Case Studies in Superconducting Magnets-Y.iwasa
Case Studies in Superconducting Magnets-Y.iwasa
Case Studies in Superconducting Magnets-Y.iwasa
Superconducting Magnets
Design and Operational Issues
Second Edition
Cover Design The blue and red shades of the covers symbolize, respectively,
low- and high-temperature superconducting (LTS/HTS) magnets. The drawings
of two magnets, by Wooseok Kim (SNU), represent, left to right, a 1.1 GHz and
a 1.3 GHz LTS/HTS NMR magnet, both to be sequentially completed in the
2010s at the M.I.T. Francis Bitter Magnet Laboratory under the sponsorship
of the National Institutes of Health. The magnet drawings, in-scale and in
proportion to each other, are based on designs of a 500 MHz and a 700 MHz
LTS magnet (both in mauve) by Masatoshi Yoshikawa (JASTEC) and that of a
600 MHz HTS insert (in light pink) by Seungyong Hahn (FBML).
ISBN 978-0-387-09799-2
Yukikazu Iwasa
Case Studies in
Superconducting Magnets
Design and Operational Issues
Second Edition
Yukikazu Iwasa
Francis Bitter Magnet Laboratory
Massachusetts Institute of Technology
Cambridge, MA
USA
springer.com
PREFACE TO THE SECOND EDITION
The Second Edition, as was the First Edition, is based on Superconducting Mag-
nets, a graduate course in the Department of Mechanical Engineering at the Mas-
sachusetts Institute of Technology that I started teaching in 1989, after the dis-
covery of high-temperature superconductors (HTS); in this book Im classifying
MgB2 , a superconductor with a critical temperature of 39 K, as an HTS. The book,
intended for graduate students and professional engineers, covers the basic design
and operational issues of superconducting magnet technology.
As prefaced in the 1st Edition, the student in the course was assigned many tu-
torial problems to review lecture materials, to discuss topics in more depth than
covered in the lecture, or to teach subjects not presented in the lecture at all.
Because the use of tutorial problems accompanied, a week later, by solutions has
been successful in the course, this format, adopted in the 1st Edition, is maintained
in the 2nd Edition. In the new edition, most problems in the 1st Edition have been
retained, either as PROBLEMS or converted to DISCUSSIONS, to which more topics
in the form of PROBLEMS & DISCUSSIONS have been added. Because the principal
magnet projects at the Francis Bitter National Magnet Laboratory (FBNML), un-
til 1995, and Francis Bitter Magnet Laboratory (FBML) thereafter, have chiey
been high-eld DC solenoidal magnets, problems directly related to other appli-
cations, with a few exceptions, are not represented. However, important topics
covered in this book, particularly on eld distribution, magnets, force, thermal
stability, dissipation, and protection, are suciently basic and generic in concept
that solenoidal magnets are suitable examples.
The 2nd Edition is more than 200 pages longer than the 1st Edition for four reasons:
1) included now are topics that should have been covered but were omitted in the
1st Edition; 2) new materials, relevant and applicable chiey to HTS magnets, are
added; 3) some concepts are given fuller explanations; and 4) more homework and
quiz problems, created since 1995, are incorporated. This time the presentation
should be signicantly improved and aicted with far fewer (but alas, not free
of) errors and typos than previously. Encountering the reader of the 1st Edition,
I often had to warn him/her of many errors, stating, though, that these errors
would in most part be obvious if the book was read with eyes wide open.
It took less than 18 months to complete the 1st Editionthe project began in May
of 1993, and the textbook came out in October 1994; by contrast, it has taken
more than 10 years to complete the 2nd Edition, starting in 1997 with the rewriting
and expansion of CHAPTER 3. In preparing both editions, I relied heavily on the
magnet projects with which I was personally involved, and am indebted to my
colleagues, the members of the Magnet Technology Division, past (FBNML) and
recent (FBML): the past members included John Williams, Mat Leupold, Emanuel
Bobrov, David Johnson, Vlad Stejskal, Andy Szczepanowski, Mel Vestal, Robert
Weggel, and Alex Zhukovsky; recent members include Juan Bascunan, my friend
and colleague the late Emanuel Bobrov (19362008), David Johnson, Haigun Lee
(now at the Korea University, Seoul), Seungyong Hahn, and Weijun Yao.
v
vi PREFACE TO THE SECOND EDITION
During the more than 10 years writing this new edition, many people helped me
to complete it. In our Division, I would particularly thank: Juan going through
the entire book more than once, checking carefully axial force equations involving
complete elliptic integrals, and providing many cryogenic data; Emanuel for his
suggestion that axial force formulas be included in the new edition and that they
are derivable from equations given in a paper by Milan Wayne Garrett; Haigun
for thoroughly going through and nding many errors in early drafts; Weijun for
key inputs in early chapters; and most of all Seungyong for his incisive critiques
on many key issues and helpful suggestions for improvement. Many recent visitors
and postdoctoral fellows made contributions to the new edition, particularly: Min
Cheol Ahn for many graphs and help in preparing the Index; Wooseok Kim (now
at Seoul National University) and Frederic Trillaud (now at Lawrence Berkeley
National Laboratory) for graphs; and Ryuya Ando (Hitachi Ltd.) for cryogenic
data. Also my former student Benjamin Haid (now at Lawrence Livermore Na-
tional Laboratory) contributed to this new edition. Robert Weggel went through
the entire text several times not only looking for typos and errors but also, more
importantly, making critical comments and suggestions, and checking several ver-
sions of the Index. To them, my deepest gratitude.
From members of the Plasma Science and Fusion Center, across the street from
the Magnet Lab, I received many valuable inputs: Joe Minervini (who taught the
course while it was also a graduate course in the Department of Nuclear Engi-
neering), Joel Schultz, Makoto Takayasu, Chen-Yu Gung, Brad Smith, and Alex
Zhukovsky. I thank them for their contributions.
From my colleagues and friends outside M.I.T. I also received valuable contri-
butions and I thank them: Dr. Luca Bottura of CERN for gures and data;
Dr. Masaki Suenaga of Brookhaven National Laboratory and Prof. Kazuo Funaki
of Kyushu University on AC losses; Drs. Hans Schneider-Muntau, Mark Bird, and
John Miller (then) of the National High Magnetic Field Laboratory (NHMFL)
on their water-cooled magnets and the 45-T hybrid magnet; Dr. Chris Rey of
Oak Ridge National Laboratory on magnetic separation; Prof. Hiroyuki Fujishiro
of Iwate University for electrical resistivity data for solder materials (CHAPTER 7);
Prof. Atsushi Ishiyama of Waseda University, Dr. Robert Duckworth of Oak Ridge
National Laboratory, and Dr. Xiaorong Wang of NHMFL for their experimental
results on normal zone propagation in HTS test samples used in CHAPTER 8.
I also thank the following friends for reviewing selected chapter drafts and making
valuable suggestions: Dr. Michael Gouge of Oak Ridge National Laboratory, Dr.
Francois-Paul Juster of the Commissariat a lEnergie Atomique (CEA), Saclay,
Prof. John Pfotenhauer of the University of Wisconsin, Dr. Soren Prestemon of
Lawrence Berkeley National Laboratory, and Dr. Martin Wilson.
Since 1995, our projects have focused principally on HTS magnets, their specic de-
sign and operation issues, e.g., protection, and their applications to advanced NMR
and MRI magnets. These magnet projects have been inspired by our Divisions
close association with Prof. Robert Grin, Director of FBML and M.I.T. De-
partment of Chemistry; Prof. Gerhard Wagner of Harvard Medical School; and
Prof. David Cory of M.I.T. Department of Nuclear Science & Engineering.
CASE STUDIES IN SUPERCONDUCTING MAGNETS vii
You know nothing till you prove it! FLY! Jonathan Livingston
To the Memory of My Parents,
Seizaburo and Shizuko Iwasa
and
viii
CONTENTS
CHAPTER 1 SUPERCONDUCTING MAGNET TECHNOLOGY . . . . . 1
1.1 Introductory Remarks . . . . . . . . . . . . . . . . . . . . . . 1
1.2 Superconductivity . . . . . . . . . . . . . . . . . . . . . . . . 3
1.2.1 Meissner Eect . . . . . . . . . . . . . . . . . . . . . . . 3
1.2.2 Londons Theory of Superconductivity . . . . . . . . . . . . . 5
1.2.3 Type I and Type II Superconductors . . . . . . . . . . . . . 5
DC and AC Responses . . . . . . . . . . . . . . . . . . . . . 6
Magnetic Behavior . . . . . . . . . . . . . . . . . . . . . . . 7
Examples of Superconductors . . . . . . . . . . . . . . . . . . 7
1.2.4 Critical Surfaces of Type II Superconductor . . . . . . . . . . 9
Critical Current Density, Jc . . . . . . . . . . . . . . . . . . . 9
1.3 Magnet-Grade Superconductors . . . . . . . . . . . . . . . . . . 9
1.3.1 Materials vs. Magnet-Grade Superconductors . . . . . . . . . 10
1.3.2 Laboratory Superconductor to Magnet-Grade Superconductor . 10
1.4 Magnet Design . . . . . . . . . . . . . . . . . . . . . . . . 11
1.4.1 Requirements and Key Issues . . . . . . . . . . . . . . . . 11
1.4.2 Eect of Operating Temperature . . . . . . . . . . . . . . 12
1.5 Numerical Solutions . . . . . . . . . . . . . . . . . . . . . . 13
1.5.1 Ballpark Solutions . . . . . . . . . . . . . . . . . . . . 13
1.5.2 Code Solutions . . . . . . . . . . . . . . . . . . . . . . 13
1.6 The Format of the Book . . . . . . . . . . . . . . . . . . . . 14
PROBLEM 1.1: Thermodynamics of Type I superconductors . . . . . . . 15
Solution to PROBLEM 1.1 . . . . . . . . . . . . . . . . . . . . 16
Values of Hc : Equation 1.6 and Experimental . . . . . . . . . . 17
PROBLEM 1.2: A superconducting loop . . . . . . . . . . . . . . . . 20
Solution to PROBLEM 1.2 . . . . . . . . . . . . . . . . . . . . 21
PROBLEM 1.3: Magnetic resonance imaging (MRI) . . . . . . . . . . . 22
Solution to PROBLEM 1.3 . . . . . . . . . . . . . . . . . . . . 22
REFERENCES . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
Other Sources of Information . . . . . . . . . . . . . . . . . . 24
CHAPTER 2 ELECTROMAGNETIC FIELDS . . . . . . . . . . . . . 25
2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 25
2.2 Maxwells Equations . . . . . . . . . . . . . . . . . . . . . . 25
2.2.1 Gauss Law . . . . . . . . . . . . . . . . . . . . . . . 26
2.2.2 Amperes Law . . . . . . . . . . . . . . . . . . . . . . 26
2.2.3 Faradays Law . . . . . . . . . . . . . . . . . . . . . . 26
ix
x CONTENTS
Stability
Figure 6.1 Spectra of gd (t) compiled for LTS magnets. . . . . . . . . . . . 356
Figure 6.15 General energy margin vs. transport current plot for CIC conductor. 381
Table 6.1 Concepts Derived from Power Density Equation (Eq. 6.1) . . . . . 352
Table 6.4 Selected values of Top , Top , and eh for LTS and HTS . . . . . 358
AC Losses
Table 7.3A Hysteresis Energy Density, ehy [J/m3 ]
Bean Slab (Width: 2a) No Transport Current . . . . . . . . . 439
Table 7.3B Hysteresis Energy Density, ehy [J/m3 ]
Bean Slab (Width: 2a) With DC Transport Current . . . . . . 440
Table 7.4 Self-eld Hysteresis Energy Density, esf [J/m3 ] Bean Slab (Width: 2a) 440
Table 7.5 Hysteresis Energy Density, ehy [J/m3 ]No Transport Current
Wire (Diameter: df ) & Tape (Width: w; Thickness: ) . . . . . . 441
Table 7.6 Self-Field Energy Density, esf [J/m3 ]Wire & Tape . . . . . . . . 442
Table 7.7 Energy Density, eih Bean Slab In-Phase Sinusoidal Current & Field
It = Im sin(2f t); He = Hm sin(2f t) . . . . . . . . . . . . . . 442
Table 7.8 Coupling Energy Density Over One Period, ecp [J/m3 ]
Wire of Outermost Diameter, Dmf , Enclosing Multilaments . . . 443
Table 7.9 Eddy-Current Energy Density, eed and < eed > [J/m3 ] . . . . . . . 443
Protection
Figure 8.2 Z(T ) plots. Ag1000 (RRR: 1000); Ag100; Cu200; Cu100; Cu50;
Al (grade 1100); Brass (70Cu-30Zn)also dashed line
(Eq. 8.10b, with m = 5.5108 m) . . . . . . . . . . . . . . 473
Figure 8.6 Y (T ) plots. Left-hand vertical scale: Ag (100 to 1000); Cu (50 to 200)
Al (grade 1100). Right-hand vertical scale: brass (70Cu-30Ni)
also dashed line (Eq. 8.32b with m = 5.5108 m . . . . . . 479
Table 8.1 Hot Spot Volume Fraction, fr , vs. Hot-Temperature, Tf
Solenoid ( = 1.5, = 2.0) of B 1.5 T30 T . . . . . . . . . . . 469
Table 8.2 Permissible Limits of Tf for Selected Materials . . . . . . . . 470
Table 8.3 Mean Linear Thermal Expansion Data . . . . . . . . . . . . . 471
Table 8.4 Data of Minimum Arcing Voltage, Vmn , and P d at Vmn for
Gases at Room Temperature . . . . . . . . . . . . . . . . . 481
Table 8.5 Selected Measured U for LTS and HTS, Bare and Composite . . . 489
Thermal, Electrical, Mechanical Properties
Figure 4.23 Thermal conductivity vs. temperature data of selected Ag-Au alloys . 284
Figure 4.24 Electrical resistivity vs. temperature data of selected Ag-Au alloys . 284
Table 4.18 Structural Data for G-10, Stainless Steel 304, Brass, and Copper . 307
Table 5.2 Diusivities of Stainless Steel and Copper at 4 K and 80 K . . . . 337
Table 6.2 Heat Capacities of Substances in Superconducting Magnets . . . . 354
Table 6.3 Thermal Conductivities of Substances in Superconducting Magnets
Electrical Resistivities of Copper and Stainless Steel . . . . . . . 355
Table 7.1 Solder Contact Resistances at 4.2 K . . . . . . . . . . . . . . 409
Table 7.2 Electrical Resistivities of Solder Alloys at Selected Temperatures . . 410
Temperature
Figure 4.19 Temperature spectrum covering 19 orders of magnitude . . . . . . 265
Table 4.11 Approximate Values of Signal Levels (V ) and Sensitivities (V /T ) of
Selected Thermometers of Diodes; Resistances; and Thermocouples . 266
Legendre and Associated Legendre Functions
Table 2.2 Legendre and Associated Legendre Functions . . . . . . . . . . . 35
Table 2.4 Solutions of Equation 2.32 in Cartesian Coordinates . . . . . . . . 36
xxxiv CONTENTS
TRIVIA (ANSWER) . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
TRIVIA 1.1 (ANSWER) . . . . . . . . . . . . . . . . . . . . . . . . . 12 (20)
TRIVIA 1.2 (ANSWER) . . . . . . . . . . . . . . . . . . . . . . . . . 16 (21)
TRIVIA 2.1 (ANSWER) . . . . . . . . . . . . . . . . . . . . . . . . . 29 (34)
TRIVIA 2.2 (ANSWER) . . . . . . . . . . . . . . . . . . . . . . . . . 35 (44)
TRIVIA 2.3 (ANSWER) . . . . . . . . . . . . . . . . . . . . . . . . . 49 (53)
TRIVIA 2.4 (ANSWER) . . . . . . . . . . . . . . . . . . . . . . . . . 59 (65)
TRIVIA 3.1 (ANSWER) . . . . . . . . . . . . . . . . . . . . . . . . . 85 (90)
TRIVIA 3.2 (ANSWER) . . . . . . . . . . . . . . . . . . . . . . . . 129 (137)
TRIVIA 3.3 (ANSWER) . . . . . . . . . . . . . . . . . . . . . . . . 146 (161)
TRIVIA 3.4 (ANSWER) . . . . . . . . . . . . . . . . . . . . . . . . 168 (172)
TRIVIA 3.5 (ANSWER) . . . . . . . . . . . . . . . . . . . . . . . . 176 (203)
TRIVIA 4.1 (ANSWER) . . . . . . . . . . . . . . . . . . . . . . . . 221 (230)
TRIVIA 4.2 (ANSWER) . . . . . . . . . . . . . . . . . . . . . . . . 235 (249)
TRIVIA 4.3 (ANSWER) . . . . . . . . . . . . . . . . . . . . . . . . 253 (261)
TRIVIA 4.4 (ANSWER) . . . . . . . . . . . . . . . . . . . . . . . . 272 (276)
TRIVIA 4.5 (ANSWER) . . . . . . . . . . . . . . . . . . . . . . . . 291 (297)
TRIVIA 4.6 (ANSWER) . . . . . . . . . . . . . . . . . . . . . . . . 302 (306)
TRIVIA 5.1 (ANSWER) . . . . . . . . . . . . . . . . . . . . . . . . 331 (337)
TRIVIA 6.1 (ANSWER) . . . . . . . . . . . . . . . . . . . . . . . . 371 (375)
TRIVIA 6.2 (ANSWER) . . . . . . . . . . . . . . . . . . . . . . . . 388 (393)
TRIVIA 7.1 (ANSWER) . . . . . . . . . . . . . . . . . . . . . . . . 405 (428)
TRIVIA 7.2 (ANSWER) . . . . . . . . . . . . . . . . . . . . . . . . 441 (451)
TRIVIA 8.1 (ANSWER) . . . . . . . . . . . . . . . . . . . . . . . . 483 (495)
TRIVIA 8.2 (ANSWER) . . . . . . . . . . . . . . . . . . . . . . . . 492 (503)
TRIVIA 8.3 (ANSWER) . . . . . . . . . . . . . . . . . . . . . . . . 521 (524)
TRIVIA 9.1 (ANSWER) . . . . . . . . . . . . . . . . . . . . . . . . 549 (555)
TRIVIA 9.2 (ANSWER) . . . . . . . . . . . . . . . . . . . . . . . . 556 (566)
TRIVIA 9.3 (ANSWER) . . . . . . . . . . . . . . . . . . . . . . . . 578 (586)
CHAPTER 1
SUPERCONDUCTING MAGNET TECHNOLOGY
1.2 Superconductivity
The complete absence of electrical resistivity for the passage of direct current
below a certain critical temperature (usually designated with the symbol Tc ) is
the basic premise of superconductivity. In addition to Tc , the critical eld Hc and
critical current density Jc are two other parameters that dene a critical surface
below which the superconducting phase can existsee 1.2.4. Tc and Hc are
thermodynamic properties that for a given superconducting material are invariant
to metallurgical processing; Jc is not. Indeed the key contribution of Kunzler and
others in 1961 was to demonstrate that for certain superconductors it is possible
to enhance Jc dramatically by means of metallurgy alone. No formal theories
of superconductivity, phenomenological or microscopic, will be presented in this
book to explain relationships among Tc , Hc , or Jc ; however, the magnetic behavior
of superconductivity, which plays a key role in superconducting magnets, will be
briey reviewed by means of simple theoretical models.
0 T
0 D Tc A
(a)
SUPERCONDUCTOR (Sc)
A = B = C = D
(b)
A = B = C = D
(c)
A = D = C = B
(d)
Fig. 1.1 (a): H-T phase diagram for two spheres, one a superconductor (Sc) and the
other a perfect conductor (Pc), respectively, when T < Tc . (b): eld prole of the Sc
sphere for a sequence of H-T environments ABCD. (c): eld proles of Pc for the
same sequence of environments. (d): eld proles of Sc or Pc for sequence ADCB.
At points C (b) for the Sc sphere, D (c) for the Pc sphere, and C (d) for both the Sc and
Pc spheres, a surface current is induced in the sphere to meet either the requirement of
B = 0 or B/ t = 0. The eld proles in (b)(d) are schematic.
SUPERCONDUCTING MAGNET TECHNOLOGY 5
where m and e are the electron mass (9.111031 kg) and charge (1.601019 C);
is the permeability of free space (4 107 H/m). Here, the density of super-
conducting electrons, nse , is dierent from that of the free electrons, nfe , being
all fully superconducting at T = 0 and zero at Tc . For a qualitative estimation in
PROBLEM 2.2 (CHAPTER 2 ) we may equate nse to nfe electrons, given by:
NA
nse nfe = (1.2)
WA
DC and AC Responses
Figure 1.2 shows schematic drawings of three superconducting rodsType I (Fig.
1.2a) and Type II (Figs. 1.2b, 1.2c)each carrying a transport current that is less
than its critical current. In the Type I rod (Fig. 1.2a), the current, DC or AC,
ows only at the surface (within the London penetration depth) and generates
no dissipation. In the Type II rod, DC current (Fig. 1.2b), though owing over
the entire rod, also generates no dissipation despite the presence of many normal-
state (resistive) regions. We may picture superconducting electrons in the rod
gliding through the material, dodging the normal-state regions. In terms of a cir-
cuit model, we may picture these resistive islets to be electrically short-circuited
by the surrounding superconducting sea. With AC current through the Type II
rod (Fig. 1.2c), however, there is dissipation (heat generation); i.e., the mixed-
state rod is electrically resistive, though its eective resistivity is still orders of
magnitude less than those of highly conductive normal metals. Each normal-state
region contains ux bundles, the so-called uxoids or vortices, and these ux bun-
dles owunder time-varying conditions of magnetic eld, current, or both. This
dissipative ux ow is the major source of AC losses in Type II superconductors.
DC
AC Type I
(a)
DC Type II
(b)
AC Type II
Heat
(c)
Fig. 1.2 Superconducting rods carrying a transport current. (a) Type I rod, DC
or AC currentno Joule dissipation generated; (b) Type II rod, DC currentno
dissipation; (c) Type II rod, AC currentJoule dissipation generated.
SUPERCONDUCTING MAGNET TECHNOLOGY 7
M
Hc
Mixed State
0 H
0 Hc1 Hc Hc2
Fig. 1.3 Schematic M vs. H plots for Type I (solid) and Type II (dashed) super-
conductors. Hc is the critical eld for Type I; Hc1 and Hc2 are, respectively, the lower
and upper critical elds of Type II. The slant-lined area represents the mixed state.
Magnetic Behavior
Exposed to a magnetic eld, a Type I superconductor is perfectly diamagnetic
up to the bulk thermodynamic critical eld, Hc ; above Hc , it behaves as a nor-
mal nonmagnetic material. The magnetic behavior of a Type II superconductor is
identical to that of Type I up to the lower critical eld, Hc1 ; in the range between
Hc1 and Hc2 , the upper critical eld, the Type II is in the mixed state (p. 5). Fig-
ure 1.3 shows schematic (not in scale) plots of magnetization (M ) vs. magnetic
eld (H) of Type I (solid curve) and Type II (dashed). (Because the supercon-
ductor is fundamentally diamagnetic, its magnetization is negative, and thus both
curves are plotted as M vs. H.) The slant-lined region in Fig. 1.3 represents the
mixed state of the Type II superconductor. Caveat: the magnetization curves of
all Type II magnet-grade-superconductors, as studied in CHAPTER 5, unlike that
shown in Fig. 1.3, are irreversible, resulting in hysteretic magnetization curves.
The hysteretic nature of magnetization is an important source of AC losses in
Type II superconductors (CHAPTER 7 ).
Examples of Superconductors
Table 1.2 shows selected superconductors, Type I and Type II, with their critical
temperatures in zero eld, Tc , and critical magnetic inductions Hc for Type I
and Hc2 for Type II. All Type I superconductors are metals, with low critical
elds. It is no wonder that Kamerlingh Onnes attempt, in 1913, to build the
rst superconducting coil, wound with lead wire, failed. Exposed to the eld of
its own coil (self eld), lead could not remain superconducting. Even at that time
0.3 T was about the minimum eld that an electromagnet needed to generate to
be considered useful. Table 1.2 clearly shows that a superconducting magnet must
employ Type II conductor. Unlike Type I, Type II comes in a variety of types:
alloys; metalloids; metallic compounds; and even oxides. Note that all the oxides
of Table 1.2 are HTS; MgB2 , a metalloid, is considered HTS. Figure 1.4 shows
Tc vs. year data of selected LTS and HTS, together with boiling temperatures
(horizontal lines) of important cryogens. The solid lines connect oxide supercon-
ductors, while the dashed line, beginning and ending, respectively, with Hg and
MgB2 , connects metallic superconductors over a 90-year span.
8 CHAPTER 1
200
C2 H6
180
160
CF4
140
120
CH4
Tc [K]
100
Ar
80
N2
60
40
Ne
20 H2
He
0
1900 1910 1920 1930 1940 1950 1960 1970 1980 1990 2000
Year
Fig. 1.4 Tc vs. Year data for selected superconductors, metallic (LTS) and oxide (HTS).
Solid lines and open circles: HTS; Dashed line and solid circles: LTS, except MgB2 (Tc =
39 K; Year 2000), considered an HTS. Dotted horizontal lines: boiling temperatures of
cryogens, from bottom to top: He (4.22 K); H2 (20.39 K); Ne (27.09 K); N2 (77.36 K);
Ar (87.28 K); CH4 (methane/111.6 K); CF4 (Freon-14/145.4 K); C2 H6 (ethane/184.6 K).
SUPERCONDUCTING MAGNET TECHNOLOGY 9
f1 (J, H, T = 0)
f2 (J, T, H = 0)
H
f3 (H, T, J = 0)
T
Fig. 1.5 Critical surface of a typical Type II superconductor.
Diculty or Cost
Protection;
Conductor
Mechanical Integrity
Stability;
Cryogenics
Top [K]
0 4 100
TRIVIA 1.1 It is said that the most intense magnetic eld is in a neutron star. Of the
large numbers below, which is closest to the ratio of its magnetic eld to the earths?
i) 1023 ; ii) 1018 ; iii) 1013 ; iv) 108 .
SUPERCONDUCTING MAGNET TECHNOLOGY 13
To derive Eq. 1.7, note that volumetric free energy at T in zero eld in the
normal state, Gn (T ), and superconducting state, Gs (T ), are related by:
Gn (T ) Gs (T ) = 12 Hc2 (T ) (1.8)
From Eq. 1.8 and S(T ) = G(T )/ T , Eq. 1.6 may be derived.
c) The dierence between the internal energy densities, Un Us , in zero eld is
maximum at Tux . By noting that U (T ) = C(T ) dT in zero eld, show:
Tc
Tux = (1.9)
3
d) Suppose that a magnetic eld is applied very slowly and adiabatically to the
superconductor initially at Ti , 0 < Ti < Tc . The eld is applied to a level
just above the critical value; this transition to the normal state lowers the
superconductors temperature to Tf (< Ti ). Derive an expression relating Tf
and Ti . Also, show the process in a thermodynamic phase diagram.
e) To the same superconductor initially at Ti , 0 < Ti < Tc , in zero eld, a
magnetic eld He is applied suddenly. If He exceeds a critical level, Hec , the
metal will be heated. Show that Hec , which depends on Ti , is given by:
2
1 + 3 Ti
Tc
Hec (Ti ) = Hc (Ti )
2 (1.10)
Ti
1
Tc
16 CHAPTER 1PROBLEMS
Because, as noted above, Sn (Tc )Ss (Tc ) = 0 at H = 0, we have from Eq. S1.3:
= 13 (a b)Tc2 (S1.4)
Equating Eqs. S1.5 and S1.6, and integrating each side with respect to T , we have:
4
T
2 Hc (T ) = 2 T 4 Tc
1 2 1 2 1 2
+A (S1.7)
Tc
where A is a constant. With Hc (T = 0) Hc , we have: A = Hc2 /2.
At T = Tc , because Hc (Tc ) = 0, Eq. S1.7 becomes:
0 = 12 Tc2 14 Tc2 12 Hc2 (S1.8)
TRIVIA 1.2 Of the following early students of magnetism (born c. 640 B.C.1777 A.D.)
below, who showed that garlic did not destroy magnetism?
i) Gilbert; ii) Oersted; iii) Peregrinus; iv) Thales.
SUPERCONDUCTING MAGNET TECHNOLOGYPROBLEMS 17
Combining Eq. S1.7 (with A = Hc2 /2) and Eq. S1.9, we have:
2 4
T T
2 Hc (T ) = Hc
1 2 2
2 Hc
1 2
12 Hc2 (S1.10)
Tc Tc
Us (T ) = 14 aT 4 (S1.13b)
The dierence between the two free energies is given by:
Un (T ) Us (T ) = 14 (b a)T 4 + 12 T 2
ab 2
= T T
2 4
(S1.14)
4 ab
Thus:
4
Tux 4Tux
3
=0 (S1.16)
ab
Tc
Ss (T ) Sn (T )
Ti e)
Tf
d)
0 S
0 Ss(Ti ) Sn(Tf ) Sn(Tc )
Fig. 1.7 Schematic T -S plots, superconducting (solid curve) and normal (dashed)
states. The solid vertical line corresponds to d); the dotted horizontal line to e).
Inserting Eq. S1.9 into Eq. S1.22, and applying S1.19, we obtain:
2 2
T i Ti
1 2 1 2 2
2 Hec (Ti ) = 2 Hc (Ti ) + 2 Hc T 1 (S1.23)
c Tc
Hence:
2
1 + 3 Ti
Tc
Hec (Ti ) = Hc (Ti )
2 (1.10)
Ti
1
Tc
Note that Hec (Ti ) Hc (Ti ) and that Hec (Tc ) = 0 because Hc (Tc ) = 0.
20 CHAPTER 1PROBLEMS
I(t) Is (t)
L Ls
AC
M
Answer to TRIVIA 1.2 i). The English physician and physicist William Gilbert
(15401603), a pioneer of experimentation, refuted many superstitions by direct
testing; his experiments on magnetism earned him the title father of electricity.
The Danish physicist Hans Christian Oersted (17771851) discovered in 1819 that
a compass needle aligned at right angles to a current-carrying conductor, thus
initiating the science of electromagnetism. The French crusader Petrus Peregrinus
(c. 1240?), discoverer of magnetic poles, showed it is impossible to separate the
poles by breaking a magnet in two, and was rst to describe in detail the pivoted
compass. The Greek philosopher Thales (640 B.C.546 B.C.), considered the rst to
study magnetism, is now known chiey for having predicted an eclipse of the sun
that occurred in Asia Minor on May 28, 585 B.C.
22 CHAPTER 1PROBLEMS
REFERENCES
[1.1] J.K. Hulm and B.T. Matthias, Overview of superconducting materials develop-
ment, in Superconductor Materials ScienceMetallurgy, Fabrication, and Appli-
cations, Eds., S. Foner and B.B. Schwartz (Plenum Press, New York, 1981).
[1.2] A.A. Abrikosov, On the magnetic properties of superconductors of the second
type (English translation), Zh. Eksp. Teor. Fiz. (Soviet Union) 5, 1174 (1957).
[1.3] U. Essmann and H. Trauble, The direct observation of individual ux lines in
Type II superconductors, Phys. Lett. 24A, 526 (1967).
[1.4] Y.B. Kim, C.F. Hempstead, and A.R. Strnad, Magnetization and critical super-
currents, Phys. Rev. 129, 528 (1963).
[1.5] J.K. Hoer, The Initiation and propagation of normal zones in a force-cooled
tubular superconductor, IEEE Trans. Mag. 15, 331 (1979).
[1.6] C. Marinucci, M.A. Hilal, J. Zellweger, and G. Vecsey, Quench studies of the
Swiss LCT conductor, Proc. 8 th Symp. on Eng. Prob. Fus. Res., 1424 (1979).
[1.7] V.D. Arp, Stability and thermal quenches in force-cooled superconducting ca-
bles, Proc. of 1980 Superconducting MHD Magnet Design Conference, MIT, 142
(1980).
[1.8] J. Benkowitsch and G. Kraft, Numerical analysis of heat-induced transients in
forced ow helium cooling systems, Cryogenics 20, 209 (1980).
[1.9] E.A. Ibrahim, Thermohydraulic Analysis of Internally Cooled Superconductors,
Adv. Cryo. Eng. 27, 235 (1982).
[1.10] C. Marinucci, A numerical model for the analysis of stability and quench char-
acteristics of forced-ow cooled superconductors, Cryogenics 23, 579 (1983).
[1.11] M.C.M. Cornellissen and C.J. Hoogendoorn, Propagation velocity for a force
cooled superconductor, Cryogenics 25, 185 (1985).
[1.12] A.F. Volkov, L.B. Dinaburg, and V.V. Kalinin, Simulation of helium pressure
rise in hollow conductor in case of superconductivity loss, Proc. 12 th Int. Cryo.
Eng. Conf. (Southampton, UK), 922 (1988).
[1.13] R.L. Wong, Program CICC ow and heat transfer in cable-in-conduit conduc-
tors, Proc. 13 th Symp. Fus. Tech. (Knoxville, TN), 1134 (1989).
[1.14] L. Bottura and O.C. Zienkiewicz, Quench analysis of large superconducting mag-
nets. Part I: model description, Cryogenics 32, 659 (1992).
[1.15] C.A. Luongo, C.-L. Chang, and K.D. Partain, A computational model applicable
to the SMES/CICC, IEEE Trans. Mag. 30, 2569 (1994).
[1.16] A. Shajii and J.P. Freidberg, Quench in superconducting magnets. I. Model and
implementation, J. Appl. Phys. 76, 3149 (1994); Quench in superconducting
magnets. II. Analytic Solution, J. Appl. Phys. 76, 3159 (1994).
[1.17] R. Zanino, S. De Palo, and L. Bottura, A two-uid code for the thermohy-
draulic transient analysis of CICC superconducting magnets, J. Fus. Energy 14,
25 (1995).
[1.18] L. Bottura, A numerical model for the simulation of quench in the ITER mag-
nets, J. Comp. Phys. 125, 26 (1996).
[1.19] L Bottura, C. Rosso, and M. Breschi, A general model for thermal, hydraulic,
and electric analysis of superconducting cables, Cryogenics 40, 617 (2000).
[1.20] Q. Wang, P. Weng, and M. Hec, Simulation of quench for the cable-in-conduit-
conductor in HT-7U superconducting Tokamak magnets using porous medium
model, Cryogenics 44, 81 (2004).
24 CHAPTER 1REFERENCES & OTHER SOURCES OF INFORMATION
Why sir, there is a very good chance that you will soon be able to tax it.
Michael Faradays reputed reply to William Gladstone, the Prime
Minister, who, after being shown by Faraday a demonstration of
the rst dynamo, asked But, after all, what use is it?
CHAPTER 2
ELECTROMAGNETIC FIELDS
2.1 Introduction
In this chapter we review electromagnetic theory by presenting Maxwells equa-
tions. This review is intended to refresh the readers understanding of electromag-
netic theory so as to allow the main subject matter of this booksuperconducting
magnetsto be approached in a quantitative manner. After a presentation of
Maxwells equations and simple solutions that are not only tractable analytically
but also useful in most magnet applications, specic cases familiar to most magnet
engineers will be presented and studied.
Equation 2.10 states that the surface integral of the B eld over the surface S is
zero; i.e., there are no point sources of B eld. In dierential form, Eq. 2.10 is:
=0
B (2.11)
Boundary Condition: The normal component of the B eld is always contin-
uous in passing through a surface from Region 1 (B1 ) to Region 2 (B 2 ). Namely:
2 B
n12 (B 1) = 0 (2.12)
As discussed below in 2.2.6, in a magnetic medium B is the sum of the magnetic
and magnetization M
eld H . It means that continuity of the normal component
through two dierent magnetic media is preserved despite any dierence in
of B
magnetization from one medium to the other.
The approximation process can continue indenitely, but for the low-frequency
cases of interest discussed in the PROBLEMS & DISCUSSIONS of this chapter, we
need to solve for only the 0th - and 1st -order elds.
ELECTROMAGNETIC FIELDS 29
Sinusoidal Case
When dealing with a sinusoidally time-varying electric eld E of complex am-
jwt jwt
plitude E , i.e., E = E e , and therefore J = (E /e )e , the time-averaged
dissipation power density < p > is expressed by:
J = 1 |E|2 = e |J|2
< p >= 12 E (2.21)
2e 2
where J is the complex conjugate of J.
=0
E (2.27)
2 = 0 (2.28)
Equation 2.28, known as Laplaces equation, expresses scalar potentials from which
elds can be derived.
physically realizable E
Similarly, in the absence of free current and under the DC condition, any magnetic
because H
eld H, = 0, is derivable from scalar potentials that satisfy the
= . Besides electromagnetic elds, there are other
Laplaces equation; i.e., H
well-known cases in engineering where Eq. 2.28 is applicable for time-independent
variables: temperature (T ) in a source-free, isotropic conductive medium; volume
expansion; the sum of linear strains in the x-, y-, and z-axes in a force- and
moment-free, isotropic elastic medium.
Selected solutions of Laplaces equation in two-dimensional cylindrical coordinates
and three-dimensional spherical coordinates are presented below.
The standard technique to solve Eq. 2.29 is to express as the product of two
functions, each a function of only one of the two coordinates:
= R(r) () (2.30)
Special Cases
n = 0: The simplest form of eld derivable from under this condition is one
whose spatial dependence is 1/r. Examples are the electric eld due to a line charge
( = 2 ) and the magnetic eld associated with a current lament (I = 2). Thus
= (1/r)r ; with a potential [0 ]H = , we
with a potential [0 ]E = ln r, we have: E
have: H = (1/r) . Note that each eld decays as 1/r from its source.
Special Cases
n = m = 0: This case gives rise to the the simplest solution, 0 1/r. A well-
= 1/r2 r valid for r > 0, that emanates from
known solution is the electric eld, E
a point charge of magnitude 4 .
n = 1, m = 0: There are two solutions, 1 cos /r2 and 1 r cos . 1 results
in a dipole eld outside of a sphere the surface of which is distributed with a
source that generates this dipole eld, while 1 results in a uniform eld within
the sphere. The dipole eld of a magnetic moment is also derivable from 1 .
Cartesian
The three orthogonal axes are: x, y, z.
U U U
grad U = U = x + y + z (2.34a)
x y z
=A
= Ax Ay Az
divA + + (2.34b)
x y z
=A
= A z A y Ax Az
curl A x + y
y z z x
Ay Ax
+ z (2.34c)
x y
2U 2U 2U
div grad U = 2 U = + + (2.34d)
x2 y2 z2
Cylindrical
The three orthogonal axes are: r, , z.
U 1 U U
grad U = U = r + + z (2.35a)
r r z
=A
= 1 (rAr ) 1 A Az
div A + + (2.35b)
r r r z
1 Az A Ar Az
curl A = A = r +
r z z r
1 (rA ) 1 Ar
+ z (2.35c)
r r r
1 U 1 2U 2U
div grad U = U =
2
r + 2 + (2.29)
r r r r 2 z2
ELECTROMAGNETIC FIELDS 33
Spherical
The orthogonal axes are: r, , and .
U 1 U 1 U
grad U = U = r + + (2.36a)
r r r sin
2
= 1 (r Ar ) + 1 (A sin ) + 1 A (2.36b)
=A
div A
r2 r r sin r sin
=A
= 1 (A sin ) 1 A
curl A r
r sin r sin
1 Ar 1 (rA )
+
r sin r r
1 (rA ) 1 Ar
+ (2.36c)
r r r
1 2U 1 U
div grad U = 2 U = r + sin
r2 r r r2 sin
1 2U
+ (2.32)
r2 sin2 2
* Dover Publications, Inc., New York, 1948. For many years until his retirement in 1960,
Dirk J. Struik (18942000) taught mathematics at M.I.T.
34 CHAPTER 2
dm Pn (u)
Pnm (u) = (1 u2 )m/2 (2.38b)
dum
(1 u2 )m/2 dm+n (u2 1)n
= (2.38c)
2n n! dum+n
Some useful recurrence formulas for Legendre functions, Pn (u), and Associated
Legendre functions, Pnm (u), are:
2(m+1)u m+1
Pnm+2 (u)
Pn (u)+(nm)(n+m1)Pnm (u) = 0 (2.39c)
(1u2 )
Table 2.2 lists Legendre functions, Pn (u), for n up to 8; and Associated Legendre
functions, Pnm (u), for m (1 m n) up to 4. Table 2.3 presents Pnm (0) for
combinations of n of 2, 4, 6, 8, 10 and m of 0, 2, 4, 6, 8, and 10. Table 2.4 presents
solutions of Eq. 2.32 in Cartesian coordinates.
Answer to TRIVIA 2.1 iv). The Danish astronomer Olaus Roemer (16441710)
used two reference objects, Jupiter and earth, to clock light at 2.27 108 m/s,
which is 76% of the modern accepted value: 2.99792458108 m/s.
ELECTROMAGNETIC FIELDS 35
TRIVIA 2.2 Identify the location for which each of the elements below is named.
i) Copper; i) Lutetium; ii) Magnesium; iv) Yttrium.
36 CHAPTER 2
Legendre Functions (m = 0)
n
n r Pn (u)
0 1
1 z
2 z 2 12 (x2 +y 2 )
3 z 3 32 (x2 +y 2 )z
4 z 4 3(x2 +y 2 )[z 2 18 (x2 +y 2 )]
5 z 5 5(x2 +y 2 )[z 2 38 (x2 +y 2 )]z
6 z 6 52 (x2 +y 2 )[3z 4 94 z 2 (x2 +y 2 )+ 18 (x2 +y 2 )2 ]
7 z 7 16
7
(x2 +y 2 )[24z 4 30z 2 (x2 +y 2 )+5(x2 +y 2 )2 ]z
8 z 8 128
7
(x2 +y 2 )[256z 6 480z 4 (x2 +y 2 )+160z 2 (x2 +y 2 )2 5(x2 +y 2 )3 ]
It was absolutely marvelous working for Pauli. You could ask him anything.
There was no worry that he would think a particular question was stupid,
since he thought all questions were stupid. Victor F. Weisskopf
ELECTROMAGNETIC FIELDSPROBLEMS & DISCUSSIONS 37
= H0 ( cos r + sin )
H (2.40)
1 ) and inside
a) Show that expressions for the magnetic inductions outside (B
2 ) the sphere are given by:
(B
1 = H0 ( cos r + sin )
B
3
R
+ H0 (2 cos r + sin ) (2.41a)
2 + r
2 = 3 H0 ( cos r + sin )
B (2.41b)
2 +
z H0
R
Boundary Conditions
1) At r = R, the tangential component ( ) of H is continuous, since there is no
surface current. This is equivalent to equating the potentials at r = R (1 = 2 ):
A
H0 + =C (S1.5)
R3
is continuous:
2) At r = R, the normal component (r ) of B
A
H0 + 2 3 = C (S1.6)
R
From Eqs. S1.5 and S1.6, we can solve for the constants C and A:
3H0
C= (S1.7)
2 +
A = (C H0 )R = H0 3
R3 (S1.8)
2 +
ELECTROMAGNETIC FIELDSPROBLEMS & DISCUSSIONS 39
1 = H0 ( cos r + sin )
B
3
R
+ H0 (2 cos r + sin ) (2.41a)
2 + r
2 = 3 H0 ( cos r + sin )
B (2.41b)
2 +
Case 1: / = 0
With = 0 inserted into Eqs. 2.41a and 2.41b, we obtain:
3
1 = H0 ( cos r + sin ) + H0
B
R
(2 cos r + sin ) (S1.9a)
2 r
2 = 0
B (S1.9b)
Case 2: / = 1
The problem reduces to the trivial case, equivalent to the absence of the sphere.
Note that with = into Eqs. 2.41a and 2.41b, the equations become identical.
Case 3: / =
This case is that of a perfectly ferromagnetic material; soft iron comes close. The
magnetic eld is drawn into the sphere. With = in Eqs. 2.41a and 2.41b:
3
R
B1 = H0 ( cos r + sin ) H0 (2 cos r + sin ) (S1.10a)
r
2 = 3 H0 ( cos r + sin )
B (S1.10b)
The important point to note is that the B eld within the sphere of / =
is three times the applied B eld. Note that if the spheres magnetization is
saturated, no longer is . The eects of eld saturation, present in all magnetic
materials, will be discussed in PROBLEM 2.3.
40 CHAPTER 2PROBLEMS & DISCUSSIONS
/ = 0.1
/ = 100
Fig. 2.2 Field lines inside and near spheres of / = 0.1 (top)
and / = 100 (bottom).
For the top sphere, the line spacing at r = is 2.6(= 2.1/0.3) times denser than
inside the sphere, while for the bottom sphere, the inside line spacing is 1.7(=
that
300/101) times that at r = . These ratios are valid in both the plane of the paper
and that perpendicular to it. A cylinder of / = 100 would have a ratio of 200/101, but
only in the plane of the paper. Note that the eld lines entering and leaving the sphere
are nearly perpendicular to the surface for / = 100perpendicular for / = .
ELECTROMAGNETIC FIELDSPROBLEMS & DISCUSSIONS 41
where H0 = 0.08 T. Initially the rod is in the normal state. The eld thus is
everywhere the same, including inside the rod. The rod is then gradually cooled
until it becomes superconducting.
a) Show that an expression for the eld outside the superconducting rod, H 1,
after transient eects of the eld change have subsided, is given by:
2
R
H1 = H0 ( cos r + sin ) + H0 (cos r + sin ) (2.42)
r
1 for a magnetic sphere.)
(Note that Eq. 2.41a corresponds to B
f [A/m],
b) Show that an expression for the surface current (free) density, K
owing within a penetration depth R, is given by:
f = 2H0 sin z
K (2.43)
c) Convert the magnitude of the surface current density to that of current den-
sity, Jf [A/m2 ], and conrm that its value is consistent with that for lead
derivable from Londons theory of superconductivity.
x H0
R
A = R 2 H0 (S2.6)
The eld outside the superconducting rod (Region 1) is thus given by:
2
1 = H0 ( cos r + sin ) + H0 R (cos r + sin )
H (2.42)
r
Note that at r = R, = 90 , |H
1 | = 2H0 ; the eld amplitude is twice the far-eld
amplitude. Physically, the eld originally inside when the rod is in the normal
state is now pushed outside and compressed at and near = 90 .
b) Because of the discontinuity at r = R of 2H0 sin in the tangential ( ) com-
there must be a surface current density K
ponent of H, f owing in the rod, as
given by Eq. 2.6. We thus have:
f = r 2H0 sin
K
= 2H0 sin z (2.43)
This sine (more commonly known as cosine) current distribution is the basis
for most of the dipole magnets in the so-called atom smashers used in high-
energy physics facilities. An ideal dipole magnet is studied in PROBLEM 3.8.
ELECTROMAGNETIC FIELDSPROBLEMS & DISCUSSIONS 43
3.31022 particle/cm3 = 3.31028 electron/m3
With v 200 m/s roughly equal to the drift velocity of superconducting electrons,
we obtain an order of magnitude for Js :
11012 A/m2
With m = 9.11031 kg and nse from Eq. S2.7, Eq. 1.1 becomes:
(9.11031 kg)
=
(4107 H/m)(1.61019 C)2 (3.31028 m3 )
3108 m
Because Kf = Jf :
2H0 2( H0 ) 2(0.08 T)
Jf = = = (S2.9)
(4107 H/m)(3108 m)
41012 A/m2
Far away from center, r , the eld outside the sphere, H 1 , is zero. Near the
sphere, the eld is derivable from a scalar potential, i.e., 1 = (A/r2 ) cos . The
operation on 1 leads to a dipole eld in spherical coordinates of the form:
1 = A (2 cos r + sin )
H (r R) (2.44b)
r3
At r = R, the normal component of B must be continuous. Because at = 0 there
is only a normal component of eld both inside and outside of the sphere:
A
H0 = 2
R3
Therefore A =H0 R3 /2, and thus:
3
1 = H0 R (cos r +
H 1
sin ) (2.44c)
r 2
= 32 H0 sin (2.45)
Note that the sin distribution could be approximated with a thin coil wound
on the surface of the sphere with a uniform turn density in the the z-axis direction.
Answer to TRIVIA 2.2 Cu: Latin cuprum, from Cyprus island; Lu: Lutetia,
ancient name of Paris; Mg: Magnesia, district in Thessaly, Greece; Y: Ytterby,
village in Sweden. Ytterby is also used for three other rare earth elements:
erbium (Er); terbium (Tb); and ytterbium (Yb).
ELECTROMAGNETIC FIELDSPROBLEMS & DISCUSSIONS 45
b) Show that in the limits of /
1 and d/R 1, the ratio Hss /H0 given by
Eq. 2.46 reduces to:
Hss 3 R
2 (2.47)
H0 d
c) Next, obtain Eq. 2.47 through a perturbation approach. First, solve the eld
in the shell (R d r R) with = . Then, use a perturbation approach
for the case /
1 and obtain Eq. 2.47.
d) In reality the magnetic ux in the shielding material must be kept below the
materials saturation ux, Msa . Show that an expression for d/R to keep
the shell unsaturated is given by:
d 3H0
(2.48)
R 2Msa
e) Draw eld lines for the case /
1.
x
r
z H0
R
A
1 = H0 r cos + cos (S3.1a)
r2
D
2 = Cr cos + cos (S3.1b)
r2
3 = Hss r cos (S3.1c)
Boundary Conditions
(H ) is continuous: 1 = 2 .
1) At r = R, the tangential ( ) component of H
2) Similarly, at r = Rd, H is continuous: 2 = 3 .
(Br ) is continuous.
3) At r = R, the normal (r ) component of B
4) Similarly, at r = Rd, Br is continuous.
The above boundary conditions give rise to the following four equations:
A D
H0 + 3
=C+ 3 (S3.3a)
R R
D
C+ = Hss (S3.3b)
(R d)3
2A 2D
H0 + 3 = C + 3 (S3.3c)
R R
2D
C + = Hss (S3.3d)
(R d)3
A 1 1
+ D Hss = H0 (S3.4)
R3 (R d)3 R3
ELECTROMAGNETIC FIELDSPROBLEMS & DISCUSSIONS 47
2A 1 1
+2 D 3 + Hss = H0 (S3.7)
R3 (R d)3 R
Combining Eqs. S3.4S3.7 and expressing Hss in terms of H0 , we obtain:
Hss 9
= 3 (2.46)
H0 d
9 + 2( ) 2
1 1
R
b) We may simplify Eq. 2.46 by dividing top and bottom by 2 and applying
the limits /
1 and d/R 1:
Hss 9/
2
(S3.8)
H0 d
9 +2 1 13
R
3
(S3.9)
d
3+2
R
c) The same result given by Eq. 2.47 for Hss /H0 can be obtained directly by a
perturbation approach for the case /
1 and d/R 1.
We proceed by assuming that of the shell material is innite. This requires that
the B lines be perpendicular to the shell at r = R. This is because H 1 has only
a radial (r ) component at r = R, because H = 0 in the shell and H must be
continuous at r = R. (This can be seen quite readily by noting that when = ,
C = D = 0.) From Eq. S3.3a, A = R3 H0 , and thus at r = R:
1 = 3H0 cos r
H (S3.10)
48 CHAPTER 2PROBLEMS & DISCUSSIONS
d
d
z R
Fig. 2.5 Flux entering into the spherical shell over the surface bounded by .
The B lines stay inside the shell without spilling into Region 3; that is, B inside
the shell has only an component. Let us now apply magnetic ux continuity,
= 0, and solve for B
i.e., B 2 when = . Once B 2 is solved for this case, an
approximate expression for H3 can be deduced for = but /
1.
First, we calculate the total magnetic ux entering into the shell over the surface
area bounded by (Fig. 2.5). This surface area, as indicated in the gure, is
given by a dierential area (a ring of radius R sin times Rd) integrated from 0
to . Thus, we have:
= 1 dA
H = 3H0 cos 2R2 sin d
0 0
2
= 3 R2 H0 sin (S3.11)
This must be equal to the total ux owing in the -direction in the shell at .
Because the shells cross-sectional area, A2 , at is given, for d R, by the shell
thickness d times the circumference of a ring of radius R sin , we have:
A2
d2R sin (S3.12)
We thus have:
= 3 R2 H0 sin2
B2 A2 = B2 d2R sin (S3.13)
ELECTROMAGNETIC FIELDSPROBLEMS & DISCUSSIONS 49
TRIVIA 2.3 The permanent magnet is ubiquitous in daily life (DISCUSSION 2.3 ).
From the gures below, choose the one that is closest to the number of permanent
magnets in a modern passenger car.
i) 20; ii) 100; iii) 500; iv) 2,500.
50 CHAPTER 2PROBLEMS & DISCUSSIONS
e) Field lines for the case / = 100 are shown below in Fig. 2.6.
Fig. 2.6 Field distribution with a spherical shell of / = 100 in a uniform eld. Note
that the eld lines entering and leaving the shell are nearly perpendicular to the shell.
ELECTROMAGNETIC FIELDSPROBLEMS & DISCUSSIONS 51
As in the spherical shell studied in PROBLEM 2.3, the cylindrical shell cannot be
arbitrarily thin; it must be thick enough to keep it from saturating:
R
Hcs = 2 H0 Msa
d
From the above, we obtain:
d 2H0
(2.50)
R Msa
where rj is measured from the center of each dipole and j in each dipole is dened
such that the eld inside the winding points in the rj -direction when j = 0 . Figure
2.7 indicates the direction of the eld inside each dipole. Also dened in Fig. 2.7
are r- coordinates and z-x coordinates common to all the dipoles. Note that for
r
d , we have 1 = +180 , 2 = 90 , 3 = , and 4 = +90 .
Show that an approximate expression for the far eld (B for r/d
1) of the
combined system is given by:
2
3rd d B d ( sin 2r + 1 cos 2 )
B (2.52)
r4 2
Neglect end eects of each dipole, i.e., consider only the plane y = 0.
2d 2rd z
2 4
3
r1
r d sin (S4.1a)
r2
r + d cos (S4.1b)
r3
r + d sin (S4.1c)
r4
r d cos (S4.1d)
With Eq. S4.1 into Eq. 2.51 for each dipole and j expressed in terms of :
1
rd2 d B
B ( cos r 1
sin ) (S4.2a)
2(r d sin )3 2
2
rd2 d B
B (sin r 1
2 cos ) (S4.2b)
2(r + d cos )3
3
rd2 d B 1
B (cos r + 2 sin ) (S4.2c)
2(r + d sin )3
4
rd2 d B 1
B ( sin r + cos ) (S4.2d)
2(r d cos )3 2
rd2 d B d
B1
1+3 sin ( cos r 12 sin ) (S4.3a)
2r3 r
rd2 d B d
B2
13 cos (sin r 12 cos ) (S4.3b)
2r3 r
2
3
rd d B 1 3 d sin (cos r + 1 sin )
B (S4.3c)
2r3 r 2
rd2 d B d
B4
1+3 cos ( sin r + 12 cos ) (S4.3d)
2r3 r
Combining each eld given by Eq. S4.3, we obtain:
2
=B
B 1 + B
2 + B 4
3rd d B d ( sin 2r +
3 + B 1
cos 2 ) (2.52)
r4 2
Answer to TRIVIA 2.3 ii). A modern military aircraft, however, may contain
up to 2,000 pieces of permanent magnetJ.D. Livingstone, Driving Force:
The Natural Magic of Magnets (Harvard University Press, 1996).
54 CHAPTER 2PROBLEMS & DISCUSSIONS
m (r, ) = mA cos r +
H 1
sin (2.53)
A 2
r3
From symmetry H m (r, ) is axisymmetric with respect to the z-axis and it is
A
derivable from a scalar potential cos /r2 . This derivation has been used in
PROBLEMS 2.1, 2.3, and 2.4, and DISCUSSION 2.1.
Hint Solve the z-axis center eld due to a single magnetic dipole moment, mA,
located on the pole perimeter at the boundary between the tapered and untapered
pole, as indicated by in Fig. 2.8.
z
r
tp
2R1
mA
2R2
y
(t)
H
R
x
Fig. 2.9 Axial view of a long thin cylinder of radius R of perfect conductor
with a narrow gap of at = 0 exposed to a sinusoidally time-varying
magnetic eld in the z-direction.
ELECTROMAGNETIC FIELDSPROBLEMS & DISCUSSIONS 57
Equations 2.55 and S6.3 demonstrate that voltages across the cylinder at the
same axial location depend on azimuthal locations of the voltage taps. This is an
important point to be remembered when measuring voltage in the presence of a
time-varying magnetic eld, applied externally as in the present case or generated
by a current owing in the system. Electrical measurement of AC losses in a
superconductor is a good example where one must be careful.
(t)
H
1
E 1
E 1
E 1
E
R x=0 R
r
H0
x
R
H0 d
1 = j rH0
E (2.56)
2
j RH0
J1 (2.57)
2e
b) Show that the resulting 1st -order magnetic eld, H 1 , in the region r R d
can be expressed by:
1 = j RdH0 z
H (2.58)
2e
Method 2
E 1 , J1 , and H
1 , derived by Method 1, each increasing with , are valid only for
frequencies well below fsk . We now demonstrate a new technique that enables us
to derive the total eld, H T = (H
0 + H R ), in the bore, valid for the entire range
of frequencies. HT is the total net eld, H 0 is the original eld, and H R is the
reaction eld of the system in the bore. In this approach, rst nd the reaction
eld H R in the bore by treating H T = (H0 +H R ) as a 0th -order eld and solve for
H R as the usual 1st -order magnetic eld response.
R, H
d) Show that expressions for H T , and J in the shell valid for d R are:
R = j RdH0 z
H (2.60)
2e + j Rd
T = 2e H0
H z (2.61)
2e + j Rd
j RH0
J = (2.62)
2e + j Rd
For r R,
E1 2r = j H0 r2 (S7.2)
1 (r R)
E j RH0
J1 = (2.57)
e 2e
1 , by multi-
For d R, we may treat the current as a 1st -order surface current, K
plying J1 by d:
1 = j RdH0
K (S7.4)
2e
1 = r [H
K 0 (H 1 )] = j RdH0
0 + H
2e
1 = j RdH0
= r H (S7.5)
2e
1 (r R d) with d R, we have:
Solving Eq. S7.5 for H
1 = j RdH0 z
H (2.58)
2e
ELECTROMAGNETIC FIELDSPROBLEMS & DISCUSSIONS 61
1| = Rd|H0 | 0|
|H |H (S7.6)
2e
From Eq. S7.6, we can obtain the frequency limit, often called the skin-depth
frequency, fsk , below which the quasi-static approximation is valid:
e
fsk = (2.59)
Rd
Note that fsk depends not only on the metals electrical resistivity but also on the
size of the object in the sinusoidally time-varying applied magnetic eld.
d) In the second approach for computing the shells reaction eld, we set H 1
HR , and substitute H0 + HR for H0 in the expression for H1 given in Eq. 2.58:
R = j Rd(H0 + HR )
H (S7.7)
2e
R , we obtain:
Solving Eq. S7.7 for H
R = j RdH0 z
H (2.60)
2e + j Rd
J and H
R are related by H
= J,
which, for K
= Jd,
reduces to:
1
J = HR (S7.8)
d
j RH0
= (2.62)
2e + j Rd
Note that in the low frequency limit, H R given by Eq. 2.60 reduces, as expected,
1 given by Eq. 2.58. In the high frequency limit, H
to H R reduces, also as expected,
to H0 and HT becomes 0. Similar observations apply to J.
62 CHAPTER 2PROBLEMS & DISCUSSIONS
e 2 2 R3 d
= |H0 |2 (2.63)
42e + 2 2 R2 d2
Method 2
The same complex power supplied to the cylinder may also be viewed as a ow of
Poynting power ux entering the cylinder at r = R from a source located at r > R.
f ) Show that the surface integral (per unit cylinder length) of the 1st -order
1 entering into the cylinder at r = R is given by:
complex Poynting vector S
S 1 dA = 1 (2R)E1 H0
2
S
je R2
= |H0 |2 (2.64)
2e + j Rd
Rd
ec = (2.65)
2
From Eq. 2.65, we note that for a given combination of resistivity (ec ) and
sample size (R, d), there exists an optimal frequency that maximizes the
heating: this is equal to the skin depth frequency, fsk , given by Eq. 2.59.
j) Compute fsk for a copper tube of 10-mm radius (R), 0.5-mm wall thickness
(d), and e = 2 1010 m (roughly the electrical resistivity of copper at
liquid helium temperatures).
ELECTROMAGNETIC FIELDSPROBLEMS & DISCUSSIONS 63
The time-averaged total dissipation power (per unit length) in the shell, < P >, is
obtained by multiplying < p > by the cross sectional area of the shell:
e 2 2 R3 d
= |H0 |2 (2.63)
42e + 2 2 R2 d2
e R
e Rd : < P > |H0 |2 e (excellent conductor) (S7.10a)
d
2 2 R3 d 1
e
Rd : < P > |H0 |2 (poor conductor) (S7.10b)
4e e
1 is given by:
where from Eq. 2.62, E
1 = e J = je RH0
E (S7.13)
2e + j Rd
We thus have:
1 dA = 1 (2R)E1 H
S 2 0
S
je R2
= |H0 |2 (2.64)
2e + j Rd
64 CHAPTER 2PROBLEMS & DISCUSSIONS
21010 m
= 10 Hz
(4107 H/m)(0.010 m)(0.0005 m)
<P >
1
e
e
e
0 ec
Fig. 2.12 Power dissipation vs. resistivity for induction-heated cylindrical shell.
b) Show that the spatially-averaged power dissipation density p (over the unit
strip volume) can be expressed by:
(bB0 )2
p = (2.67)
12e
y
b B0
2
2b
Answer to TRIVIA 2.4: iv). The magnetostriction eect of a magnetic eld on iron
causes cyclic changes in the iron sheet dimensions, creating hum of frequency twice
that of the current. For 60-Hz current the hum is about an octave below middle C.
E1x = y B0 (2.66)
b) The local power density dissipated in the strip, p(y), is given by E 1 J1 . The
total power dissipation (per unit strip length), P , is thus given by:
b/2 b/2
2a(B0 )2 ab(bB0 )2
P =a p(y) dy = y 2 dy = (S8.2)
b/2 e 0 12e
P (bB0 )2
p = = (2.67)
ab 12e
Note that p and < p > are proportional, respectively, to (bB0 )2 and (b B0 )2 ; i.e.,
both depend not only on the square of time rate of change of magnetic induction
but also on the square of conductor width.
N c2
(t) I(t) (2.69)
2R
where 1 (t) is the total ux linked to one loop.
b) Show that an exact expression for (t) is given by:
(t) = N R R2 c2 I(t) (2.70)
Place one of the N loops centered on the x-y coordinates to compute (t).
c) Show that in the limit (c/R)4 1, Eq. 2.70 reduces to Eq. 2.69.
d) Show that (t) given by Eq. 2.70 is valid even if the Rogowski coils axis is
o-center of the current axis.
e) For a Rogowski coil with N = 3600; c = 3 mm; R = 0.5 m, compute the
volt-seconds generated across its terminal for I(t) = 1 MA.
N loops
y
R
c
I(t) x
(a) (b)
Fig. 2.14 (a) Rogowski coil comprised of N circular loops, each of diam-
eter 2c, encircling a time-varying current, I(t), to be measured; (b) Cross-
sectional view of one loop (radius c), centered on the x-y coordinates, with
the loop center located radially R from the center of the current.
68 CHAPTER 2PROBLEMS & DISCUSSIONS
I(t) c
c2 y 2
= dy (S9.3)
c R+y
Equation S9.3 may be solved in a closed form with a new variable, R+y (note
that d = dy). Thus, we have:
I(t) R+c c2 R2 + 2R 2
1 (t) = d
Rc
= R R2 c2 I(t) (S9.4)
The total ux linked to a Rogowski coil having N loops is given by (t) = N 1 (t):
(t) = N R R2 c2 I(t) (2.70)
1 c 1 c
(t) N R R 1 2 2 + 8 4 I(t) (S9.6)
R R
N c2
(t) I(t) (2.69)
2R
ELECTROMAGNETIC FIELDSPROBLEMS & DISCUSSIONS 69
Point A
2c
R r
x
(0. 0) s
i
Current
Fig. 2.15 Cross sectional view in the plane normal (x, y) to the azimuthal
(current) direction of a current/Rogowski coil set with the Rogowski coil at the
center (0, 0) of the plane and the current center o downward by distance i .
70 CHAPTER 2PROBLEMS & DISCUSSIONS
N I(t) R+c
c2 (r R)2
= 0+ dr (S9.11b)
Rc r
Note that the integration remarkably eliminates i . Now, Eq. S9.11b is integrated
over the radial extent of one loop, from r = Rc to r = R+c, resulting in:
N I(t)
(t) = (R R2 c2 ) (S9.11c)
Thus, a Rogowski coil measures I(t) accurately regardless of its concentricity to
the current that the coil encircles.
e) For N = 3600, c = 3 mm; R = 0.5 m; and I = 1 MA, Eq. 2.69 is applicable
because (c/R)4 = 1.3109 1. Thus, from Eq. S9.2:
N c2 I
V (t) dt =
2R
(4107 H/m)(3600)(3103 m)2 (1106 A)
=
2(0.5 m)
41 mV s
. . .as we know, there are known knowns; these are things we know we
know. We also know there are known unknowns; that is to say we know there
are some things we do not know. But there are also unknown unknowns
the ones we dont know we dont know. Donald Rumsfeld (2002)
At the present time, eld and force computations are generally performed with
computer codes that for a given magnet conguration give accurate numerical so-
lutions at any location. These codes can also compute the self and mutual induc-
tances of coils comprising the magnet and Lorentz forces acting on the coils [3.1].
Analytical expressions derived in this chapter give eld values only at specialized
locations such as the magnet center; however, they illustrate subtle relationships
among elds, forces, and magnet parameters.
In this introductory section, we rst study the law of Biot-Savart that is basic to
computation of a magnetic eld generated by a current-carrying element in the
absence of magnetic materials. Also presented in this section are rather extensive
treatments of: 1) eld analysis; 2) axial forces for rings and thin solenoids; 3)
stresses and strains in solenoids; and 4) self and mutual inductances.
z=0
a
With cos = a/r and r2 = a2 +z 2 , Hz (z, ) on the z-axis ( = 0), Hz (z, 0), becomes:
a2 I a2 I
Hz (z, 0) = 3 = (3.3a)
2r 2(a2 + z 2 )3/2
In terms of the center (z = 0, = 0) eld, Hz (0, 0), Hz (z, 0) may be given by:
Hz (0, 0)
Hz (z, 0) = (3.3b)
[1 + (z/a)2 ]3/2
Equation 3.3a may be used to derive an expression for the eld along the axis of a
solenoid of arbitrary winding cross section having any current distribution that is
invariant in the -direction. PROBLEM 3.1 and DISCUSSION 3.1 are good exam-
ples. From Eq. 3.3b, we note that for z a or far from the center, the axial eld
of a ring coil decreases as 1/(z/a)3 this is further studied in PROBLEM 3.11.
a2 K dz
Bz = = K B (3.5)
2 (a2 + z 2 )3/2
The integration is taken over the entire range of z because the current rings
extend from z = to z = . The application of Amperes law gives that,
for this innitely long solenoid, B outside the solenoid (r > a) is zero and that
within the solenoid bore (r < a) is B , uniform in both z- and r-directions. (From
symmetry of the current distribution, the eld is also independent of . Also note
that Eq. 3.3a is valid only for such cases.) That is, the magnetic induction within
the bore of an innitely long solenoid is completely uniform and directed only in
the z-direction. A more detailed discussion of axial eld analysis is given next.
The Bz eld just inside the winding is B and that just outside the winding is zero,
decreasing linearly with r over the winding thickness . The average induction,
Bz , to which the current element in the winding is exposed, is thus B /2, resulting
in an r-directed average Lorentz force density fLrr acting on the winding:
K K B
fLrr = Bzr = r (3.6)
2
An r-directed Lorentz force, FLrr acting on the winding volume element, dened
in Fig. 3.2, is equivalent to a magnetic pressure, pmr , acting on the winding
surface element, also dened in Fig. 3.2. Thus:
B2
pm = (3.8)
2
That is, the magnetic pressure is equal to the magnetic energy density. For B
equal to 1 T, Eq. 3.8 gives a magnetic pressure of 3.98105 Pa or 4 atm, from
which it follows that for B = 50 T, a magnetic pressure of 1 GPa is reached.
pm fLr
Fig. 3.2 Axial view of a dierential element for a thin-walled solenoid (thickness )
of average diameter 2a. The element is z high in the z-direction (out of the paper).
74 CHAPTER 3
2a2
2a1
2b
Fig. 3.3 Cross sectional view of a solenoidal coil of winding i.d., o.d., and length
(width), respectively, of 2a1 , 2a2 , and 2b. Field lines are drawn to indicate that
the magnetic eld generated by the coil is chiey axial in the coil bore, diverging,
except along the axial midplane, to the radial direction outside the bore.
MAGNETS, FIELDS, AND FORCES 75
For solenoidal systems for which all current densities are invariant in , i.e., sym-
metric with respect to the axis, only the m = 0 terms remain. Along the z-axis
(r = z, = 0), we may simplify Eq. 3.9 for Hz (z) to:
Hz (z) = z n (n + 1)A0n (3.10a)
n=0
Along the x- (or y-) axis (x, = 90 ), at midplane, Eq. 3.9 for solenoids becomes:
n
Hz (x) = xn (n + m + 1)Pnm (0)Am
n (3.10b)
n=0 m=0
Equation 3.10a has no Legendre functions because Pn0 (1) = Pn (1) = 1. For Eq. 3.10b
note that Pnm (0) = 0 when n is even and m is odd. Table 2.3 in CHAPTER 2 gives
values of Pnm (0) for even numbers of m up to m = 10. Using values of Pnm (0) for n
and m = 0, 2, and 4, we may express Hz (z) and Hz (x) in Cartesian coordinates:
A00 is the eld at the magnet center (0,0,0): A00 H0 . For an ideal solenoidal
coil, coecients are zero for m > 0: Am
n = 0 for m > 0. Coecients for n > 0, m = 0,
A0n , are functions of coil parametersspecically, and . Introducing hz ()
Hz (z)/H0 , where z/a1 , and hz () Hz (x)/H0 , where x/a1 , we obtain:
hz () = 1 + E2 (, ) 2 + E4 (, ) 4
+ E6 (, ) 6 + E8 (, ) 8 + E10 (, ) 10 + (3.12a)
hz () = 1 12 E2 (, ) 2 + 38 E4 (, ) 4
5
16 E6 (, )
6
+ 35
128 E8 (, )
8
63
256 E10 (, )
10
+ (3.12b)
76 CHAPTER 3
Note that the 2 term coecient is only half (and of opposite sign) of the 2
coecient, and the 4 term is 3/8 of the 4 term and of the same sign. Indeed,
every coecient in the plane direction is numerically smaller than that in the z-
direction, so that eld inhomogeneity tends to be smaller in the x- and y-directions
than that in the z-direction.
The center eld H0 ( A00 ) is given by:
+ 2 + 2
H0 = Ja1 F (, ) = Ja1 ln (3.13a)
1+ 1+ 2
From Eq. 3.13a, we note that:
+ 2 + 2
F (, ) = ln (3.13b)
1+ 1+ 2
This concise format is used below for terms of the 6th , 8th , and 10th order:
8 r=
r3 8r +44r6 2 +99r4 4 +28r2 6 +280 8
F (, )E6 (, ) =
240 5 (r2 + 2 )5.5 r=1
(3.14c)
r=
16r12 +120r10 2 +390r8 4 +715r6 6
r3 +1080r4 8 1008r2 10 +1344 12
F (, )E8 (, ) = 7
896 (r2 + 2 )7.5 r=1
(3.14d)
F (, )E10 (, ) =
r=
128r16 +1216r14 2 +5168r12 4 +12920r10 6 +20995r8 8
r3 +19976r6 10 +49632r4 12 46464r2 14 +21120 16
11520 9 (r2 + 2 )9.5 r=1
(3.14e)
MAGNETS, FIELDS, AND FORCES 77
In general:
2k hz ()
= (2n)(2n1)( )(2n2k+1)E2n (, ) 2(nk) (3.16b)
2k
n=k
At the origin, = 0, Eqs. 3.16a and 3.16b reduce to simpler expressions because
only the rst term is nonzero. Thus:
2 hz ()
= 2E2 (, ) (3.17a)
2 0
2k hz ()
= (2k)! E2k (, ) (3.17b)
2k 0
Nested-Coil Magnet
For a magnet comprised of
nested (coaxial and concentric) coils, each of J, a1 ,
, , Eq. 3.12a, given only up to the 2 term, is generalized:
j=1 (J)j a1j F (j , j )E2 (j , j ) 2
hz () = 1 + + (3.18)
j=1 (J)j a1j F (j , j )
Before discussing in 3.4.2 errors arising in a two-coil nested magnet, we shall rst
consider a few special cases for single solenoidal coils.
78 CHAPTER 3
Short Coil
For a short coil ( 0), such as a pancake, F (, ) may be simplied as:
F (, 0) = ln (3.19)
Although extremely tedious, expressions for E2 (, 0), . . . , E10 (, 0) may be derived
from Eq. 3.14. In the limit 0, the denominator of each term in fn (, ), e.g.,
(1+ 2 )n0.5 , may be expanded in a power series of 2k up to the n term, where
the integer k runs from 1 to n/2.
1 2 4
= 1(n0.5) +(n0.5)(n+0.5)
(1+ 2 )n0.5 2!
6
(n0.5)(n+0.5)(n+1.5) +
3!
2k
(1)k (n0.5)(n+0.5)(n+1.5) (n+k1.5) (3.20)
k!
In the limit of 0, the terms higher than n in the right-hand side of Eq. 3.20
become negligible compared with n . Also, remarkably, all the terms below n
cancel out, leaving only the n term in the numerator within the brackets of the
right-hand side of Eq. 3.14 with a numerical coecient, e.g., 3/4 for n = 2.
En (, 0) terms, derivable from Eqs. 3.14 and 3.19, may be given by:
3 (2 1) 3(2 1)
E2 (, 0) = = (3.21a)
22 2 ln 42 ln
35 (4 1) 15(4 1)
E4 (, 0) = = (3.21b)
25 4 ln 324 ln
57 (6 1) 35(6 1)
E6 (, 0) = = (3.21c)
25 3 6 ln 966 ln
579 (8 1) 315(8 1)
E8 (, 0) = = (3.21d)
210 8 ln 10248 ln
7911 (10 1) 693(10 1)
E10 (, 0) = = (3.21e)
29 5 10 ln 256010 ln
Thus, Eq. 3.12a becomes:
3(2 1) 2 15(4 1) 4 35(6 1) 6
hz () = 1 +
42 ln 324 ln 966 ln
315(8 1) 8 693(10 1) 10
+ + (3.22)
10248 ln 256010 ln
MAGNETS, FIELDS, AND FORCES 79
Thin-Walled Coil
For a thin-walled coil ( = 1), F (, ) becomes:
lim F (, ) = (3.23a)
1 1 + 2
( 1)
= (3.23b)
1 + 2
Combining Eqs. 3.14 and 3.23, we may obtain expressions for En (1, ):
3
E2 (1, ) = (3.24a)
2(1+ 2 )2
5(34 2 )
E4 (1, ) = (3.24b)
23 (1+ 2 )4
7(520 2 +8 4 )
E6 (1, ) = (3.24c)
24 (1+ 2 )6
9(35280 2 +336 4 64 6 )
E8 (1, ) = (3.24d)
27 (1+ 2 )8
3 5(34 2 ) 4 7(520 2 +8 4 ) 6
hz () = 1 2
+
2(1+ 2 )2 8(1+ 2 )4 16(1+ 2 )6
9(35280 2 +336 4 64 6 ) 8
+
128(1+ 2 )8
As may be inferred from Eq. 3.26, note that in the limiting case of ,
En (1, ) = (n+1)/2 (n+2) . As expected, homogeneity improves with coil length.
80 CHAPTER 3
Ring Coil
For a ring coil ( = 1, = 0), En (, ) is derivable from Eq. 3.21 in the limit of
= 1, or from Eq. 3.25 in the limit = 0. In Eq. 3.21,
n 1
lim =n (3.27)
1 n ln
Thus, we may derive the same expressions by combining Eqs. 3.21 and 3.27 or
simply inserting = 0 in each expression of Eq. 3.24. Either way, we obtain:
3
E2 (1, 0) = = 1.5 (3.28a)
2
35 35
E4 (1, 0) = = = 1.875 (3.28b)
23 24
57 357
E6 (1, 0) = = 2.188 (3.28c)
24 246
579 3579
E8 (1, 0) = = 2.461 (3.28d)
27 2468
7911 357911
E10 (1, 0) = 8
= 2.707 (3.28e)
2 246810
Thus, for a ring coil, hz () is given by:
hz () = 1 32 2 + 15 4
8 35 6
16 + 315 8
128 693 10
256 + (3.29)
Values of E12 (1, 0), E14 (1, 0), E16 (1, 0), E18 (1, 0), and E20 (1, 0) are given, respec-
tively, by Eqs. 3.124a, 3.124b, 3.124c, 3.124d, and 3.124e in PROBLEM 3.4 (p. 140).
where [Hz (z)]1 and [Hz (z)]2 may be given, from Eq. 3.11a:
Note that the total axial eld at the center: Hz (0) = [A00 ]1 + [A00 ]2 H0 .
{6[A02 ]2 z + 20[A04 ]2 z3 + }z
{20[A04 ]2 z + }z 3
+ {5[A04 ]1 + 5[A04 ]2 }z 4 (3.32b)
As may be observed from Eq. 3.32b an axial misalignment of Coils 1 and 2 by z not
only perturbs the coecient of each even-numbered z-term but, more importantly,
gives rise to odd-numbered z-terms which make Hz (z) axially asymmetric.
82 CHAPTER 3
FzA () is +z-directed or towards Coil B; i.e., FzA () is attractive. If the two cur-
rents ow in opposite directions, the sign changes to minus, and FzA () becomes
repulsive. In Eq. 3.34, K(k) and E(k) are the complete elliptic integrals, respec-
tively, of the rst and second kinds, dened below.
/2
d
K(k) = (3.35a)
0 1 k 2 sin2
/2
E(k) = 1 k 2 sin2 d (3.35b)
0
The modulus, k, of the elliptic integrals for this system is given by:
4aA aB
k2 = (3.36)
(aA + aB )2 + 2
2aA 2aB
z
Fig. 3.5 Ring coils A and B coaxially aligned and separated by a distance .
84 CHAPTER 3
Table 3.1: Complete Elliptic Integrals of the First and Second Kinds:
K(k) and E(k) for Selected Values of k2 and k
k2 k K(k) E(k) k2 k K(k) E(k)
0 0 /2 /2 0.7 0.8367 2.0754 1.2417
0.1 0.3162 1.6124 1.5308 0.8 0.8944 2.2572 1.1785
0.2 0.4472 1.6596 1.4890 0.90 0.9487 2.5781 1.1048
0.3 0.5477 1.7139 1.4454 0.95 0.9747 2.9083 1.0605
0.4 0.6325 1.7775 1.3994 0.98 0.9899 3.3541 1.0286
0.5 0.7071 1.8541 1.3506 0.99 0.9950 3.6956 1.0160
0.6 0.7746 1.9496 1.2984 1 1 1
Table 3.1 gives K(k) and E(k) for selected values of k 2 and k. Note that K(0) =
E(0) = /2; also obvious are K(1) = and E(1) = 1, i.e., K(k) increases with k,
whereas E(k) decreases with k. For example, if k 2 = 0.5, K(k = 0.7071) = 1.8541.
K(k) and E(k) may be expressed by power series in k 2 :
2 2 2 2
1 2 13 4 135 6 1357 8
K(k) = 1+ k + k + k + k + (3.37a)
2 2 24 246 2468
2 2 4 2 2
1 2 13 k 135 k 6 1357 k 8
E(k) = 1 k (3.37b)
2 2 24 3 246 5 2468 7
For k 2 1, the two integrals and their dierence may be approximated by:
K(k) 1 + 14 k 2 + 64
9 4 25 6
k + 256 1225 8
k + 16384 k (3.38a)
2
E(k) 1 14 k 2 64 k 256
3 4 5
k 6 16384
175
k8 (3.38b)
2
2 3 4 15 6
K(k) E(k) k + 8 k + 64 k + 1024 175 8
k (3.38c)
4
TRIVIA 3.1 A legend has it that one of the mathematicians below advanced
his argument for the existence of God by using an irrelevant algebraic equation:
Sir, (a+b)n/n = x; hence God exists, answer please! Who (and to whom)?
i) Euler; ii) Gauss; iii) Laplace; iv) Legendre.
86 CHAPTER 3
The elliptic integral moduli, kb , kb+ , and k2b , are given by:
4a2 4a2 4a2
kb2 = ; kb2+ = ; 2
k2b =
4a2 + (b z)2 4a2 + (b + z)2 4a2 + 4b2
(aR +aS )2 +2 2 K(kR )E(kR ) kR2 K(kR ) (3.42)
NR IR
NS IS
2aR
2aS z
2bS
Fig. 3.6 Thin-walled solenoid and a ring coil coaxially aligned and separated by .
88 CHAPTER 3
where aT = aA +aB and bT = bA +bB . The moduli, kA , kB , kT , and k , are given by:
4aA aB 4aA aB
kA2 = ; kB2 =
a2
T + (2bA +)2 a2 +(2b
T B + )
2
4aA aB 4aA aB
kT2 = 2 ; k2 =
aT +(2bT + )2 aT2 + 2
NB IB
NA IA
2aA 2aB
2bB
2bA
Fig. 3.7 Thin-walled Solenoids, A and B, with the two nearest ends separated by .
MAGNETS, FIELDS, AND FORCES 89
(c2, k) is the complete elliptic integral of the third kind, dened by:
/2
d
(c2, k) = (3.46)
0 (1 c2 sin2 ) 1 k 2 sin2
As evident from Eq. 3.46, (c2, k) has two moduli, c2 1 and k 1. The modulus
c2 in the cases considered here is given by:
4aA aB 4aA aB
c2 = = (3.47)
aT2 (aA + aB )2
(0, k) = K(k) and (1, k) = . (c2, k) may be expressed in a power series in c2
and k 2 :
m
(2m)! (2j)! k 2j c2(mj)
(c2, k) = (3.48)
2 m=0 j=0 4m 4j (m!)2 (j!)2
Low-order terms are:
(c2, k) = 1+ 12 c2 + 14 k 2 + 38 c4 + 16
3 2 2 9 4
c k + 64 k+
2
5 6
+ 16 5 4 2
c + 32 15 2 4
c k + 128 c k + 256 k +
25 6
(3.49a)
(c2, 0) = 1+ 12 c2 + 233 c4 + 254 c6 + 57 8 79 10 3711 12
27 c + 28 c + 210 c
2
+ 31113
211 c14
+ 591113 16
215 c + 5111317 18
216 c + 11131719 20
218 c + (3.49b)
Note that when c2 = 0, Eq. 3.49a becomes identical to Eq. 3.38a, because as noted
above, (0, k) = K(k). However, for most problems of interest c2 is generally
close to 1, in which case neither expansion of Eq. 3.49 is very useful, since rapid
convergence requires c2 1. Later, in discussing mutual inductance, we shall use
Eq. 3.49b. Table 3.2 gives (c2, k) for selected values of c2 and k 2 . Mathcad and
a number of other software programs can evaluate K(k), E(k), and (c2, k).
c2 k=0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 0.999
0.1 1.6558 1.6600 1.6732 1.6961 1.7307 1.7803 1.8509 1.9541 2.1173 2.4295 4.8804
0.2 1.7562 1.7609 1.7752 1.8002 1.8380 1.8923 1.9696 2.0829 2.2625 2.6077 5.3514
0.4 2.0279 2.0336 2.0513 2.0822 2.1290 2.1963 2.2925 2.4343 2.6604 3.1001 6.7100
0.6 2.4836 2.4913 2.5148 2.5561 2.6187 2.7090 2.8389 3.0315 3.3418 3.9550 9.2511
0.8 3.5124 3.5246 3.5622 3.6283 3.7290 3.8751 4.0867 4.4042 4.9246 5.9821 16.070
0.9 4.9673 4.9863 5.0448 5.1480 5.3056 5.5355 5.8710 6.3796 7.2263 8.9943 27.895
0.92 5.5536 5.5754 5.6426 5.7610 5.9421 6.2069 6.5939 7.1824 8.1667 10.239 33.280
0.94 6.4127 6.4387 6.5186 6.6597 6.8758 7.1921 7.6557 8.3633 9.5535 12.086 41.737
0.95 7.0248 7.0537 7.1429 7.3002 7.5414 7.8948 8.4135 9.2071 10.547 13.414 48.138
0.96 7.8540 7.8869 7.9886 8.1681 8.4435 8.8475 9.4417 10.353 11.897 15.227 57.268
0.97 9.0690 9.1079 9.2280 9.4403 9.7662 10.245 10.951 12.036 13.884 17.905 71.507
0.98 11.107 11.156 11.308 11.575 11.986 12.592 13.486 14.867 17.233 22.440 97.397
0.99 15.708 15.780 16.002 16.395 17.001 17.894 19.219 21.277 24.832 32.789 163.12
0.991 16.558 16.634 16.869 17.286 17.927 18.874 20.279 22.462 26.239 34.711 176.15
0.992 17.562 17.643 17.894 18.338 19.022 20.033 21.532 23.864 27.903 36.986 191.85
0.993 18.775 18.862 19.131 19.609 20.345 21.431 23.045 25.557 29.914 39.735 211.22
0.994 20.279 20.374 20.667 21.185 21.985 23.167 24.923 27.658 32.409 43.151 235.80
0.995 22.214 22.319 22.642 23.214 24.096 25.400 27.339 30.362 35.623 47.552 268.26
0.996 24.836 24.954 25.318 25.962 26.956 28.426 30.613 34.027 39.978 53.523 313.54
0.997 28.679 28.816 29.239 29.989 31.147 32.860 35.412 39.400 46.365 62.286 382.16
0.998 35.124 35.294 35.817 36.745 38.178 40.300 43.464 48.416 57.088 77.009 502.08
0.999 49.673 49.916 50.666 51.996 54.050 57.096 61.643 68.776 81.309 110.30 787.66
where aT = aA +aB and the moduli k2b and k4b are given by:
2 4aA aB 2 4aA aB
k2b = 2 2
; k4b = 2
aT + 4b aT + 16b2
Because c2 1, though always c2 < 1, for most problems of our interest, Eq. 3.50
cannot be approximated even for long coils ( 1).
Derivation of Eq. 3.50 from Eq. 3.44
If we make Solenoids A and B identical in diameter and surface current density but
each one half the length of the original, for = 0 the two solenoids are transformed
into one solenoid of length 2b. Next, if the following substitutions, i.e., 2aA = 2aB =
2a, NA IA /2bA = NB IB /2bB = N I/2b, and 2bA = 2bB = b, are made in Eq. 3.44, the
modied equation is reduced to Eq. 3.50.
Answer to TRIVIA 3.1 i). The Swiss mathematician Leonhard Euler (17071783)
to the French encyclopedist Denis Diderot (17131784). This anecdote is disputed
by Dirk Struik, who states in A Concise History of Mathematics (Dover Publica-
tions, 1948) that . . . no reason exists to think that the thoughtful Euler would
have behaved in the asinine way. The great Pierre Simon de Laplace (17491827),
once asked by Napoleon why his monumental ve-volume Celestial Mechanics had
no mention of God, replied, I had no need of that hypothesis.
MAGNETS, FIELDS, AND FORCES 91
2b
[aT2 +4b2 ] K(k2b )E(k2b ) (c2, k2b ) (3.51)
aT2 +4b2
where aT = aA +aB . The moduli, kb and k2b , are given by:
4aA aB 4aA aB
kb2 = ; 2
k2b =
aT2 + b2 a2
T + 4b
2
FzAA (0), the midplane axial force by Subsolenoid A on itself, is given by Eq. 3.41a
with (N I/4b) in place of (N I/2b). Similarly, FzB (0) is the total midplane axial
force on Subsolenoid B comprised of FzBB (0) and FzBA (0):
Solenoid B
(N I)/2
2aA 2aB
b
Solenoid A
(N I)/2
2b
Thus:
2
N I
FzA (0) =
2 4b
2b 4aA2 +b2 K(kbA)E(kbA) 2b 4aA2 +4b2 K(k2bA )E(k2bA )
2b
+ [aT2 +b2 ] K(kb )E(kb ) (c2, kb )
aT2 +b2
2b
[aT2 +4b2 ] K(k2b )E(k2b ) (c2, k2b ) (3.53a)
aT2 +4b2
Again c2 is given by Eq. 3.47. The moduli kbA , k2bA , kbB , and k2bB are given by:
In this case, because 2bA = 2bB = 2b and NA IA = NB IB , FzAB (0) = FzBA (0). Thus:
2
N I
FzB (0) =
2 4b
2b 4aB2 +b2 K(kbB )E(kbB ) 2b 4aB2 +4b2 K(k2bB )E(k2bB )
2b
+ (aT2 +b2 ) K(kb )E(kb ) (c2, kb )
aT2 +b2
2b
(aT2 +4b2 ) K(k2b )E(k2b ) (c2, k2b ) (3.53b)
aT2 +4b2
The total midplane axial force of a solenoid FzT (0), divided into two thin-walled
subsolenoids, is the sum of FzA (0) and FzB (0). We thus obtain:
2
N I
FzT (0) =
2 4b
2b 4aA2 +b2 K(kbA )E(kbA ) 2b 4aA2 +4b2 K(k2bA )E(k2bA )
+ 2b 4aB2 +b2 K(kbB )E(kbB ) 2b 4aB2 +4b2 K(k2bB )E(k2bB )
4b
+ (aT2 +b2 ) K(kb )E(kb ) (c2, kb )
aT2 +b2
4b
(aT +4b ) K(k2b )E(k2b ) (c , k2b )
2 2 2
(3.54)
aT2 +4b2
MAGNETS, FIELDS, AND FORCES 93
Note that when a solenoid is divided into two thin-walled subsolenoids, FzT (0)
requires computation of four terms in Eq. 3.54. When a solenoid is divided into
m > 2 thin-walled subsolenoids, it will require computation of 2(m!)/[(m 2)!]
terms to obtain FzT (0). For m = 3 an expression of FzT (0) thus will contain twelve
terms and 18 moduli, clearly too tedious for hand computation.
To create a computer code that will accurately compute the midplane axial force
of a practical solenoidal coil, i.e., a thick-walled solenoid, one has to extend
Eq. 3.54 to a set of expressions applicable to m thin-walled subsolenoids. To
ensure that each subsolenoid is thin-walled, m may have to be 10 or even greater.
Note that because in most cases c2 is close to 1, it is not possible to approximate
the (c2, k) term contained in each (c2, k) even for long ( 1) solenoids.
Note that when aA = aB , Eq. 3.55 reduces to Eq. 3.41b. In the second form of
Eq. 3.55, the last term within the braces may be considered a correction term.
NA IA NB IB 2b
+ B [aT2 +bB2 ] K(kB )E(kB ) (c2, kB )
2bA 2bB aT2 +bB2
bD
[aT2 +bD2 ] K(kD )E(kD ) (c2, kD )
aT2 +bD2
bT
[aT +bT ] K(kT )E(kT ) (c , kT )
2 2 2
aT2 +bT2
(3.56)
Note that aT = aA +aB , bD = bB bA , and bT = bA +bB . kbA , k2bA , kA , kB , and kAB are:
4aA2 4aA2 4aA aB 4aA aB 4aA aB
kb2A = ; 2
k2b = ; kB2 = ; kD2 = ; kT2 =
4aA2 + bA2 A
4aA2 + 4bA2 aT2 + bB2 aT2 + bD2 aT2 + bT2
Solenoid B
NB IB
2aA 2aB
Solenoid A
NA IA 2bA
2bB
The second term in the right-hand side of Eq. 3.57a represents FzAB . Physically
this is due to Solenoid Bs Br on Solenoid A at the axial location near each end of
Solenoid B. Note that as bA and bB , the second term becomes zero.
NB IB NA IA 2b
+ A [aT2 +bA2 ] K(kA )E(kA ) (c2, kA )
2bB 2bA aT2 +bA2
bD
+ [aT2 +bD2 ] K(kD )E(kD ) (c2, kD )
aT2 +bD2
bT
[aT2 +bT2 ] K(kT )E(kT ) (c2, kT )
aT2 +bT2
(3.58)
96 CHAPTER 3
Here too c2 is given by Eq. 3.47. The moduli k2bB and kA are given by:
2 aB2 4aA aB
k2b = ; kA2 =
B
4aB2 + 4bB2 aT2 + bA2
The other moduli are the same as those given above for Eq. 3.56.
As in Special Case 11, the second term in Eq. 3.59a represents FzBA , which is
nonnegligible because Solenoid As Br impinges on Solenoid B at the axial location
near each end of Solenoid A.
bD +
+ [aT2 +(bD +)2 ] K(kD + )E(kD + ) (c2, kD + )
aT2 +(bD +)2
bT +
[aT2 +(bT +)2 ] K(kT + )E(kT + ) (c2, kT + )
aT2 +(bT +)2
bD
[aT +(bD ) ] K(kD )E(kD ) (c , kD )
2 2 2
aT2 +(bD )2
(3.60)
Equation 3.61 is later used to derive an expression of MAB (), the mutual
induc-
tance between Solenoids A and B misaligned by a small distance ( aT2 +bD2 ).
1 r r
n
bk rk+2
uTr = J (3.69)
Er 2 (k + 2)2
k=0
100 CHAPTER 3
E # $
= + (1+) (3.72b)
1 2 T
+ a1 (1+)T (3.76a)
C2 1 2 JB1 a21 2+ (3+)
(1 + )C1 (1 ) 2 = ( ) (1 ) 2
E 1 3 8
+ a1 (1+)T (3.76b)
Thin-Walled Coil
Figures 3.10a and 3.10b show, respectively, plots, for a thin coil, here = 1.2,
of normalized stresses, hoop, /(JB1 a1 ), and radial, r /(JB1 a1 ),
as functions of normalized radial distance, r/a1 , at selected eld ratios,
B2 /B1 . Note that = 0.1 is appropriate for a stand-alone solenoid; = 0 for
an innitely long coil; and > 0 for a coil exposed to a uniform background eld.
For each eld ratio () decreases with , while r , being 0 at r = a1 and r = a2 ,
has an extremum (maximum or minimum) radially midway through the winding.
1.2
1.1 0.95
0.9
1.0 0.8
0.7
0.9
0.5
0.8
0.7 0.2
0.6
0
0.5 0.1
0.4
1.0 1.04 1.08 1.12 1.16 1.20
(a)
0.010 0.95
0.9
0.005 0.8
0.7
0
0.5
0.005
r
0.001 0.2
0.015
0
0.020
0.1
0.025
1.0 1.04 1.08 1.12 1.16 1.20
(b)
Fig. 3.10 Plots, at selected B2 /B1 values, for a thin-walled coil ( = 1.2): (a)
normalized hoop stress, /(JB1 a1 ), vs. normalized radial distance, r/a1 ; (b)
normalized radial stress, r /(JB1 a1 ), vs. . In each graph , from the bottom
trace to the top trace: 0.1 (bottom); 0; 0.2; 0.5; 0.7; 0.8; 0.9; and 0.95 (top).
MAGNETS, FIELDS, AND FORCES 103
For > 0.5, the winding throughout develops a positive radial stress, which tends
to separate turns, a condition that generally should be avoided. The positive eect
of winding tension in reducing r is discussed shortly (3.6.3).
Medium-Walled Coil
Figure 3.11 shows plots similar to those of Fig. 3.10 for a medium-walled coil,
here = 1.8. In this medium-walled coil, r > 0 for 0.2; furthermore, r exceeds
0.1 for > 0.8. That is, an insert coil in the bore of a high-eld background
magnet should be thin-walled, or otherwise it should be subdivided.
2.0
1.6
0.95
0.9
0.8
1.2 0.7
0.5
0.2
0.8
0
0.1
0.4
0.2
1 1.2 1.4 1.6 1.8
(a)
0.14
0.95
0.9
0.10 0.8
0.7
0.06 0.5
r
0.02 0.2
0
0.02
0
0.1
0.06
1 1.2 1.4 1.6 1.8
(b)
Fig. 3.11 Plots, at selected B2 /B1 values, for a medium-walled coil ( = 1.8):
(a) /(JB1 a1 ) vs. ; (b) r r /(JB1 a1 ) vs. . In each graph , from the
bottom trace to the top trace: 0.1 (bottom); 0; 0.2; 0.5; 0.7; 0.8; 0.9; and 0.95 (top).
104 CHAPTER 3
Thick-Walled Coil
Figure 3.12 shows plots for a thick-walled coil, here = 3.6. Note that is
roughly twice that for the medium-walled coil. Most signicantly, though, in the
thick-walled coil normalized radial stress r > 0 unless is suciently negative.
There are two practical ways to make r nearly 0 or negative: 1) wind the coil with
conductor in tension; and 2) wrap the coil at its outermost layer with a banding
wire of high modulus of elasticity. Also, subdividing the coil into thinner coils
reduces not only r but also .
4.5
4.0
3.0
0.95
0.9
0.8
2.0 0.7
0.5
0.2
0
1.0
0.1
0
1.0 1.4 1.8 2.2 2.6 3.0 3.4
(a)
0.9 0.95
0.9
0.8
0.8
0.7
0.6
0.5
r 0.4
0.2
0.2
0
0 0.1
0.1
1.0 1.4 1.8 2.2 2.6 3.0 3.4
(b)
Fig. 3.12 Plots, at selected B2 /B1 values, for a thick-walled coil ( = 3.6): (a)
/(JB1 a1 ) vs. ; (b) r r /(JB1 a1 )vs. . In each graph , from the bottom
trace to the top trace: 0.1 (bottom); 0; 0.2; 0.5; 0.7; 0.8; 0.9; and 0.95 (top).
MAGNETS, FIELDS, AND FORCES 105
14
12 = 0
10
40 N
8
r [MPa]
6 80
4
120
2
0 160
2
200
4
40 45 50 55 60 65 70 75 80
r [mm]
Fig. 3.13 r vs. r plots of a solenoidal coil of = 1.8, with winding tensions, , from 0
(no tension) to 200 N ( 20 kg). With = 0, r has a peak, at r 57.5 mm, of 11.1 MPa.
Only at winding tensions above 160 N ( 16 kg), r 0 everywhere.
106 CHAPTER 3
= LI (3.78)
The proportionality constant L is the self inductance of the coil. Note that is a
eld concept quantity, while I is a circuit concept quantity: L links these two
concepts, and because the eld concept quantity necessarily involves a volume, L
is a geometry-dependent quantity. L is also related to the stored magnetic energy,
Em , of the coil through:
Em = 12 LI 2 (3.79)
For systems containing no magnetic materials, Em may be computed by integrating
(1/2) H 2 over all space, i.e., Em is another eld-concept quantity; L connects
Em to I through Eq. 3.79.
In the right-hand side of each equation above, the rst term is the self inductance
of the loop due to the ux linked within the circular area (0 r R a) of
the loop and the second term represents, as will be studied in PROBLEM 3.18
(p. 208), the self inductance of the interior of the wire, 2R long. Because the
ux linkage outside the wire is frequency independent, the rst term is valid for
all frequencies, while the second term is frequency dependent. Thus, (1/4)R is
valid only at low frequencies; the second term reduces to 0 at high frequencies.
Derivation of the rst term of the right-hand side of Eq. 3.80 is quite complex,
involving, among others, elliptic integrals.
5
0.04 0.05 0.06 0.08 0.1
4.5
0.2
4
0.4
3.5 0.6
0.8
0.04 1
3
0.1 1.5
L(, )
2.5 0.2 2
2 0.4 3
0.6 4
1.5 0.8 5
1
7
1 1.5 10
2
3
0.5 4
5
7
10
0
1 1.5 2 2.5 3 3.5 4 4.5 5
Fig. 3.14 L(, ) for solenoidal coil of parameters and .
108 CHAPTER 3
Wire
The interior of a wire of radius a per unit length, L [H/m]:
L= (3.83)
8
Long Coils
a. A long ( > 0.75), thin coil of i.d. 2a1 with N total turns [3.7]:
(1 + )2 2 (1 + )2
L = a1 N 2
1 + (3.84a)
8 3 32 2
Thus, as noted above, for a very long and thin coil, L(, ) = /2.
Short Coil
A short ( < 0.75), thin ( 1) coil of i.d. 2a1 with N total turns [3.7]:
+1 2( + 1) 2
L a1 N 2
ln 1+
2 2(1 + 2 )
2
2 1+
1
(3.85a)
4(1 + 2 )
Note that for a ring (N = 1), i.e., a1 (+1) = 2R (ring diameter) and 2a1 = 2a
or a1 (1) = 2a (ring wire diameter), both Eqs. 3.85b and 3.86b simplify to:
4R R
L R ln 0.5 = R ln + 0.886 (3.86c)
a a
Equation 3.86c applies to a ring having a rectangular cross sectional area, while
Eq. 3.80a to a ring of circular cross sectional area. For a really fat ( 1) pancake,
Eq. 3.86a further simplies to:
25 101
L 0.5 a1 N ln2
= 0.538 a1 N 2
6 288
L 0.5 a2 N 2 (3.86d)
Note that it is a2 , not the usual a1 , that appears in Eq. 3.86d.
L = 18 N 2 (3.87)
Note that for an ideal dipole or quadrupole magnet, the inductance is independent
of winding radius, but instead is proportional to length.
110 CHAPTER 3
Coupling Coecient
Mutual inductance M12 is related to L1 and L2 by:
M12 = k L1 L2 (3.95a)
M12
k= (3.95b)
L1 L2
k is called the coupling coecient; k = 0 when the coils are uncoupled, and k = 1
when fully coupled. For a tightly nested solenoidal coil pair, in which one coil
is in the bore of the other coaxially and concentrically (with the two midplanes
coinciding), k ranges 0.30.6, being close to 0.6 when their s and s are similar
and towards 0.3 when they are not.
112 CHAPTER 3
As long as , though much greater than aR and aS , is still not much greater than bS ,
the correction terms, bS / and that with (c2, k)s, must be retained in Eq. 3.99.
2bS
NS
NS
2aR 2aS
Inserting Eq. 3.92, with the subscripts 1 and 2 replaced, respectively, by A and B
in the right-hand side of Eq. 3.104a, and noting that FAB in the z-direction (with
variable ) is FzR (), given by Eq. 3.60, we obtain:
EAB MAB ()
FzR () = = IA IB (3.105b)
For small ( aT2 +bD2 ), we may integrate FzR () given by Eq. 3.61:
For small , MAB () thus decreases as 2 , the square of the o-center distance.
MAGNETS, FIELDS, AND FORCESPROBLEMS & DISCUSSIONS 115
2b r
Bz (0, 0)
2a1
2a2
10 76 5 4
3.5
4
3
2.5
3 2
1.8
1.6
F (, )
1.4
2 1.2
1
0.8
0.6
1
0.4
0.2
0.1
0
1 2 3 4 5 6 7 8
Fig. 3.17a F (, ) vs. for constant lines (dashed). The heavy curve
represents the minimum conductor volume for a solenoid coil of given F (, ).
MAGNETS, FIELDS, AND FORCESPROBLEMS & DISCUSSIONS 117
200 100 50 20 15 10
7
9
6 8
7
5 6.5
6
F (, ) 4 5.5
5
4.5
3
4
3.5
2 3
2.5
1 2
1.5
0 1.1
0 1 2 3 4 5 6 7 8
Fig. 3.17b F (, ) vs. for constant lines (dashed). The heavy curve
represents the minimum conductor volume for a solenoid coil of given F (, ).
7
F (, ) 5.5
6
5
5 4.5
4
3.5
4 3
2.5
3 2
1.5
1
2 0.5
0.25
1
0
1 2 3 4 5 6 7 8
Fig. 3.17c vs. for constant F (, ) lines (dashed). The heavy curve
represents the minimum conductor volume for a solenoid coil of given F (, ).
118 CHAPTER 3PROBLEMS & DISCUSSIONS
Thus:
a2
a2
r2 b dr
Bz (0, 0) = J = Jb ln r + r2 + b2
2 r 2 + b2
a1 r a1
= Jb ln a2 + a22 + b2 ln a1 + a21 + b2
b a2 /a1 + (a2 /a1 )2 + (b/a1 )2
= Ja1 ln (S1.1)
a1 (a1 /a1 ) + (a1 /a1 )2 + (b/a1 )2
Thus:
b) Ring Coil For a ring coil ( 1; 0), the logarithmic term in the
right-hand side of Eq. 3.110 becomes:
+ 2 + 2
lim ln = ln
0 1 + 1 + 2
And thus:
P = cd J 2 a31 2(2 1) (3.112)
where J is the current density in the conductor only. From Eq. 3.112, we can solve
for J in terms of P and other parameters:
P 1
J= (S1.4)
cd a1 a1 2(2 1)
Thus:
P
Bz (0, 0) = G(, ) (3.113a)
cd a1
+ 2 + 2
G(, ) = ln (3.113b)
2(2 1) 1 + 1 + 2
Equation 3.113a implies that in resistive solenoids the required power P for a given
set of and , increases quadratically with the central magnetic eld:
cu a1 Bz2 (0, 0)
P = (3.113c)
2 G2 (, )
Fig. 3.18 Silhouette of two nested Florida-Bitter plates, with an outer plate of 140 mm
in diameter, in water magnets at the National High Magnetic Field Laboratory [3.9].
MAGNETS, FIELDS, AND FORCESPROBLEMS & DISCUSSIONS 123
and combine this with Eqs. 3.115 to relate [Bz (0, 0)]B and PB :
B PB
[Bz (0, 0)]B = [G(, )]B (3.116a)
cu a1
where
1
[G(, )]B = [F (, )]B (3.116b)
4 ln
As in uniform-current-density solenoids, [Bz (0, 0)]B increases as the square root
of PB ; PB is a quadratic function of eld. The maximum of [G(, )]B occurs
when 6.42 and 2.15: [G(6.4, 2.15)]B 0.166. [G(, )]B is at least 99%
of its peak value for 5 9 and 1.8 2.6. That is, PB for a given eld is
within 2% of minimum throughout this range of and . However, because eld
homogeneity for a given value of a1 improves with 2b or , most Bitter magnets
have values greater than 2.5, as in the magnet considered below.
Illustration We may apply Eq. 3.116 to compute PB for a Bitter magnet of
2a1 = 6 cm, 2a2 = 40 cm, 2b = 22 cm, B = 0.8, and cu = 2106 cm, generating
[Bz (0, 0)]B = 20 T. With = (a2 /a1 ) = 40/6 = 6.67; = (b/a1 ) = 22/6 = 3.67,
1 3.67+ 1+(3.67)2
[G(6.67, 3.67)]B = ln 6.67 0.159
43.67 ln(6.67) 3.67+ (6.67)2 +(3.67)2
We have an expression of PB , similar to Eq. 3.113c, in terms of [Bz (0, 0)]B and
other parameters:
cu a1 [Bz (0, 0)]2B
PB =
2 B [G(, )]2B
Kelvin Coil
The current density that gives the best eld eciency is known as the Kelvin
distribution, JK (r, z):
r
JK (r, z) 2
(r + z 2 )3/2
Its unique feature is that every portion of the Kelvin coil produces the same eld
per unit power. In comparison with a Kelvin coil, a uniform-current-density coil
for the same total power produces 66% of eld at the magnet center; for a Bitter
coil, the ratio is 77%. It is not feasible to fabricate coils having the Kelvin current
density.
Gaume
The Gaume distribution, JG (r, z), also gives a good eld eciency:
1 1 1
JG (r, z) 2
r a21 + z 2 a2 + z 2
The Gaume coils make each turn produce the same eld per unit power as every
other turn. A Gaume coil produces 85% of the eld of a Kelvin coil. The current
distribution of Bitter coils often approximates to a degree the Gaume distribu-
tion. This is achieved by using thicker Bitter plates axially away from the magnet
midplane: JB (r, z) 1/r(z), where (z) is the z-dependent plate thickness.
Polyhelix
A polyhelix coil consists of many nested single-layer coils, in which the current
density of each layer is adjusted to maximize the eld eciency and/or to match
the stress in each layer to its conductor strength:
JP (r, z) f (r)
A polyhelix coil of maximum eciency, with JP (r, z) 1/r2 , generates 92% of a
Kelvin coil eld. In practice, because of the need to have many electrodes at both
ends, polyhelix coils are considered more dicult to manufacture than Bitter coils.
126 CHAPTER 3PROBLEMS & DISCUSSIONS
3 2 15(34 2 ) 4 35(520 2 +8 4 ) 6
hz () = 1 + + +
4(1+ 2 )2 64(1+ 2 )4 256(1+ 2 )6
315(35280 2 +336 4 64 6 ) 8
+
16384(1+ 2 )8
b) Also, show that for a short coil ( = 0) an expression for hz () is given by:
3(2 1) 2 45(4 1) 4 175(6 1) 6
hz () = 1 + + +
82 ln 2564 ln 15366 ln
11025(8 1) 8 43659(10 1) 10
+ + + (3.117b)
1310728 ln 65536010 ln
3 2 15 4 35 6 315 8 693 10
hz () 1 + 4
6
+ 8
10
+
4 16 32 256 512 12
(3.117c)
d) Compute an approximate value of Hm /Hz (0, 0) hm for a thin-walled
( = 1) and relatively short coil with = 0.4.
e) Calculate hm for a pancake coil of = 2 and 0.
f ) Determine hm for a thin-walled and long coil of = 1 and = 2.
Note that for a single solenoid of a set of any and values, Hm /Hz (0, 0) may
be given by hz () obtained by combining Eqs. 3.12b and 3.14.
MAGNETS, FIELDS, AND FORCESPROBLEMS & DISCUSSIONS 127
Ic (B, T )
Ic (0, T )
Load lines
(no background eld)
Iop (Bmx , T )
Load lines
(in center eld of Bb )
Iop (Bmxb , T )
0 B
0 Bmx Bb Bmxb
A. End Field
The axial end eld of an axially symmetric solenoid is half the center eld of a
solenoid of identical composition but twice the length of the original. We may
visualize this by considering an axially symmetric solenoid comprised of two iden-
tical sections, each of the original length 2b (Fig. 3.20). The axial center eld of
the new solenoid is the sum of two equal elds, generated by the two sections:
Because F (, 2) > F (, ) (Fig. 3.17b), it follows that H(z = b) > 0.5H(0). That
is, the end axial eld of a solenoid is always greater than half its central eld. Note
that H(z = b) 0.5H(0) in the limit .
1 b+z bz
(Case 1: z < b) H(z) = 2 Ja1
F , = +F , = (3.120a)
a1 a1
b+z zb
1
(Case 2: z > b) H(z) = 2 Ja1 F , = F , = (3.120b)
a1 a1
= 2
1
2a1
2b 4b
Fig. 3.20 In a superposition technique, the end eld of an axially symmetric solenoid
of length 2b may be computed as half of the center eld of a solenoid of the same
construction having length 4b.
MAGNETS, FIELDS, AND FORCESPROBLEMS & DISCUSSIONS 131
ORIGINAL COIL
b+z bz
2b
=
2
1
2b+2z
+
2
1
2b2z
Case 2: z > b
z
ORIGINAL COIL
2b
b+z
=
1
2
2b+2z
1
2
2z 2b
Fig. 3.21 Superposition technique for o-center axial points. Case 1: z < b; Case 2:z > b.
132 CHAPTER 3PROBLEMS & DISCUSSIONS
1232 mm
Magnet Center
Fig. 3.22 Cross sectional view of the 45-T hybrid magnet at NHMFL [3.31].
134 CHAPTER 3PROBLEMS & DISCUSSIONS
Table 3.3: Parameters of 14-T SCM for NHMFL 45-T Hybrid Magnet [3.31]
Coils (each wound with CIC conductor) Coil A Coil B Coil C
Strand composite superconductor Nb3 Sn NbTi
Type of winding Layer Pancake
Number of layers/double pancakes 6 7 29
Total number of turns 306 378 1015
Winding i.d., 2a1 [mm] 710 908 1150
Winding o.d., 2a2 , [mm] 888 1115 1680
Winding length, 2b [mm] 869 868 992
Operating current, Iop [kA] 10*
J @ Iop [MA/m2 ] 39.6* 44.3* 38.6*
Field contribution at center @ Iop [T] 3.3* 3.6* 7.4*
Bpeak @ Iop with water magnet idle [T] 15.7* 11.7* 8.5*
Combined inductance [H] 1.96
Stored energy @ Iop [MJ] 98*
Water Magnet
In DISCUSSION 3.2, it has been shown that the power requirement PB for
Bitter magnets is proportional to a1 and [Bz (0, 0)]2B (from Eq. 3.116a) , with
PB typically in the 630 MW range. A Bitter magnet is power hungry,
and it is best to minimize its overall volume: use it as an insert in a hybrid.
However, because the stronger the eld strength, the greater the magnetic
stresses in the conductor, the conductor materials
must be stronger, which are
generally
more resistive. Thus, Bz (0, 0) P (Eq. 3.113a) and [Bz (0, 0)]B
PB (Eq. 3.116a) both become invalid at higher elds.
Normal metals such as copper have no intrinsic eld limit above which they
cannot be used to build a magnet. However, as stated above, because those
made of stronger materials require greater power, and equally demanding, the
cooling that must match this increased Joule dissipation, the range 3040 T
is considered the limit for practical magnet facilities.
Superconducting Magnet
Superconductors have fairly well-dened upper eld limits above which they
cannot remain superconducting. Thus it is best to place this magnet in the
lower-eld part of a hybrid: place it outside the water magnet.
The total energy storage increases with magnet size, but the power required,
chiey for cryogenics, remains insignicant. A 100-MJ magnet does not
require a 100-MW power supply; typically 10100 kW supplies suce.
Combination of these features makes it natural for a hybrid magnet to consist of
a water-cooled insert surrounded by a superconducting magnet.
A
C
Fig. 3.23 Pictorial view of a double-pancake coil, with the top and bottom pancakes
separated axially for clarity. The pancakes in this drawing are wound with a tape
conductor. Points A and B indicate the ends of a continuous conductor, with Point
C marking the approximate midpoint.
MAGNETS, FIELDS, AND FORCESPROBLEMS & DISCUSSIONS 137
Answer to TRIVIA 3.2 ii). Corresponds to a net temperature rise of 25 C. Note that
this 17-MW dissipation must obviously be matched by a cooling power of 17 MW. It has
been suggested, half in jest, that a water-cooled magnet is a zero-eciency machine.
138 CHAPTER 3PROBLEMS & DISCUSSIONS
a) Idealizing the two coils by two rings each of radius a, show that when d = a,
dHz2 (0, z)/dz 2 = 0 at the magnet center. The solid curve in Fig. 3.24b gives
Hz (0, z) of a coil with d that does not satisfy Eq. 3.121.
b) Show that with coils of opposing polarity, a gradient eld is generated at the
magnet center. Evaluate this dHz /dz at z = 0.(Note that d3 Hz (0, z)/dz 3 = 0
when d = a; d3 Hz (0, z)/dz 3 = 0 requires d = 3a.) This conguration with
reverse current is called a Maxwell coil. The dotted curve in Fig. 3.24b gives
Hz (z) of a gradient coil.
z
z = d/2
a
d z =0
(a)
Hz(z)
z
d d / 2 0 d/2 d
(b)
Fig. 3.24 (a) Ideal Helmholtz coil arrangement; (b) Hz (0, z) for a uniform
eld case (solid) and Hz (0, z) for a gradient eld case (dotted).
a2 I
Hz (0, z) =
2[a2 + (z + d/2)2 ]3/2
Adding the eld from the top coil, located at z = d/2, we have:
a2 I 1 1
Hz (0, z) = + 2 (S3.1)
2 [a2 + (z + d/2)2 ]3/2 [a + (z d/2)2 ]3/2
Note that from symmetry dHz (0, z)/dz = 0 at z = 0 for any value of a.
The second derivative of Eq. S3.1 is given by:
d2 Hz (0, z) 3a2 I a2 4(z + d/2)2 a2 4(z d/2)2
= 2 2
dz 2 2 [a + (z + d/2)2 ]7/2 [a + (z d/2)2 ]7/2
This method of locating two identical coils having opposite currents to achieve a
eld gradient is the basic principle used in magnets requiring a gradient eld at
the midplane. A pulsed magnet used in an MRI system to produce a gradient eld
(to extract spatial information for imaging) is one example.
140 CHAPTER 3PROBLEMS & DISCUSSIONS
From the above, it is clear that d2 hz (0)/d 2 |1/2 0 as more higher terms are
added. Note that for a single ring coil placed at = 0: d2 hz (0)/d 2 = 3.
142 CHAPTER 3PROBLEMS & DISCUSSIONS
106.44
100.00
112.89
100.00 148.21
Coil B
289.24
112.89
100.00
177.55
Coil A
Coil C
Fig. 3.26 The magnet of Fig. 3.25 represented by Coils A, B, and C, with each
center at the magnet center. Note that the minus sign signies that Coil C carries
current in the direction opposite from those in Coils A and B. Dimensions are in mm.
144 CHAPTER 3PROBLEMS & DISCUSSIONS
When all coils have the same a1 and Ja1 , Eq. 3.18 becomes:
k
j=1 F (j , j )E2 (j , j ) 2
h () = 1 + k + (S5.1)
j=1 F (j , j )
And so the design of the rst powerful water-cooled coils at M.I.T. gradually
evolved. An interesting side light at this point was that a thorough search of the
literature revealed that actually we were not the rst to attempt to build coils along
these lines. During World War I two Frenchmen tried to do the same sort of
thing. They designed water-cooled coils and found that the only place they could
conveniently get the power to operate them was in the private power station of one
of the big department stores in Paris. But they had to stop work because of the
war, and for about twenty years it was forgotten. My designs were quite dierent
from theirs, and I decided to follow my own plans. Now I was faced with the same
problem as the Frenchmen. If we did build some coils, where could we test them?
Throughout this period and up to the completion of the rst magnets the chief
person responsible for the realization of my ambitions was Vannevar Bush, at that
time the Vice-President of M.I.T. He was interested in what I was doing, and gave
it the needed backing. To begin with, after I had made some preliminary designs
that looked reasonable, he suggested that I try them out at one of the substations
of the Boston Edison Company. They had old-fashioned DC power stations which
had spare power during the early hours of the morning. He arranged to have space
made available for me to set up a magnet, and to get water cooling from the city
water mains.
And so the rst magnet was constructed in the basement of the physics building.
It was to dissipate about 1000 kilowatts in a magnet having a volume of about one
cubic foot (30 liters). If the water cooling should fail, the 1000 kilowatts dissipated
in this volume would melt everything in it in a few seconds. Our water-cooling tests
had indicated that the heating would not fail, that steam layers would not form and
stop the cooling processes. But it was still an exciting time. When we set up the
magnet at the Scotia Street substation of the Edison Company, the engineers were
frankly skeptical. But because of Van Bushs backing they were willing enough to
give me a chance to prove my ideas.
146 CHAPTER 3PROBLEMS & DISCUSSIONS
x2 + y 2
s = sin = (S6.1c)
x2 + y 2 + z 2
y
sin = (S6.1d)
x2 + y2
x
cos = (S6.1e)
x2 + y 2
Equations 3.9 and S6.1aS6.1e are combined for cases n = 0, 1, and 2.
n=0:
0
Hz (x, y, z) = r0 (1 + 0)P00 (u)(A00 cos 0 + B00 sin 0) (S6.2a)
m=0
= (1)(1)(1)(A00 ) (S6.2b)
For n = 0, we thus have:
Hz (x, y, z) = A00 (3.126a)
Note that A00 represents the magnet center eld: Hz (0, 0, 0).
n=1:
1
Hz (x, y, z) = r1 (2+m)P1m (Am m
1 cos m+B1 sin m) (S6.3a)
m=0
x 2 +y 2 A1
x+B 1
y
+ 3 x2 +y 2 +z 2
1 1
x2 +y 2 +z 2 x2 +y 2
n=2:
2
Hz (x, y, z) = r2 (3+m)P2m (Am m
2 cos m+B2 sin m) (S6.4)
m=0
= 32 A02 (2z 2 x2 y 2 )
Note that when evaluated for n = 0, 1, and 2, Hz (x, y, z) contains terms varying
as x, y, z, z 2 , x2 , y 2 , zx, zy, and xy.
Ignorance is like a delicate exotic fruit; touch it and the bloom is gone.
Lady Bracknell
148 CHAPTER 3PROBLEMS & DISCUSSIONS
C L D H F
I G K
b2
a3 a2
b1
A B E
a1
z1
Because of the use of ln(, ), Eq. 3.128b is a dimensionless eld expression slightly
dierent from that derivable from Eq. 3.13a.
2a2
2a1
w
2N
(2N 1)
+
2N (w+)
P2 + w
= + 2w+ +
P1 + w
Two-Pancake Magnet Coil 2 Coil 2
P4 + w
P3 + w
Pancakes 3&4 Coil 4 Coil 4
The total dimensionless axial eld due to all 4 coils, 4 (z) is:
4 (z) = [()]2 [ ()]2 + [()]4 [ (z)]4 (3.132)
where
n
[()]4 = 4 ln(, 4 ) 1 + E2j (, 4 ) 2j (3.133a)
j=1
n
[ ()]4 = 4 ln(, 4 ) 1+ E2j (, 4 ) 2j (3.133b)
j=1
In this analysis it is assumed that the spacing between pancake coils within each
double-pancake coil and those between adjacent double-pancake coils are equal.
MAGNETS, FIELDS, AND FORCESPROBLEMS & DISCUSSIONS 153
+
P 2N
= + 2bN + 2bN
P (2N 1)
+
where
2kw + (2k 1)
2k =
2a1
2(k 1)w + (2k 1)
2k =
2a1
Equation 3.136 may also be expressed as:
2N ()
= 1 + [E2 ]2N 2 + + [E2n ]2N 2n (3.137)
2N (0)
where 2N (0) is the dimensionless center eld and [E2n ]2N is the nth overall error
coecient. 2N ( = 0) and [E2n ]2N are given by:
N
2N (0) = 2k ln(, 2k ) 2k ln(, 2k ) (3.138a)
k=1
N
k=1 2k ln(, 2k )E2n (, 2k )2k ln(, 2k )E2n (, 2k )
[E2n ]2N = N
(3.138b)
k=1 2k ln(, 2k )2k ln(, 2k )
Thus Eq. 3.138b gives an expression for computation of nth error coecient com-
prised of 2N identical pancake coils, each of 2a1 i.d., 2b = w, , and , and
separated from each other by .
154 CHAPTER 3PROBLEMS & DISCUSSIONS
c) Show that an expression for the Lorentz force density per unit length, f L
[N/m2 ], acting on a current-carrying element of the shell, is given by:
f L = H02 sin 2 (3.141)
y
x
R
4R H02
FLx = (3.142)
3
e) Show that an expression for the total magnetic energy stored (per unit dipole
length), Em [J/m], is given by:
R2 B02
Em = (3.143)
To reduce the eld outside the dipole, an iron yoke ( = ) of radial thickness d
is placed outside the dipole, as shown in Fig. 3.33.
f ) Show that the new K eld inside the dipole
f 1 needed to generate the same H
is exactly one-half that given by Eq. 3.140. Explain this current reduction.
g) In reality, the iron yoke cannot maintain its high for an unlimited value of
H0 . Show that an expression for the minimum dm to keep the yoke unsatu-
rated is given by:
H0
dm = R (3.144)
Msa
where Msa is the yoke materials saturation magnetization. Compute dm for
the following set: H = 5 T; Msa = 1.2 T; R = 20 mm.
y
r
x
R R+d
K d2 H
f = r (H d1 ) = r 2H0 cos
y y
(a)
x x
(b) (c)
f L = K
f H0 sin r (S8.1)
= 2 H02 cos sin
= H02 sin 2 (3.141)
Note that f L has no r-component; it has only a -component (Fig. 3.34c). Also
the force density is maximum at = /4+n/2
and zero at = 0 + n/2, where
n = 0, 1, 2, 3. The cumulative force, fL () d, is maximum at = 0 and 180 .
d) It is clear from Fig. 3.34c that the net Lorentz force per unit length [N/m]
acting on the right-hand segment of the shell is +x-directed. Thus:
/2
FLdx = f L x dx = R fL sin d (S8.2a)
/2
/2 /2
= 2R fL sin d = 4R H02 cos sin2 d (S8.2b)
0 0
2 2 2 2
= H R + H R (S8.3b)
2 0 2 0
R2 B02
= R2 H02 = (3.143)
From Eq. S8.3b it is clear that the total stored magnetic energy is divided equally
inside and outside the dipole shell. We may imagine that one-half of the current
d1 and the other half is used to create H
owing in the dipole is used to create H d2 .
Inserting H0 = B0 = 5 T and R = 0.02 m into the above expression, we obtain:
2(250103 J)
L=
(5000 A)2
= 20 mH
In 3.7.3 the inductance per unit length of an ideal dipole is given by:
L = 18 N 2 (3.87)
f 1 = H0 cos
K (S8.5)
d) Show that an expression for the magnetic spring constant, kLx , in the x-
direction for a proton traveling in the +z-direction along the center of the
magnet with a speed nearly equal to that of light, c, is given by:
qc H0
kLx (3.148)
R
e) Similarly, show that an expression for the magnetic spring constant, kLy ,
in the y-direction for a proton traveling in the +z-direction along the center
of the magnet with a speed nearly equal to that of light, c, is given by:
qc H0
kLy (3.149)
R
f ) By stating whether kLx and kLy are unstable or restoring, describe the func-
tion of quadrupoles for charged particle accelerators.
K q2 H
f = r (H q1 ) = r 2H0 cos 2
f vectors change directions four times around the magnet shell (Fig. 3.35b).
The K
y (a) y
x x
(b) (c)
Fig. 3.35 a) Quadrupole elds inside and outside the bore; b) surface current
density vectors in the magnet; c) force directions in the magnet.
MAGNETS, FIELDS, AND FORCESPROBLEMS & DISCUSSIONS 161
Answer to TRIVIA 3.3 iii). Despite his friendship with the royal family, having been
lionized by Marie Antoinette, it is said that Joseph Louis Lagrange (17361813) survived
the Terror because of the general respect for his accomplishments in mathematics and
his foreign birth (Torino). He was appointed to head this commission in 1793.
162 CHAPTER 3PROBLEMS & DISCUSSIONS
2a2
2a1
Racetrack 1 (N turns) y
x 2c
Racetrack 2 (N turns)
Fig. 3.36 Cross section of a magnet comprised of two ideal racetrack coils.
1+ = K d
dH (3.150)
2r1
where the eld direction is shown in the gure and K = N I/(a2 a1 ). The y-
component of eld, Hy1+ , contributed by the entire +z-surface current, from =
a1 to = a2 , is given by integration of Eq. 3.151 (below) from = a1 to = a2 :
a2
K cos 1 d
Hy1+ = (3.151)
2 a1 r1
With r1 = ( x)2 +(cy)2 and cos 1 = ( x)/r1 into Eq. 3.151, we obtain:
K a2
( x) d K (a2 x)2 +(cy)2
Hy1+ = = ln (3.152a)
2 a1 ( x)2 +(cy)2 4 (a1 x)2 +(cy)2
Similarly, the contributions from the remaining current sheets are given by:
K a2 ( +x) d K (a2 +x)2 +(cy)2
Hy1 = = ln (3.152b)
2 a1 ( +x)2 +(cy)2 4 (a1 +x)2 +(cy)2
K a2 ( x) d K (a2 x)2 +(c+y)2
Hy2+ = = ln (3.152c)
2 a1 ( x)2 +(c+y)2 4 (a1 x)2 +(c+y)2
K a2 ( +x) d K (a2 +x)2 +(c+y)2
Hy2 = = ln (3.152d)
2 a1 ( +x)2 +(c+y)2 4 (a1 +x)2 +(c+y)2
r1
1 c
(x, y)
z
(0, 0) 1
1+
dH
2a
2 1
y NI
x 2c
(0, 0)
4 3
The quantity K(a22 a21 ) = K(a2 +a1 )(a2 a1 ) in the second term of the right-hand
side of the rst line of Eq. 3.156a becomes 2aN I. Thus:
K(a22 a21 )[3c4 +(a22 +a21 )c2 a22 a21 ] 2 2 2aN I[3c4 +2a2 c2 a4 ] 2 2
2 2 (x y ) (x y )
(a2 +c2 )2 (a1 +c2 )2 (a2 +c2 )4
4
3c +2a2 c2 a4 2 2
= Hy (0, 0) (x y )
(a2 +c2 )3
where F1 |2 , F1 |3 , and F1 |4 are the force vectors on element 1 by elements, respec-
tively, 2, 3, and 4.
The force, F1 |2 , on element 1 carrying I1 = N I by element 2 carrying I2 = N I, is
+x-directed and given by:
I1 I2 N 2 I 2
F1 |2 = x = x (3.159a)
4a 4a
MAGNETS, FIELDS, AND FORCESPROBLEMS & DISCUSSIONS 167
I well remember the occasion for the rst trial. We were to have the power at
some time in the middle of the night, around 1 A.M. Shortly before the appointed
hour Van arrived to see how things went. Then, as usual, there were seemingly
endless delays. First of all, for some unforeseen reason, the power was not available,
and we were told to wait half an hour. Then another hour. Then we went out to
get some coee. I dont remember just how long, in the end, we had to wait.
But nally the moment arrived. Sometime near dawn, when we were all worn
out with waiting, we stood around the corner of a wall and watched as the power
began to be turned up. At rst everything was all right. Then there were slight
hissing sounds. They got louder; nally there was a bang. Whereupon the power
was shut o. When we went to examine the magnet, we could nd nothing much
wrong. One of the bolts in the anges of the case holding the parts together had
mysteriously exploded. The magnet had failed for some reason quite dierent from
those we had expected.
168 CHAPTER 3PROBLEMS & DISCUSSIONS
I I
r
B a
B
Fig. 3.39 Ideal toroidal magnet comprised of N loops, each carrying current I.
TRIVIA 3.4 Of the four pressures below, three are close to the magnetic pressure
corresponding to a 1.5-T eld, while the other is not. Identify this odd man out.
i) Compressive, on the toes of a ballerina performing a pirouette;
ii) Internal, of an uncorked champagne bottle;
iii) Lift, on the wings of a 747 jet at cruising speed;
iv) Thermal, of the plasma in an experimental Tokamak.
MAGNETS, FIELDS, AND FORCESPROBLEMS & DISCUSSIONS 169
N I
B (r) = (3.161)
2r
Outside the torus, no net current is enclosed when the above integral is performed
over the entire circumference. Therefore H (r) = 0 and B (r) = 0.
b) Figure 3.40 shows a single loop in which a dierential force dFL acts on a
dierential element ds with dierential force dFLr in the r-direction.
dFL is given by:
dFL = I ds B (r) (S10.2)
where B (r) is the average eld acting on the surface current. Because this eld
varies from that given by Eq. 3.161 just inside the surface current to zero just
outside, the average eld is one half of that given by Eq. 3.161. Thus:
N I 2 ds
dFL = I ds B (r) = (S10.3)
4r
where the vector points in the direction of FL (Fig. 3.40). The r-component of
this dierential force is given by:
N I 2 cos ds
dFLr = (S10.4)
4r
R
L
dF
ds
dFr
d
r
a
N I 2 a cos d
dFLr = (S10.5)
4(R + a cos )
where H0 is the central eld. We may compute Re by equating the dipoles far
eld (r Re ) along the z-axis (r-direction at = 0), given by Eq. 3.163, and that
(z a) of a ring of current I and radius a, given by Eq. 3.3a. Thus:
H0 Re3 = 12 a2 I (3.164)
a) For a solenoidal coil having total ampere-turns of N I, i.d. of 2a1 and o.d. of
2a2 , show, by using a weighted-average, a2 , that Eq. 3.164 is modied to:
When Eq. 3.164 is applied to each layer of a winding comprising n layers, each
layer having the same total turns per layer, nt/ , we have, given without derivation,
an expression, which the reader with extra time may derive:
3 1 2 (n +1) 2 (n +1)(2n +1)
H0 Re = 2 a1 N I 1+(1) +(1) (3.166)
n 6n2
Note that N = nt/ n . For n 1, Eq. 3.166 may be approximated by Eq. 3.165.
Because Re is proportional to the cube root of the right-hand side of each equa-
tion, in most cases Eq. 3.165 is a good approximation to Eq. 3.166. In any case
these equations are valid only for r Re , and thus Re computed by Eq. 3.165 is
independent of magnet length, 2b, and N I. For a nested-coil magnet comprised
of k coils, Eq. 3.165 may be generalized to:
k
H0 Re3 = 16 I (a21j + a22j + a1j a2j )Nj (3.167)
j=1
b) By using Eq. 3.167 and parameter values given in Table 3.3, show that
Re = 0.67 m for the 45-T SCM. (Because the water magnets volume is much
smaller than the SCMs, the water magnet, despite its central eld of 31 T,
contributes little to the fringing elds and may be neglected here.)
c) For reasons of safety, people and equipment associated with operation and
experiment of this magnet should be outside the 100-gauss contour prolate
spheroid of the magnet. Determine the radial distance rm , at z = 2.75 m, at
f | is 100 gauss.
which the fringing eld magnitude | H
Thus:
b) Applying Eq. 3.167 to the 45-T SCM (Table 3.3) with H0 = 14 T/ , we have:
(14 T)
R3 = 16 (104 A)
(4107 H/m) e
[(0.355 m)2 +(0.444 m)2 +(0.355 m)(0.444 m)]306
Re = 0.667 m
Solving Eqs. S11.2a, S11.2b, and S11.2c for x, we nd: x = 6.52 m. For x = 6.52 m,
we nd from Eq. S11.2b that R 7.05 m and 67.5 .
Note that this 100-gauss exposure to the experimenters, presumably over a period
of a half day or at most a day, is dierent from the long-term exposure level of
5 gauss sanctioned by the FDA.
Answer to TRIVIA 3.4 iii) 747 (0.06, in atm); 1.5-T eld (9)
champagne (6); ballerina toes (8); plasma (10).
MAGNETS, FIELDS, AND FORCESPROBLEMS & DISCUSSIONS 173
A. Spatial Homogeneity
As discussed in 3.4, the spatial eld homogeneity of a solenoidal magnet is deter-
mined completely by and . Therefore, the full-scale magnet will have the same
eld homogeneity (over the scaled volume) as the model magnet (over the original
volume) if and are kept the same.
J (2a )a (1)
N = (3.168a)
I
174 CHAPTER 3PROBLEMS & DISCUSSIONS
(J/)(2a )a (1)
(for I = I ) N = = N (3.168b)
I
(J/)(2a )a (1)
(for I = I ) N = = N (3.168c)
I
(for I = I ) = N (a + a ) = N a ( + 1) = (3.169a)
(for I = I ) = N a ( + 1) = 2 (3.169b)
I = 2 I (3.170)
Illustrative Example
We have a model magnet of the following parameters: Hz (0, 0) = 1.53 T; 2a1 =
80 mm; 2a2 = 130 mm; 2b = 220 mm; total number of turns, N = 2976. The
magnet has a self inductance of 0.301 H and generates a center eld of 1.53 T at
an operating current, I , of 100 A. Now let us consider a full-scale magnet with
its dimensions scaled by 10 from the model magnet with remaining the same.
We may compute a few parameters for the full-scale magnets.
Inductance Because L = a N2 L(, ), 1) L = L (if I = I ); or 2)
L = 3 L (if I = I ). Thus: 1) L 3.01 H if I = 1000 A; and 2) L = 301 H if
I = 100 A.
Let us rst compute :
a (+1)
= N
2
(0.04 m)(1.625+1)
= (2976) 156 m
2
Thus: 1) 1.56 km if I = 1000 A; or 2) = 15.6 km if I = 100 A.
MAGNETS, FIELDS, AND FORCESPROBLEMS & DISCUSSIONS 175
TRIVIA 3.5 If an electron is to circle around the earth at the equator, should it
travel eastward or westward? (Note that it will travel at nearly the speed of light.)
MAGNETS, FIELDS, AND FORCESPROBLEMS & DISCUSSIONS 177
r
1.0 ()
0.1 (2b)
2.1 1.5 z
(2a2 ) (2a1 ) (0.5,0) (0,0) (0.5,0)
Coil A Coil B
Fig. 3.41 Cross sectional view of a 2-coil magnet consisting of two identical coils,
A and B, energized in the same polarity. Dimension in meters.
178 CHAPTER 3PROBLEMS & DISCUSSIONS
0.05
1.0
0.1
0.5
0.25
) [T]
0.5
0.75
Bz (
1.0
1.5
0
0.5
1.0
0 0.2 0.4 0.6 0.8 1.0
Radial Displacement, r [m]
(a)
1.0
z = 0.05
0.8 0.1
) [T]
0.6
0.25
Br (
0.4
0.2 0.5
0.75
1.0
0 1.5
0 0.2 0.4 0.6 0.8 1.0
Radial Displacement, r [m]
(b)
Fig. 3.42 (a) Bz (r, z) and (b) Br (r, z) plots for one coil, A or B, at I = 65 A,
with the other coil unenergized, at axial (z) locations of the energized coil.
MAGNETS, FIELDS, AND FORCESPROBLEMS & DISCUSSIONS 179
b) Total conductor length, , is given by: = N (a2 +a1 ) 50.3 km. This 2-coil
magnet thus contains a total conductor length of 100 km.
c) We may approximate Coil A (or Coil B) as a pancake coil with = 1.4. By
applying Eq. 3.111f , we obtain BzA , the center axial eld of Coil A:
N I ln
BzA Bz (0, 0) = (3.111f )
2a1 1
(4107 H/m)(8900)(65 A) 0.336
= = 0.4 T
(1.5 m) 0.4
d) From Fig. 3.14, we have L( = 1.4, = 0.067) 2.8. From Eq. 3.81:
L = a1 N 2 L(, ) (3.81)
= (4107 H/m)(0.75 m)(8900)2 (2.8) = 209 H
where K(k) and E(k) are, respectively, the complete elliptic integrals of the rst
and second kinds.
4aA aB
k2 = (3.36)
(aA + aB )2 + 2
MAGNETS, FIELDS, AND FORCESPROBLEMS & DISCUSSIONS 181
14 12
10 10
25 25
(a) ( b)
Fig. 3.43 (a) Solenoid with 2a1 = 10 cm; 2a2 = 14 cm; 2b = 25 cm. (b) Solenoid subdivided
into subsolenoids A and B, with 2aA = 10 cm; 2aB = 12 cm; 2b = 25 cm.
MAGNETS, FIELDS, AND FORCESPROBLEMS & DISCUSSIONS 183
4(0.125 m)
+ [(0.11 m)2 +(0.125 m)2 ](1.800991.38373)
(0.11 m)2 +(0.125 m)2
(0.05 m0.06 m)2 (22.37011.8010)
4(0.125 m)
[(0.11 m)2 +4(0.125 m)2 ](1.640411.50558)
2
(0.11 m) +4(0.125 m) 2
[(0.05 m0.06 m)2 (18.72271.6404)
= (5.655107 N/m2 ) 14.680103 m2 7.705103 m2
+20.634103 m2 11.009103 m2
+34.739103 m2 6.177103 m2
18.413103 m2 +3.127103 m2
Note that it is assumed that for a long coil (k 2 1 or 1), for which
Eq. 3.41b is valid, Bz (0, 0) and Bz (0, b) may be assumed constant over the circular
plane of area a2 . For this long solenoid, it is also true that Bz (0, b) 0.5Bz (0, 0)
(see DISCUSSION 3.4 ), thus Eq. S14.2 may be given by:
b
a2
2aBr (z) dz Bz (0) (S14.3)
0 2
For a long coil, Bz (0, 0) = N I/2b (Eq. 3.111), thus:
b
a2 N I
2aBr (z) dz (S14.4)
0 2 2b
Combining Eqs. S14.1 and S14.4, we have:
2
N I a2 N I NI
Fz (0) a2 (3.41b)
2b 2 2b 2 2b
MAGNETS, FIELDS, AND FORCESPROBLEMS & DISCUSSIONS 187
Because z is close to b, kb 2
= 4a2 /[4a2 + (b z)2 ]
/ 1 and thus K(kb ) E(kb )
2
cannot be approximated in kb . Let us simply guess the location of z near b at
which point Fz (z b) is close to that given by Eq. 3.41b. If we guess z = b2a, then:
2
NI
Fz (z = b2a) 2a 4a2 +4a2 K(kb )E(kb ) (S14.6)
2 2b
2
where kb = 4a2 /8a2 = 0.5, which gives, from Table 3.1, K(kb = 0.7071) = 1.8541
and E(kb ) = 1.3506. Inserting these values into Eq. S14.6, we obtain:
2
NI
Fz (z = b2a) 4a2 2(0.5035)
2 2b
2
NI
= 2.85a2 (S14.7)
2 2b
It was too late for us to do any more about it that night, but we had made some
progress although we had not reached the maximum power input into that magnet by
a big factor. Then came a period of taking the magnet apart, nding out what had
gone wrong, putting it together, trying again. Finally it became clear that there was
nothing fundamentally wrong with the design, but that many little details hard to
foresee had to be looked after. The usual expression for this sort of thing is getting
the bugs out of the apparatus. When we got the bugs out, the magnet behaved just
as had been calculated. It was a success.
188 CHAPTER 3PROBLEMS & DISCUSSIONS
2bB
SOLENOID B
SOLENOID A
2bA
z 2aA 2aB
where
4aA aB 4(5 cm)(7 cm)
c2 = 2
= = 0.972222
(aA + aB ) (5 cm+7 cm)2
4aA aB 4(5 cm)(7 cm)
kB2 = 2 2
= = 0.052950 = kB 0.230109
aT + bB (12 cm)2 +(50 cm)2
4aA aB 4(5 cm)(7 cm)
kD2 = = = 0.182055 = kD 0.426679
aT2 + bD2 (12 cm)2 +(25 cm)2
4aA aB 4(5 cm)(7 cm)
kT2 = = = 0.024268 = kT 0.155781
aT2 + bT2 (12 cm)2 +(75 cm)2
190 CHAPTER 3PROBLEMS & DISCUSSIONS
where c2 , kD2 , and kT2 are given above; for this case, because bA = bD , kA2 = kD2 . Thus:
2
(4107 H/m) 8106 A
Fz B (0.07 m)2
2 1.0 m
8106 A 3106 A
+ [(0.05 m)2 +(0.07 m)2 ]
1.0 m 0.5 m
(0.05 m0.07 m)2 (210.2692+10.26929.5245)
Dov
Dcd
Asc 1 Acu
fsc = = 2.1 Atr = 0.230 (S16.3b)
Atr
Ain Acu + Asc
fin = =1 = 1 0.484 0.230 = 0.286 (S16.3c)
Atr Atr
c) The Youngs modulus for the composite, E, may be given from the parallel
mixture rule:
E = fcu Ecu + fsc Esc + fin Ein (S16.4)
= (0.48)(100 GPa) + (0.23)(85 GPa) + (0.29)(30 GPa)
76 GPa
We may calculate the stress of each component at the innermost winding radius:
Ecu 100 GPa
cu = i = (105 MPa) 137 MPa (S16.5a)
E 76 GPa
Esc 85 GPa
sc = i = (105 MPa) 117 MPa (S16.5b)
E 76 GPa
Esc 30 GPa
in = i = (105 MPa) 41 MPa (S16.5b)
E 76 GPa
These values ignore residual stresses, which in fact may be very large.
MAGNETS, FIELDS, AND FORCESPROBLEMS & DISCUSSIONS 193
Table 3.10: Stresses () & Strains () in HTS Insert Midplane Pancake Coil [3.32]
(Operating at 86.7 A in a Background Field of 14.1 T)
Average Silver Steel Silver Silver Tape
r [mm] r [MPa] [MPa] [MPa] z [MPa] T [MPa] [%]
39.10 0 62.3 54.3 152.9 2.1 56.4 0.030
51.19 2.6 44.9 39.2 110.3 2.1 41.3 0.028
63.29 0 33.6 29.3 82.4 2.1 31.4 0.026
* Tresca stress: T = z .
Table 3.10 gives values of stresses and hoop strains at three radial locations, in-
nermost, midpoint, and outermost, in one of the midplane pancakes [3.32]. The
midplane axial force is 16 kN. Note that the highest stresses occur at the in-
nermost (a1 = 39.10 mm) winding. The maximum combined stress, Tresca stress,
T (= z ), in silver is 56.4 MPa, which is less than the allowable stress,
allow = 2/3 (0.2%): for silver at 77 K, allow 120 MPa.
Stresses and strains may be computed by solving Eqs. 3.62 and 3.171. However,
some of the equations derived in 3.5 may be applied to estimate the midplane axial
force. As noted above, the HTS magnet at its midplane will have a computed axial
force of 16.2 kN (i.e., compressive), which as discussed in 3.5, arises from the
self eld of the HTS magnet and from the interaction of the HTS and the LTS
magnet. Here, we compute approximate values of the midplane force in the HTS
magnet by its own eld alone. The above analysis gives a value of 8.6 kN [3.32].
a) Treating the HTS magnet as thin-walled, apply Eq. 3.41a to compute
Fz (0). Choose a = 39.1 mm, b = 203.3 mm, and N I = 5.91106 A.
b) Apply Eq. 3.41b to compute Fz (0) with 1) a = 39.1 mm, the a1 of the HTS
magnet and 2) a = 51.2 mm, which is the average winding radius.
c) Now treat the HTS magnet as two thin-walled and long subsolenoids, A
and B, of the same winding thickness, and apply Eq. 3.55 to compute FzT (0).
Choose aA = 39.1 mm and aB = 51.2 mm.
d) The 14.1-T LTS magnet that surrounds this HTS insert consists of nested
coils. Model the LTS magnet as a thin-walled solenoid of aB = 191.3 mm;
bB = 337.5 mm; NB IB = 7.535 MA; the HTS insert as a thin-walled solenoid
of aA = 51.2 mm; bA = 203.3 mm; NA IA = 0.591 MA. By applying Eq. 3.57a,
compute the total midplane axial force for the HTS insert magnet: FzA (0),
which is 16.2 kN. Explain why Eq. 3.57a does not give the correct value.
e) Using the same models for both magnets and applying Eq. 3.60, compute
an axial restoring force on the HTS insert when it is o-center by 5 mm
( = 5 mm in Eq. 3.60). A code gives a value of 2.1 kN.
In the following pages, appropriate elliptic integral values are given, Tables 3.11
through 3.14.
MAGNETS, FIELDS, AND FORCESPROBLEMS & DISCUSSIONS 195
where kb and k2b are the moduli determined by a and b. For this case, in which
a = 39.1 mm and b = 203.3 mm, we have:
Appropriate complete elliptic integral values are given in Table 3.11. By applying
Eq. 3.38c up to the k 6 term, we may compute K(k)E(k) for k 2 1. Thus:
K(k2b )E(k2b ) (0.128888)+ 38 (0.128888)2 + 15
64 (0.128888)3
4
= 0.106515
= 6.2 kN
The moduli, kb , k2b , and c2, with aT = aA + aB , aA = 39.1 mm, aB = 51.2 mm, b =
203.3 mm, aT = 90.3 mm, are given by:
4aA aB 4(39.1 mm)(51.2 mm)
c2 = 2
= = 0.982045
aT (90.3 mm)2
4aA aB 4(39.1 mm)(51.2) mm
kb2 = 2 2
= = 0.161820 kb = 0.402269
aT +b (90.3 mm)2 +(203.3 mm)2
4aA aB 4(39.1 mm)(51.2) mm
2
k2b = 2 2
= = 0.046160 k2b = 0.214848
aT +4b (90.3 mm)2 +4(203.3 mm)2
Thus:
2
7 (0.591106 A)
Fz (0) = (410 H/m) [(0.0391 m)2 + (0.0512 m)2 ]
(0.8132 m)
(0.0391 m 0.0512 m)2
1 [2(12.66889) 11.96934]
[(0.0391 m)2 + (0.0512 m)2 ]
= 7.4 kN
This underestimates the code by 15%.
MAGNETS, FIELDS, AND FORCESPROBLEMS & DISCUSSIONS 197
Parameter values other than those given in question d) are: aT = aA+aB = 242.5 mm;
bT = bA +bB = 540.8 mm; bD = bA bB = 134.2 mm; and
4aA aB 4(51.2 mm)(191.3 mm)
c2 = 2
= = 0.666226
aT (242.5 mm)2
4aA aB 4(51.2 mm)(191.3) mm
kD2 = 2 2
= = 0.510028 kD = 0.714162
aT +bD (242.5 mm)2 +(134.2 mm)2
4aA aB 4(51.2 mm)(191.3) mm
kT2 = 2 2
= = 0.111533 kT = 0.333965
aT +bT (242.5 mm)2 +(540.8 mm)2
4aA aB 4(51.2 mm)(191.3) mm
kB2 = = = 0.226841 kB = 0.476278
aT2 +bB2 (242.5 mm)2 +(337.5 mm)2
This is 2.7 times the code value of 16.2 kN. The major source of error comes from
the approximation of Eq. 3.57a involving the interaction force, FzAB . In order
for Eq. 3.57a to be valid, as remarked in its derivation, the conditions bA2 4aA2 ,
bB2 aT2 , bD2 aT2 , must be satised. In this particular case, we have: bA2 /4aT2 = 0.176;
bB2 /aT2 = 1.94; and bD2 /aT2 = 0.306. That is, none of the conditions necessary for
Eq. 3.57a to be valid is satised in this particular case.
198 CHAPTER 3PROBLEMS & DISCUSSIONS
c2 k (c2, k) c2 k (c2, k)
0.666226 0.331412 2.820467 0.666226 0.336555 2.823887
0.666226 0.707891 3.364664 0.666226 0.720364 3.400959
Parameter values other than those given in question e) are: aT = aA+aB = 242.5 mm;
bT = bA +bB = 540.8 mm; bD = bA bB = 134.2 mm; and
bT + (0.5458 m)
0.913860
2 (0.2425 m)2 +(0.5458 m)2
aT +(bT +)2
aT2 +(bT +)2 (0.2425 m)2 +(0.5458 m)2 0.356704 m2
K(kT + )E(kT + ) 1.6168151.526733 0.090082
(aA aB ) [(c , kT + )K(kT + )] (0.019628 m )(2.8204671.616815) 0.023625 m2
2 2 2
bD (0.1392 m)
0.497833
aT +(bD )2
2 (0.2425 m)2 +(0.1392 m)2
aT2 +(bD )2 (0.2425 m)2 +(0.1392 m)2 0.078183 m2
K(kD )E(kD ) 1.8550161.350085 0.504931
(aA aB ) [(c , kD )K(kD )] (0.019628 m )(3.3646641.855016) 0.029632 m2
2 2 2
FzR () = 2.2 kN
This overestimates the code value (2.1 N) by 4%; the sign implies that the
force is restoring. The spring constant between the magnets is 400 kN/m.
In Eqs. 3.174 and 3.175, it is assumed that the impinging eld is uniform for
energy density computation, but nonuniform for force density computation.
z
fmr
y0.1g
45-T Hybrid Platform fmy
fm
1m IRON SPHERE
r
Iop
45-T HYBRID MAGNET
2Re
Combining the above expression with Eq. 3.174 and using the grad operator in
spherical coordinates (Eq. 2.36a) on Eq. 3.173, we obtain:
6
3 H02 2 1 2 Re
fm (r, ) = (cos + 4 sin ) r
2 r r
6
1 Re 2 1 2
+ cos + 4 sin
r r
7
3 H02 Re
= 6(cos2 + 14 sin2 )r 3
2 sin cos
2Re r
2
7
9 H0 Re
fm (r, ) = [(1 + 3 cos2 )r + sin cos ] (3.176a)
4Re r
Note that fm (r, ) varies as 1/r7 and, as expected for any ferromagnetic object,
the r-component of fm (r, ) is directed towards the magnet center.
The magnetic force thus varies as 1/r4 when the iron sphere is saturated. Note
also that because it is r-directed, as in the unsaturated case, the iron sphere is
attracted to the magnet center.
202 CHAPTER 3PROBLEMS & DISCUSSIONS
Combining the above expressions and Eq. 3.178, and with fmy = 0.1g we have:
3.5
9( H0 )2 Re6 1 4z 2 y
0.1g = 2 1+ 2 0.1g
4 y0.1g + z2 y0.1g + z 2 y 2 + z2
0.1g
z2
Coil 2 z1
Coil 1
Fig. 3.47 Two nested solenoidal coils, one displaced radially from the other.
Answer to TRIVIA 3.5 Eastward. The current must point westward to make
a resultant magnetic force on the electron directed toward the earth center.
204 CHAPTER 3PROBLEMS & DISCUSSIONS
The Edison Company was getting rather fed up with the time-consuming exper-
iments. While they were willing enough to make it possible to test out a brand-new
piece of equipment, they were not exactly eager to go on interrupting their own
schedules and having their people work overtime on a project of no interest to them.
So we called a halt to these tests, and Van set out to nd some money for me to
get a power station of my own at M.I.T. and to construct some new magnets us-
ing the experience gained in the preliminary tests. In due course the money was
made available. The sum was perhaps a tenth of what was required ten or fteen
years later to duplicate the installation, but luckily in those days (the mid-1930s)
secondhand equipment could be had.
MAGNETS, FIELDS, AND FORCESPROBLEMS & DISCUSSIONS 205
brr cur
s0 sr
br0 cu0
Ebr Ecu Es
Abr Acu As
Fig. 3.48 Schematic strain states in the Nb3 Sn composite after cooldown.
Equation 3.180a states that the net internal force is zero. Equations 3.180b 3.180d
give the strain compatibilities for, respectively, the bronze/Nb3 Sn, copper/Nb3 Sn,
and copper/bronze. Be aware that Eq. 3.180 implicitly assumes that each con-
stituent is in its elastic range, which usually is not the case.
B. Residual Strains
From Eqs. 3.180b and 3.180c, we may obtain expressions for sr and cur :
sr = br0 s0 brr
cur = brr br0 + cu0
Combining the above expressions with Eq. 3.180, we obtain:
brr Abr Ebr + (brr br0 + cu0 )Acu Ecu + (brr br0 + s0 )As Es = 0
Note that both matrix materials are in tension, while Nb3 Sn is in compression; sr
of 0.72% is too severe and would certainly damage the conductor. However, when
the magnet is energized, the conductor is subjected mostly to a tensile stress, which
tends to place sr towards zero strain; usually, the Lorentz stresses are sucient
to put Nb3 Sn in tensile strain when the magnet is energized.
The yield stresses of annealed bronze, bry , and annealed copper, cuy , are both
only 100 MPa; both bronze and copper thus yield plastically during cooldown.
b) Equation 3.84c, for a very long ( ) and thin coil with N total turns:
2
L = a1 N (3.84c)
2
c) Equation 3.87, for the unit length of an ideal dipole with N total turns:
L = 18 N 2 (3.87)
g) Equation 3.90a, for an ideal toroid of major radius R and rectangular section
of width (r-axis) 2a, height (z-axis) 2b, and N total turns.
2 1 R+a
L = bN ln (3.90a)
Ra
b) From Eq. 3.111d, the center eld, Bz (0, 0), of a long solenoid ( ) is given
by N I/2b. As may be inferred from Eq. 3.117c, the eld in a thin-walled and
long solenoid is uniform in both axial and radial directions and equal to the
center eld. Thus the total ux linked by N turns of the solenoid is given by:
a1 2
a1 N I
= 2rHz (0, 0) dr = N (S18.4)
0 2b
Combining Eqs. 3.78 ( = LI) and S18.4, we obtain:
L = a1 N 2 (3.84c)
2
where R is the dipole radius. Combining Eqs. 3.140 (for K f ) and S18.5, we have:
/2
NI = 2H0 R cos d = 4H0 R (S18.6)
/2
The total magnetic energy Em per unit length of the dipole is given by Eq. 3.143:
R2 B02
Em = (3.143)
From S18.6, we compute H0 R = N I/4. Now combining this, Eqs. 3.143 and 3.79,
and solving for L (L per unit length), we have:
L = 18 N 2 (3.87)
210 CHAPTER 3PROBLEMS & DISCUSSIONS
Because the dipole has a uniform eld of H0 , the total ux linkage by the N turns
of the dipole is given by:
/2 /2
N = n() H0 (2R cos ) d = N H0 R cos 2 d
/2 /2
= 12 N H0 R = 18 N 2 I (S18.8)
L = 18 N 2 (3.87)
The total magnetic energy Em per unit length of the quadrupole is obtained by
integrating |H(r, )|2 /2 over the entire surface. |H(r, )|2 , from Eq. 3.145, is
independent of , because sin2 2+cos2 2 = 1.
2 R
1 H0 3 1 2 6 2 dr
Em = 2 2r dr + 2 H0 R
R 0 R r5
= 12 H02 R2 (S18.11)
Therefore:
/4
N = N H0 R cos2 2 d
/4
2
= 14 N H0 R = 1
16 N I (S18.14)
Combining Eqs. 3.78 and S18.14, we obtain:
1 2
L = 16 N (3.88)
The equation of the circle that denes the cross section is given by:
z 2 + (R r)2 = a2 (S18.17)
Solving z from Eq. S18.17 for the limits of z in the integral of Eq. S18.16 in terms
of r and constants, and after combining with Eq. S18.15, we have:
a2 (Rr)2
N 2 R+a dz dr
=
2 Ra a2 (Rr)2 r
N 2 I R+a
a2 R2 +2Rrr2
= dr (S18.18)
Ra r
212 CHAPTER 3PROBLEMS & DISCUSSIONS
N 2 I a2 R2 a 2
= R + = N RI 1 1
2
(S18.19)
R2 a2 R
Combining Eqs. 3.78 and S18.19, we obtain:
a 2
L = RN 2
1 1 (3.89a)
R
f) For a R, we have:
a 2 a 2 a 4 a 6
1 = 1 12 18 16
1
(S18.20)
R R R R
From Eqs. 3.89a and S18.20,
a 6
a 2 1 a 4
L = RN 2 12 +8 + 161
R R R
Thus:
a2 N 2 2 4
1 a 1 a
L= 1+ 4 +8 (3.89b)
2R R R
g) The eld within the rectangular section of this ideal toroid is the same as that
in the circular section of the toroid studied above. Thus:
N 2 I R+a z=+b 1
N = dz dr
2 Ra z=b r
N 2 bI R+a
= ln (S18.21)
Ra
Combining Eqs. 3.78 and S18.21, we obtain:
bN 2 R+a
L= ln (3.90a)
Ra
After inquiring here and there, I found that the place to look for big secondhand
generators was in New Jersey, out beyond Jersey City. It was a peculiar feeling,
going shopping for big secondhand electrical machinery, about which I knew very
little. But I found something that looked suitable. It had a central motor, and on
each end of the shaft a generator capable of delivering up to 5000 amps at 170 volts.
It was an impressive objectmuch bigger than any magnet I had seenabout 12
feet (4 m) high and 20 feet (6 m) long. When I reported my ndings to Van, he
suggested that we employ a rm of consulting engineers to check on the condition of
the motor generator and then to design a proper installation . . . and before long we
had a 1.7 megawatt (million-watt) motor generator which could deliver power at any
voltage from zero to 170 volts. This was very valuable, since gradual starting and
stopping were necessary. By means of the voltage control we were able to connect
the magnet to the power line without drawing current, and then slowly build up the
current and increase the power drawn from the power mains.
For about three years we had three or four magnets in more or less continuous
operation, providing facilities for a considerable range of experiments. There were
rst of all some low-temperature experiments which, as I pointed out in Chapter III,
might be expected to lead to particularly interesting results in high magnetic elds . . .
. . .Some years after the war was over, and we were continuing work with our
battered twenty-year old magnets, a colleague asked what I felt about the possibility
of making even better and stronger magnets. I told him that it seemed to me that
our experiments with heat transfer were inconclusive, . . . That is, it seemed possible
to remove heat from the copper not at the rate of 200 watts per square centimeter,
as I had designed it, but at approximately 2000 watts per square centimeter. The
design of such a magnet would lead to even more critical cooling conditions, and in
the event that something went wrong the magnet might literally explode. But such
magnets are being built. The bugs are being taken out of the new designs, and more
and more powerful magnets are becoming available for scientic investigations.*
* In the 1930s Francis Bitter (19021967) built 10-T Bitter magnets [3.65] to perform
research in high magnetic elds [3.66]. His magnet laboratory in the basement of an M.I.T.
physics building led to the establishment in 1960 of the National Magnet Laboratory
(NML) at M.I.T., the rst major facility in the world dedicated to high magnetic eld
sciences and technologies. The NML in turn led to the creation of high magnetic eld
facilities in Europe (England, Netherlands, France, Germany, Poland, Russia) and Asia
(Japan, China). Currently, the largest and most extensive is the National High Magnetic
Field Laboratory, in operation since 1995 on the campus of Florida State University. In
1995 the NML, by then the Francis Bitter National Magnet Laboratory, named in his honor
in 1968, became the Francis Bitter Magnet Laboratory, as before, an M.I.T. laboratory.
MAGNETS, FIELDS, AND FORCESREFERENCES 215
REFERENCES
[3.1] Many codes are available, created by individuals and institutions, e.g., SOLDE-
SIGN (M.I.T.), and by commercial outts, e.g., COMSOL, ANSYS, ANSOFT.
[3.2] D. Bruce Montgomery, Solenoid Magnet Design (Robert Krieger Publishing, New
York, 1980).
[3.3] R.J. Weggel (personal communication, 1999).
[3.4] Francis Bitter, Magnets: The Education of a Physicist (Doubleday, New York,
1959).
[3.5] Milan Wayne Garrett, Calculation of elds, forces, and mutual inductances of
current systems by elliptical integrals, J. Appl. Phys. 34, 2567 (1963).
[3.6] Based on a reformulation with new materials by Emanuel Bobrov (FBML) in
2005, with additional contribution by Seung-Yong Hahn (FBML), of a paper by
E.S. Bobrov and J.E. Williams, Stresses in superconducting solenoid Mechanics
of Superconducting Structures, F.C. Moon, Ed. (ASME, New York, 1980), 1341.
[3.7] Standard Handbook for Electrical Engineers, Ed. Archer E. Knowlton (McGraw-
Hill Book, 1949).
[3.8] Benjamin J. Haid (personal communication, 2003).
[3.9] Hans-J. Schneider-Muntau and Mark Bird (personal communication, 2004).
[3.10] John Peter Voccio,Qualication of Bi-2223 high-temperature superconducting
(HTS) coils for generator applications (Ph.D. Thesis, Department of System De-
sign Engineering, Keio University, 2007).
[3.11] D.B. Montgomery, J.E.C. Williams, N.T. Pierce, R. Weggel, and M.J. Leupold,
A high eld magnet combining superconductors with water-cooled conductors,
Adv. Cryogenic Eng. 14, 88 (1969).
[3.12] M.J. Leupold, R.J. Weggel and Y. Iwasa, Design and operation of 25.4 and 30.1
tesla hybrid magnet systems, Proc. 6 th Int. Conf. Magnet Tech. (MT-6) (ALFA,
Bratislava), 400 (1978).
[3.13] M.J. Leupold, J.R. Hale, Y. Iwasa, L.G. Rubin, and R.J. Weggel, 30 tesla hybrid
magnet facility at the Francis Bitter National Magnet Laboratory, IEEE Trans.
Magn. MAG-17, 1779 (1981).
[3.14] M.J. Leupold, Y. Iwasa and R.J. Weggel, 32 tesla hybrid magnet system, Proc.
8 th Int. Conf. Magnet Tech. (MT-8) (J. Physique Colloque C1, supplement to 45),
C1-41 (1984).
[3.15] M.J. Leupold, Y. Iwasa, J.R. Hale, R.J. Weggel, and K. van Hulst, Testing a
1.8 K hybrid magnet system, Proc. 9 th Int. Conf. Magnet Tech. (MT-9) (Swiss
Institute for Nuclear Research, Villigen), 215 (1986).
[3.16] M.J. Leupold, Y. Iwasa, and R.J. Weggel, Hybrid III system, IEEE Trans. Magn.
MAG-24, 1070 (1988).
[3.17] Y. Iwasa, M.J. Leupold, R.J. Weggel, J.E.C. Williams, and Susumu Itoh, Hybrid
III: the system, test results, the next step, IEEE Trans. Appl. Superconduc. 3,
58 (1993).
[3.18] Y. Iwasa, M.G. Baker, J.B. Con, S.T. Hannahs, M.J. Leupold, E.J. McNi, and
R.J. Weggel, Operation of Hybrid III as a facility magnet, IEEE Trans. Magn.
30, 2162 (1994).
[3.19] K. van Hulst and J.A.A.J. Perenboom, Status and development at the High Field
Magnet Laboratory of the University of Nijmegen, IEEE Trans. Magn. 24, 1397
(1988).
216 CHAPTER 3REFERENCES
[3.20] Jos A.A.J. Perenboom, Stef A.J. Wiegers, Jan-Kees Maan, Paul H. Frings, First
operation of the 20 MW Nijmegen High Field Magnet Laboratory, IEEE Trans.
Appl. Superconduc. 14, 1276 (2004).
[3.21] S.A.J. Wiegers, J. Rook, J.A.A.J. Perenboom, and J.C. Maan, Design of a 50 mm
bore 31+ T resistive magnet using a novel cooling hole shape, IEEE Trans. Appl.
Superconduc. 16, 988 (2006).
[3.22] Y. Nakagawa, K. Noto, A. Hoshi, S. Miura, K. Watanabe and Y. Muto, Hybrid
magnet project at Tohoku University, Proc. 8 th Int. Conf. Magnet Tech. (MT-8)
(Supplement au Journal de Physique, FASC. 1), C1-23 (1984).
[3.23] K. Watanabe, G. Nishijima, S. Awaji, K. Takahashi, K. Koyama, N. Kobayashi,
M. Ishizuka, T. Itou, T. Tsurudome, and J. Sakuraba, Performance of a cryogen-
free 30 T-class hybrid magnet, IEEE Trans. Appl. Superconduc. 16, 934 (2006).
[3.24] H.-J. Schneider-Muntau and J.C. Vallier, The Grenoble hybrid magnet, IEEE
Trans. Magn. MAG-24, 1067 (1988).
[3.25] G. Aubert, F. Debray, J. Dumas, K. Egorov, H. Jongbloets, W. Joss, G. Martinez,
E. Mossang, P. Petmezakis, Ph. Sala, C. Trophime, and N. Vidal, Hybrid and
giga-NMR projects at the Grenoble High Magnetic Field, IEEE Trans. Appl.
Superconduc. 14, 1280 (2004).
[3.26] A. Bonito Oliva, M.N. Biltclie, M. Cox, A. Day, S. Fanshawe, G. Harding, G. How-
ells, W. Joss, L. Ronayette, and R. Wotherspoon, Preliminary results of nal test
of the GHMFL 40 T hybrid magnet, IEEE Trans. Appl. Superconduc. 15, 1311
(2005).
[3.27] K. Inoue, T. Takeuchi, T. Kiyoshi, K. Itoh, H. Wada, H. Maeda, T. Fujioka,
S. Murase, Y. Wachi, S. Hanai, T. Sasaki, Development of 40 tesla class hybrid
magnet system, IEEE Trans. Magn. 28, 493 (1992).
[3.28] John R. Miller, The NHMFL 45-T hybrid magnet system: past, present, and
future, IEEE Trans. Appl. Superconduc. 13, 1385 (2003).
[3.29] M. Bird, S. Bole, I. Dixon, Y. Eyssa, B. Gao, and H. Schneider-Muntau, The 45T
hybrid insert: recent achievement, Phys. B, 639 (2001).
[3.30] J.R. Miller, Y.M. Eyssa, S.D. Sayre and C.A. Luongo, Analysis of observations
during operation of the NHMFL 45-T hybrid magnet systems, Cryogenics 43, 141
(2003).
[3.31] J.R. Miller (Personal communication, 2003).
[3.32] E.S. Bobrov (Personal communication, 2003).
[3.33] Juan Bascunan, Emanuel Bobrov, Haigun Lee, and Yukikazu Iwasa, A low- and
high-temperature superconducting (LTS/HTS) NMR magnet: design and perfor-
mance results, IEEE Trans. Appl. Superconduc. 13, 1550 (2003).
[3.34] Haigun Lee, Juan Bascunan, and Yukikazu Iwasa, A high-temperature supercon-
ducting (HTS) insert comprised of double pancakes for an NMR magnet, IEEE
Trans. Appl. Superconduc. 13, 1546 (2003).
[3.35] J. Allinger, G. Danby, and J. Jackson, High eld superconducting magnets for
accelerators and particle beams, IEEE Trans. Magn. MAG-11, 463 (1975).
[3.36] A.D. McIntur, W.B. Sampson, K.E. Robins, P.F. Dahl, R. Damm, D. Kassner,
J. Kaugerts, and C. Lasky, ISABELLE ring magnets, IEEE Trans. Magn. MAG-
13, 275 (1977).
[3.37] W.B. Fowler, P.V. Livdahl, A.V. Tollestrup, B.F. Strauss, R.E. Peters, M. Kuchnir,
R.H. Flora, P. Limon, C. Rode, H. Hinterberger, G. Biallas, K. Koepke, W. Hanson,
and R. Borcker, The technology of producing reliable superconducting dipoles at
MAGNETS, FIELDS, AND FORCESREFERENCES 217
[3.52] Bruce Gamble, David Cope, and Eddie Leung, Design of a superconducting mag-
net system for Maglev applications, IEEE Trans. Appl. Superconduc. 3, 434
(1993).
[3.53] Kenji Tasaki, Kotaro Marukawa, Satoshi Hanai, Taizo Tosaka, Toru Kuriyama,
Tomohisa Yamashita, Yasuto Yanase, Mutuhiko Yamaji, Hiroyuki Nako, Motohiro
Igarashi, Shigehisa Kusada, Kaoru Nemoto, Satoshi Hirano, Katsuyuki Kuwano,
Takeshi Okutomi, and Motoaki Terai, HTS magnet for Maglev applications (1)
coil characteristics, IEEE Trans. Appl. Superconduc. 16, 2206 (2006).
[3.54] A. den Ouden, W.A.J. Wessel, G.A. Kirby, T. Taylor, N. Siegel, and H.H.J. ten
Kate, Progress in the development of an 88-mm bore 10 T Nb3 Sn dipole magnet,
IEEE Trans. Appl. Superconduc. 11, 2268 (2001).
[3.55] G. Ambrosio, N. Andreev, S. Caspi, K. Chow, V.V. Kashikhin, I. Terechkine,
M. Wake, S. Yadav, R. Yamada, A.V. Zlobin, Magnet design of the Fermilab 11
T Nb3 Sn short dipole model, IEEE Trans. Appl. Superconduc. 10, 322 (2000).
[3.56] A.R. Hafalia, S.E. Bartlett, S. Caspi, L. Chiesa, D.R. Dietderich, P. Ferracin,
M. Goli, S.A. Gourlay, C.R. Hannaford, H. Higley, A.F. Lietzke, N. Liggins, S.
Mattarri, A.D. McIntur, M. Nyman, G.L. Sabbi, R.M. Scanlan, and J. Swanson,
HD 1: design and fabrication of a 16 tesla Nb3 Sn dipole magnet IEEE Trans.
Appl. Superconduc. 14, 283 (2004).
[3.57] J.E.C. Williams, L.J. Neuringer, E.S. Bobrov, R. Weggel, and W.G. Harrison,
Magnet system of the 500 MHz spectrometer at the FBNML: 1. Design and
development of the magnet, Rev. Sci. Instrum. 52, 649 (1981).
[3.58] Haigun Lee, Emanuel S. Bobrov, Juan Bascunan, Seung-yong Hahn and Yukikazu
Iwasa, An HTS insert for Phase 2 of a 3-phase 1-GHz LTS/HTS NMR magnet,
IEEE Trans. Appl. Superconduc. 15. 1299 (2005).
[3.59] Juan Bascunan, Wooseok Kim, Seungyong Hahn, Emanuel S. Bobrov, Haigun Lee,
and Yukikazu Iwasa, An LTS/HTS NMR magnet operated in the range 600
700 MHz, IEEE Tran. Appl. Superconduc. 17, 1446 (2007).
[3.60] D.S. Easton, D.M. Kroeger, W. Specking, and C.C. Koch, A prediction of the
stress state in Nb3 Sn superconducting composites, J. Appl. Phys. 51, 2748 (1980).
[3.61] J.W. Ekin, Strain scaling law for ux pinning in practical superconductors. Part
1: Basic relationship and application to Nb3 Sn conductors, Cryogenics 20, 611
(1980).
[3.62] J.W. Ekin, Four-dimensional J-B-T - critical surface for superconductors, J.
Appl. Phys. 54, 303 (1983).
[3.63] S.L. Bray, J.W. Ekin, and C.C. Clickner, Transverse compressive stress eects
on the critical current of Bi-2223/Ag tapes reinforced with pure Ag and oxide-
dispersion-strengthened Ag, J. Appl. Phys. 88, 1178 (2000).
[3.64] J.W. Ekin, S.L. Bray, N. Cheggour, C.C. Clickner, S.R. Foltyn, P.N. Arendt,
A.A. Polyanskii, D.C. Larbalestier and C.N. McCowan, Transverse stress and
fatigue eects in Y-Ba-Cu-O coated IBAD tapes, IEEE Trans. Appl. Supercon-
duc. 11, 3389 (2001).
[3.65] F. Bitter, The design of powerful electromagnets Part IV. The new magnet lab-
oratory at M.I.T., Rev. Sci. Inst. 10, 373 (1939).
[3.66] George R. Harrison and Francis Bitter, Zeeman eects in complex spectra at elds
up to 100,000 gauss, Phys. Rev. 57, 15 (1940).
CHAPTER 4
CRYOGENICS
4.1 Introduction
Cryogenics is essential for superconductivity. This reliance of superconductivity
on cryogenics has been a major hindrance to the widespread use of superconduc-
tivity for applications such as electric power. It is, however, important to put
cryogenics in perspective and not overemphasize its role. From cryogenics alone,
it is clearly more ecient to operate a superconducting magnet at the highest
permissible temperature, but if this superconducting magnet is a part of a system,
the impact of this operating temperature on the overall system must be evaluated.
The question of What is the best operating temperature for a superconducting
magnet? becomes a real, and extremely important, design/operation issue, par-
ticularly with HTS magnets. To generate the same eld an HTS magnet operating
at 77 K, for example, unquestionably requires considerably more superconductor
than one at 20 K; the savings in cryogenics may be insucient to oset the in-
creased superconductor cost.
Also not to be overlooked is the mandatory requirement of thermal insulation for
every superconducting system: the best thermal insulation is vacuum. A thermally
eective vacuum is relatively easy to achieve in a cryostat when the cold temper-
ature is below 20 K, at which point hydrogen condenses. Hydrogen, outgassed from
the evacuated cryostat surfaces, is the primary heat transfer medium within the
cryostat. Thus, for the overall HTS magnet system it might be more cost eec-
tive to operate the magnet below, rather than above, 20 K. Alternatively, if we
can select an operating temperature high enough, e.g., above 70 K, arguably
most peoples preferred option, we may be able to eliminate vacuum insulation
altogether from the cryogenic system, making it one step closer to less intrusive.
In this introductory section, the cryogenic design/operation issues of superconduct-
ing magnets are briey discussed : 1) two cooling methods for superconducting
magnets, wet and dry; 2) cooling sources, heating sources, and cryogenic mea-
surements; 3) cryogens for wet magnets; and 4) solid cryogens that may become
useful for dry magnets. Details of some of these topics are further studied and
covered in PROBLEMS & DISCUSSIONS which follow this section.
Table 4.1: Cooling Methods for Wet and Dry Superconducting Magnets
Wet Magnets
Cooling Method Cooling-Conductor Coupling Heat Transfer
Bath-cooled, cryostable Good; entire conductor Convective
Bath-cooled, adiabatic Essentially nonexistent Conductive*
Force-cooled, cryostable Good; entire conductor Convective
Force-cooled, quasi-stable Close proximity, but indirect Conductive
Dry Magnets
Cryocooled, quasi-stable Indirect Conductive
* To the surfaces of the magnet winding pack.
CRYOGENICS 221
4.3.3 Measurement
In the operation of superconducting magnets, cryogenic parameters generally mea-
sured include: 1) temperature; 2) pressure (cryostat vacuum or cryogen pressure);
3) cryogen ow rate (force-cooled wet magnets); and 4) vapor ow rate (current
leads in wet magnets). In this book, only temperature measurement is discussed
briey in DISCUSSION 4.13 (p. 264).
TRIVIA 4.1 Of the important events below for helium, which is said to
have been achieved for the rst time on February 29?
i) Discovery; ii) Liquefaction; iii) Solidication; iv) Superuidity.
222 CHAPTER 4
gq
qpk
qf m
T (= T Ts )
Tpk
Fig. 4.1 Boiling heat transfer ux vs. temperature (dierence between the
surface temperature and liquid saturation temperature, Ts ) for a typical liquid.
CRYOGENICS 223
2.5
Phase transition (35.61 K)
8.2 J/cm3
2.0
SNe
Cp [J/cm3 K]
1.5
SN2
Pb
1.0
Ag
SAr Cu
0.5
0
0 10 20 30 40 50 60 70 80 90
T [K]
Fig. 4.2 Heat capacity, Cp , vs. temperature, T , plots for SNe, SN2 and SAr (solid
lines); and Pb, Ag, and Cu (dashed lines). Note that SN2 has a solid-to-solid phase
transition at 35.61 K, absorbing an energy density of 8.2 J/cm3.
224 CHAPTER 4
In the PROBLEMS & DISCUSSIONS that follow this introductory section, cryogenic
topics relevant to wet and dry magnets are studied; they are followed by a brief
discussion on temperature measurement. In the last section, a rather extensive
treatment is given, by means of PROBLEMS & DISCUSSIONS, on current leads, all
vapor-cooled, except one dry lead. This is because large magnet systems,
particularly those operating at currents above 1 kA, are still dominated by wet
LTS magnets that require vapor-cooled current leads.
The perpetual motion? Nonsense! It can never be discovered. It is a dream that may
delude men whose brains are mystied with matter, but not me.Owen Warland
CRYOGENICSPROBLEMS & DISCUSSIONS 225
c) Show that for a Carnot refrigerator with Twm = 300 K, Wca /Q 70 for
Top = 4.2 K and Wca /Q 3 for Top = 77 K.
Twm
Swm Qwm
Wca
Scl Q
Top
Wca is the work input to the refrigerator, equal to the area enclosed on the T -
S diagram. (When the directions of Wca , Q, and Qwm in Fig. 4.3 are reversed,
this Carnot cycle represents an ideal working machine and the area enclosed on
the T -S diagram represents the work output.) Because each process is reversible,
Q = Top (Swm Scl ) and Qwm = Twm (Swm Scl ), and thus:
Q
Swm Scl = (S1.2a)
Top
Qwm
Swm Scl = (S1.2b)
Twm
Equating Eqs. S1.2a and S1.2b, we obtain:
Q Qwm
= (S1.3)
Top Twm
0 S
0 Scl Swm
Fig. 4.4 Temperature vs. entropy plot for the Carnot refrigerator.
CRYOGENICSPROBLEMS & DISCUSSIONS 227
10,000
1,000
Wcp
Q 100
1W
10 W
10 100 W
1 kW
100 kW
1
0 10 20 30 40 50 60 70 80
Top [K]
Fig. 4.5 Wcp/Q vs. Top plots for cryocoolers at specied levels of Q [W] (inset) [4.19].
The symbol is for the Carnot cycle at selected Top , computed for Twm = 300 K.
228 CHAPTER 4PROBLEMS & DISCUSSIONS
30 60 80
15
20
8
10
4
1
0
0
15 20 30 40 50 60 70
1st -Stage Temperature [K]
Thus, 1-W cryocoolers (solid circles) have Wcp/Q ratios, for example, of 7500 for a
unit designed for Top = 4.2 K and 600 for a 1-W unit with Top = 20 K. As discussed
below, its Wcp/Q ratios at dierent levels of Top are dierent from a cryocooler
optimized at specic temperatures of 4.2 K and 20 K.
Answer to TRIVIA 4.1 iii). In 1928 (a leap year), by a colleague of Kamerlingh Onnes,
the Dutch physicist Willem Keesom (18761956), who in 1927 also discovered the phe-
nomenon of superuidity in liquid helium. The element was rst observed in 1868 by
the French astronomer Pierre Jules Cesar Janssen (18241907) as a spectral line in the
sun (helios) during a total eclipse in India. The Scottish chemist Sir William Ramsay
(18521916) in 1895 discovered the element on earth in America from samples of gas
obtained from a uranium mineral.
CRYOGENICSPROBLEMS & DISCUSSIONS 231
The minus sign for Qr (T ) in Eq. 4.4 signies that the cryocooler is providing
refrigeration: the magnet (copper mass) is cooling, i.e., dT /dt < 0.
100
80
60
Qr [W]
40
20
0
0 100 200 300
T [K]
Table 4.4 lists Qr (T ), Ccu (T ), and K(T ) Ccu (T )/Qr (T ) at selected temperatures.
We may integrate Eq. 4.4 to solve for cn for a set of Ti , Top , Qr (T ), and Mcu :
Mcu Ti Ccu (T ) Mcu Ti
cn = dT = K(T ) dT (4.5a)
cu Top Qr (T ) cu Top
cu cn
Mcu = Ti
(4.5b)
K(T ) dT
Top
A. Transport Properties
Because of its extremely high thermal conductivity superuid helium is some-
times used as a coolant for superconducting magnets, generally operated at 1.8 K
(< T ). (Although we use the classical denition of thermal conductivity here, i.e.,
heat conduction temperature dierence, in He II the smaller the temperature
dierence, the greater becomes the equivalent conductivity, which also varies
with heat ux.) The high thermal conductivity of He II does not allow a tem-
perature gradient in the liquid sucient for creation of vapor. Thus, unlike the
winding of a 4.2-K He I cooled cryostable magnet, the bubbleless winding of
a 1.8-K He II cooled cryostable magnet does not require ventilation. However,
this does not mean that He II can transport unlimited heat uxes through narrow
channels. Analogous to the critical current density in superconductors, He II has
a critical heat ux.
10000
Solid
1000
Liquid
Pressure [kPa]
(He I)
100
Liquid Vapor
(He II)
10
1
0 1 2 3 4 5 6 7
Temperature [K]
Fig. 4.8 Phase diagram of ordinary helium (He4 ) [4.20].
234 CHAPTER 4PROBLEMS & DISCUSSIONS
where Tcl [K] is the cold-end temperature and Twm [K] is the warm-end tempera-
ture. q [W/cm2 ] is the heat ux through a channel L [cm] long between the two
ends, one end at Tcl and the other at Twm . Equation 4.6a is applicable for the
case with no additional heating introduced to the liquid from the channel itself.
Under normal operating conditions, Tcl = Tb , where Tb is the bath temperature;
Twm is the liquid temperature adjacent to a heated region within the winding,
which cannot exceed T . When Twm = T , from Fig. 4.9, X(Twm ) = 0, and we
can simplify Eq. 4.6a to:
X(Tb ) = qc3.4 L (4.6b)
600
500
400
X(T ) 300
T T
200
q He II
100 L
T
0
1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2
T [K]
Heated Channel
When heating is uniformly introduced over the entire length L of the channel,
rather than at the hot end as discussed above, Eq. 4.6b is modied [4.22]:
qc3.4
X(Tb ) = L (4.6c)
4.4
mh
4 5
Pump
Vapor-cooled current leads
Poppet valve
Relief valve Transfer tube
Filter vertical motion of tube
Reservoir (4.2 K)
Pumping
line Hydraulic
1 communication
Feed
valve Current links
Magnet vessel
Evaporator (1.8 K)
2 3
Vacuum
boundary
J-T valves (coarse & ne needle valves)
Qin Qout
mh mh
2 3
c.v.
J-T valve
Q1.8
Fig. 4.11 Heat balance for the evaporator. See Fig. 4.10
for locations of Points 2 and 3 in the cryostat.
238 CHAPTER 4PROBLEMS & DISCUSSIONS
= Cp /Cv for a perfect gas is 5/3; v4 is the specic volume at Point 4, which for
helium at 300 K and 12.3 torr is 371 m3/kg. With P4 = 12.3 torr = 1.64 103 Pa,
P5 /P4 = 61.8, and mh = 0.001 kg/s (1 g/s), we obtain: Ps = 6400 W (or 9 hp,
horsepower). Note that this is for an ideal case; the power requirement for a real
pump is roughly 20 kW.
CRYOGENICSPROBLEMS & DISCUSSIONS 239
* Although for many years until his death a faculty member of Boston University, the biochemist
Isaac Asimov (19201992) devoted his life writing for the general public on virtually every
subject conceivable. In the sci- genre his books include the Galactic Empire series (Pebble in
the Sky, 1950) and the Robot series (I, Robot, 1950).
242 CHAPTER 4PROBLEMS & DISCUSSIONS
9
6
7
4
8
11 10
6
5
4
3
13
2
14
1.5 m
12
1
75-mm
RT bore
a
a j j
Heat Exchanger 1
b i
1st Stage
Q1 c i
Heat Exchanger 2
Cryocooler
Q2 d h
e
2nd Stage f g
J-T Valve
Fig. 4.13 Schematic drawing of a cryocooler-based mini helium liqueer.
Figure 4.13 shows schematic details of the mini liqueer, comprising the warm-to-
cold stream (mass ow rate mhe ; pressure 10 atm) and cold-to-warm helium vapor
stream (mv ; 1 atm), and the 1st and 2nd stages of a cryocooler as cooling sources.
The processes are assumed ideal, e.g., no pressure drop; perfect heat exchange
with no temperature dierence between the two streams.
Warm-to-Cold Helium StreamIndicated by solid arrows in Fig. 4.13
Point a. Mini Liqueer entrance, from Cold trap (13 in Fig. 4.12): Ta = 295 K;
Pa = 10 atm (1 MPa).
Point a. Heat Exchanger 1 entrance: thermodynamically identical to Point a.
Point b. Heat Exchanger 1 exit (1st Stage entrance): Tb .
Point c. Heat Exchanger 2 entrance (1st Stage exit): Tc . Between Points b and
c, the 1st Stage extracts Q1 from the helium stream.
Point d. Heat Exchanger 2 exit (2nd Stage entrance): Td .
Point e. 2nd Stage exit (J-T Valve entrance): Te . Between Points d and e, the 2nd
Stage extracts Q2 from the helium stream.
Point f. J-T Valve exit: Tf = 4.22 K; Pf = 1 atm; mv ; mp (liquefaction mass
rate). LHe to Point g, indicated by the dotted arrow.
Cold-to-Warm Helium StreamIndicated by dashed arrows in Fig. 4.13
Point g. To LHe transfer line (9 in Fig. 4.12): Tg = 4.22 K; Pg = 1 atm; m .
Point h. Heat Exchanger 2 entrance: Th = 4.22 K; Ph = 1 atm; mv and mr =
mp m (LHe return mass rate).
Point i. Heat Exchanger 2 exit: Ti = Tc ; Pi = 1 atm.
Point i. Heat Exchanger 1 entrance: thermodynamically identical to Point i.
Point j. Heat Exchanger 1 exit: Tj = 295 K; Pj = 1 atm.
Point j. To warm helium exit line (10 in Fig. 4.12).
244 CHAPTER 4PROBLEMS & DISCUSSIONS
c) The performance data of a 2-stage cryocooler shown in Fig. 4.6, indicate that
this cryocooler with a refrigeration capacity of 1 W at 4.2 K delivers a cooling
power of 4 W at 7 K. Thus a cryocooler with a capacity of 7.44 W at 7 K would
most likely deliver a cooling power of 2 W at 4.2 K. Cryocoolers with a cooling
power of 2 W at 4.22 K are expected to be commercially available by 2010.
d) An equation similar to Eq. S2.3 may be applied between Points b and c:
mhe hhe (46 K,10 atm) = Q1 + mhe hhe (30 K,10 atm) (S2.4)
From Eq. S2.4 we nd:
Q1 = mhe [hhe (46 K,10 atm) hhe (30 K,10 atm)]
= (1 g/s)(252 J/g 168 J/g) 84 W
The performance data of Fig. 4.6 indicate that Q1 84 W is not achievable with a
cryocooler having the performance data of Fig. 4.6 even if its 4.22-K performance
is increased to 2 W. A cryocooler with an enhanced 2nd stage refrigeration power
will be required.
e) The total enthalpy decrease of the warm-to-cold stream from Points c to d
must be equal to the total enthalpy increase of the cold-to-warm stream from
Points h to i:
mhe [hhe (30 K,10 atm) hhe (8 K,10 atm)]
= (mv + mr )hv (30 K,1 atm) mv hv (4.22 K,1 atm) + mr hl (4.22 K,1 atm)
(S2.5)
246 CHAPTER 4PROBLEMS & DISCUSSIONS
mhe [hhe (295 K,10 atm) hhe (46 K,10 atm)] = (1 g/s)(1550.0 J/g 253.9 J/g)
1296 W (S2.7a)
(mv + mr )[hhe (295 K,1 atm) hhe (30 K,1 atm)]
(0.798 g/s + 0.144 g/s)(1547.0 J/g 170.2 J/g)
1297 W/s (S2.7b)
Equations. S2.7a and S2.7b are equal, within round-o errors of 0.1 %.
g) Applying Eq. 4.10, we obtain an ideal compressor power requirement, Ps :
1
Pa
Ps = mh (Pj vj) 1 (S2.8)
1 Pj
= Cp /Cv for a perfect gas is 5/3; vi is the specic volume at the compressor inlet
of Pj = 1 atm, which for helium at 295 K is 6 m3/kg. With Pj = 1 atm 1105 Pa,
Pa /Pj = 10, and mh = 0.001 kg/s (1 g/s), we obtain: Ps = 2.3 kW (or 3 hp,
horsepower). Note that this is for an ideal case; the power requirement for a real
pump would be roughly 7 kW.
Of course in a real mini helium liqueer, the liquid yield, because chiey of imper-
fection of the heat exchangers and pressure drops in both the warm-to-cold and
cold-to-warm helium streams, would be smaller than that of an ideal liqueer:
mp (and thus m ) will be o by a factor of perhaps as much as two, respectively,
from 0.202 g/s (6 liters/h) and 0.058 g/s (1.7 liter/h) computed above.
CRYOGENICSPROBLEMS & DISCUSSIONS 247
MAGNET CRYOCIRCULATOR
CIRCULATOR
RADIATION
1st STAGE
SHIELD LOAD
Top/Top 0
CIRCULATOR
MAGNET 2nd STAGE
CHAMBER LOAD
Answer to TRIVIA 4.2 The French chemist Georges Claude (18701960) showed that
electric discharge through inert gases could produce light, the start of neon lights, making
him rich; now more well-known as the developer of an eponymous refrigeration cycle.
250 CHAPTER 4PROBLEMS & DISCUSSIONS
VACUUM GAUGE
TURBO- MECHANICAL
MOLECULAR CRYOSTAT PUMP
PUMP
Table 4.10 lists approximate values of Dth and corresponding sd for sd = 10 mm for
solid neon (SNe), solid nitrogen (SN2 ), and copper (Cu) in the range 460 K, based
on (T ), cp (T ), and k(T ) data [4.334.35]. Over this temperature range thermal
diusivities of SN2 are 3 to 5 orders of magnitude less than those of copperheat
penetrates into SN2 much more slowly than into copper. For example, at 30 K, as
given in Table 4.10 for sd = 10 mm, sd = 46 s in SN2 , whereas it is only 1.3 ms for
Cu. However, SN2 can absorb much more heat per unit volume than Cu.
Slow Heating
If the heat required to be absorbed by a volume of solid cryogen of a reason-
able thermal diusion distance can be absorbed gradually, over a period of time
much longer than sd , the entire solid will remain nearly uniform in temperature.
Indeed, as noted above, solid cryogen appears best applied to magnets that are
generally operated in persistent mode such as those for MRI and NMR. As stud-
ied in PROBLEM 4.3 below, in a solid-cryogen-cooled magnet which has a heating
time of hours over a diusion distance of 12 cm, the entire solid-nitrogen volume
may be assumed to be at a uniform temperature.
Transient Heating
Under rapid transient conditions only a very thin layer of solid cryogennote that
sd sd (from Eq. 4.21)is eective in absorbing heat that would otherwise
heat the magnet winding. Even with this limitation, and despite the probable
occurrence of thermal dry-out (below), SN2 has proven eective in suppressing
temperature rises in HTS test samples subjected to transient heating [4.36, 4.37].
Recently, a group in Kyoto University has shown that a phenomenon they term
thermal dry-out occurs, creating a large temperature gap across the contact
between the transiently heated surface and solid nitrogen [4.384.40]. Thermal
dry-out and a solution to overcome this phenomenon are discussed below.
256 CHAPTER 4PROBLEMS & DISCUSSIONS
29
Temperature [K] 28 SN2 ONLY
27
SN2 -LNe MIXTURE
26
25
0 100 200 300 400 500 600
Time [s]
Fig. 4.17 Temperature vs. time plots of an HTS strip under over-current disturbances.
Solid curve: with the HTS strip cooled by SN2 only; dashed: with the strip cooled by
an SN2 -LNe mixture; dotted: initial temperature at 25.1 K [4.384.40].
C. Thermal Dry-Out
Nakamura and his colleagues in Kyoto University have experimentally demon-
strated that when the surface in thermal contact with SN2 is subjected to a large
heating ux, thermal dry-out occurs [4.384.40]. For solid nitrogen at 60 K, for ex-
ample, a thermal dry-out begins at a power ux of 1.5 W/cm2 . Apparently, a
thin vapor layer at the interface is responsible for this temperature discontinuity.
Figure 4.17 shows temperature vs. time plots for an HTS strip under over-current
disturbances, each disturbance lasting up to 600 s. The solid curve corresponds to
a typical run for the strip in contact with solid nitrogen only. The dotted horizontal
line, at 25.1 K, indicates the initial temperature of both the strip and solid nitrogen.
Beyond 400 s, at which the heat ux is 14.3 W/cm2 with a temperature dierence
(T ) in excess of 3 K, a thermal runaway occurs that, were the over-current to
persist, would lead to conductor damage.
60
50
Cu Pb SN2
Temperature [K]
40
30
20
SNe
10
LHe
0
0 5 10 15 20
Time [h]
Fig. 4.18 T (t) plots with an initial temperature of 4.2 K and a nal temperature of
60 K (25 K, SNe) for a 1-liter volume each of Cu, Pb, SNe, and SN2 subjected to a
constant heat input of 1 W. The duration required to boil o 1 liter of liquid helium
(LHe), 0.7 h, is indicated by the dotted horizontal line at 4.2 K.
260 CHAPTER 4PROBLEMS & DISCUSSIONS
Cp (10 K) + Cp (15 K)
h(15 K) h(10 K) (15 K 10 K)
2
(0.175 J/cm3 K + 0.475 J/cm3 K)
(5 K)
2
1.625 J/cm3
Thus the 1015 K warm-up time for 15-liter solid nitrogen with 0.25-W heat input,
[t(10 K15 K)]0.25 W
15liter may be given by:
c) The volume of solid nitrogen, VN2 , in a cylinder of i.d. D, length , and layer
thickness rN2
D, is:
VN2 = DrN2 (S3.1)
With VN2 = 15000 cm3 , D = 90 cm, = 30 cm inserted in Eq. S3.1, we have:
We may equate rN2 = 1.8 cm to sd of Eq. 4.21 and with Dth 55102 cm2/s, a
rough average value between 10 K and 15 K, solve Eq. 4.21 for sd :
2
1 sd
sd = (4.21)
Dth
2
1 1.8 cm
= 0.6 s
(55102 cm2/s)
The diusion time of 0.6 s is much less than the 30-h warm-up time. Thus, the
assumption of uniform temperature within this 15-liter SN2 is valid.
d) We reverse the order of the two approaches to the solution given in b). Thus,
using the Cp (T ) data for solid neon given in Fig. 4.2 and by taking the area
underneath the Cp (T ) curve for SNe between 10 K and 15 K, we have:
Cp (10 K) + Cp (15 K)
h(15 K) h(10 K) (15 K 10 K)
2
(0.400 J/cm3 K + 0.875 J/cm3 K)
(5 K)
2
3.2 J/cm3
CRYOGENICSPROBLEMS & DISCUSSIONS 261
7.6 liter 1
2 15 liter
From Fig. 4.18 we note that for 1-liter solid neon, T (t 0.33 h) = 10 K and T (t
liter 0.89 h for SNe; for 1-liter SN2 :
1.22 h) = 15 K, resulting in [t(10 K15 K)]11 W
[t(10 K15 K)]1 liter 0.486 h. We thus have:
1W
(0.486 h)
VNe (10 K15 K) = (15 liter) 8.2 liter 1
2 (15 liter)
(0.89 h)
e) From Fig. 4.18 we note that for 1-liter solid nitrogen, T (t 0.6 h) = 15 K and
T (t 17.75 h) = 60 K, or [t(15 K60 K)]11 W liter 17.15 h. For an average heat
input of 3.3 W and 15-liter SN2 , we have:
3.3 W 15 liter 1W
[t(15 K60 K)]15liter = [t(15 K60 K)]11 liter
W
1 liter 3.3 W
(15)(0.303)(17.15 h)
= 78 h 80 h
Answer to TRIVIA 4.3 ii). The German physicist Otto von Guericke (16021686)
of Magdeburg hemispheres fame was the rst with a pump, though the Italian
physicist Evangelista Torricelli (16081647) achieved, in 1643, the rst man-made
vacuum by using a glass tube and a dish full of mercury.
262 CHAPTER 4PROBLEMS & DISCUSSIONS
Thus the contribution of the extra enthalpy at 35.61 K more than doubles (2.1
times) the warm-up time for the same temperature increment of 5 K.
i) Equation 4.21 is used to compute sd for Dth = 2.4 103 cm2/s and sd =
0.04 cm and demonstrate that this sd indeed is 0.1 s. Thus:
2
1 sd
sd = (4.21)
Dth
2
1 0.04 cm
= 0.067 s
(2.4103 cm2/s)
0.1 s
j) The maximum power ux, psd , that may inject energy in 0.067 s into a 0.04-
mm thick layer of SN2 from 30 K to 35 K may be given by:
sd
psd = [h(35 K) h(30 K)]
sd
(0.04 cm)
= (7.0 J/cm3 ) = 4.2 W/cm2
(0.067 s)
CRYOGENICSPROBLEMS & DISCUSSIONS 263
This linear change occurs in three dimensions. If the windings were to expand
equally in all directions, there would be no deterioration of homogeneity. In reality,
because every winding is anisotropic, a degradation of the eld homogeneity is
expected but to what extent depends on how much the medium is anisotropic,
and consequently this amplitude cannot accurately be predicted.
Note that (L/L ) in all materials increases not only with Top but also with
initial temperature. Therefore, because of this potential eld homogeneity degra-
dation during an operating temperature excursion, for solid cryogen cooled mag-
nets applied to NMR and MRI, it may be prudent to keep their initial operating
temperatures to below 20 K and Top no greater than 10 K.
264 CHAPTER 4PROBLEMS & DISCUSSIONS
Temperature Spectrum
Figure 4.19 shows a temperature spectrum over the range from 100 picokelvins
(100 pK) to 1 gigakelvin (1 GK). Note that superconducting magnets operate over
a temperature span that extends less than 2 log scales, 1 K100 K.
T [K]
1G Interior of the hottest star
100M Thermonuclear fusion
Suns interior
10M
1M
Nuclear ssion (300,000 K)
100k
10k
Luminous nebulae; Tungsten melts (3683 K)
1k Superconductors reacted (1000 K)
Ice point (273.15 K=0 C)
100 Liquid nitrogen boils (77.3 K) HTS magnets
10
Liquid helium boils (4.2 K) LTS magnets
1 3
Pumped He (0.5 K)
100m
10m Dilution refrigerators
3
1m He becomes superuid (2 mK)
100
10 16 K, by nuclear adiabatic demagnetization (1956)
1
100n 150 nK, by laser cooling (1995)
10n
1n
450 pK, by laser cooling (2003)
100p 100 pK, by nuclear adiabatic demagnetization (2001)
Capacitance Thermometers
Though not included in the above discussion, capacitance thermometers such as
strontium titanate sensors are used because of their very low magnetic eld de-
pendence. Several other types based on glass or plastic materials are popular for
temperatures below 1 K.
Fahrenheit Scale
Passage from Linus Paulings College Chemistry [4.46]
The Fahrenheit scale was devised by Gabriel Daniel Fahrenheit (16861736), a
natural philosopher who was born in Danzig (now Gdansk) and settled in Holland.
He invented the mercury thermometer in 1714; before then alcohol had been used as
the liquid in thermometers. As the zero point on his scale he took the temperature
produced by mixing equal quantities of snow and ammonium chloride. His choice of
212 o for the boiling point of water was made in order that the temperature of his body
should be 100 oF. The normal temperature of the human body is 98.6 oF (37.0 C);
perhaps Fahrenheit had a slight fever while he was calibrating his thermometer.
268 CHAPTER 4PROBLEMS & DISCUSSIONS
z +z
(T )I2
A
mI cp (T )T Ak(T ) dT
dz z
The power input QI into the liquid boils o the liquid at a rate mI given by:
QI
mI = (4.31a)
hL
where hL is the liquid heliums latent heat of vaporization [J/kg]. Combining
Eqs. 4.30 and 4.31a and solving for mI , we obtain:
k0 0
mI = I (4.31b)
cp0 hL
Inserting mI given by Eq. 4.31b into Eq. 4.30 and solving for QI /I , we derive:
QI hL k0 0
= (4.32a)
I cp0
Note that QI depends neither on , the lead active length between the bottom
end (z = 0) and top end (z = ), nor on A, the lead conductor cross section. It
is, however, directly proportional to I , the leads rated current. Note that if the
lead carries I < I , the cold-end heat input will not be (I/I )QI .
Throughout this discussion T0 is the cold-end temperature of a vapor-cooled lead,
assumed below to be T0 = 6 K. However, in a real vapor-cooled lead, this cold end
(z = 0) may be located high above the top of the magnet (and minimum liquid
level)in some cases 25 cm or even higher, because it is often desirable to provide
a reservoir of liquid helium above the magnet. The cold end is then electrically
and thermally connected to a superconductor-shunted copper extension, the other
end (z < 0) of which penetrates into the liquid. The superconductor shunt (often of
copper/NbTi composite) carries I to the magnet, and the copper conducts QI to
the liquid. The copper extension must have enough cross sectional area to conduct
QI , even when exposed out of liquid helium over its entire span, without raising
T0 too high, thus enabling the superconductor shunt to carry I superconductively.
270 CHAPTER 4PROBLEMS & DISCUSSIONS
d2 T dT
Ak 2
m0 cp =0 (4.36)
dz dz
where k is the average thermal conductivity and cp is the average specic heat of
helium, both in the temperature range, T0 to T .
T
1
k = k(T ) dT (4.37)
T T0 T0
Table 4.18 (p. 307) presents values of k for four materials (G-10, stainless steel
304, brass, and copper) over three common temperature ranges in cryogenic ap-
plications: 480 K; 4300 K; and 80300 K.
With the boundary conditions, T (z = 0) = T0 and Ak(dT/dz)z=0 = m0 hL , T (z)
may be given by:
hL m0 cp z
T (z) = T0 + exp 1 (4.38a)
cp Ak
Thus:
hL m0 cp
T () T = T0 + exp 1 (4.38b)
cp Ak
Solving Eq. 4.38b for m0 cp /Ak, we have:
m0 cp cp (T T0 )
= ln +1 (4.39a)
Ak hL
Equation 4.44 assumes a linear temperature gradient along the current lead. Com-
bining Eqs. 4.43b, where cu is given in the unit of m, and Eq. 4.35b, we have:
V 2.3107 cu V (4.45)
That is, Vot is the same among leads optimized for any rated current.
For copper, cu is 2.51010 m (below 50 K) and varies linearly with tem-
perature above 50 K, being 1.75 108 m at 273 K. From Eq. 4.44, we have:
cu 0.4108 m. The voltage drop across an optimum lead at its rated current
is independent of rated current; for a copper lead it is 100 mV. Thus, V 100 mV
at 100 A for a 100-A lead and at 10 kA for a 10-kA lead; when the warm end is
cool, which often happens in vapor-cooled leads, V can be less than 100 mV.
This conclusion agrees, within an order of magnitude, with experimental data for
vapor-cooled copper leads in the 1 kA30 kA range [4.54].
Dierentiating Eq. 4.53 with respect to /A and setting it to 0 for (I /A)dr , we
obtain (I /A)dr for an optimal dry lead that minimizes [QI ]dr (Eq. 4.12, p. 240):
I 2k(T T0 )
= (4.12)
A dr
From Eqs. 4.12 and 4.53, we obtain an expression for [QI ]dr :
[QI ]dr = I 2k (T T0 ) (4.54)
With k = 460 W/m K and = 1 108 m (for copper in the range 80300 K),
Eq. 4.54 gives: [QI ]dr /I 45 mW/A; for brass, Eq. 4.54 gives 32 mW/A [4.19].
That is, for a conduction-cooled lead, brass is preferable to copper.
B. HTS Extension
After the discovery of YBCO, HTS lead extensions were built for the cold portion
(below 80 K) of a normal metal lead [4.57, 4.58]. The heat input to the magnet
environment at T0 by an HTS extension, [QIhts ]dr , of average thermal conduc-
tivity khts , cross sectional area Ahts , length hts , and warm-end temperature Twhts
is ideally by conduction, for which:
(Twhts T0 )
[QIhts ]dr = khts Ahts (4.55)
hts
In a real normal-metal/HTS lead, the normal metal heat at its cold end, given by
Eq. 4.54 with T0 Twhts , generally is absorbed by the 1st stage of a cryocooler,
the 2nd stage of which maintains the magnet at T0 . Also, the HTS lead, whether
of bulk or tape, must be protected; protection of Bi2223 tape current lead is
discussed in DISCUSSION 8.5. Besides the HTS lead considered here, there also
are vapor-cooled HTS versions, as studied in PROBLEMS 4.44.6.
It is assumed here that heat transfer between the lead and helium vapor is perfect,
with zero temperature dierence between the two.
In Eq. 4.56, [Am ]fs is the total cross sectional area in the FSV lead of HTS matrix
(normal) metal, e.g., Ag-Au in Bi2223 tape: [Am ]fs = Nfs am , where am is the
matrix cross sectional area of an individual tape; Nfs is the number of parallel
HTS tapes; km (T ) is the thermal conductivity of the matrix metal; [mh ]fs is the
helium ow rate; and cp (T ) is the specic heat of helium. We linearize Eq. 4.56
by making km (T ) = k and cp (T ) = cp , both constants, respectively, temperature-
averaged km (T ) and cp (T ) over the range from T0 at z = 0 to T at z = , Thus:
d2 T dT
[Am ]fs k 2
[mh ]fs cp =0 (4.57)
dz dz
Equation 4.57 may be solved for T (z):
T T0 []fs (z/)
T (z) = T0 + e 1 (4.58)
e[]fs 1
[mh ]fs cp
[]fs = (4.59)
k[Am ]fs
a) Show that [Qin ]fs , the heat input to the helium bath (z = 0), is given by:
k[Am ]fs hL cp (T T0 )
[Qin ]fs = ln +1 (4.60)
cp hL
where hL is the latent heat of vaporization of helium. Note that [Qin ]fs , unlike
QI I (from Eq. 4.32a) for the vapor-cooled copper lead, is proportional
not to current but to Nfs through [Am ]fs = Nfs am .
b) Show that the dierence between [Qin ]fs and [Q ]fs , the heat conducted into
the FSV lead from the copper lead at z = , is equal to the dierence in
enthalpy energy of the euent helium vapor at z = (T ) and z = 0 (T0 ):
Answer to TRIVIA 4.4: Pluto (40, in K); MgB2 (39); CrBr3 (37); N2 (35.6).
CRYOGENICSPROBLEMS & DISCUSSIONS 277
Evaluating dT/dz|0 from Eq. 4.58 and inserting []fs given by Eq. 4.59, we obtain:
dT T T0 []fs
k[Am ]fs = k[Am ]fs
dz 0 e[]fs 1
T T0
= [mh ]fs cp = [mh ]fs hL (S4.2)
e[]fs 1
Solving Eq. S4.2 for e[]fs , we have:
cp (T T0 )
e[]fs = +1 (S4.3a)
hL
cp (T T0 )
[]fs = ln +1 (S4.3b)
hL
Combining Eqs. 4.59 and S4.3b, we have:
k[Am ]fs cp (T T0 )
[mh ]fs = ln +1 (S4.4)
cp hL
Substituting Eq. S4.4 into S4.1, we obtain:
k[Am ]fs hL cp (T T0 )
[Qin ]fs = ln +1 (4.60)
cp hL
Strictly, Eq. 4.60 is valid only when the system bath is continuously replenished so
that its liquid level is kept constant; if it is not, a correction factor of (1v / ) is
needed for [mh ]fs hL [4.63], where v and are respectively the densities of vapor
and liquid at saturation. At 4.2 K the density ratio is 0.135: (1v / ) = 0.865.
b) Conduction heat into the lead at z = from above, [Q ]fs , is given by:
dT T T0
[Q ]fs = k[Am ]fs = [mh ]fs cp e[]fs (S4.5a)
dz e[]fs 1
cp (T T0 )
= [mh ]fs hL +1
hL
= [mh ]fs cp (T T0 ) + [mh ]fs hL (S4.5b)
Combining the outboard terms of Eq. S4.1 and Eq. S4.5b, we note that:
[Q ]fs [Qin ]fs = [mh ]fs cp (T T0 ) (4.61)
Equation 4.61 shows that the three power terms are balanced. Namely, the dier-
ence in heat entering the lead at z = and that leaving the lead at z = 0 is exactly
equal to the gain in thermal power of the euent helium vapor.
278 CHAPTER 4PROBLEMS & DISCUSSIONS
Advantage of CSV
For the same rated transport current, the CSV lead contains fewer (Ncs ) expensive
HTS tapes than its FSV counterpart (Ncs < Nfs ). Furthermore, as discussed in
PROBLEM 4.6 for 6-kA leads, the cold-end heat input of the CSV lead, despite
its Joule dissipation near the warm end, is less than that of the FSV counterpart.
Ic (T )
Ic0
I
Ic
I Ic
0 T
0 T0 Tcs T Tc
Fig. 4.22 Ic (T ) plot (solid line), approximated to be linear with temperature for
HTS. The dotted horizontal line is the leads rated current, I . The dashed line
indicates the current through the matrix, IIc (T ), in the current-sharing region.
CRYOGENICSPROBLEMS & DISCUSSIONS 279
1/(2cs )
[ ]cs cp (Tcs T0 )
e 2 = +1 (4.70a)
hL
1/2
[ ]cs cs cp (Tcs T0 )
e 2 = +1 (4.70b)
hL
We obtain:
1+ cs
1 hL cp (Tcs T0 ) cp (Tcs T0 ) 2cs
cs
ln +1 +1 =
cs cp (T Tcs ) hL hL sin cs (1cs )
(4.71)
Note that for a set of parameters of a CSV leadx , k, [Am ]cs , I , Ic the
only unknown parameter in Eq. 4.71 is cs , because the only unknown in cs
(Eq. 4.68c) is also cs : cs must satisfy both Eqs. 4.68c and 4.71.
d) Show that over the fully superconducting section (0 z cs ) of this CSV
lead, the three power terms are balanced:
where [Qcs ]cs is the conduction heat entering the fully superconducting sec-
tion at z = cs .
e) Show that in the current-sharing region, the four power terms[Q ]cs , the
conduction into the CSV lead at z = ; Qj , the total Joule dissipation gen-
erated within the region; [Qcs ]cs , the conduction out of the lead at z = cs ;
and convective cooling, mh cp (T Tcs )are balanced.
f ) Show that:
[Q ]cs + Qj [Qin ]cs = [mh ]cs cp (T T0 ) (4.74)
Tcs T0 e[ ]cs cs T0 Tcs
T (0) = 1+
e[ ]cs cs 1 Tcs T0
Tcs T0 T0 (e[ ]cs cs 1)
= = T0 (S5.1)
e[ ]cs cs 1 Tcs T0
Tcs T0 [ ]cs cs e[ ]cs cs T0 Tcs
T (cs ) = e +
e ]cs cs 1
[ Tcs T0
Tcs T0 Tcs (e[ ]cs cs 1)
= = Tcs (S5.2)
e[ ]cs cs 1 Tcs T0
b) We have:
k[Am ]cs dT
[Qin ]cs = (S5.3)
d =0
cp (Tcs T0 )
e[ ]cs cs = +1 (4.65a)
hL
1 cp (Tcs T0 )
[ ]cs = ln +1 (4.65b)
cs hL
Finally, by combining Eqs. S5.3 and 4.65b, and noting cs = cs , we have:
k[Am ]cs hL cp (Tcs T0 )
[Qin ]cs = ln +1 (4.64)
cp cs hL
Although cs < , [Qin ]cs < [Qin ]fs , given by Eq. 4.60 for FSV, because [Am ]cs /cs is
generally less, at most 80% of [Am ]fs / for FSV. Also note that Tcs T0 < T T0 ,
which also contributes to making [Qin ]cs < [Qin ]fs .
282 CHAPTER 4PROBLEMS & DISCUSSIONS
With [ ]cs given by Eq. 4.63 and e[ ]cs cs derivable from Eq. 4.70b, we obtain:
cp (Tcs T0 )
[mh ]cs cp hL + 1 (Tcs T0 )
[Qcs ]cs = k[Am ]cs
k[Am ]cs c (T T )
p cs 0 hL
cp (Tcs T0 )
= [mh ]cs hL +1 (S5.8)
hL
Because [Qin ]cs = [mh ]cs hL , combining this with Eq. S5.8, we have:
That is, within the fully superconducting section of a CSV lead, power is balanced.
e) [Q ]cs , the conduction into the lead at z = may be evaluated from Eq. 4.67:
k[Am ]cs dT k[Am ]cs (T Tcs )
[Q ]cs = = [ ]
d e cs sin cs (1cs )
e[ ]cs 12 [ ]cs sin cs (1 cs ) + cs cos cs (1 cs )
By inserting T (), given by Eq. 4.67 (with = z/), into the last term of the
left-hand side of Eq. 4.66 and integrating from = cs to = 1, we evaluate Qj :
1
x I (I Ic ) [ ]cs
Qj = [ ]cs
e 2 sin cs ( cs ) d (S5.11)
[Am ]cs e 2 sin cs (1cs ) cs
1cs
[ ]cs [ ]cs
e 2 x sin cs x cs cos cs x (S5.12a)
2 0
4x I (I Ic )
= [ ]cs
[Am ]cs e 2 sin cs (1cs ) ([ ]2cs +4cs
2)
[ ]cs [ ]cs [ ]cs [ ]cs cs
e 2 sin cs (1 cs ) cs e 2 cos cs (1 cs ) + cs e 2
2
(S5.12b)
From Eq. 4.68b we may derive:
4x I (I Ic )2
[ ]2cs + 4cs
2
= (S5.13)
k[Am ]2cs (T Tcs )
1000
Pure Ag
Thermal Conductivity [W/cm K]
100
0.085at%Au
10
0.22at%Au
1at%Au
1
2.9at%Au
11at%Au
0.1
5 10 20 50 100 200 300
Temperature [K]
Fig. 4.23 Thermal conductivity vs. temperature data of selected Ag-Au alloys [4.78].
10
11at%Au
1
2.9at%Au
Resistivity [ cm]
1at%Au
0.22at%Au
0.1
0.085at%Au
0.01
Pure Ag
0.001
1 2 5 10 20 50 100 200 300
Temperature [K]
Fig. 4.24 Electrical resistivity vs. temperature data of selected Ag-Au alloys [4.78].
CRYOGENICSPROBLEMS & DISCUSSIONS 285
kn and [An ]fs are, respectively, the thermal conductivity and total cross sectional
area of the normal metal tapes paralleled to the HTS tapes. Note that for the sake
n
of protection, heat input to the liquid helium is increased because [Qin ]fs > [Qin ]fs ,
as given by Eq. 4.60. []fs , given by Eq. 4.59, is modied to:
n
n
[mh ]fs cp
[ ]fs = (4.77a)
n
[kA] fs
cp (T T0 )
= ln +1 (4.77b)
hL
n
where [mh ]fs is the euent helium mass ow rate for an FSV lead paralleled with
normal metal tapes.
[Q ]cs + Qj [Qin ]cs = mh cp (T T0 ) (4.74)
CSV For a CSV lead, [Qin ]cs , given by Eq. 4.64, is modied to [Qin ]ncs :
n hL cp (Tcs T0 )
[kA]
n cs
[Qin ]cs = ln +1 (4.78)
cp cs hL
n = k[Am ]cs + kn [An ]cs
[kA] (4.79)
cs
where [An ]cs is the total cross sectional area of normal metal tapes used in a CSV
lead. Note that cs remains the same whether the CSV lead is paralleled with
normal metal tapes or not. Note also that because [Am ]fs and [Am ]cs are dierent,
[An ]fs and [An ]cs can be dierent. [ ]cs also is dierent:
[mh ]ncs cp
[ ]ncs = (4.80a)
n
[kA] cs
1 cp (Tcs T0 )
= ln +1 (4.80b)
cs hL
where [mh ]ncs and cs (= cs /) are, respectively, the helium mass ow rate and
dimensionless length from z = 0 at which the current-sharing region starts for a
CSV lead paralleled with normal metal tapes.
286 CHAPTER 4PROBLEMS & DISCUSSIONS
where
n = x [An ]cs + n [Am ]cs
[A] (4.82)
cs
Electricity
Passage from Helen Davis The Chemical Elements [4.79]
Among the special uses for certain elements, none is more remarkable than the
dependence of electrical industry upon copper. Its softness and ductility allow the
metal to be drawn into wire which carries the electric current, and its electrical
conductivity is greater than that of any other suitable metal except silver [and now,
obviously, superconductors].
Alessandro Volta (17451827) in 1800 wrote to Sir Joseph Banks,* describing a
piece of apparatus he had built: The apparatus of which I speak to you, and which
without doubt will surprise you, is only the assemblage of a number of good conduc-
tors of dierent sorts, arranged in a certain number, 30, 40, 60 pieces, preferably,
of copper, or better silver, each touching a piece of tin, or, which is better, of zinc,
and an equal number of layers of water, or some other liquid which should be a
better conductor than simple water like salt water, lye solution, etc., or pieces of
cardboard, leather, etc., well soaked in these liquids; of which pads interposed be-
tween each couple or combination of the two dierent metals, alternating with each
set, and always in the same order, of the three sorts of conductors; that is all there
is to my new instrument.
Volta drew sparks from his battery, felt light shocks in the wrists of both hands
when he touched the top and bottom plates with wet ngers, and admired the way
it would, after each discharge, re-establish itselfan inexhaustible charge, a per-
petual eect. He did not notice the slight corrosion of the metal which accounts
for the source of the electric current.
* The English botanist Sir Joseph Banks (17431820) was that convenient but rare phe-
nomenon, a scientist of great independent wealth who spends that wealth liberally in the
support of science. While still a student at Oxford he nanced a lectureship in botany,
which is how the subject came to be taught there for the rst time. [4.27]
CRYOGENICSPROBLEMS & DISCUSSIONS 287
In Eq. 4.84 kcu and cfl are, respectively, the T -averaged copper thermal conduc-
tivity and the uid specic heat; cu and are, respectively, the resistivity tem-
perature coecient and resistivity at the cold-end temperature, T (here 77 K).
Because over most of the 77293 K range the last term in the left-hand side of
Eq. 4.84 is negligible compared with the third term, Eq. 4.84 may be simplied to:
2
d2 T mfl cfl cu dT 2
I cu cu
+ (T T ) = 0 (4.85)
d 2 kcu Acu d kcu A2cu
Combining Eqs. 4.86 and 4.90, and then invoking Eq. 4.87c, we obtain:
I2 cu cu (Tcu T ) 1 cu
Qj = e sin cu d
Acu ecu sin cu 0
I2 cu cu (Tcu T )
= 2 + 2 )
(cu cu cot cu )
Acu (cu cu
Note that, as expected, Eq. 4.92 states that the Joule dissipation generated over
the entire length of the copper lead is balanced by the cooling power of uid
introduced at the cold end of the lead.
CRYOGENICSPROBLEMS & DISCUSSIONS 289
where
mfl cfl (cu q )
cu = (4.95a)
2kcu Acu
I2 cu (cu q )2 )2
cu = (cu (4.95b)
kcu A2cu
There is a unique combination of mfl and q that satises both Eqs. 4.93 and 4.94.
100Xwt %B WA
Xat %B =
(100 Xwt %B)WB + Xwt %B WA
290 CHAPTER 4PROBLEMS & DISCUSSIONS
As studied in DISCUSSION 4.14 (Eq. 4.32b ), a copper vapor-cooled 6-kA lead would
have had Qin 6 W; thus, 0.08 W is indeed an impressive improvement.
c) With [Qin ]fs = 0.0798 W, we obtain [mh ]fs = [Qin ]fs /hL = 0.00385 g/s. Heat
input at z = is given by the rst form of Eq. S4.5b:
cp (T T0 )
[Q ]fs = [mh ]fs hL +1
hL
(5.28 J/g K)(77.3 K4.2 K)
= (0.00385 g/s)(20.7 J/g) +1
(20.7 J/g)
= 1.566 W
Thus: [Q ]fs[Qin ]fs = 1.486 W, which essentially equals [mh ]fs cp (TT0 ) = 1.487 W.
b) Verify numerically that the power terms are balanced for this lead too.
n hL c (T T )
[kA]
n fs p 0
[Qin ]fs = ln +1 (4.75)
cp hL
n
We have thus: [Q ]fs [Qin ]fs
n
= 8.106 W [mh ]fs
n
cp (T T0 ) = 8.105 W.
CRYOGENICSPROBLEMS & DISCUSSIONS 293
n
sin cs (1cs ) = 1 (4.96a)
n
cs (1 cs ) = /2 (4.96b)
n
You may determine cs iteratively, rst guessing a value of cs , compute cs
n
from Eq. 4.96b, then insert cs and cs into Eq. 4.83 to see if Eq. 4.83 is indeed
satised. Because ic (T ) is assumed a linear function of T , an appropriate
starting value of cs into Eq. 4.96b may be 0.9 ( 69.3/77.3).
n n
c) Verify that cs found in b) agrees with cs computed by Eq. 4.81.
d) Compute [Qin ]ncs , the heat input to liquid helium at z = 0, given by Eq. 4.78.
e) Numerically compute each term appearing in the following power balance
equation and demonstrate that power is balanced.
[Q ]ncs + [Qj ]ncs [Qin ]ncs = [mh ]ncs cp (T T0 ) (4.97)
[Q ]ncs is the heat into the lead at z = ; [Qj ]ncs is the Joule dissipation within
the current-sharing region; and [Qcs ]cs is the heat out of the lead at z = cs .
[Q ]ncs and [Qj ]ncs are given by modied forms of Eq. S5.9 and Eq. S5.14:
n (T Tcs )
[kA]
[Q ]ncs = cs 1
2 [ ]ncs + cs
n n
cot cs (1 cs ) (4.98)
[ n (T Tcs )
kA]
2 [ ]cs cs cot cs (1 cs )
cs
[Qj ]ncs = 1 n n n
[ ]n
cs cs
n e 2
+ cs [ ]ncs (4.99)
e 2 sin cs n (1 )
cs
n
Note that because cs (1cs ) = /2 here, Eqs. 4.98 and 4.99 are simplied:
n (T Tcs )
[kA]
[Q ]ncs = cs
12 [ ]ncs (4.100)
n (T Tcs )
[kA] [ ]n
cs (cs 1)
n cs 1 n n
[Qj ]cs = 2 [ ]cs + cs e
2 (4.101)
294 CHAPTER 4PROBLEMS & DISCUSSIONS
n n
Table 4.14: Determination of cs and cs
Guess Value of Eq. 4.96b Left-Hand Side
n
cs cs * Eq. 4.83
0.9 15.708 32.3975
0.95 31.4159 28.2120
0.94 26.1799 28.9759
0.945 28.5599 28.5898
0.9451 28.6120 28.5821
0.94505 28.5859 28.5860
* sin cs
n
(1cs ) = 1 and thus cs
n
is equal to the left-hand side of Eq. 4.83.
CRYOGENICSPROBLEMS & DISCUSSIONS 295
d) With cs = 0.94505 and = 20 cm: cs = 18.9 cm. The current-sharing region
for this CSV lead thus spans 1.1 cm, from z = 18.9 cm to z = 20 cm. From Eq. 4.78:
n hL cp (Tcs T0 )
[kA]
n cs
[Qin ]cs = ln +1 (4.78)
cp cs hL
(0.698 W cm/K)(20.7 J/g) (5.28 J/g K)(69.3 K4.2 K)
= ln +1
(5.28 J/g K)(18.9 cm) (20.7 J/g)
n (T Tcs ) [ ]n
[kA] [ ]n
cs (cs 1)
cs cs
[Qj ]ncs = n
+ cs e 2 (4.101)
2
n (T Tcs )
3.035
[kA] 3.035(0.9451)
cs
= + (28.586)e 2
2
(0.698 W cm/K)(77.3 K69.3 K) 3.035
= + 26.30 = 7.7658 W
(20 cm) 2
With [mh ]ncs cp (T T0 ) = (0.02006 g/s)(5.28 J/g K)(77.3 K4.2 K) = 7.7743 W and
[Qin ]ncs = 0.4153 W, we obtain:
[Q ]ncs + [Qj ]ncs [Qin ]ncs = [mh ]ncs cp (T T0 ) (4.97)
n (T Tcs )
[kA]
[Q ]ncs = cs 1
[ ]
2 cs
n
+ n
cs cot n
cs (1 cs ) (4.98)
c) The values of cu and cu determined in a) are also valid for gaseous nitrogen
entering the cold-end (z = 0) at 77.3 K. The only dierence between helium and
nitrogen is cfl , which for the latter is 1.04 J/g K. Thus: mfl = 1.812 g/s.
This ratio is 0.77 mW/A for copper leads (Eq. 4.32b ). Note that for the brass
lead, despite its much greater resistivityby a factor of 100than coppers
at 4.2 K, [QI /I ]br is only 60% greater than QI /I (for copper).
c) Show that [Q0 /QI ]br , the ratio of cold-end heat input of a brass lead with
no current (I = 0) to that of the lead at I is given by:
Q0
0.21 (4.104)
QI br
Use kbr = 55 W/m K (Table 4.18); cp = 5.2 kJ/kg K; and cp0 = 6.0 kJ/kg K.
CRYOGENICSPROBLEMS & DISCUSSIONS 301
Inserting (t) = T (t) T to Eq. 4.106, we have d(t)/dt and solution (t):
d(t) 1
= 2br cp AI[1 + br (I I )] br I 2 (t)
dt A2 Cbr
+ 2br cp AT I[1 + br (I I )] [0 ]br I 2
(4.107a)
(t) = (I) 1 et/j (I) (4.107b)
(I) and j (I), both I-dependent constants, are, respectively, the temperature
rise and response time, which is assumed to be much longer than the period during
which the condition I > I is established. Solving for (I) and j (I), we have:
[0 ]br I 2 2br cp AT I[1+ br (I I )]
(I) = (4.108a)
2br cp AI[1+ br (I I )] br I 2
A2 Cbr
j (I) = (4.108b)
2br cp AI[1+ br (I I )]br I 2
302 CHAPTER 4PROBLEMS & DISCUSSIONS
e) Show that in the overcurrent region, [QI ]br , the heat input is given by:
TRIVIA 4.6 List the metals below in the descending order of their electrical resistivities.
i) Aluminum 1100 at 4.2 K; ii) Copper (RRR 20, 000) at 100 K;
iii) Copper (RRR 100) at 50 K; iv) Copper (RRR 100) at 4.2 K in a 20-T eld.
CRYOGENICSPROBLEMS & DISCUSSIONS 303
T
I cp0 [k0 ]br [0 ]br dT
= (S7.1)
A br hL T0 br (T )
With cp0 = 6.0 kJ/kg K and other property values inserted into Eq. S7.1, we have:
I (6.0 kJ/kg K)(22 W/m K)(21 n m)
= (0.971010 K/ m)
A br (20.7 kJ/kg)
Thus:
I
[ ]br 0.35107 A/m (4.102)
A br
QI hL [k0 ]br [0 ]br (20.4 kJ/kg)(22 W/m K)(21 n m)
= = (S7.2)
I br cp0 (6.0 kJ/kg K)
Thus:
QI
1.25 mW/A (4.103)
I br
c) From Eq. 4.41 with kbr = 55 W/m K (Table 4.18) and other values, including
ln[cp (T T0 )/hL + 1] 4.32, we have:
Q0 kbr hL cp0 cp (T T0 )
= ln +1 (S7.3)
QI br cp [ ]br
[k0 ]br [0 ]br hL
(55 W/m K) (20.7 kJ/kg)(6.0 kJ/kg K)
7
(4.32)
(5.2 kJ/kg K)(0.3610 A/m) (22 W/m K)(21 n m)
Thus:
Q0
0.21 (4.104)
QI br
f) With [I /A]br from Eq. 4.109 and [QI ]br from Eq. 4.110 at I = I :
I 2cp T
= br (S7.5a)
A br [0 ]br
Inserting [QI /I ]br from Eq. S7.2 into Eq. S7.6, we obtain:
I 2cp T [k0 ]br [0 ]br
= (4.111)
A br [0 ]br hL cp0
We may determine the value of T , the only unknown parameter in Eq. 4.111:
[0 ]br I hL cp0
T = (S7.7)
2cp A br [k0 ]br [0 ]br
(21 n m) (20.7 kJ/kg)(6.0 kJ/kg K)
= (0.358107 A/m)
2(5.19 kJ/kg K) (22 W/m K)(21 n m)
120 K
At I = 25 kA, we have, from Eq. 4.103, [QI ]br = 31.5 W. When carrying no
current, this 25-kA brass lead has, from Eq. 4.104, [Q0 ]br 0.21[QI ]br 6.6 W.
Thus, a total energy injected into the cold bath, Ebr , over a 300-s period at 75 kA
and over a 1800-s period at 25 kA, is given by:
Over this 2100-s period, a copper lead would have required a cold-end heat input of
33.8 W, to match the total energy Ebr = 72 kJ. For a vapor-cooled copper lead op-
erating continuously at I , this translates to a current rating of 35 kA (<75 kA).
Although no detailed study is available for vapor-cooled copper leads in overcur-
rent mode, it is quite possible that a copper lead too can be operated safely with
an overcurrent up to a reasonable value of I/I .
Figure 4.25 shows a measured heat input vs. I plot of this brass lead in the current
ranges of 0 to its nominally rated current of 40 A, and 40 A to 90 A (overcurrent).
As expected, the results show that the heat input is linear with I up to 40 A,
while beyond 40 A, the results agree well with the solid curve given by Eq. 4.110.
Figure 4.26 shows V (I) vs. t plots for three constant overcurrent levels, 130 A,
150 A, and 203 A. The solid curves are experimental, while the dotted curves are
analytical, given by Eq. 4.113. The experimental curve at 203 A shows that at this
level of overcurrent (I 5I ), the lead eventuallyafter 400 senters into an
unstable (overheated) region. The results indicate that it is safe to operate the
lead up to an overcurrent level 4 times the rated current.
306 CHAPTER 4PROBLEMS & DISCUSSIONS
200
150
Heat Input [mW]
100
50
0
0 20 40 60 80 100
I [A]
Fig. 4.25 Measured heat input vs. I plot for a vapor-cooled brass lead (I = 40 A) [4.87].
Solid curve: based on Eq. 4.110, with br = 6.13108 kg/s A; and br = 0.0138 A1 .
900
I2 =203 A
800
700
V (I) [mV]
600
I2 =150 A
500
I2 =130 A
400
300
200
0 100 200 300 400 500
Time [s]
Fig. 4.26 V (I) vs. t plots for over-current levels of 130, 150, and 203 A [4.87]. Solid
curves: experimental; Dotted curves: based on j given by Eq. 4.108b and V (I) by
Eq. 4.113, with A = 0.0613 cm2 ; = 54 cm; cp = 5.19 J/g K; and b = 7.4 n cm/K.
Table 4.18: Structural Data for G-10, Stainless Steel 304, Brass, and Copper
REFERENCES
[4.1] E. Trifon Laskaris, Kenneth G. Herd and Bijan Dorri, A compact 0.8 T super-
conducting MRI magnet, Cryogenics 34, 635 (1994).
[4.2] Toru Kuriyama, Masami Urata, Takashi Yazawa, Kazutaka Yamamoto, Yasumi
Ohtani, Kei Koyanagi, Tamaki Masegi, Yutaka Yamada, Shunji Nomura, Hideaki
Maeda, Hideki Nakagome and Osamu Horigami, Cryocooler directly cooled 6
T NbTi superconducting magnet system with 180 mm room temperature bore,
Cryogenics 34, 643 (1994).
[4.3] Lisa Cowey, Peter Cetnik, Kevin Timums, Peter Daniels, John Mellors and Ian Mc-
Dougall, Cryogen free Nb3 Sn magnet, operated at 9.5 K with high Tc BiSrCaCuO
(2212) current leads, IEEE Trans. Appl. Superconduc. 5, 825 (1995).
[4.4] Weijun Shen, M. Coey, W. McGhee, Development of 9.5 T NbTi cryogen-free
magnet, IEEE Trans. Appl. Superconduc. 11, 2619 (2001).
[4.5] C.H. Chang, F.Z. Hsiao, C.S. Hwang, M.H. Huang, and C.T. Chen, Design of a
7.5 T superconducting quadrupole magnet for magnetic circular dichroism exper-
iments, IEEE Trans. Appl. Superconduc. 12, 718 (2002).
[4.6] Achim Hobl, Detlef Krischel, Michael Poier, Ruediger Albrecht, Ralf Bussjaeger,
and Uwe Konopka, Design, manufacture, and test of a large bore cryogen-free
magnet, IEEE Trans. Appl. Superconduc. 13, 1569 (2003).
[4.7] R. Hirose, S. Hayashi, S. Fukumizu, Y. Muroo, H. Miyata, Y. Okui, A. Itoki,
T. Kamikado, O. Ozaki, Y. Nunoya, and K. Okuno, Development of 15 T cryogen-
free superconducting magnets, IEEE Trans. Appl. Superconduc. 16, 953 (2006).
[4.8] Yingming Dai, Luguang Yan, Baozhi Zhao, Shousen Song, Yuanzhong Lei, and
Quiliang Wang, Tests on a 6 T conduction-cooled superconducting magnet, IEEE
Trans. Appl. Superconduc. 16, 961 (2006).
[4.9] M.A. Daugherty, J.Y. Coulter, W.L. Hults, D.E. Daney, D.D. Hill, D.E. McMurry,
M.C. Martinez, I.G. Phillips, J.O. Willis, H.J. Boenig, F.C. Prenger, A.J. Roden-
bush and S. Young HTS high gradient magnetic separation system, IEEE Trans.
Appl. Superconduc. 7, 650 (1997).
[4.10] K. Watanabe, S. Awaji, K. Takahashi, G. Nishijima, M. Motokawa, Y. Sasaki,
Y. Ishikawa, K. Jikihara, J. Sakuraba, Construction of the cryogen-free 23 T
hybrid magnet, IEEE Trans. Appl. Superconduc. 12, 678 (2002).
[4.11] J. Good and R. Mitchell, A desktop cryogen free magnet for NMR and ESR,
IEEE Trans. Appl. Superconduc. 16, 1328 (2006).
[4.12] G. Snitchler, S.S. Kalsi, M. Manlief, R.E. Schwall, A. Sidi-Yekhlef, S. Ige and R.
Medeiros, High-eld warm-bore HTS conduction cooled magnet, IEEE Trans.
Appl. Superconduc. 9, 553 (1999).
[4.13] K. Sato, T. Kato, K. Ohkura, S. Kobayashi, K. Fujino, K. Ohmatsu and K. Hayashi,
Performance of all high-Tc superconducting magnets generating 4 T and 7 T at
20 K, Supercond. Sci. Technol. 13, 18 (2000).
[4.14] Hitoshi Kitaguchi, Hiroaki Kumakura, Kazumasa Togano, Michiya Okada, Katsu-
nori Azuma, Hiroshi Morita, Jun-ichi Sato, Cryocooled Bi-2212/Ag solenoid mag-
net system generating 8 T in 50 mm room temperature bore; design and prelimi-
nary test, IEEE Trans. Appl. Superconduc. 10, 495 (2000).
[4.15] R. Musenich, P. Fabbricatore, S. Farinon, C. Ferdeghini, G. Grasso, M. Greco,
A. Malagoli, R. Marabotto, M. Modica, D. Nardelli, A.S. Siri, M. Tassisto, and
A. Tumino, Behavior of MgB2 react & wind coils above 10 K, IEEE Trans. Appl.
Superconduc. 15, 1452 (2005).
[4.16] Lubomr Kopera, Pavol Kovac, and Tibor Melisek, Compact design of cryogen-
CRYOGENICSREFERENCES 309
free HTS magnet for laboratory use, IEEE Trans. Appl. Superconduc. 16, 1415
(2006).
[4.17] Richard McMahon, Stephen Harrison, Steve Milward, John Ross, Robin Staord
Allen, Claude Bieth, Said Kantas, and Gerry Rodrigues, Design and manufacture
of high temperature superconducting magnets for an electron cyclotron resonance
ion source, IEEE Trans. Appl. Superconduc. 14, 608 (2004).
[4.18] Kaoru Nemoto, Motoaki Terai, Motohiro Igarashi, Takeshi Okutomi, Satoshi Hira-
no, Katsuyuki Kuwano, Shigehisa Kusada, Tomohisa Yamashita, Yasuto Yanse,
Toru Kuriyama, Taizo Tosaka, Kenji Tasaki, Kotaro Marukawa, Satoshi Hanai,
Mutsuhiko Yamaji, and Hiroyuki Nakao, HTS magnet for Maglev applications
(2)magnet structure and performance, IEEE Trans. Appl. Superconduc. 16,
1104 (2006).
[4.19] Juan Bascunan (Personal communication, 2002).
[4.20] Luca Bottura (Personal communication, 2004).
[4.21] G. Bon Mardion, G. Claudet, and P. Seyfert, Practical data on steady state heat
transport in superuid helium at atmospheric pressure, Cryogenics 29, 45 (1979).
[4.22] Steven W. Van Sciver (personal communication, 1993).
[4.23] G. Claudet, C. Mwueia, J. Parain, and B. Turck, Superuid helium for stabilizing
superconductors against local disturbances, IEEE Trans. Magn. MAG-15, 340
(1979).
[4.24] Steven W. Van Sciver, Helium Cryogenics, (Plenum Press, New York, 1986), 182.
[4.25] See, for example, C. Taylor, R. Althaus, S. Caspi, W. Gilbert, W. Hassenzahl,
R. Meuser, J. Reschen, R. Warren, Design of epoxy-free superconducting dipole
magnets and performance in both helium I and pressurized helium II, IEEE Trans.
Magn. MAG-17, 1571 (1981).
[4.26] M.J. Leupold and Y. Iwasa, A subcooled superuid helium cryostat for a hybrid
magnet system, Cryogenics 26, 579 (1986).
[4.27] Isaac Asimov, Asimovs Biographical Encyclopedia of Science and Technology
(Doubleday, New York, 1964).
[4.28] Takashi Noguchi, Vacuum insulation for a cryostat, Cryogenic Engineering (in
Japanese) 28, 355 (1993).
[4.29] Randall F. Barron, Cryogenic Systems 2nd Ed., (Clarendon University Press, Ox-
ford, 1985).
[4.30] Yukikazu Iwasa, A permanent HTS magnet system: key design & operational
issues, Advances in Superconductivity X (Springer-Verlag, Tokyo, 1998), 1377.
[4.31] Benjamin J. Haid, A permanent high-temperature superconducting magnet op-
erated in thermal communication with a mass of solid nitrogen, Ph.D. thesis,
Department of Mechanical Engineering, M. I. T., Cambridge, MA (June, 2001).
[4.32] Benjamin J. Haid, Haigun Lee, Yukikazu Iwasa, Sang-Soo Oh, Young-Kil Kwon,
and Kang-Sik Ryu, Design analysis of a solid heat capacitor cooled Permanent
high-temperature superconducting magnet system, Cryogenics 42, 617 (2002).
[4.33] L.A. Koloskova, I.N. Krupskii, V.G. Manzhelii, and B.Ya. Gorodilov, Thermal
conductivity of solid nitrogen and carbon monoxide, Sov. Phys. Solid State 15,
1278 (1973).
[4.34] T.A. Scott, Solid and liquid nitrogen, Physics Reports (Section C of Physics
Letters 27, 89 (1976).
[4.35] V.A. Rabinovich, A.A. Vasserman, V.I. Nedostup, L.S. Veksler, Thermophysical
310 CHAPTER 4REFERENCES
Y. Iwasa, Case Studies in Superconducting Magnets: Design and Operational Issues, 313
DOI: 10.1007/b112047_5,
Springer Science + Business Media, LLC 2009
314 CHAPTER 5
Hs (x)
Hp Hp
He He
0 0
0 x a x 2a
Fig. 5.1 Slab of Type II superconductor exposed to an external eld.
to the slab, we obtain the magnetic eld within the superconductor, Hs (x):
0 (x x x ) (5.2a)
Hs (x) = He Jc x (0 x x) (5.2b)
H + J (x 2a)
e c (x x 2a) (5.2c)
Note that the slope of Hs (x) is equal to Jc , positive where Jc is positive (z-directed,
out of the paper) and negative where Jc is negative. x (and 2a x ) gives the
extent of the eld penetration in the slab; in terms of He and Jc :
He
x = (5.3a)
Jc
At He = Hp Jc a, x = x = a, and the entire slab is in the critical state; Hp is
known as the penetration eld:
Hp Jc a (5.3b)
The average magnetic induction within the slab, Bs , is thus given by
2a
Bs = Hs (x) dx = <shaded area in Fig. 5.1> (5.4a)
2a 0 2a
He x He2
=2 = (5.4b)
2a 2 2aJc
He2
= (5.4c)
2Hp
From the denition, M = Bs / He , we have:
He2
M = He (0 He Hp ) (5.5)
2Hp
The superconductor is diamagnetic; M is the magnitude of its magnetization.
As the external eld is increased further, the eld eventually penetrates the slab
completely (He Hp ), and Bs = He Hp /2, and thus:
M = 12 Hp = 12 Jc a (He Hp ) (5.6)
The dashed magnetization prole in Fig. 5.1 corresponds to He = Hp .
MAGNETIZATION 315
Hs (x)
Hm Hm
2Hp 2Hp
He He
Hm Hp Hm Hp
Hp Hp
He He= Hm 2Hp
He He= 0
0 a 2a
The dotted lines in Fig. 5.2 represent Hs (x) at He = Hm > 2Hp , where Hm is the
maximum external eld applied in this eld sweep sequence.
As He is reduced from Hm towards 0, Hs (x) changes as indicated by the solid
lines in Fig. 5.2. When He = Hm 2Hp , M becomes Hp /2, as may be inferred
from Fig. 5.2, because Hs (x) is a perfect mirror image of that at Hm . It can be
shown that for the return eld sweep from Hm to He= 0, M (He ) is given by:
(Hm He )2
M (He ) = 12 Hp (Hm He ) + (He= Hm Hm 2Hp ) (5.7a)
4Hp
= 12 Hp (He= Hm 2Hp 0) (5.7b)
1
H
2 p
He
0 Hp Hc2
12 Hp
Fig. 5.3 Magnetization vs. eld traces for a hard superconducting slab subjected to an
external eld sequence of 0 Hc2 0. The solid curve presents the case Jc = constant;
the dashed curve qualitatively presents the case Jc (He ), with Jc = 0 at Hc2 [5.25.5].
Figure 5.4 shows the current distribution within the slab corresponding to the eld
distribution of Fig. 5.1. Note that Jc = Hp /2a. The net current per unit length in
the y-direction [A/m] owing through the slab in the z-direction is given by:
2a
I= J(x) dx = 0 (5.8)
0
As expected, I = 0 in the absence of transport current.
J(x)
Hp/a Hp/a
Jc = 12 Hp/a
0
x a x 2a
Jc
Hp/a Hp/a
Hs (x)
Hp Hp
1
I
2 t
0
x a x 2a
12 It
Hp Hp
12 It + Jc x = 0 (5.9a)
Jc (x 2a) + 12 It = 0 (5.9b)
It It
x = and x = 2a (5.9c)
2Jc 2Jc
Figure 5.6 shows current distribution J(x) in the slab. By integrating J(x) across
the slab width, we can show that the net current owing in the slab is indeed It :
2a
I= J(x) dx = Jc x + Jc (2a x ) (5.10a)
0
= 12 It + 12 It = It (5.10b)
As expected, the net current in the slab is the current supplied by the external
source. Note that the presence of an external eld He y , when applied after It , does
not fundamentally change the distributions shown in Figs. 5.5 and 5.6; but if He y
is applied before It , dierent Hs (x) and J(x) would emerge. In DISCUSSION 5.1
the eects of transport current on magnetization are studied in detail.
J(x)
Hp/a Hp/a
Jc Jc = 12 Hp/a
0
x a x 2a
Hp/a Hp/a
dHe
Vsc (t) = Nsc Asc (5.11b)
dt
sc
The subscripts pc and sc refer to the primary and secondary search coils, respec-
tively. N is the number of turns of each search coil and A is the eective area of
each turn in the coil through which He (t) is coupled. He is the space-averaged
eld over each search coil.
The bridge output voltage Vbg (t) is given by:
Vbg (t) = (k 1)Vpc (t) + kVsc (t) (5.12)
where k is a constant (from 0 to 1) that expresses the fraction of the potentiome-
ter resistance (R) appearing in the primary side of the search coil in Fig. 5.7.
Combining Eqs. 5.11 and 5.12, we obtain:
dM
Vbg (t) = (k 1) Npc Apc
dt
dHe dHe
+ (k 1) Npc Apc + k Nsc Asc (5.13)
dt dt
pc sc
dHe dHe
(k 1) Npc Apc + k Nsc Asc =0 (5.14a)
dt dt
pc sc
dM
Vbg (t) = (k 1) Npc Apc (5.14b)
dt
Although in practice the condition required by Eq. 5.14a is not always satised
over a wide frequency range, Eq. 5.14b is a good approximation for most cases.
Generally k is close to 0.5. Vbg (t) is fed into an integrator, and its output, Vmz (t),
is proportional to M . Specically, if the test sample is in the virgin state (M = 0)
and He (t) is increased (He) from 0 (at t = 0) to He (at t = t1 ), then we have:
t1
1
Vmz (He) = Vbg (t) dt
it 0
(k 1) Npc Apc
= M (He ) (5.15)
it
The factor fm is the ratio of magnetic material volume to the total test sample
volume. This factor is needed because generally a test sample does not consist
entirely of magnetic material for which magnetization is measured; in the case
of a multilamentary conductor, for example, the test sample consists not only
of superconducting laments but also of a matrix metal and other nonmagnetic
materials, e.g., insulator. If an external eld excursion is 0 Hm > Hp He <
Hm 2Hp , then we have:
(k 1) Npc Apc Jc a
Vmz (He < Hm 2Hp ) = fm (5.16b)
it 2
Vmz = Vmz (He > Hp ) Vmz (He < Hm 2Hp ) is thus proportional to the
width of magnetization curve at He :
(k 1) Npc Apc
Vmz = fm Jc a (5.16c)
it
80
60
40
Magnetization [emu/cm3 ]
20
20
40 10 K
20 K
60 30 K
80
2 1.5 1 0.5 0 0.5 1 1.5 2
Magnetic Field [T]
Fig. 5.8 Magnetization vs. field (in tesla) traces of MgB2 at 10 K, 20 K, and 30 K [5.7].
Figure 5.8 shows magnetization vs. field (here given in the unit of tesla) traces of
MgB2 at 10 K, 20 K, and 30 K, each completing a field cycle, from 0 to +1.7 T,
back to 0, to 1.7 T, and finally to 0 [5.7]. Note that unlike Fig. 5.3 which shows
M (He ) plots, these are +M (He ) plots. Because the traces are not tilted along
the x (field) axis, we may conclude that the primary and secondary coils used in
this measurement are well balanced.
The hysteretic nature of its magnetization clearly indicates that MgB2 is a Type
II superconductor. Note that the diamagnetic nature of this superconductor is
evident in the first part of each trace when the field is increased from 0 towards
1.7 T, magnetization enters into the negative quadrant.
As may be inferred from Hp = Jc a of Beans model, magnetization is directly
proportional to the superconductors Jc . However, unlike in Beans model, Jc is
a decreasing function of not only field but temperature. Dependence on Jc and T
are clearly evident in Fig. 5.8. Here, the magnetization is in emu/cm3 , a non-SI
unit. In DISCUSSION 5.1, we shall apply Beans model to the magnetization data
of Fig. 5.8 to compute the materials Jc at 10 K in zero field.
Hs (x)
3.5Hp 3.5Hp
3Hp 3Hp
2.5Hp 2.5Hp
2Hp 2Hp
1.5Hp 1.5Hp
0
0 x x a 2a
Fig. 5.9 Field proles in the presence of He = 2.5Hp , rst with It = 0 (dotted
lines), then It = Jc a = Ic /2 (solid), and nally It = 2Jc a = Ic (dashed).
Hs22 (x) = He Jc x (x x x )
Hs23 (x) = He + 12 It + Jc (x 2a) (x x 2a)
and
Hs22 (x ) = Hs23 (x )
He Jc x = He + Hp i + Jc (x 2a) = x = a 1 12 i
Hs22 (x ) = He Hp + 12 Hp i
Hs (x)
He + 12 It
He He
He 12 It
A1 A2 A3
He Hp
0
0 x x a 2a
Fig. 5.10 Field prole with transport current (solid) for computation of mag-
netization. The vertical dashed lines partition three areas, A1 , A2 , and A3 .
MAGNETIZATION PROBLEMS & DISCUSSIONS 323
= 12 a(1 i)(2He Hp )
= a He He i 12 Hp + 12 Hp i
Shaded area = A1 + A2 + A3
= a 12 He i 38 Hp i2 + He He i 12 Hp + 12 Hp i
+ He + 12 He i 12 Hp 14 Hp i + 34 Hp i + 38 Hp i2
= a(2He Hp + Hp i)
1
M (i) = He (Shaded area)
2a
= He He + 12 Hp 12 Hp i
= 12 Hp (1 i) (5.17a)
Hs (x)
He + 12 It
He He (= 2Hp )
He 12 It
Hp Hp
1
I
2 t
0
0 0.5a a 1.5a 2a
12 It
Fig. 5.11 Field proles, with current only (dotted) and with current and eld (solid).
MAGNETIZATION PROBLEMS & DISCUSSIONS 325
A2 = 12 (2a a + ai)(He + Hp i + He Hp )
= a(1 + i) He 12 Hp + 12 Hp i
= a He + He i 12 Hp + 12 Hp i2
Shaded area = A1 + A2
= a(2He Hp + Hp i2 )
Once the shaded area is known, we have M :
M (i) = He 12 (2He Hp + Hp i2 )
= 12 Hp (1 i2 ) (5.18a)
Hs (x)
He + 12 It
He He
He 12 It
A1 A2
0
0 x a 2a
Fig. 5.12 Field prole with transport current and eld for computation
of magnetization. The vertical dashed line partitions areas A1 and A2 .
326 CHAPTER 5PROBLEMS & DISCUSSIONS
Hs2 (x) = He + Hp i Jc x (x x a)
Hs3 (x) = He + Hp i + Jc (x 2a) (a x x )
Hs2 (x) = He = He + Hp i Jc x
Hp i
x = = ai
Jc
Hs3 (x ) = He = He + Hp i + Jc (x 2a)
Hp
x = 2a i = 2a ai
Jc
Hs (x)
He +Hp i He +Hp i (= He + 12 It )
He He
He Hp i He Hp i (= He 12 It )
0
0 x a x 2a
He +Hp i He +Hp i
base
He
height
He Hp i
A1 A2 A3 A4
0
0 x a x 2a
0
0 0.5 1
i
Magnetization FunctionsSummary
Figure 5.15 presents three normalized magnetization functions, f1 (i), f2 (i), and
f3 (i), where i = It /Ic . It is interesting to note how dierent sequences of transport
current and external eld applications aect M (i). These f (i) functions were val-
idated with experimental results [5.3, 5.4], thereby making Beans model accepted
quite quickly after its formulation.
A. Initial State
For He Hp , where Hp is the critical-state eld, the eld within the lament,
Hs (r), is zero from r = 0 to r = (df/2He /Jc ) and varies as Jc r from r to df /2:
r r
Hs (r) = He (r r) (5.21)
df/2 r
Note that r = 0 when He = Hp , where Hp is the critical-state eld:
Hp = 12 Jc df (5.22)
Using steps similar to those taken with Eq. 5.4, we may compute the average
magnetic induction within the lament, Bs :
df/2
4 r r
Bs = He (2r) dr
2
df r df /2 r
8 He 1
= 3
24 df 18 df2 r + 16 r3 (5.23)
df2 (df/2 r)
Unlike in the case of a slab, where the integration may be performed geometrically
from Hs (x), here the area integration must be performed mathematically. By
inserting r = (df/2He /Jc ) into Eq. 5.23 and noting that Hp = Jc df/2, we obtain:
Bs 2He2 4He3 He2 He3
= = (5.24)
df Jc 3(df Jc )2 Hp 3Hp2
From the denition M = Bs / He , we have:
He2 H3
M = He + e2 (0 He Hp ) (5.25)
Hp 3Hp
Note that Eq. 5.25 is similar to, but clearly dierent from, Eq. 5.5 for the slab.
The conductor volume per unit length in the z-direction is df2 /4. Thus:
4mA 4 df df
M= 2 = 3 Jc 2 0.424Jc 0.5Jc a (5.29a)
df 2
8 df
Hp = Jc (5.29b)
3 2
This is nearly the same (8/3 1) as that for a Bean slab of thickness df .
y
He He He He
x
2a
df
(a) (b)
Fig. 5.16 Induced current distributions in a) Bean slab of width 2a and b) an innitely
long lament of diameter dd , both subjected to an external eld He in the y-direction.
MAGNETIZATIONPROBLEMS & DISCUSSIONS 331
6M (0 T; 10 K)
Jc (0 T; 10 K) =
df
6(240103 A/m)
=
(0.531103 m)
= 2.7109 A/m2
TRIVIA 5.1 Fill in the blank in a poetic couplet about one of the
poets contemporaries below. God said, Let be! and all was light.
i) Bach; ii) Halley; iii) Newton; iv) Wren.
332 CHAPTER 5PROBLEMS & DISCUSSIONS
38 100
Be
38
(a) (b)
42
Secondary Coil
20 (280 turns)
107 B
e
Primary Coil
40 (500 turns)
70
Secondary Coil
20 (280 turns)
70
(c)
(k 1) Npc Apc
Vmz = fm Jc a (5.16c)
it
where for NbTi, B0 0.3 T. J0 is the zero-eld critical current density, which is
usually dicult to measure. Thus from the Jc value at 5 T and B0 = 0.3 T, we
can rst solve for J0 B0 :
J0 B0
2.0109 A/m2 = = J0 B0 = 10.6109 A T/m2
5 T + 0.3 T
10.6109 A T/m2
Jc (2.5 T) = = 3.8109 A/m2
2.8 T
(0.5)(4107 H/m)(500)(3.38103 m2 )
Vmz = 0.25
1s
(3.810 A/m )(50106 m)
9 2
50 mV
MAGNETIZATIONPROBLEMS & DISCUSSIONS 335
dBe dBe
Npc Apc = Nsc Asc (S1.2)
dt dt
pc sc
Because Apc = Asc , we have: Npc [Be ]pc = Nsc [Be ]sc . From symmetry, we consider
only the upper half (the unit mm is omitted in the following equations):
2
20
Be (0) z
[Be ]pc = 1c dz (S1.3a)
20 0 z0
2
80
Be (0) z
[Be ]sc = 1c dz (S1.3b)
20 60 z0
2 2
20 80
250 z 280 z
1c dz = 1c dz (S1.4)
20 0 z0 20 60 z0
c (20)3 c (80)3 c (60)3
250 20 = 280 80 60 +
3 (75)2 3 (75)2 3 (75)2
5000 118.5c = 22400 8495.4c 16800 + 3584c
600
c 0.125
4793
336 CHAPTER 5PROBLEMS & DISCUSSIONS
B
Faradays law: =
E (2.8)
t
For the slab (width 2a) geometry, we can express Eqs. 2.5 and 2.8 as:
Hy Ez
Amperes law: = Jz = (5.31)
x e
Ez By Hy
Faradays law: = = (5.32)
x t t
where e is the materials electrical resistivity. From Eqs. 5.31 and 5.32, we obtain:
2Hy Hy
e =
x2 t
e 2Hy 2Hy Hy
2
D mg 2
= (5.33)
x x t
Equation 5.33 is a magnetic diusion equation, for which:
e
Dmg = (5.34)
Similarly, the one-dimensional thermal diusion equation having constant thermal
properties is given by:
2T T
k 2 =C (5.35a)
x t
where k and C are, respectively, the materials thermal conductivity and heat
capacity. Dividing both sides of Eq. 5.35a by C, we obtain:
k 2T 2T T
D th = (5.35b)
C x2 x2 t
Equation 5.35b is a thermal diusion equation, for which:
k
Dth = (5.36)
C
Note that Eq. 5.36 and 4.20 are equivalent, because C = cp .
Table 5.2 presents approximate values of electrical and thermal properties and
corresponding diusivities at 4 K and 80 K for stainless steel and copper. From
Table 5.2 we can clearly see that stainless steel, a stand-in for normal-state su-
perconductors, and copper are opposite with respect to magnetic and thermal
diusivities. Specically, changes in magnetic eld propagate quickly through
stainless steel, whereas temperature gradients are relatively slow to propagate;
hence, large nonuniform temperature distributions can be created in stainless steel
during changing magnetic elds. Physically, it means that magnetic heating hap-
pens essentially adiabatically in Type II superconductors. In copper, the reverse
is true: the magnetic eld diuses very slowly, while any nonuniformity in temper-
ature is quickly evened out. Therefore, copper in intimate contact with Type
II superconductor can alleviate eld-motion-induced instability in Type II super-
conductors. This thinking is the essence of dynamic stability, one of the stability
criteria developed during the 1960s and 1970s [5.8] and applied also to HTS in the
late 1980s [5.9].
Answer to TRIVIA 5.1 iii). The Principia Mathematica of Sir Isaac Newton
(16421727) is regarded as a dominant force ushering in the Age of Reason.
The complete couplet of the English poet Alexander Pope (16881744): Nature
and Natures laws lay hid in night/God said, Let Newton be! and all was light.
338 CHAPTER 5PROBLEMS & DISCUSSIONS
Jc |Jc |a2
e = (5.37)
3
Note that the entire slab is in the critical state with its surface (a) exposed
to an external eld of He y .
b) Derive Eq. 5.37 by computing the Poynting energy ow into the slab at x = a
and equating it with the change in magnetic energy storage and dissipation
energy E in the positive half of the slab.
c) To relate Jc to an equivalent temperature rise in the conductor, we may
assume a linear temperature dependence for Jc (T ):
Tc T
Jc (T ) = Jc (5.38)
Tc Top
where Jc is the critical current density at the operating temperature Top . Tc
is the critical temperature at a given magnetic induction B . From Eq. 5.38,
Jc in Eq. 5.37 may be related to an equivalent temperature rise T :
T
Jc = Jc (5.39)
Tc Top
Dissipation power density, p(x), is given by Ez (x)Jc ; the total energy density per
unit length dissipated in the slab or per unit slab surface area in the y-z plane, E
[J/m2 ], is given by:
a
E = p(x)t dx
0
a
x2 Jc |Jc |a3
= Jc |Jc | ax dx = (S2.5)
0 2 3
Jc |Jc |a2
e = (5.37)
3
340 CHAPTER 5PROBLEMS & DISCUSSIONS
Jc2 T a2
e = (S2.15)
3(Tc Top )
Ts Jc2 a2
< (S2.17)
T 3Cs (Tc Top )
c) Estimate the temperature rise for ux jump A. Assume the heat capacity of
Nb-Zr to be independent of temperature and equal to 6 kJ/m3 K.
3 Hf
A
2
1
M [Kilogauss]
0 5 10 15 20 25 30 35 40
Ambient Field [Kilogauss]
Fig. 5.19 Magnetization vs. ambient field trace for a 0.5-mm dia. Nb-Zr monofilament.
df 2w df
y
E
x 2l
E
H0z H0z
2
Icp = Jcu dy = y dy = (5.43)
0 cu 0 2cu
c) At a critical length
c , the net current Icp given by Eq. 5.43, becomes equal
to Jc df , the slabs critical current (per unit conductor depth). Show that
the critical length
c is:
2cu Jc df
c = (5.44)
H0z
b) Once the E eld is known, the current density Jcu in the copper slab is given
by: Jcu = E1x /cu . The net current owing in the copper from one superconduct-
ing slab to the other over half the conductor length is given by:
H0z H0z
2
Icp = Jcu dy = y dy = (5.43)
0 cu 0 2cu
= 1.1104 m = 110 m
In typical submicron strands, the twist pitch length is thus 100 m. This means
that the diameter of such strands, by mechanical requirements, should be 1 m;
actually a thermal-magnetic stability criterion, similar to the ux jump criterion,
requires it to be even smaller than 1 m. This is because the strands, to reduce
coupling losses, use Cu-Ni alloys as the matrix materials, resulting in a magnetic
diusion time constant that is smaller than the thermal diusion time constant.
e) Critical current (Ic ), critical current density (Jc ), lament number (Nf ) and
diameter (df ) in a multilamentary conductor are related by:
d2f
Ic = Nf Jc (S4.2)
4
Solving for Nf from Eq. S4.2 with appropriate values of parameters, we obtain:
4Ic 4(100 A)
Nf = =
d2f Jc (0.2106 m)2 (2109 A/m2 )
= 1.6106
In submicron strands, the number of laments may approach ten million.
MAGNETIZATIONPROBLEMS & DISCUSSIONS 347
B3
SWEEP RATES
B2 110 Oersted/s
320
M [Arbitrary Linear Unit]
A 900
B1
0 5 10 15 20 25 30 35 40
Ambient Field [Kilo-oersted]
Fig. 5.21 Magnetization traces for Conductors 1, 2, and 3 [5.12].
= 2.7102 m = 27 mm
This value is close enough to the actual twist pitch of 10 mm. Because the magneti-
zation of Conductor 1 (Trace C) at a sweep rate of 320 oersted/sec is considerably
smaller than that of Conductor 2 for the same eld sweep rate, we conclude that
where hs (Tc ) and hs (Top ) are the superconductor enthalpies, respectively, at Tc and
Top . Because e (Top ) = [ Hp (Top )]2 /6 , and for Bean slab, Hp (Top ) = aJc (Top ),
the conductor size criterion, ac , for suppressing complete ux jumping is given by:
6[hs (Tc ) hs (Top )]
ac = (5.46)
Jc2 (Top )
Comparing these two size criteria (Eqs. 5.40 and 5.46), we may conclude that under
adiabatic conditions, a ux jump may initiate if the conductor size is greater than
that specied by Eq. 5.40, but it may be only partial if the size does not exceed
that of Eq. 5.46. Thus, ux jumps will be only partial in a superconductor that
either violates the size criterion of Eq. 5.40 but not Eq. 5.46, or the process is not
adiabatic. Note that even those ux jumps seen in Fig. 5.19 are, strictly speaking,
not complete, most probably because the process was not perfectly adiabatic.
REFERENCES
[5.1] C.P. Bean, Magnetization of hard superconductors, Phys. Rev. Lett. 8, 250
(1962).
[5.2] Y.B. Kim, C.F. Hempstead, and A.R. Strnad, Magnetization and critical super-
currents, Phys. Rev. 129, 528 (1963).
[5.3] M.A.R. LeBlanc, Inuence of transport current on the magnetization of a hard
superconductor, Phys. Rev. Lett. 11, 149 (1963).
[5.4] Ko Yasukochi, Takeshi Ogasawara, Nobumitsu Usui, and Shintaro Ushio, Mag-
netic behavior and eect of transport current on it in superconducting Nb-Zr wire,
J. Phys. Soc. Jpn. 19, 1649 (1964).
Ko Yasukochi, Takeshi Ogasawara, Nobumitsu Usui, Hisayasu Kobayashi, and
Shintaro Ushio, Eect of external current on the magnetization of non-ideal Type
II superconductors, J. Phys. Soc. Jpn. 21, 89 (1966).
[5.5] H.T. Coey,Distribution of magnetic elds and currents in Type II superconduc-
tors, Cryogenics 7, 73 (1967).
[5.6] W.A. Fietz, Electronic integration technique for measuring magnetization of hys-
teretic superconducting materials, Rev. Sci. Instr. 36, 1621 (1965).
[5.7] Figure 5.8, modied for use here by Mohit Bhatia (Ohio State University), from the
work of M.D. Sumption, E.W. Collings, E. Lee, X.L. Wang, S. Soltanina, S.X. Dou,
and M.T. Tomsic, Real and apparent loss suppression in MgB2 superconducting
composite, Physica C, 98 (2002).
[5.8] H.R. Hart, Jr., Magnetic instabilities and solenoid performance: Applications of
the critical state model, Proc. 1968 Summer Study on Superconducting Devices
and Accelerators, (Brookhaven National Laboratory, Upton, NY, 1969), 571.
[5.9] T. Ogasawara, Conductor design issues for oxide superconductors. Part 2: exem-
plication of stable conductors, Cryogenics 29, 6 (1989).
[5.10] See, for example, M.S. Lubell, B.S. Chandrasekhar, and G.T. Mallick, Degrada-
tion and ux jumping in solenoids of heat-treated Nb-25% Zr wire, Appl. Phys.
Lett. 3, 79 (1963).
[5.11] Superconducting Applications Group (Rutherford Laboratory), Experimental and
theoretical studies of lamentary superconducting composites, J. Phys. D3, 1517
(1970).
[5.12] Y. Iwasa, Magnetization of single-core, multi-strand, and twisted multi-strand
superconducting composite wires, Appl. Phys. Lett. 14, 200 (1969).
[5.13] A. Nabialek, M. Niewczas, H. Dabkowska, A. Dabkowski, J.P. Castellan, and B.
D. Gaulin, Magnetic ux jumps in textured Bi2 Sr2 CaCu2 O8+ , Phys. Rev. B
67, 024518 (2003).
[5.14] Igor S. Aranson, Alex Guerevich, Marco S. Welling, Rinke J. Wijngaarden, Vitalii
K. Vlasko-Vlasov, Valerii M. Vinokur, and Ulrich Welp, Dendritic ux avalanches
and nonlocal electrodynamics in thin superconducting lms, Phys. Rev. Lett. 94,
037002 (2005).
CHAPTER 6
STABILITY
6.1 Introduction
Reliability is one of the major requirements that all devices must meet, super-
conducting magnets included. Historically, reliability has been one of the most
troubling, and therefore most challenging, aspects of superconducting magnet
technology. As illustrated in Fig. 1.5, superconductivity exists within a phase
volume bounded by three parameters: current density (J), magnetic eld (H),
and temperature (T ).
Of these parameters, as examined in CHAPTER 3, the designer can dene, and more
importantly control, current density and magnetic eld quite well, at least under
normal operating conditions. Even under a complex fault-mode condition such as
that involving more than one solenoid in a hybrid or nested multi-coil magnet,
the current density and magnetic eld are tractable: the magnet designer has rm
control of these two parameters. This is not strictly the case with temperature: it is
the least tractable of the three. The magnitude of its excursion from the operating
point could vary unpredictably in time and, more intractably, in space within
the winding. The energy stored in the magnet, both magnetic and mechanical,
can easily be converted into heat, raising the conductor temperature to above its
critical value at one or more locations in the winding. Indeed, virtually every
stability problem of a superconducting magnet may be traceable to the magnet
designers inability to keep the winding temperature intact at its operating point.
In this chapter we shall consider: 1) basic physics controlling temperature in
a superconducting winding; and 2) stability evaluation methods to quantify the
likelihood of an unscheduled temperature rise within the winding. CHAPTERS 7
and 8 also deal with temperature rise in the winding under dierent contexts:
CHAPTER 7 on causes or sources of temperature rises; and CHAPTER 8 on methods
to protect magnets subsequent to unscheduled temperature rises. First, there is a
striking dierence in this stability issue between LTS and HTS magnets.
In Eq. 6.1, the left-hand side represents the time rate of change of thermal energy
density of the conductor, where Ccd (T ) is the heat capacity per unit volume of the
conductor, which, after the development in 1964 by Stekly [6.9] of composite su-
perconductors, consists of superconductor and normal-metal matrix. For complete
steady-state stability, this term must remain zero at all times; in practice a modest
temperature excursion, Top from the operating point, Top , is permitted during
operation in most windings, even adiabatic ones. Because this permissible Top
is generally much greater in HTS magnets than in LTS magnets, as stated at the
outset, stability is almost a non-issue for HTS magnets. This point is elaborated
a bit more below.
In the right-hand side, each on a per unit volume basis, the rst term describes
thermal conduction into the composite superconductor element, where kcd (T ) is
the thermal conductivity of the composite. The second term is Joule heating,
where cd (T ) is the composites electrical resistivity (zero in the superconducting
state), and Jcd (t) is the current density at operating current Iop (t), which can
depend on time. gd (t) describes non-Joule heat generation, primarily magnetic and
mechanical in origin. The last term represents cooling, where fp is the fraction
of the composite perimeter, PD , exposed to cryogen, Acd is the composite cross
sectional area, and gq (T ) is the convective heat transfer ux for the cryogen.
The history of the development of theories and concepts for stability (and protec-
tion to be discussed in CHAPTER 8 ) has evolved around solutions to simplications
of Eq. 6.1. Table 6.1 lists various concepts derived from Eq. 6.1 under special con-
ditions. In the table, a parameter
labeled 0 signies that it is negligible or not
considered in the equation; signies inclusion. Before discussing each term of
Eq. 6.1, we briey discuss the concepts listed in Table 6.1.
Table 6.1: Concepts Derived from Power Density Equation (Eq. 6.1)
Ccd (T )( T/ t) [kcd (T )T ] 2
cd (T )Jcd (t) gd (t) gq (T ) Concept
0 0 0 Flux jump
0 0 0 Cryostability
0 Dynamic stability
0 0 Equal area
0 0 0 MPZ*
0 0 0 Protection
0 0 Adiabatic NZP
* Minimum propagating zone.
Normal zone propagation.
STABILITY 353
Flux Jumping
As examined in CHAPTER 5, criteria have been developed to eradicate most in-
stances of ux jumping, which generally aicts LTS.
Cryostability
The basic concept of cryostability was developed in the mid 1960s as an engi-
neering solution to achieve reliable magnet operation [6.9]. In a cryostable
composite conductor, a superconductor is co-processed with a highly conductive
matrix metal [6.10], and a large portion of the conductor surface is exposed to
cryogen to ensure local cooling. As shown by Table 6.1, the terms other than
the Joule heating and cooling terms may be neglected. Many successful magnets in
the 1970s are cryostable [6.11, 6.12]; it is now applied only to large LTS magnets.
As we shall see later, it is not applied to HTS magnets. The cryostable concept is
further studied in this chapters PROBLEMS & DISCUSSIONS.
Dynamic Stability
As studied in CHAPTER 5, when the magnetic diusivity is much greater than
the thermal diusivity, as in Type II superconductors, ux jumping can occur
when the conductor size criterion to suppress it cannot be readily met, e.g., a tape
conductor. By loading the superconductor with a high thermal diusivity material,
e.g., copper, we may balance the two diusivities and achieve stable operation free
of ux jumping. Tape-LTS is now rarely used; ux jumping is unlikely in HTS
tapes (DISCUSSION 5.7). This criterion is not discussed further in this chapter.
Equal Area
The equal-area criterion is a special case of cryostability in which the thermal
conduction term ( [kcd (T )T ]) in Eq. 6.1 is included to improve the overall
current density at which the magnet may be cryostable. This will be further
discussed in the PROBLEMS & DISCUSSIONS.
MPZ
The concept of MPZ (minimum propagating zone) considers the eect on coil
performance of a local disturbance, gd (t) in Eq. 6.1, in the winding [6.13]. The
MPZ concept shows that it is possible for a magnet to remain superconducting
even in the presence of a small normal-state region in its winding, provided that the
normal-zone volume is smaller than a critical size dened by the MPZ theory. Its
importance in adiabatic magnets was recognized by Wilson in the late 1970s [1.27],
and it has since become an indispensable concept for analyzing the stability of
adiabatic magnets. The MPZ concept will be further studied in the PROBLEMS
& DISCUSSIONS.
Nonsteady Cases
The last two cases in Table 6.1 concern the non-steady-state thermal behavior of
the winding. Both are treated in CHAPTER 8.
354 CHAPTER 6
Unlike its heat capacity, the thermal conductivity of each substance in the table
varies much less dramatically with temperature. Copper has the most remarkable
thermal conductivity. It is much greater than those of the rest, making it, par-
ticularly for stability (this chapter) and protection (CHAPTER 8), unquestionably
indispensable in any winding, LTS or HTS. As discussed in CHAPTER 7, because of
Joule dissipation from eddy currents in copper windings subjected to time-varying
electromagnetic elds, the presence of copper can present operational diculties,
which are generally circumvented or minimized by a complex conguration for the
composite superconductor.
101
AC Losses
Wire
102 Motion
Flux Particle
Jumping Showers
gd [J/cm3 ]
103
Nuclear
Heat
104 Heat
Leaks
105 6
10 105 104 103 102 0.1 1 10
Time [s]
Note that eh depends not only on Ccd (T ), Top , and Tcs (Iop ) or [Top (Iop )]st but
also on Iop relative to Ic , i.e., iop Iop /Ic . Specically, for the simple straight
line approximation of Ic (T ) of Fig. 6.2, [Top (Iop )]st is given by:
Table 6.4: Selected Values of Top , Top , and eh for LTS and HTS
LTS HTS
Top [K] [Top (Iop )]st [K] eh [J/cm3 ] Top [K] [Top (Iop )]st [K] eh [J/cm3 ]
2.5 0.3 1.2104 4.2 25 1.6
4.2 0.5 0.6103 10 20 1.8
4.2 2 4.3103 30 10 3.7
10 1 9103 70 5 8.1
LTS Comparison of energy margins in Table 6.4 with disturbance energy den-
sities in Fig. 6.1 clearly indicates that LTS magnets are very much susceptible to
quenching induced by a disturbance, the energy density of which is represented by
gd (t) in Eq. 6.1. Aptly, techniques have been developed over the years to suppress
these disturbances, e.g., wire motion and ux jumping, or minimize AC losses for
most DC LTS adiabatic magnets to make them operate stably most of the time.
Techniques to minimize or eradicate mechanical disturbances, e.g., wire motion,
important only to LTS magnets, discussed more extensively in the 1st Edition, are
briey discussed in CHAPTER 7 of this Edition.
HTS With the exception of AC losses, the disturbance energy spectra for HTS
magnets should be those given in Fig. 6.1. Referring to eh values in Table 6.4,
we may conclude that HTS magnets, at least under DC conditions, are absolutely
stable: every DC HTS magnet should thus be designed to operate adiabatically.
6.2.7 Cooling
Although cooling is required for operation of every superconducting magnet, as
discussed in CHAPTER 4, only bath-cooled, cryostable magnets require cryogen
cooling within the winding. The gq (T ) term in Eq. 6.1 thus refers to cooling
present only within the winding; cooling exterior to the winding, which every su-
perconducting magnet requires, is literally peripheral; it does not enter Eq. 6.1.
As discussed above, HTS magnets can, and really should always, be operated adi-
abatically. Furthermore most HTS magnets, except those coupled to LTS magnets
and which therefore must operate at liquid helium temperature, operate at tem-
peratures above 20 K. Therefore, liquid helium heat transfer data are no longer
as essential for the design of HTS magnets as for bath-cooled cryostable LTS mag-
nets; even liquid nitrogen heat transfer data are not essential for HTS magnets,
which are adiabatic, because liquid nitrogen is not present in the winding.
Bottura has summarized helium heat transfer coecient, hq , vs. T plots (Fig. 6.3)
for various cooling regimes [6.14]; here T is the temperature dierence between
the heated surface and the helium at 4.2 K, except where noted otherwise. The
plots include: 1) nucleate boiling and lm boiling, including peak nucleate boiling
point (hpk ) of 1.23 W/cm2 K; 2) transient nucleate; 3) Kapitza at 1.8 K (super-
uid) and at 4.2 K; 4) forced-ow at 3.5 atm, 4.5 K, and Reynolds numbers of 104
and 105 . Of these plots, the nucleate boiling plot, including hpk = 1.23 W/cm2 K, is
for bath-cooled and cryostable magnets; the Kapitza (1.8 K) plot is for superuid-
cooled cryostable magnets; forced-ow plots are for cryostable magnets wound
with cable-in-conduit (CIC) conductor.
STABILITY 359
10
Kapitza (4.2 K)
Transient nucleate boiling
hpk Kapitza
(1.8 K)
(1.23 W/cm2 K)
1
hq [W/cm2 K]
Forced-flow
(P = 3.5 atm;
Nucleate boiling (4.2 K) T = 4.5 K;
Re =105 )
0.1
Forced-flow Film boiling
(P = 3.5 atm; (4.2 K)
T = 4.5 K;
Re =104 )
0.01
0.01 0.1 1 10 100
T [K]
Superconductor
Asc Composite Structure (AS )
Superconductor &
Matrix Metal
(Acd ) Others (Ain )
Am
Non-Matrix Metals
Coolant
Am
(Aq )
(a) (b)
Fig. 6.4 Schematic cross sectional area drawings of: (a) composite superconductor;
and (b) winding pack, showing their components and respective area symbols.
360 CHAPTER 6
Rn
+ Vcd
Matrix Rm
It Im It
Is
0 Is
0 Ic
(a) (b)
Gj (T ) = Rm It2 (T Tc ) (6.13c)
Assume Rm to be temperature-independent.
b) Make a plot of Eq. 6.13 for the temperature range from Top to T > Tc .
c) Give a physical explanation of Gj (T ) given by Eq. 6.13b.
d) Discuss qualitatively how Eq. 6.13b must be modied above 30 K, at which
point Rm becomes temperature-dependent, Rm (T ), as may be the case with
a composite HTS.
Ic (T )
I c
B (CONSTANT)
It
0 T
Top Tcs Tc
Fig. 6.6 Linear approximation of Ic vs. T (Eq. 6.12) for the superconductor.
Setting It = Ic (Tcs ) and inserting this into Eq. 6.12, we can solve for Ic :
Tc Top
Ic = It (S1.2)
Tc Tcs
Gj (T ) = Rm It2 (T Tc ) (6.13c)
Tcs T Tc :
T Tcs
Gj (T ) = Rm (T )It2 (6.14a)
Tc Tcs
T Tc :
Gj (T ) = Rm (T )It2 (6.14b) 0 T
Top Tcs Tc
m Ic2
1 (6.18)
fp PD Am hq (Tc Top )
The Stekly stability parameter, sk , is given by Eq. 6.18:
m Ic2
sk = (6.19)
fp PD Am hq (Tc Top )
Note that sk , a dimensionless number, expresses the ratio of Joule dissipation
density to cooling density. Operation is thus stable when sk 1 (sucient
cooling) and unstable when sk > 1 (insucient cooling.)
366 CHAPTER 6PROBLEMS & DISCUSSIONS
A. Monolith
The superconductor and the normal metal form one entity, achieved chiey through
metallurgical processes. By visual inspection, it is impossible to distinguish, except
through the conductor cross section, the existence of more than one constituent in
a monolith. Most round composite superconductors are monolithic. For values of
m/s above 10, however, it is dicult to manufacture monolithic superconductors
without breaking laments in the metal forming processes, particularly for those
having laments less than 100-m diameter.
B. Built-Up
A built-up conductor is comprised of a monolithic superconductor having a m/s
close to 1 and normal-metal stabilizer parts that are generally soldered to the
monolith after the monolith has been prepared. Mechanical properties of the sta-
bilizer parts are therefore unaected by manufacturing processes of the monolith,
making it sometimes easier to satisfy conductor specications. The CIC conductor
is a variant of built-up conductors.
q(T ), gj (T )
q(T )
qpk
Acd
qfm gj (Tc ) = Rm Ic2
fp PD
Acd
gj (Tc ) = Rm It2 (It < Ic )
fp PD
T
Top Tcs Tc
Fig. 6.8 Qualitative plots of q(T ) for liquid helium and gj (T ) for the case Iop = Ic
(solid straight lines) and Iop < Ic (dashed straight lines).
STABILITYPROBLEMS & DISCUSSIONS 369
Figure 6.9 shows one example of gj (T ) curves satisfying the equal area cri-
terion. In this example, Eq. 6.23 is satised by having two cross-lined areas
in Fig. 6.9 equal, the one bounded by the gq (T ) > gj (T ) and gj (T ) curves
and the other bounded by the gj (T ) >
q(T ), gj (T )
gq (T ) and gq (T ) curves. Physically,
the excess heating in the warmer
region where the conductor tempera-
ture ranges from Tc to Teq is con- qpk
gj (T )
ducted to the cooler region where
the conductor temperature ranges be-
tween Top and Tc ; this cooler region gq (T )
thus provides excess cooling. Figure qfm
6.9 clearly shows that the Joule dissi-
pation line, gj (T Tc ), for the com-
posite satisfying the equal-area crite-
rion is greater than that given by the T
Top Tc Teq
Stekly cryostability criterion.
Fig. 6.9 Example of gj (T ) curves sat-
Wilson extended this 1-D equal-area
isfying the equal area criterion. The
criterion to the 2-D equal-area crite- cross-lined area bounded by the gq (T ) >
rion [6.16]; the 2-D criterion was ver- gj (T ) and gj (T ) curves is equal to the
ied by an experiment conducted on other cross-lined area bounded by the
pancake test coils. gj (T ) > gq (T ) and gq (T ) curves.
370 CHAPTER 6PROBLEMS & DISCUSSIONS
Vs
n3 =
n2 = 50
n1 = 5
Vc
n1
n2
0 Is
0 Ic
Im
Matrix Rm
It It
Rdif
+
VS
Is
Fig. 6.11 Circuit model for a composite superconductor with a superconductor of the Vs
vs. Is characteristic (Eq. 6.25a), shunted by a matrix resistance Rm . The superconductor
consists of VS , an ideal voltage source, in series with Rdif , the dierential resistor.
50
300
40 mV 40
Voltage [mV]
200 200 A
Current [A]
30 mV 30
20 mV 20
100
10 mV 10
0 0A 0
Fig. 6.12 (a) Schematic drawing of the cross section of a 10-mm wide composite YBCO;
the Cu/Ag side is exposed to boiling LN2 . (b) Transport current (dashed) and voltage
(solid) vs. time traces recorded with an over-current pulse applied to the composite.
STABILITYPROBLEMS & DISCUSSIONS 375
10
gq [W/cm2 ]
1 Film Boiling
1 10 100
T (= T Ts ) [K]
Fig. 6.13 Typical heat transfer ux data for liquid nitrogen boiling at 77.3 K. The
data point corresponding to c) and d) is indicated by the solid circle in the gure.
378 CHAPTER 6PROBLEMS & DISCUSSIONS
Cable
The cable basically consists of many strands, each a composite of diameter 1 mm
or smaller containing many laments of 10100 m diameter superconductor, NbTi
or Nb3 Sn; often copper strands of the same diameter as that of either NbTi or
Nb3 Sn strands are substituted for the superconducting strands either to enhance
the overall matrix metal cross sectional area, reduce the cable cost or both. Gen-
erally three or seven strands are bundled to form what we call here a basic
cable. The 3-strand basic cables may be deciphered from each of the CIC con-
ductors illustrated in Fig. 6.14. A typical CIC conductor is at least one cm or
two acrossfor fusion magnets this can exceed 5 cmand therefore operates at
a high current, with an operating current, Iop , of at least 10 kA and, for some
fusion magnets, close to 100 kA. To meet this high Iop requirement, the bundling
process continues, in which generally 3, 5, or 7 basic cables are bundled to form
the 2nd -step cable, and so forth to additional steps, as required.
Table 6.6 lists three CIC conductors: Coils A and C of the 45-T Hybrid Magnet
at NHMFL; and a proposed CIC conductor for an ITER toroidal-eld coil design.
The 45-T Hybrid Magnet is operating, while the ITER coil is still in the design
stage, though many model versions have been operated. The rst part of Table
6.6 lists parameters of the cables for these coils.
Helium
Helium in supercritical state, i.e., its pressure above the critical pressure, 227.5 kPa
(2.25 atm), is forced through the windings at a nominal operating pressure Pop 3
5 atm. The operating temperature, Top , though, is generally maintained below its
critical temperature (5.2 K), by having the circulating helium cooled to typically
4.34.5 K, before it enters the winding. In the 45-T Hybrid the helium is subcooled
superuid, at a nominal Pop of 1 atm and a nominal Top of 1.8 K. The remarkable
properties of superuidity (high thermal conductivity and low viscosity) enable
this system to rely simply on natural convection, rather than forced ow, to
transport dissipation to a heat exchanger located outside the winding. The helium
parameters for the three coils are in the middle part of Table 6.6.
For forced-ow helium, its heat transfer coecient, hhe [W/cm2 K], is based on the
so-called Dittus-Boelter-Giarratano-Yaskin correlation [6.23]:
0.716
khe 0.8 0.4 The
hhe = 0.0259 Re Pr (6.29)
Dhy Tcd
In Eq. 6.29 khe , Re, and Pr are, respectively, the thermal conductivity of helium,
Reynolds number, and Prandtl number; Dhy is the hydraulic diameter; and Tcd
is the conductor temperature. Heat transfer uxes, gq , at 3.5 atm and 4.5 K, for
Re = 104 and Re = 105 are given in Fig. 6.3.
380 CHAPTER 6PROBLEMS & DISCUSSIONS
Conduit
The conduit encases a bundle of cabled strands and provides space for coolant,
which is generally forced supercritical helium but can be stagnant if the coolant is
superuid helium, like the superconducting coils of the 45-T Hybrid Magnet. For
composite Nb3 Sn strands, which are brittle and must not be strained more than
0.3%, the nal heat treatment for Nb3 Sn reaction must be performed after the
unreacted strands are rst encased in the conduit and wound into the coil. In the
1980s a nickel-iron based superalloy, Incoloy 908, was the preferred conduit metal,
but recently most conduits use stainless steel grade 316LNthe designation LN
indicates 316 stainless steel with low carbon and high nitrogen contents.
STABILITYPROBLEMS & DISCUSSIONS 381
eh
DUAL STABILITY
WELL-COOLED
ILL-COOLED
0 It /Ic
0 0.5 1
Fig. 6.15 General energy margin vs. normalized transport current plot
for a CIC conductor [6.25].
382 CHAPTER 6PROBLEMS & DISCUSSIONS
Joint
A CIC conductor handles both strands that carry current and the conduit that
carries coolant. Therefore, joining of two CIC conductors is much more dicult
than joining two composites unconned in a conduit. Several techniques have been
developed to deal with this issue [6.516.57].
Ramp-Rate Limitation
A conductor such as CIC conductor that is comprised of cabled composite strands
sometimes suers from a phenomenon known as ramp-rate limitation, a form
of instability that quenches the conductor below its designated operating current.
The instability phenomenon occurs only when the current ramp rate exceeds a
critical rate or for a conductor carrying a constant current but exposed to a rapid
change in the background magnetic eld. Clearly, nonuniform sharing of transport
current within the cabled strands is due chiey to inequalities in inductances and
resistances of strands in the cable. The problem has been studied extensively in
the past decade [6.586.62]. Because the ramp-rate-limitation does not occur in
cabled strands that operate at currents below Ilim , perhaps it is another reason
for these large magnets to operate at the conservative level set by Ilim .
In this very spot there are whole forests which were buried millions of years
ago; now they have turned to coal, and for me they are an inexhaustible mine.
Captain Nemo (c. 1870)
Rm (I Ic )
V = (6.31)
Rm IIc
1
fp Pcd hq (Tc Top )
c) Condition 1: fp = 1 (sk = 0.1). For Tc = 5.2 K (and Top = 4.2 K), compute v
at i = 1, 1.1, 1.5, and 2.
d) Condition 2: fp = 0.1 (sk = 1). Show that v is indeterminate at i = 1.
e) Condition 3: fp = 0.05 (sk = 2). Here, the surface area is nearly insulated
from liquid helium, the composite will behave unstably, and as observed by
Stekly in his experiment using a real supply, the supply momentarily drops
its current from the designated level, matching the positive load voltage,
before settling back to the original current and the corresponding voltage
on the v = i line [6.63]. Although Eq. 6.32, derived from Eq. 6.31 based on
the premise that v = 0 for i < 1, is not valid for i > 1 under the steady-state
conditions, you may use the equation to nd values of v for for i = 1, 0.9,
0.8, 0.75, and 0.707.
f ) Plot v(i) traces for the three conditions studied above. Plot v = i with a solid
line. Label sk = 0.1 for the curve corresponding to Condition 1; sk = 1 for
the Condition 2 curve; sk = 2 for the Condition 3 curve.
Rm2
Ic I(I Ic )
= Rm (I Ic ) + (S5.3b)
fp Pcd hq (Tc Top ) Rm Ic I
1.5
v=i
v 1.0
sk = 1
0.5
sk = 2 sk = 0.1
0
0 0.5 1.0 1.5 2.0
i
Fig. 6.16 Normalized voltage vs. normalized current traces for sk = 0.1, 1, and 2.
386 CHAPTER 6PROBLEMS & DISCUSSIONS
SPACER STRIPS
0.5-mm THICK
a INSULATOR
DOUBLE PANCAKE
SPACER STRIPS
. . .there must be discipline. For many things are not as they appear.
Discipline must come from trust and condence. Robert Jordan
388 CHAPTER 6PROBLEMS & DISCUSSIONS
4
Ic [kA]
Iop
2
1
Top Tcs Tc
0
0 1 2 3 4 5
Temperature [K]
Fig. 6.18 Ic vs. T plot (solid) for a Hybrid III Nb-Ti conductor at 10 T. The
intersection of the line at Ic = 0 determines Tc = 4.7 K. It = 2100 A is given
by the dashed line, which intersects the solid line at Tcs = 3.7 K.
TRIVIA 6.2 List the items below in the descending order of their surface heat fluxes.
i) Bottom wet surface, of a heated teapot keeping the water at 373 K;
ii) LED bulb, of a lit Christmas tree miniature decoration light;
iii) Normal-state composite HTS, cryostabilized by boiling liquid nitrogen at 77 K;
iv) Sunlight over Mars.
STABILITYPROBLEMS & DISCUSSIONS 389
200
150
gj (T ) & q(T ) [kW/m2 ]
100
50
0
0 1 2 3 4 5 6
T [K]
300
CRYOSTABLE (HF)
200 CRYOSTABLE (LF)
Stress [MPa]
100 QUASI-ADIABATIC
100
200
300 350 400 450 500 550
r [mm]
Fig. 6.20 Hoop stress vs. radius plots for cryostable and quasi-adiabatic windings.
STABILITYPROBLEMS & DISCUSSIONS 391
If we insert values of kwd , Tc , Top , m , and Jm typical for an LTS magnet oper-
ating at liquid helium temperature, Rmz is 0.110 mm. (If kwd is anisotropic but
orthotropic, as within a real winding, the MPZ shape will be an ellipsoid, not a
sphere.) The MPZ volume is minute compared with the winding volume, but it
is possible to sustain a minute amount of gd (t) even within an adiabatic winding,
provided its extent is limited to less than Rmz .
Although (Tc Top ) in Eq. 6.33 can be an order of magnitude greater for an HTS
winding than an LTS winding, because m increases with Top and kwd remains
relatively constant, the MPZ size of an
y
HTS winding at Top = 77.3 K is about
the same as that for an LTS magnet op-
erating at Top = 4.2 K. A parameter that
dramatically increases with Top is the 2
enthalpy densities of the winding mate-
rials, making it virtually impossible, as 1
stated at the outset of this chapter, to x
Rmz
drive any HTS magnet above its criti-
cal temperature with disturbance ener-
gies such as those originating from me-
chanical events, e.g., wire motion and
epoxy cracking, that still, though rarer
than in the early 1980s, continue to af-
Fig. 6.21 Isotropic winding where in re-
ict high-performance LTS magnets for gion 1 (r Rmz ) the winding is gener-
high-energy physics accelerators, NMR, ating Joule heating and in region 2 (r
and MRI. Rmz ) the winding is superconducting.
c) Show that the critical dissipation density, gdc , at which the maximum tem-
perature rise within the winding from Top , Tmx Tmx Top , is given by:
4kwd Tmx
gdc = 2 (6.37a)
1 2
1
a21 1+ ln 1
2 ln 2 ln
4
dc () 2 (6.37c)
12
1
1+ ln 1
2 ln 2 ln
b) Dierentiating Eq. 6.35 with respect to and setting the resultant expression
equal to zero, we obtain:
2
dT gd a21 1 1
= 2 + =0 (S7.3)
d 4kwd ln
Solving for , we have:
2 1
mx = (6.36)
2 ln
c) At = mx , T (mx ) = Tmx . Thus, from Eq. 6.35, Tmx is given by:
2
a21 gd 1
T (mx ) Tmx = (1 mx ) +
2
ln mx + Top (S7.4)
4kwd ln
Noting that Tmx = Tmx Top and gd = gdc at Tmx , and combining Eqs. 6.36 and
S7.4, we obtain:
2
gdc a21 2 1 1
Tmx = 1+ ln 1 (S7.5)
4kwd 2 ln 2 ln
Answer to TRIVIA 6.2 Teapot (50, in W/cm2 ); HTS (10); LED bulb (0.2); Mars (0.05).
394 CHAPTER 6PROBLEMS & DISCUSSIONS
200
100
dc 20
10
2
1
0.3
1 2 3 4 5
Fig. 6.22 dc () for the range of 1.255.
Illustration Let us consider a few examples that may approximate real cases.
Case 1 Here we consider = 2; a1 = 10 cm; kwd = 0.01 W/cm K (a typical
winding at 30 K, consisting of metal and insulator); and Tmx = 3 K. Because
dc (2) = 7.9, applying Eq. 6.37b, we obtain:
kwd Tmx
gdc = dc () (6.37b)
a21
(0.01 W/cm K)(3 K)
= (7.9) 2.4103 W/cm3 2.4103 W/m3
(10 cm)2
For this HTS solenoid, of a1 = 10 cm and a2 = 20 cm (and of innite axial length),
in which the winding is kept at 30 K and cooled by conduction from its i.d. and
o.d., a uniform dissipation density, for example by AC losses, up to 2.4 kW/m3
(2.4 mW/cm3 ) within the winding will not destroy superconductivity, provided the
superconductor has a temperature margin, Tmx , of at least 3 K. Note that this
upper limit of dissipation density increases proportionately with Tmx . Of course,
even if the HTS can remain fully superconducting under a dissipation density of
2.4 kW/m3 , the system cryogenic load may be prohibitively high.
As studied in CHAPTER 7, for an HTS tape of width w and thickness exposed
to a cyclic magnetic eld of (amplitude Hm ), directed parallel to the tapes width
(see Fig. 7.1d), hysteresis energy density, ehy , one of three major AC losses in mul-
tilamentary superconductor composite, for example, is given by: ehy Jc Hm
(Eq. 7.29d), where Jc is the superconductor critical current density. For YBCO at
30 K and 2 T, Jc 2109 A/m2 and 1106 m; thus for Hm = Bm = 2 T, we
nd: ehy = 4 kJ/m3 = 4 mJ/cm3 . This dissipation energy density translates to a
power density, at 60 Hz, of 240 kW/m3 , 100 times greater than the maximum per-
missible level computed above! To limit the hysteresis dissipation power density
to 2.4 kW/m3 , for the same Jc , Bm has to be limited to 0.02 T. Note that this is
for the eld parallel to the tape.
Case 2 Here, we consider a thinner solenoid: = 1.25, with the other parameters
the same. With dc (1.25) 128, we have: gdc 38 kW/m3 .
STABILITYPROBLEMS & DISCUSSIONS AND REFERENCES 395
REFERENCES
[6.1] E.W. Collings, Design considerations for high Tc ceramic superconductors, Cryo-
genics 28, 724 (1988).
[6.2] Y. Iwasa, Design and operational issues for 77-K superconducting magnets, IEEE
Trans. Mag. MAG-24, 1211 (1988).
[6.3] L. Ogasawara, Conductor design issues for oxide superconductors Part I: criteria
of magnetic stability, Cryogenics 29, 3 (1989).
[6.4] L. Dresner, Stability and protection of Ag/BSCCO magnets operated in the 20
40 K range, Cryogenics 33, 900 (1993).
[6.5] L.Y. Xiao, S. Han, L.Z. Lin, and H.M. Wen, Stability study on composite con-
ductors for HTSC superconducting magnets, Cryogenics 34, 785 (1994).
[6.6] J.W. Lue, L. Dresner, S.W. Schwenterly, D. Aized, J.M. Campbell, and R.E.
Schwall, Stability measurements on a 1-T high temperature superconducting mag-
net, IEEE Appl. Superconduc. 5, 230 (1995).
[6.7] Yu. A. Ilyin, V.S. Vysotsky, T. Kiss, M. Takeo, H. Okamoto and F. Irie, Stability
and quench development study in small HTS magnet, Cryogenics 41, 665 (2001).
[6.8] T. Obana, K. Tasaki, T. Kuriyama, and T. Okamura, Thermal stability analysis
of conduction-cooled HTS coil, Cryogenics 43, 603 (2003).
[6.9] A.R. Kantrowitz and Z.J.J. Stekly, A new principle for the construction of stabi-
lized superconducting coils, Appl. Phys. Lett. 6, 56 (1965). Also see, Z.J.J. Stekly,
R. Thome, and B. Strauss, Principles of stability in cooled superconducting mag-
nets, J. Appl. Phys. 40, 2238 (1969).
[6.10] J. Wong, D.F. Fairbanks, R.N. Randall, and W.L. Larson, Fully stabilized su-
perconducting strip for the Argonne and Brookhaven bubble chambers, J. Appl.
Phys. 39, 2518 (1968).
[6.11] G. Bogner, C. Albrecht, R. Maier, and P. Parsch, Experiments on copper- and
aluminum-stabilized Nb-Ti superconductors in view of their application in large
magnets, Proc. 2 nd Intl Cryo. Eng. Conf. (Ilie Science and Technology Publi-
cation, Surrey, 1968), 175.
[6.12] J.R. Purcell, The 1.8 tesla, 4.8 m i.d. bubble chamber magnet, Proc. 1968 Sum-
mer Study on Superconducting Devices and Accelerators (Brookhaven National
Laboratory, Upton New York, 1969), 765.
[6.13] A.P. Martinelli and S.L. Wipf, Investigation of cryogenic stability and reliability
of operation of Nb3 Sn coils in helium gas environment, Proc. Appl. Superconduc.
Conf. (IEEE Pub. 72CHO682-5-TABSC), 331 (1977).
[6.14] L. Bottura [Private communication, 2004].
[6.15] B.J. Maddock, G.B. James, and W.T. Norris, Superconductive composites: heat
transfer and steady state stabilization, Cryogenics 9, 261 (1969).
396 CHAPTER 6REFERENCES
[6.16] M.N. Wilson and Y. Iwasa, Stability of superconductors against localized distur-
bances of limited magnitude, Cryogenics 18, 17 (1978).
[6.17] J.E.C. Williams, E.S. Bobrov, Y. Iwasa, W.F.B. Punchard, J. Wrenn, A. Zhukov-
sky, NMR magnet technology at MIT, IEEE Trans. Magn. 28, 627 (1992).
[6.18] Y. Iwasa and V.Y. Adzovie, The index number (n) below critical current in
Nb-Ti superconductors, IEEE Trans. Appl. Superconduc. 5, 3437 (1995).
[6.19] K. Yamafuji and T. Kiss, Current-voltage characteristics near the glass-liquid
transition in high-Tc superconductors, Physica C 290, 9 (1997).
[6.20] Kohei Higashikawa, Taketsune Nakamura, Koji Shikimachi, Naoki Hirano, Shigeo
Nagaya, Takenobu Kiss, and Masayoshi Inoue, Conceptual design of HTS coil for
SMES using YBCO coated conductor, IEEE Trans. Appl. Superconduc. 17, 1990
(2007).
[6.21] Mitchell O. Hoenig and D. Bruce Montgomery, Dense supercritical-helium cooled
superconductors for large high eld stabilized magnets, IEEE Trans. Magn. MAG
-11, 569 (1975).
[6.22] M. Morpurgo, A large superconducting dipole cooled by forced circulation of two
phase helium, Cryogenics 19, 411 (1979).
[6.23] P.J. Giarratano, V.D. Arp and R.V. Smith, Forced convection heat transfer to
supercritical helium, Cryogenics 11, 385 (1971).
[6.24] Y. Iwasa, M.O. Hoenig, and D.B. Montgomery, Cryostability of a small supercon-
ducting coil wound with cabled hollow conductor, IEEE Trans. Magn. MAG-13,
678 (1977).
[6.25] J.W. Lue, J.R. Miller, and L. Dresner, Stability of cable-in-conduit superconduc-
tors, J. Appl. Phys. 51, 772 (1980).
[6.26] L. Bottura, Stability, protection and ac loss of cable-in-conduit conductors a
designers approach, Fusion Eng. and Design 20, 351 (1993).
[6.27] J.V. Minervini, M.M. Steeves, and M.O. Hoenig, Calorimetric measurement of
AC loss in ICCS conductors subjected to pulsed magnetic elds, IEEE Trans.
Magn. MAG-23, 1363 (1980).
[6.28] T. Ando, K. Okuno, H. Nakajima, K. Yoshida, T. Hiyama, H. Tsuji, Y. Takahashi,
M. Nishi, E. Tada, K. Koizumi, T. Kato, M. Sugimoto, T. Isono, K. Kawano,
M. Konno, J. Yoshida, H. Ishida, E. Kawagoe, Y. Kamiyauchi, Y. Matsuzaki,
H. Shirakata, S. Shimamoto, Experimental results of the Nb3 Sn demo poloidal
coil (DPC-EX), IEEE Trans. Magn. 27 2060 (1991).
[6.29] G.B.J. Mulder, H.H.J. ten Kate, A. Nijhuis and L.J.M. van de Klundert, A new
test setup to measure the AC losses of the conductors for NET, IEEE Trans.
Magn. 27, 2190 (1991).
[6.30] R. Bruzzese, S. Chiarelli, P. Gislon, and M. Spadoni, and S. Zannella, Critical
currents and AC losses on subsize cables of the NET-EM/LMI 40-kA Nb3 Sn cable-
in-conduit conductor prototype, IEEE Trans. Magn. 27, 2198 (1991).
[6.31] D. Ciazynski, J.L. Duchateau, B. Turck, Theoretical and experimental approach
to AC losses in a 40 kA cable for NET, IEEE Trans. Magn. 27, 2194 (1991).
[6.32] P. Bruzzone, L. Bottura, J. Eikelboom, A.J.M. Roovers, Critical currents and
AC losses on subsize cables of the NET-EM/LMI 40-kA Nb3 Sn cable-in-conduit
conductor prototype, IEEE Trans. Magn. 27, 2198 (1991).
[6.33] S.A. Egorov, A. Yu. Koretskij and E.R. Zapretilina, Interstrand coupling AC losses
in multistage cable-in-conduit superconductors, Cryogenics 32, 439 (1992).
STABILITYREFERENCES 397
[6.34] Naoyuki Amemiya, Takayuki Kikuchi, Tadayoshi Hanafusa and Osami Tsukamoto,
Stability and AC loss of superconducting cablesAnalysis of current imbalance
and inter-strand coupling losses, Cryogenics 34, 559 (1994).
[6.35] B.J.P. Baudouy, K. Bartholomew, J. Miller and S.W. Van Sciver, AC loss mea-
surement of the 45-T hybrid/CIC conductor, IEEE Trans. Appl. Superconduc. 5,
668 (1995).
[6.36] B. Blau, I. Rohleder, G. Vecsey, AC behaviour of full size, fusion dedicated cable-
in-conduit conductors in SULTAN III under applied pulsed eld, IEEE Trans.
Appl. Superconduc. 5, 697 (1995).
[6.37] Arend Nijhuis, Herman H.J. ten Kate, Pierluigi Bruzzone and Luca Bottura, First
results of a parametric study on coupling loss in subsize NET/ITER Nb3 Sn cabled
specimen, IEEE Trans. Appl. Superconduc. 5, 992 (1995).
[6.38] M. Ono, S. Hanawa, Y. Wachi, T. Hamajima, M. Yamaguchi, Inuence of cou-
pling current among superconducting strands on stability of cable-in-conduit con-
ductor, IEEE Trans. Magn. 32, 2842 (1996).
[6.39] K. Kwasnitza and St. Clerc, Coupling current loss reduction in cable-in-conduit
superconductors by thick chromium oxide coating, Cryogenics 38, 305 (1998).
[6.40] Toshiyuki Mito, Kazuya Takahata, Akifumi Iwamoto, Ryuji Maekawa, Nagato
Yanagi, Takashi Satow, Osamu Motojima, Junya Yamamoto, EXSIV Group, Fu-
mio Sumiyoshi, Shuma Kawabata and Naoki Hirano, Extra AC losses for a CICC
coil due to the non-uniform current distribution in the cable, Cryogenics 38, 551
(1998).
[6.41] P.D. Weng, Y.F. Bi, Z.M. Chen, B.Z. Li and J. Fang, HT-7U TF and PF conductor
design, Cryogenics 40, 531 (2000).
[6.42] Soren Prestemon, Stacy Sayre, Cesar Luongo and John Miller, Quench simula-
tion of a CICC model coil subjected to longitudinal and transverse eld pulses,
Cryogenics 40, 511 (2000).
[6.43] Kazutaka Seo, Katuhiko Fukuhara and Mitsuru Hasegawa, Analyses for inter-
strand coupling loss in multi-strand superconducting cable with distributed resis-
tance between strands, Cryogenics 41, 511 (2001).
[6.44] Yoshikazu Takahashi, Kunihiro Matsui, Kenji Nishi, Norikiyo Koizumi, Yoshihiko
Nunoya, Takaaki Isono, Toshinari Ando, Hiroshi Tsuji, Satoru Murase, and Susumu
Shimamoto, AC loss measurement of 46 kA-13 T Nb3 Sn conductor for ITER,
IEEE Trans. Appl. Superconduc. 11, 1546 (2001).
[6.45] Qiuliang Wang, Cheon Seong Yoon, Sungkeun Baang, Myungkyu Kim, Hyunki
Park, Yongjin Kim, Sangil Lee and Keeman Kim, AC losses and heat removal
in three-dimensional winding pack of Samsung superconducting test facility under
pulsed magnetic eld operation, Cryogenics 41, 253 (2001).
[6.46] S. Egorov, I. Rodin, A. Lancetov, A. Bursikov, M. Astrov, S. Fedotova, Ch. Weber,
and J. Kaugerts, AC loss and interstrand resistance measurement for NbTi cable-
in-conduit conductor, IEEE Trans. Appl. Superconduc. 12, 1607 (2002).
[6.47] A. Nijhuis, Yu. Ilyin, W. Abbas, B. ten Haken and H.H.J. ten Kate, Change of
interstrand contact resistance and coupling loss in various prototype ITER NbTi
conductors with transverse loading in the Twente Cryogenic Cable Press up to
40,000 cycles, Cryogenics 44, 319 (2004).
[6.48] S. Lee, Y. Chu, W.H. Chung, S.J. Lee, S.M. Choi, S.H. Park, H. Yonekawa,
S.H. Baek, J.S. Kim, K.W. Cho, K.R. Park, B.S. Lim, Y.K. Oh, K. Kim, J.S. Bak,
and G.S. Lee, AC loss characteristics of the KSTAR CSMC estimated by pulse
test, IEEE Trans. Appl. Superconduc. 16, 771 (2006).
398 CHAPTER 6REFERENCES
7.1 Introduction
Although the perfect conductivity of superconductors is what makes supercon-
ductivity perpetually fascinating for scientists and enticing to engineers and en-
trepreneurs, Type II superconductors, suitable for magnets, operate in the mixed
state and, as seen in CHAPTER 5, are magnetically hysteretic. They are intrinsi-
cally dissipative under time-varying conditions of magnetic eld, transport current,
or both. Furthermore, when a Type II superconductor is processed into a compos-
ite conductor in the form of multilaments embedded in a normal metal matrix,
other magnetic losses besides hysteresis come into play. These magnetic losses
are commonly known as AC losses. In addition, the magnet is subjected to other
dissipations, the sources of which include: 1) conductor splices; 2) Lorentz-force
induced conductor, and even winding, motion, which results in frictional heat-
ing; and 3) Lorentz-force induced cracking in the winding impregnants, which also
results in dissipation. Although not discussed here, there is another source of
dissipation in fusion magnets: neutron radiation.
The dissipation power density, expressed by gd in Eq. 6.1 of CHAPTER 6, lumps all
these chiey non-Joule heating dissipation densities. Generally, its size is miniscule
2
compared with the Joule dissipation density, cd (T )Jcd
(t) of Eq. 6.1. Despite its
small magnitude, it can play a critical role in adiabatic superconducting magnets,
particularly LTS, because the steady-state dissipation base line for LTS magnets is
zero or nearly so. As seen in CHAPTER 6, an adiabatic HTS magnet, on the other
hand, can remain superconducting despite a large dissipation density within
its windingas much as 400 kW/m3 in an illustrative case, though at a great
cryogenic load. For comparison, the dissipation density base line for water-cooled
magnets can be tens of GW/m3 ; dissipations other than Joule heating are com-
pletely negligible.
In CHAPTER 7 we discuss and study three types of the disturbance term gd : 1)
magnetic (AC losses); 2) electrical (splice resistance); and 3) mechanical (frictional
and epoxy cracking). For LTS magnets AC losses have proven devastating: only
LTS magnets having windings locally cooled with liquid heliumcryostable
can tolerate AC losses, limiting their applications (e.g., research and fusion)
and essentially excluding commercially relevant applications, where, for eciency,
adiabatic windings are preferred. Only those in which AC losses may be reduced
at will (i.e., for DC applications, e.g., NMR and MRI), are adiabatic LTS mag-
nets usable and applied successfully. With AC losses controllable and remedies
against mechanical disturbances in place, most adiabatic LTS NMR/MRI mag-
nets nowadays operate successfully most of the time. Remarkably, as noted on
the disturbance spectra of Fig. 6.1, there are no major intractable disturbances,
except AC losses, in HTS magnets. CHAPTER 7 thus focuses chiey on AC losses;
splice dissipation and mechanical disturbances are treated as other losses.
Y. Iwasa, Case Studies in Superconducting Magnets: Design and Operational Issues, 399
DOI: 10.1007/b112047_7,
Springer Science + Business Media, LLC 2009
400 CHAPTER 7
7.2 AC Losses
Following the basic philosophy of this book, only those cases amenable to analytical
expressions with which to compute ballpark gures of AC losses are considered;
i.e., only a few simple cases are presented and studied. Thus a complex real-
world case may have to be either simplied to an analytically solvable modela
recommended approach for every problemor computed head-on with a code at
the outset, an unattractive and much less revealing approach.
Three distinguishable AC loss energy densities [J/m3 ] in a multilamentary com-
posite conductor or strand are: 1) hysteresis, ehy ; 2) coupling, ecp ; and 3) eddy-
current, eed . AC losses in the conductor are generated by a magnetic eld and/or
transport current, with one or both varying in time; here only selected eld-current
excitations are considered. Subjected to one of these eld-current excitations, AC
losses in the conductor generally depend on 1) the conductor cross-section shape
considered here are Bean slab, cylinder, tapeand 2) the magnetic eld direc-
tion with respect to the conductor axiseither longitudinal, parallel to the broad
face (if any), or perpendicular to it.
AC losses within a magnet winding add to the system cryogenic load and, be-
cause an HTS magnet can operate at temperatures much higher than 4.2 K, it
can tolerate some AC losses. Still, for any AC superconducting magnet to
compete with a room-temperature magnet, its total AC losses multiplied by the
ratio of compressor power input to heat load at operating temperature (Wcp/Q)
must be less than those of a room-temperature counterpart. Note that, as shown
in Fig. 4.5, the range of Wcp/Q is 2508000 at 4.2 K and 1050 at 77 K. These
ratios alone make the task of enabling an AC superconducting magnet to succeed
in the marketplace daunting.
Work on AC losses for Type II superconductors as applied to magnets started in
the late 1960s and has continued to date. The bases for the work were established
in the 1970s and early 1980s [1.27, 7.17.16]. More recent articles are cited where
appropriate, including those on HTS.
df
2a df
w
w
(d) (e)
E H dA dt = 1 2
E J dt + 2 H + H dM dV (7.1)
S V
2a
E(x) He dA dt = 1 2
E(x) Jc (x) dt + 2 Hs (x) dx (7.2)
S 0
2a
1
ehy = Jc E(x) dt dx (7.3a)
2a 0
By combining Eqs. 7.2 and 7.3a, we have another expression for ehy :
2a
1
ehy = E(x) He dA dt 2
1
Hs2 (x) dx (7.3b)
2a S 0
Equation 7.3b states that the hysteresis energy density in the slab is equal to the
total energy density supplied by the Poynting energy density to the slab minus
the magnetic energy density in the slab.
AC AND OTHER LOSSES 403
When the external eld H e makes one complete cycle in which the initial and nal
elds, respectively, Hei and H e as well as magnetizations, M
(H
e ) and M (H
e ),
f i f
are identical, ehy may also be given by:
ehy = H e (H
e dM e) (7.4a)
In Eq. 7.4b, the vector symbols are dropped because each vector parameter points
in only one direction: He and M in the y direction.
Hs (x) plots for Cases 1, 2, and 4 are shown, respectively, in Figs. 7.27.4. In
the absence of transport current the magnetic behavior of a slab (of width 2a) is
symmetric about its midpoint (x = a); therefore, Hs (x) plots are shown only for
(0 x a). Hs (t) plots for selected Cases 1i6i are shown later.
404 CHAPTER 7
Hs (x) Hs (x) Hs (x)
Hm
Hm
Hp Hp Hp
Hm
He He He
0 0 0
0 x+ xm a 0 x+ xm 0 x+ xm
(a) (b) (c)
Fig. 7.2 Field proles for Case 1, He = 0 Hm , at 0 < He < Hm (solid lines) and at
Hm (heavy dashes): (a) small eld (Hm Hp ); (b) medium (Hp Hm 2Hp ); and
(c) large (Hm 2Hp ). He indicates that He is increasing. In each graph, x+ = He /Jc
and xm = Hm /Jc , which becomes xm = a for (b) and (c).
Hm
Hm
Hp Hp Hp
Hm
He He
He
0 0 0
0 x x0 xm a 0 x x0 xm 0 x xm
(a) (b) (c)
Hm
Hp Hp Hp
Hm
He He
He
x+
0 x+ 0 0
0 x0 xm a 0 xm 0 xm
x0
Hm
Hp Hp Hp
Hm
Hm
(a) (b) (c)
Fig. 7.4 Field proles for Case 4, in each graph, starting at He = 0 (heavy square-dots)
after He returns from Hm ; i.e., the slab is no longer virgin at the start of Case 4: He
(the arrow indicates He is increasing) between 0 and Hm (solid lines) and at He = Hm
(heavy dashes). (a) small eld (Hm Hp ); (b) medium (Hp Hm 2Hp ); and (c)
large (Hm 2Hp ). Note that x0 = Hm /2Jc , x+ = (Hm +He )/2Jc , and xm = a for (b)
and xm = x0 = x+ = a for (c).
2p
cp = (7.6)
8 2 ef
where p is the twist pitch length of laments and ef is the eective matrix
resistivity for inter-lament currents. Because, as stated above, the longer the
coupled currents last, the greater the energy dissipated, the greater cp is, the
greater will be ecp . As Wilson points out [1.27], ecp may be viewed as a fraction of
the total magnetic eld energy density in the composite, Hm 2
/2, and as cp 0,
ecp 0. As stated above, useful formulas for ecp are presented later in Table
7.8; it includes formulas of ecp for multilamentary wire under four time functions
of external magnetic eld: 1) sinusoidal; 2) exponential; 3) triangular; and 4)
trapezoidal. Key time parameters for these functions are dened in Fig. 7.18.
1 f
ef0 = m (7.7a)
1 + f
1 + f
ef = m (7.7b)
1 f
(a) ( b)
a sd SOLDER LAYER
CONDUCTOR A
CONDUCTOR B
sp
Contact Resistance
The splice resistance, Rsp , is the sum of three components:
Rsp = RcA + Rsd + RcB (7.9a)
Rct
= (7.9b)
Act
where RcA is the contact resistance between Conductor A and the solder layer
and RcB is that between Conductor B and the solder layer. Rct is the contact
resistance in the unit of [ m2 ], and Act = asp is the contact area. If the solder
wets the surface of each conductor, it is reasonable to assume RcA RcB 0 or
at least RcA RcB Rsd :
Rct
Rsp = Rsd (7.9c)
Act
Note that Rsp can be made arbitrarily small by making sp suciently long. Com-
bining Eqs. 7.8 and 7.9c, we obtain:
Rct sd sd (7.10)
Thus, we may achieve small Rct by selecting a low-resistivity solder and, equally
important, by minimizing sd , certainly no thicker than 1050 m. Indeed, if a
superconducting solder is used, Rct becomes zero, provided such a splice is placed
in a low eld region and enough area (asp ) provided for current passage. With
a superconducting solder, it is possible to create an essentially superconducting
joint with a resistance in the range of pico-ohms. Such a joint must be in a very
low-eld ( 1 T) and cold ( 9 K) environment, with the contact area large enough
to keep the solder superconducting as the joint carries the operating current.
Table 7.1 presents values of Rct for selected tin-lead solders. Rct is magnetic eld-
dependent, increasing linearly with B. In some data, a nonlinearity is observed
between zero and 1 T. These data are presented only as a general guide. In projects
involving large magnets and where splice resistance is an important design issue,
it is prudent to rely on actual measurements.
Table 7.2 presents zero-eld electrical resistivity vs. temperature data, by Fu-
jishiro [7.21], of indium (In) and six common solder alloys of In, lead (Pb), and
tin (Sn), listed in the table, where Tsl and Tlq are, respectively, solidus and liquidus
temperatures. (The smaller the dierences in these temperatures, the easier the
soldering; when Tsl = Tlq , an alloy has a well-dened melting point.) As indicated
in the table, these common solder alloys are superconducting above 4.2 K; Tc is
the zero-eld critical temperature [4.41, 4.42, 7.20].
Mechanical-Contact Switch
For some applications it is advantageous to use a mechanical-contact switch rather
than a heater-activated persistent-current switch (PCS). Resistances of mechani-
cal contacts between copper or indium-coated copper surfaces, though small, are
not superconducting even at 4.2 K [7.22]. Recently, mechanical contacts between
HTS bulk disks operating at 77 K have been explored by Sawa and others to build
mechanical-contact switches [7.23, 7.24]. Although obviously not superconducting
at 77 K, if operated below 10 K and with HTS bulk disks coated with a supercon-
ducting solder, it may be possible to achieve a superconducting disk-to-disk contact
for a superconducting mechanical-contact switch.
* John Wiley & Sons, Inc., New York, 1995. For many years until his retirement in 1993,
Ernest Rabinowicz (19272006) taught mechanical engineering at M.I.T. While still a
student in Cambridge University, he once handed back a crystal radio, restored by his
friend, to the great English physicist Paul Andrien Maurice Dirac (1902-1984), who
thanked and commended him, Youre a promising young man.
AC AND OTHER LOSSES 413
AE
V
1V
0 2 4 6 8 10 12 14 16 18
TIME [s]
Fig. 7.8 AE and voltage signals from an Nb3 Sn coil prematurely quenching [7.57].
414 CHAPTER 7
AE SENSOR 1
AE SENSOR 2
AE SENSOR 3
VOLTAGE
Fig. 7.9 Oscillogram traces of AE signals and voltage from a natural quench of
a superconducting magnet at its critical current (time scale: 2 ms/div) [7.57].
AE signals appear. It is likely that the signals were caused by a nonuniform tem-
perature distribution created by the quench, which, though naturally triggered,
was localized to the high-eld region.
AE signals could become useful complements to voltage signals for detecting the
onset of overheating in HTS magnets, particularly as the resistive voltage of HTS,
owing chiey to its low index, does not rise with current as sharply as that of LTS.
Wozny and others recorded the AE signals generated by the temperature rise at
the superconducting-normal transition in YBCO bulk samples [7.58]; Arai detected
heating-induced AE signals in a test pancake coil wound with Bi2223 [7.59]. Fur-
ther work towards making AE signals complement voltage signals for protection
of HTS magnets is in progress [7.60].
We can know the time, we can know a time. We can never know Time. Ada
AC AND OTHER LOSSESPROBLEMS & DISCUSSIONS 415
He2 2Hm He Hm
2
M (He ) = He + (He = Hm 0) (7.12)
4Hp
Figure 7.10 shows M (He ) as given by Eqs. 5.5 and 7.12.
a) Applying Eq. 7.3a for one half of the slab (0 x a), show that ehy for Case
1, He = 0 Hm , is given by:
3
Hm
ehy = (0 Hm Hp ) (7.13a)
6Hp
b) Similarly, applying Eq. 7.3a for one half of the slab (0 x a), show that
ehy for Case 2, He = Hm 0, is given by:
3
Hm
ehy = (0 Hm Hp ) (7.14a)
24Hp
M (He )
1
H
2 p
A
B
0 He
Hm H p
12 Hp
Fig. 7.10 M (He ) for small eld excitation (Hm Hp ). Case 1: Trace A;
Case 2: Trace B; Case 3: Traces A and B. Sequence A begins at the origin
(0,0), with a virgin Bean slab.
416 CHAPTER 7PROBLEMS & DISCUSSIONS
Applying Eqs. 7.3a for the rst half of the slab, and S1.1b, we have:
1 a
ehy = Jc E(x) dt dx (S1.2a)
a 0
Jc Hm x+
= (x+ x) dx dHe (S1.2b)
a 0 0
Note that in Eq. S1.2b the order of integration is reversed from that of Eq. S1.2a
because x+ depends on He . Equation S1.2b leads to:
Jc Hm 2 x+2
ehy = x+ dHe (S1.2c)
a 0 2
Jc Hm He2 Hm3
= dH e = (S1.2d)
a 0 2Jc2 6aJc
3
Hm
ehy = (Hm Hp ) (7.13a)
6Hp
Note that ehy Hm 3
; i.e., in a small eld excitation ehy increases as the 3rd power
of Hm ; this has been veried by many experiments.
b) Similarly in a decreasing-eld sequence, He = Hm 0, Ez (x) dt is given by:
Ez (x) dt = (x x ) dHe (S1.3)
where x = (Hm He )/2Jc , as indicated in Fig. 7.3a. Thus ehy is given by:
x
Jc 0 Jc 0 x2
ehy = (x x ) dx dHe = x2 dHe (S1.4)
a Hm 0 a Hm 2
Jc (Hm He )2
0 3
Hm
= dH e =
a Hm 8Jc2 24aJc
3
Hm
ehy = (0 Hm Hp ) (7.14a)
24Hp
Note that in the decreasing-eld sequence, ehy is 1/4 that of the increasing-eld
sequence. This is, as may be evident from Figs. 7.2a and 7.3a, because Ez induced
within the slab in Case 1 is between x = 0 and xm = Hm /Jc (Fig. 7.2a), while in
Case 2 it is between x = 0 and x0 = Hm /2Jc (Fig. 7.3a); again, ehy Hm
3
.
AC AND OTHER LOSSESPROBLEMS & DISCUSSIONS 417
3
Hm
epy 1 = (S1.6a)
3Hp
3
Hm
epy 2 = (S1.6b)
12Hp
The sign in Eq. S1.6b indicates that epy 2 was returned to the source. At the end
of the full eld sequence, He = 0 Hm 0 (Case 3), as may be inferred from
Eq. 7.12, M (0) = Hm 2
/4Hp : the slab stores magnetic (or magnetization) energy
density, emf , through Hs (x), which, with x0 = Hm /2Jc , is given by:
Hs (x) = Jc x (0 x x0 ) (S1.7a)
Hs (x) = Hm Jc x (x0 x Hm /Jc ) (S1.7b)
3
Hm
emf = (S1.8)
24Hp
418 CHAPTER 7PROBLEMS & DISCUSSIONS
Energy Flow
Here, let us check the energy ow from the source to the slab during each eld
sequence. In each sequence, the energy densities must be balanced:
epy = ehy + emf emi (S1.9)
where emf and emi are the magnetic energy densities in the slab, respectively, at
the nal and initial states. Note that Eq. S1.9 is essentially the same as Eq. 7.2.
For the rst eld sequence, epy = epy 1 (Eq. S1.6a), ehy (Eq. 7.13a), emi = 0 (because
the slab is in the virgin state), and emf 1 may be evaluated from Hs (x) = Hm Jc x:
a 2
emf 1 = Hs (x) dx
2a 0
Hm
Jc
= (Hm Jc x)2 dx
2a 0
3
Hm
emf 1 = (S1.10)
6Hp
Inserting Eqs. 7.13a and S1.10 into the right-hand side of Eq. S1.9, we obtain:
3 3 3
Hm Hm Hm
epy 1 = + = (S1.11a)
6Hp 6Hp 3Hp
epy 1 of Eq. S1.11a is identical to Eq. S1.6a, demonstrating that the energy density
ow in the rst eld sequence is balanced.
We may also check the energy balance during the second eld sequence, He =
Hm 0. During this sequence, appropriate energy densities in the right-hand side
of Eq. S1.9 are: ehy (Eq. 7.14a), emf (Eq. S1.8), and emi = emf 1 (Eq. S1.10):
3 3 3 3
Hm Hm Hm Hm
epy 2 = + = (S1.11b)
24Hp 24Hp 6Hp 12Hp
epy 2 of Eq. S1.11b indeed equals epy 2 given by Eq. S1.6b, once again demonstrating
that the energy balance of Eq. S1.9 is valid in the second eld sequence. The sign
indicates, as remarked above, a net ow of energy back to the external source during
the second eld sequence: this energy density plus the hysteresis energy density
are paid for by a net reduction in the magnetic energy storage in the slab.
AC AND OTHER LOSSESPROBLEMS & DISCUSSIONS 419
= 12 Hp (He = Hp Hm ) (5.6)
(Hm He )2
M (He ) = 12 Hp (Hm He ) + (He = Hm 0) (5.7a)
4Hp
Figure 7.11 shows M (He ) plots given by Eqs. 5.5, 5.6, and 5.7a.
a) Applying Eq. 7.3a for one half of the slab (0 x a), show that ehy for Case
1 is given by:
2Hp
ehy = 2 Hp Hm 1
1
(Hp Hm 2Hp ) (7.13b)
3Hm
b) For Case 2, explain why ehy is still given by Eq. 7.14a (p. 415).
c) Using Eq. 7.3b, show that ehy for Case 3, which is obviously the sum of
Eqs. 7.13b and 7.14a, is given by:
2
2H p H m
ehy = 12 Hp Hm 1 + 121
(Hp Hm 2Hp ) (7.15b)
3Hm Hp
d) Show that for Hm = Hp , Eqs. 7.15a (p. 415) and 7.15b give the same ehy .
M (He )
1
H
2 p
A
B
0 He
Hp Hm
12 Hp
Fig. 7.11 M (He ) for medium eld excitation. Case 1: Trace A; Case 2:
Trace B; Case 3: Traces A and B. Note that Hp Hm 2Hp = 2Jc a.
420 CHAPTER 7PROBLEMS & DISCUSSIONS
= 12 Hp Hm 13 Hp2
2Hp
ehy = 2 Hp Hm 1
1
(Hp Hm 2Hp ) (7.13b)
3Hm
2
epy 1 = 12 (Hm Hp2 ) (S2.4b)
AC AND OTHER LOSSESPROBLEMS & DISCUSSIONS 421
Again the sign indicates that epy 2 was returned to the source. At the end of
the full eld sequence (Case 3), as may be inferred from Fig. 7.11, the slab stores
magnetic (or magnetization) energy density, emf , through Hs (x), which, with
x0 = Hm /2Jc , is again given by Eq. S1.7. Using Eq. S1.7, we may evaluate emf .
Here, unlike in the previous integration (Eq. S1.8), where the integration from
x = 0 to x = x0 = Hm /2Jc is doubled to compute emf , the integration must be
performed separately for the two ranges, x = 0 x0 and x0 a (Fig. 7.3b):
Hm a
a 2 2Jc
emf = Hs (x) dx = Jc2 x2 dx + (Hm Jc x)2 dx
2a 0 2a 0 Hm
2J c
3
Hm
2
emf = 12 Hm 12 Hm Hp + 16 Hp2 (S2.7)
8Hp
Combining Eqs. 7.3b, S2.4a, S2.4b, S1.6b, and S2.7 we obtain ehy for Case 3:
ehy = epy 1 + epy 1 + epy 2 emf
3
Hm
= 13 Hp2 + 12 (Hm
2
Hp2 )
12Hp
3
Hm
1 2
2 Hm 2 Hm Hp + 6 Hp
1 1 2
8Hp
3
Hm
= 13 Hp2 + + 12 Hp Hm (S2.8)
24Hp
b) Show that ehy 2 , the hysteresis energy density in the second part of Case 2,
He = (Hm 2Hp ) 0, is given by:
2Hp
ehy 2 = 2 Hp Hm 1
1
(7.16b)
Hm
e) Show that Eqs. 7.15b (p. 419) and 7.15c agree at Hm = 2Hp .
M (He )
1
H
2 p
A
B
0 He
Hm2Hp Hp 2Hp Hm
12 Hp
Fig. 7.12 M (He ) for large eld excitation. Case 1: Trace A; Case 2:
Trace B; Case 3: Traces A and B. Note that Hm 2Hp = 2Jc a.
AC AND OTHER LOSSESPROBLEMS & DISCUSSIONS 423
b) Again, Ez (x) is given by Eq. S1.3 and ehy 2 by Eq. S1.4 with x = a:
a
Jc 0 Jc 0 2
ehy 2 = (x a) dx dHe = a dHe
a Hm2Hp 0 2a Hm2Hp
2Hp
ehy 2 = 1
2 Hp Hm 1 (7.16b)
Hm
d) We may simply add Eq. 7.13b (ehy for He = 0 Hm ) and Eq. 7.14b:
2Hp 4Hp
ehy = 12 Hp Hm 1 + 12 Hp Hm 1
3Hm 3Hm
Hp
ehy = Hp Hm 1 (Hm 2Hp ) (7.15c)
Hm
e) At Hm = 2Hp , ehy may be given by either Eqs. 7.15b (p. 419) or 7.15c:
2
2Hp Hm
ehy = 2 Hp Hm 1
1
+ 121
(7.15b)
3Hm Hp
1 4
= Hp 1 +
2
= Hp2
3 12
Hp
ehy = Hp Hm 1 (7.15c)
Hm
1
= 2 Hp 1
2
= Hp2
2
424 CHAPTER 7PROBLEMS & DISCUSSIONS
Figure 7.13 shows M (He ) plots over the eld range from Hm to Hm . The dash-
dotted, dashed, and solid lines correspond, respectively, to small, medium, and
large eld excursions, each applicable to Case 6.
M (He )
1
H
2 p
He
2Hp Hp Hp 2Hp
12 Hp
Fig. 7.13 M (He ) plots from Hm to Hm : dash-dotted; dashed; and solid lines
correspond, respectively, to small, medium, and large eld excursions.
AC AND OTHER LOSSESPROBLEMS & DISCUSSIONS 425
Case 4Medium As may be inferred from Fig. 7.4b, when He reaches such
that He + Hm = 2Hp , i.e., He = 2Hp Hm (remember here Hp Hm 2Hp ),
Hs (x) = He Jc x throughout the slab (of course, from x = 0 to x = a) and Ez (x) is
given simply by Eq. S2.2 of Solution to PROBLEM 7.2:
dHe
(a x)
Ez (x) = (S2.2)
dt
Until He = 2Hp Hm , as may be inferred from Fig. 7.4b, He penetrates only to
x+ . Thus, ehy must be computed over two eld ranges, He = 0 2Hp Hm and
He = 2HpHm Hm , with Ez (x) dt given, respectively, by Eq. S1.1b and Eq. S2.2:
2HpHm 2 Hm
Jc Hm + He 2
ehy = dHe + a dHe
2a 0 2Jc 2HpHm
2HpHm Hm
2
= (Hm + 2Hm He + He2 ) dHe + 1
2 Hp dHe
8Hp 0 2HpHm
3
Hm
= 1 2
3 Hp + ( Hp Hm Hp2 )
24Hp
Thus:
2
2Hp Hm
ehy = Hp Hm 1 1
24 (Hp Hm 2Hp ) (7.19b)
3Hm Hp
Note that Eqs. 7.19a and 7.19b, as expected, agree at Hm = Hp : ehy = 7 Hp2 /24.
426 CHAPTER 7PROBLEMS & DISCUSSIONS
2Hp
ehy = 2 Hp Hm 1 (Hp Hm 2Hp ) (7.20b)
3Hm
We may also derive Eq. 7.20b above from Eq. 7.21, the equivalent of Eq. 7.4a.
Hm
ehy = 2 [M (He )]He =0Hm [M (He )]He =Hm 0 dHe (7.21)
0
Because M (He ) is given by Eq. 7.17b for He = 0 2Hp Hm and by Eq. 5.6
for He = 2Hp Hm Hm , the M (He ) dHe integral of Eq. 7.21 consists of three
components. With Eqs. 7.17b, 5.6, and 5.7a into Eq. 7.21, we have:
2HpHm
(Hm +He )2
ehy = 2 1
2 Hp + (Hm +He ) dHe
0 4Hp
Hm
Hm
(Hm He )2
2 Hp dHe H
1 1
+ 2 H p (H m e ) + dHe
2Hp Hm 0 4Hp
2HpHm 2
Hm Hm He He2
= 2 1
2 Hp + Hm +He dHe
0 4Hp 2Hp 4Hp
Hm
H2 Hm He H2
+ Hp (Hm Hp ) 1
2 Hp Hm + He + m + e dHe
0 4Hp 2Hp 4Hp
3
Hm
= 2 1 2
3 Hp + 12 Hp Hm 12 Hm
2
+
12Hp
3
Hm
+ Hp (Hm Hp ) 1
2 Hp Hm 1 2
2 Hm +
12Hp
= 2 23 Hp2 + Hp Hm
Thus:
2Hp
ehy = 2 Hp Hm 1 (Hp Hm 2Hp ) (7.20b)
3Hm
428 CHAPTER 7PROBLEMS & DISCUSSIONS
2Hp
ehy = 2 Hp Hm 1 (Hm 2Hp ) (7.20c)
3Hm
Equation 7.20c may also be derived from Eq. 7.21. Here too, although M (He ) is
given by Eq. 5.6 for the entire eld range in the increasing-eld sequence (Case 4),
because in the decreasing-eld sequence (Case 5) M (He ) is given by Eq. 5.7a in
the range He = Hm Hm 2Hp and by Eq. 5.7b in the range He = Hm 2Hp 0,
the M (He ) dHe integral (Eq. 7.21) consists of three components:
Hm
ehy = 2 [M (He )]He =0Hm + [M (He )]He =Hm 0 dHe (7.21)
0
Hm
1
= 2 2 Hp dHe
0
Hm2Hp
Hm
(Hm He )2 1
1
2 Hp (Hm He ) + 2 Hp dHe
Hm2Hp 4Hp 0
= 2 Hp (Hm Hp )
Hm 2
Hm H m He He2
+ 2 Hp + Hm
1
He + dHe
Hm2Hp 4Hp 2Hp 4Hp
= 2 Hp (Hm Hp ) + 13 Hp2
2Hp
ehy = 2 Hp Hm 1 (Hm 2Hp ) (7.20c)
3Hm
Note that for Hm Hp , the condition which is generally met in most applications,
ehy is proportional to Hm ; because Hp = Jc a, ehy also increases with Jc and a:
Hp Hp
Hp Hp
Hm +Hp i
Hm +Hp i
Hp i
Hp i
HmHp i Hm Hp i
0 0
Hp i Hp i
0 a 2a 0 a 2a
(a) (b)
Hs (x) Hs (x)
Hm +Hp i Hm+Hp i
HmHp i HmHp i
Hp Hm
+Hp i Hp Hp
Hp i Hp i
Hm
Hp i
0 0
Hp i Hp i
0 a 2a 0 a 2a
(c) (d)
Fig. 7.14 Graphs of Hs (x) plots for a Bean slab of width 2a carrying DC transport
current of It (= 2iHp ), subjected to He (t) of Cases 1i and 2i, where Hm is the maximum
external eld. In each graph, the dotted and dashed lines correspond, respectively, to
Hs (x) at the start and the end of a eld sequence. Graphs (a) and (b), respectively Cases
1i and 2i, are for Hm < Hp (1i); graphs (c) and (d), respectively Cases 1i and 2i, are
Hm Hp (1i). The dash-dot-dot lines in (c) and (d) represent Hs (x) at the end of a
eld sequence when Hm = Hp (1i) Hm .
430 CHAPTER 7PROBLEMS & DISCUSSIONS
Hp Hp
Fig. 7.15 Hs (x) plots for Cases 4i and
5i, in which Hm Hp (1i). Graph (a) He+Hp i
is for Case 4i, i.e., He (t) = 0 Hm , after Hp i
the eld sequence from Hm to 0; graph
(b) is for Case 5i, i.e., He (t) = Hm 0.
In each graph the dotted and dashed HeHp i
lines correspond to Hs (x), respectively, Hp i
at the start and end of each sequence;
the solid lines correspond to Hs (x) for
He (Case 4i) or He (Case 5i) between Hp Hp
0 and Hm . Note that the horizonal co-
x
ordinate is either x or , measured, re- 0 x+ + 0
spectively, from the left-hand end or the
right-hand end of the slab. (b)
AC AND OTHER LOSSESPROBLEMS & DISCUSSIONS 431
Hp Hm
+Hp i
b) Show that the hysteresis energy density for Case 5i, He = Hm 0, after a
small eld excursion, i.e., Hm Hp (1i), is given by:
3
Hm
ehy = [0 Hm Hp (1 i)] (7.22b)
24Hp
Note that ehy given by Eq. 7.22c is identical to ehy given by Eq. 7.20a (p. 426)
for a slab carrying no transport current: for a small eld excursion, trans-
port current has no eect on hysteresis dissipation.
d) Show that the hysteresis energy density for Case 4i, He = 0 Hm , for a
large eld excursion, i.e., Hm 2Hp (1i), is given by:
e) Show that the hysteresis energy density for Case 5i, He = Hm 0, after a
large eld excursion, i.e., Hm 2Hp (1i), is given by:
g) Show that Eq. 7.23c with i = 0zero transport currentreduces to Eq. 7.20c
(p. 428), an expression for ehy with no transport current in the slab.
AC AND OTHER LOSSESPROBLEMS & DISCUSSIONS 433
dHe
Ez (x) = (x+ x) (S4.1a)
dt
Ez (x) dt = (x+ x) dHe (S4.1b)
Inserting x+ = (Hm +He )/2Jc into Eq. S4.2a and performing the integration, we
obtain:
3
7 Hm
ehy x = (S4.2b)
48Hp
Next, we derive ehy , the hysteresis energy density in the side of the slab,
Hs (); Hs (0) =He +Hp i, as shown in Fig. 7.15a. With + = (Hm +He )/2Jc , it is
quite clear that ehy = ehy x and hence ehy = 2ehy x . Thus:
3
7 Hm
ehy = [0 Hm Hp (1 i)] (7.22a)
24Hp
b) We consider the x side of the slab. Similar to Eq. S4.2a, ehy x is given by:
0 x+
Jc
ehy x = (x x+ ) dx dHe (S4.3a)
2a Hm 0
Again, the hysteresis energy density in the side of the slab, ehy , is the same as
ehy x . Therefore ehy for this case is twice ehy x given by Eq. S4.3c:
3
Hm
ehy = [0 Hm Hp (1 i)] (7.22b)
24Hp
434 CHAPTER 7PROBLEMS & DISCUSSIONS
Next, we consider the side of the slab. Similar to Eq. S4.2a, we have:
Jc Hm
ehy = ( ) d dHe (S4.6a)
2a 0 0
Because ehy = ehy x +ehy , combining Eqs. S4.5b and S4.6b, we obtain:
ehy = 12 Hp Hm (1 + i2 ) [Hm 2Hp (1 i)] (7.23a)
e) For a decreasing eld sequence, Case 5i, we rst consider the range between
Hm and Hm Hm 2Hp (1i). Considering the x side of the slab, we have:
Jc Hm x+
ehy x = (x x+ ) dx dHe (S4.7a)
2a Hm 0
where, in this decreasing eld sequence, x+ = (Hm He )/2Jc . Thus:
2
Jc Hm Hm He
ehy x = dHe (S4.7b)
4a Hm 2Jc
Hm
= (Hm2
He Hm He2 + 13 He3 )
16Hp Hm 2Hp (1i)
= 16 Hp2 (1 i)3
AC AND OTHER LOSSESPROBLEMS & DISCUSSIONS 435
ehy for Case 5i is given by the sum of those given by Eqs. S4.7c and S4.8c:
f ) ehy for Case 6i, Hm Hp (1i), is the sum of ehy for Case 4i and ehy for Case 5i
multiplied by 2 to include those for the eld sequences He (t) = 0 Hm and
He (t) = Hm 0:
ehy = 2 1
2 Hp Hm (1 + i2 ) + 12 Hp Hm (1 + i2 ) 23 Hp2 (1 i3 ) (S4.9)
In Eq. 7.25 Im is the amplitude of the cyclic AC current. Note that in Cases 1sf3sf
considered here the slab is no longer virgin. Figure 7.17 shows Hs (x) plots for a
Bean slab of width 2a, in which graphs (a) and (b) correspond, respectively, to
Cases 1sf, i(t) = 0 im , and 2sf, i(t) = im 0, where i I/Ic im Im /Ic 1. As
seen in each graph, Hs (x) is antisymmetric about the slabs midpoint: to derive
expressions of ehy below, consider only one half of the slab, from x = 0 to x = a.
a) Applying Eq. 7.3a (p. 402), show that esf for Case 1sf is given by:
7 2 3
esf = 24 Hp im (7.26a)
b) Applying Eq. 7.3a, show that esf for Case 2sf is given by:
1 2 3
esf = 24 Hp im (7.26b)
c) Applying Eq. 7.3a, show that esf for Case 3sf is given by:
esf = 23 Hp2 i3m (7.26c)
d) Derive Eq. 7.26c from Eq. 7.20a (p. 426), ehy valid for Case 6 (small).
Hp Hp Hp Hp
Hp im Hp im Hp im Hp im
Hp i Hp i
Hp i Hp i
Hp im Hp im Hp im Hp im
Hp x Hp Hp x Hp
0 x+ a 0 x+ a
(a) (b)
Fig. 7.17 Hs (x) plots for Cases 1sf and 2sf, in which i I/Ic im Im /Ic 1. Graph
(a) for Case 1sf; graph (b) for Case 2sf. In each graph the dotted and dashed lines
correspond to Hs (x), respectively, at the start and end of a current sequence.
AC AND OTHER LOSSESPROBLEMS & DISCUSSIONS 437
Hp im x+
Jc
esf x = (x x+ ) dx dHsf (S5.1a)
2a 0 0
Hp im
Jc 2
= (x+ ) dHsf (S5.1b)
4a 0
Substituting Hsf = Hp i into Eq. S5.1b and, with x+ = Hp (im +i)/2Jc , we obtain:
Hp2 im
esf x = (i2m + 2im i + i2 )di (S5.2a)
16 0
7 2 3
= 48 Hp im (S5.2b)
Because the same dissipation energy density is generated over the other half of the
slab, a x 2a, esf for Case 1sf is twice esf x :
7 2 3
esf = 24 Hp im (7.26a)
b) Similarly to Case 1sf, for Case 2sf we have, with x+ = Hp (im i)/2Jc :
0 x+
Jc
esf x = (x+ x) dx dHsf (S5.3a)
2a Hp im 0
Hp2 0
= (i2m 2im i + i2 )di (S5.3b)
16 im
1 2 3
= 48 p im
H (S5.3c)
Thus, esf for Case 2sf is twice esf x given by Eq. S5.3c:
1 2 3
esf = 24 Hp im (7.26b)
c) For Case 3sf ehy is the sum of those of Cases 1sf and 2sf multiplied by 2,
because Case 3sf covers a full cycle, i(t) = 0 im 0 im 0:
esf = 2 7 2 3
24 Hp im + 1 2 3
24 Hp im (S5.4)
2 (Hp im )3
=
3Hp
Triangular: Trapezoidal:
Hm Hm
t t
2m m1 m2 m1
Fig. 7.18 He (t) functions for coupling and eddy-current energy densities.
AC AND OTHER LOSSESPROBLEMS & DISCUSSIONS 439
Table 7.3A: Hysteresis Energy Density, ehy [J/m3 ]Bean Slab (Width: 2a)
No Transport Current
CASE 1 CASE 4
! !
He (t) = 0 (Virgin slab) Hm 0 Hm 0 Hm 0 Hm 0 (7.5)
! !
CASE 2 CASE 5
! !
CASE 3 CASE 6
Case 1
3
Hm
ehy = (0 Hm Hp ) (7.13a)
6Hp
2Hp
ehy = 2 Hp H m 1
1
(Hm Hp ) (7.13b)
3Hm
Cases 2 and 5
3
Hm
ehy = (0 Hm 2Hp ) (7.14a)
24Hp
4Hp
ehy = 12 Hp Hm 1 (Hm 2Hp ) (7.14b)
3Hm
Case 3
3
5 Hm
ehy = (0 Hm Hp ) (7.15a)
24Hp
2
2Hp Hm
ehy = 1
H H
2 p m
1 1
+ 12 (Hp Hm 2Hp ) (7.15b)
3Hm Hp
Hp
ehy = Hp H m 1 (Hm 2Hp ) (7.15c)
Hm
Case 4
3
7 Hm
ehy = (0 Hm Hp ) (7.19a)
24Hp
2
2Hp Hm
ehy = Hp Hm 1 1
24
(Hp Hm 2Hp ) (7.19b)
3Hm Hp
Case 6
3
2 Hm
ehy = (0 Hm Hp ) (7.20a)
3Hp
2Hp
ehy = 2 Hp Hm 1 (Hp Hm 2Hp ) (7.20b)
3Hm
2Hp
ehy = 2 Hp Hm 1 (Hm 2Hp ) (7.20c)
3Hm
ehy = 2 Hp Hm (Hm Hp ) (7.20d)
ehy = 2 Jc aHm (Hm Hp ) (7.20e)
440 CHAPTER 7PROBLEMS & DISCUSSIONS
Table 7.3B: Hysteresis Energy Density, ehy [J/m3 ]Bean Slab (Width: 2a)
With DC Transport Current It (i = It /Ic )
CASE 1i CASE 4i
! !
He (t) = 0 (Virgin slab) Hm 0 Hm 0 Hm 0 Hm 0 (7.5)
! !
CASE 2i CASE 5i
! !
CASE 3i CASE 6i
Cases 1i3i
Only Hs (x) proles for Cases 1i and 2i are studiedsee Figs. 7.14 and 7.15.
Case 4i
3
7 Hm
ehy = [0 Hm Hp (1i)] (7.22a)
24Hp
ehy = 12 Hp Hm (1 + i2 ) [Hm 2Hp (1i)] (7.23a)
Case 5i
3
Hm
ehy = [0 Hm Hp (1i)] (7.22b)
24Hp
ehy = 12 Hp Hm (1 + i2 ) 23 Hp2 (1 i3 ) [Hm 2Hp (1i)] (7.23b)
Case 6i
3
2 Hm
ehy = [0 Hm Hp (1i)] (7.22c)
3Hp
ehy = 2 Hp Hm (1 + i2 ) 43 Hp2 (1 i3 ) [Hm 2Hp (1i)] (7.23c)
CASE 1sf
!
It (t) = 0 (Virgin slab) Im 0 Im 0 Im 0 Im 0 (7.25)
!
CASE 2sf
!
CASE 3sf
im = Im /Ic
Case 1sf
esf = 7
H 2 i3
24 p m
(7.26a)
Case 2sf
esf = 1
H 2 i3
24 p m
(7.26b)
Case 3sf
esf = 23 Hp2 i3m [0 Hm Hp (1i)] (7.26c)
AC AND OTHER LOSSESPROBLEMS & DISCUSSIONS 441
Table 7.5: Hysteresis Energy Density, ehy [J/m3 ]No Transport Current
Wire (Diameter: df ) [1.27, 7.14] & Tape (Width: w; Thickness: ) [1.27, 7.13]
Wire
Field Parallel to Wire Axis (Fig. 7.1b): Hp Jc (df /2)
3 4
Hm Hm
ehy = 4
3
2
3
(0 Hm Hp ) (7.27a)
Hp Hp2
ehy = 43 Hp Hm 23 Hp2 (Hm Hp ) (7.27b)
Tape
Field Parallel to Tape Surface (Fig. 7.1d):* Hp Jc (/2)
*same as Bean Slab Case 1 (Table 7.3A)
3
Hm
ehy = 2
3
(0 Hm Hp ) (7.29a)
Hp
2Hp
ehy = 2 Hp Hm 1 (Hm Hp ) (7.29b)
3Hm
ehy 2 Hp Hm (Hm Hp ) (7.29c)
Table 7.6: Self-Field Energy Density, esf [J/m3 ]Wire & Tape [1.27, 7.14]
It (t) = V irgin 0 Im 0 Im 0
! !
HISTORY LOSS CYCLE
Table 7.7: Energy Density, eih [J/m3 ]Bean Slab (Width: 2a) [7.5]
In-Phase Sinusoidal Transport Current & Field
It (t) = Im sin(2f t); He (t) = Hm sin(2f t)
Hp2 i3m
eih = 2 Hp2 2
+ im (0 Hm Hp ) (7.34a)
2Hm
Hp4 3 3
Im Ic
2
i m 2
(0 Hm
Hp ) (7.34b)
Hm Hm
Hm Hp i2m
eih = 2 Hp2 + (0 Hm Hp ) (7.35a)
3Hp Hm
2 Hp (3+im )
2
2Hp2 (1i3m ) 6Hp3 i2m (1im )2
eih = 2 Hp 2
+ 2 (H H i )
3Hm 3Hm 3Hm m p m
3 2 2 4 2 3
6Hp im (1im ) 4Hp im (1im )
+ 2 (H H i )
2 (H H i )2
(Hm Hp ) (7.35b)
3Hm m p m 3Hm m p m
2 Hp3 I3
(3+i2m ) c (3+i2m ) (Hm Hp ) (7.35c)
3Hm Hm
AC AND OTHER LOSSESPROBLEMS & DISCUSSIONS 443
Table 7.8: Coupling Energy Density Over One Period, ecp [J/m3 ] [7.16]
Wire of Outermost Diameter, Dmf , Enclosing Multilaments
2
2 Dmf
ecp = 2 Hm 1+ 14 (7.36 )
p
2 2 m cp
Sinusoidal: = 2 + 4 2 2
(7.37a)
m cp
m 2 2 cp
(m
cp ) (7.37b); (m cp ) (7.37c)
2cp m
cp
Exponential: = (7.38 )
2(m + cp )
2cp 2cp m
Triangular: = 1 tanh (7.39a)
m m 2cp
m 2cp
(m
cp ) (7.39b); (m cp ) (7.39c)
4cp m
Trapezoidal:
m1
(m +m ) m2
cp cp 1 2
= 2+ 1 e cp e cp
e cp 2 (7.40a)
m1 m1
m
2 2cp
1e cp (m1
cp ) (7.40b); (m1 cp ) (7.40c)
m1
Trapezoidal, Cyclic With a Period of 2(m1 +m2 )
m1
1 e cp
m2
1 + e cp
2cp
= 1 (7.41a)
m1 m1 ( +
m1 m2
)
1+e cp
cp
m m1
cp 2cp m1 cp
1 1 (7.41b); (7.41c)
3cp m2 cp m1 m2
cp
Table 7.9: Eddy-Current Energy Density, eed and < eed > [J/m3 ]
e (t) = H
Wire (Diameter: d), Under H m sin(2t/m ) or H
e (t) = H
m sin(2t/m )
2 d2 ( Hm )2 2 d2 ( He )2
< eed > = (7.42a); < eed > = (7.42b)
4m m 12m m
)
Tape (Width: w; Thickness: ; Figs. 7.1d & 7.1e); Hm (Hm )2 +(wHm )2
4 2 ( Hm )2 ( Hm )2
Sinusoidal: < eed > = (7.43a); Exponential: eed = (7.43b)
24m m 24m m
( Hm )2
Triangular (time-averaged) & Trapezoidal: eed = (7.43c)
12m m
* Time-averaged.
444 CHAPTER 7PROBLEMS & DISCUSSIONS
Figure 7.19 shows a schematic drawing of a cross section, based on the original
drawing [7.87], of a calorimetric setup for measuring total AC dissipation of a
superconducting Test Coil. In this technique Calibration Heater, placed inside
Calorimeter Housing with Test Coil, is controlled to keep the rate of the boil-o
vapor, measured by Vapor Flow Meter, constant as Test Coil, carrying either AC
or DC transport current, is exposed to a time-varying external eld, generated by
Bitter Magnet. The variation in Calibration Heater input is related to the total
AC dissipation of Test Coil. To stabilize the cryostat pressure, liquid helium is
not transferred during measurement. The euent helium vapor of Test Coil and
BITTER MAGNET
BITTER MAGNET BITTER MAGNET
CONSTANT-FLOW
CURRENT CONTROL
CURRENT CONTROL CURRENT CONTROL
FEEDBACK
LHe LEVEL
CHIMNEY
FOR VAPOR
BITTER MAGNET
BITTER MAGNET BITTER MAGNET
HEATER CURRENT
CURRENT CONTROL
SUPPLY CURRENT CONTROL
SUPPLY
CALORIMETER
HOUSING TEST COIL
CALIBRATION
HEATER LHe REPLENISHMENT
HOLES
Fig. 7.19 Schematic drawing of a cross section, based on the original drawing [7.87], of
a calorimeter setup for measuring total AC dissipation of a superconducting Test Coil
carrying either AC or DC current and subjected to a time-varying magnetic eld.
446 CHAPTER 7PROBLEMS & DISCUSSIONS
0.8
1.0
0.8
0.6
0.6
0.4
0.4
0.2 0.3
M [T]
0.3
0.2 0.4
0.4 0.6
0.8
0.6
1.0
0.8
3 2 1 0 1 2 3
H [T]
Fig. 7.20 M vs. H plots, measured at 4.2 K, for Nb3 Sn/Cu composite
wires of diameters 1.0, 0.8, 0.6, 0.4, and 0.3 mm [7.114].
448 CHAPTER 7PROBLEMS & DISCUSSIONS
(0.47 T)
Jc (3 T, 4.2 K) Jc (3 T, 4.2 K)3.92
(0.12 T)
2.81010 A/m2
b) From the Bean model, for H Hp = Jc a, M (H) = Hp /2, and using a = deff /2,
where deff is the lament diameter, we have:
4 M (0 T) 4(470103 T)
deff = = 53 m
Jc (0 T) (4107 H/m)(2.81010 A/m2 )
Note that deff may be computed at any value of H at which both M (H) and Jc (H)
are known.
c) As may be inferred from the positive magnitudes of magnetization at H =
2 T, the magnitude, deff Jc , is nearly proportional to the overall wire size.
It implies that deff decreases with wire diameter though perhaps not exactly in
proportion to it. The critical size, ac , for a Bean slab is given by Eq. 5.40:
+
3Cs (Tc Top )
ac = (5.40)
Jc2
To avoid ux jumping, the product aJc must satisfy the following condition:
+
3Cs (Tc Top )
aJc (S6.1)
(75106 m)2
Ehy2 = (2500)(8100 m)
4
8 (0.14 T)(1.8 T)
(S7.3b)
3 (4107 H/m)
= (89.4103 m3 )(0.53106 J/m3 ) 48 kJ
The total hysteresis loss released into the liquid is thus 53 kJ.
c) In a rapidly changing eld, the most important losses are coupling (ecp ) and
eddy (eed ). We shall consider here only ecp , because it is sucient to drive the
conductor-helium unit volume to T , even without any contribution from eed .
We shall rst compute cp , the coupling time constant, for the NbTi conductor.
2p
cp = (7.6)
8 2 ef
For NbTi composite, ef given by Eq. 7.7b is generally used. In terms of c/s , the
copper-to-superconductor ratio, ef is given by:
1 + f
ef = m (7.7b)
1 f
c/s + 2
= m (S7.4)
c/s
AC AND OTHER LOSSESPROBLEMS & DISCUSSIONS 453
Consider a unit length (1 cm) of conductor. Because its cross section is (0.92
cm)(0.26 cm) = 0.24 cm2 , the conductor has volume Vcd of 0.24 cm3 . Over this
conductor length, helium occupies 0.4 cm length (40% lling) and 0.5 mm channel
depth (1-mm deep channel is shared by conductors of the top and bottom pan-
cakes) over the conductor width of 2.6 mm. Thus for a unit conductor length,
helium occupies a volume Vhe of 5.2 103 cm3 . The total dissipation energy over
the unit conductor length, Ecp , will thus be given by:
Ecp = ecp Vcd (S7.9)
(0.16 J/cm3 )(0.24 cm3 ) 38 mJ
The total thermal energy, Eth needed to raise the unit conductor (and accom-
panying liquid helium) from 1.8 K to T is given by:
Eth = [hcu (T ) hcu (1.8 K)]Vcd + [hhe (T ) hhe (1.8 K)]Vhe (S7.10)
Inserting hcu 0.1 mJ/cm3 and hhe 290 mJ/cm3 into Eq. S7.10, we have:
Eth = (0.1 mJ/cm3 )(0.24 cm3 ) + (290 mJ/cm3 )(5.2103 cm3 ) (S7.11)
2 mJ
Because Ecp Eth , the entire helium surrounding the unit conductor will be
heated to well above T , making it unlikely for the conductor to recover.
Oh gures! You can make gures do whatever you want. Ned Land
454 CHAPTER 7PROBLEMS & DISCUSSIONS
DIODES
L
POWER
HEATER PCS
SUPPLY
Rpc r
Thermal Insulation
Heating injected into the switch to maintain it resistive is a cryogenic load to the
cryostat. The more thermally-insulated the switch is from the cold environment,
the less the cryogenic load.
C. Index Loss
As given by Eq. 6.25b, for magnet-grade superconductors the electric eld E and
current density Js may be approximated by:
n
Js
E s = Ec (6.25b)
Jc
V2
V1
0 I
0 I1 I2
Fig. 7.23 Typical V vs. I trace from which to determine an index number.
AC AND OTHER LOSSESREFERENCES 459
REFERENCES
[7.1] Donald J. Hanrahan, A Theoretical and Experimental Study of the AC Losses of
a High Field Superconductor and the Implications for Power Applications (Naval
Research Laboratory, Washington DC, 1969).
[7.2] J.J. Rabbers, AC Loss in Superconducting Tapes and Coils (Doctoral Thesis, Uni-
versity of Twente, 1970).
[7.3] W.T. Norris, Calculation of hysteresis losses in hard superconductors carrying ac:
isolated conductors and edge of thin sheets, J. Phys. D3, 489 (1970).
[7.4] G. Ries and H. Brechna, AC Losses in Superconducting Pulsed Magnets (KFK
Report 1372, Gesellschaft fur Kernforschung m.b.H. Karlsruhe, 1972).
[7.5] Marijin Pieter Oomen, AC Loss in Superconducting Tapes and Cables (Doctoral
Thesis, University of Twente, 1972).
[7.6] W.J. Carr, Jr., ac loss in a twisted lamentary superconducting wire. II, J. Appl.
Phys. 45, 935 (1974).
[7.7] K. Kwasnitza, Scaling law for the ac losses of multilament superconductors,
Cryogenics 17, 616 (1977).
[7.8] F. Sumiyoshi, F. Irie and K. Yoshida, Magnetic eld dependence of ac losses in
multi-lamentary superconducting wires, Cryogenics 18, 209 (1978).
[7.9] J.D. Thompson, M.P. Maley, John R. Clem, Hysteretic losses of a type II super-
conductor in parallel ac and dc magnetic elds of comparable magnitude, J. Appl.
Phys. 50, 3531 (1979).
[7.10] J.P. Soubeyrand and B. Turck, Losses in superconducting composite under high
rate pulsed transverse eld, IEEE Trans. Magn. MAG-15, 248 (1979).
[7.11] T. Ogasawara, Y. Takahashi, K. Kanbara, Y. Kubota, K. Yasohama and K. Ya-
sukochi, Alternating eld losses in superconducting wires carrying dc transport
currents. Part 1: single core conductors, Cryogenics 19, 736 (1979).
[7.12] T. Ogasawara, Y. Takahashi, K. Kanbara, Y. Kubota, K. Yasohama and K. Yasu-
kochi, Alternating eld losses in superconducting wires carrying dc transport cur-
rents. Part 2: multilamentary composite conductors, Cryogenics 21, 97 (1981).
[7.13] I. Hlasnik, Review on AC losses in superconductors, IEEE Trans. Magn. MAG-
17, 2261 (1981).
[7.14] E.H. Brandt and M. Indenbom, Type-II superconductor strip with current in a
perpendicular magnetic eld, Phys. Rev. B 48, 12893 (1993).
[7.15] John R. Clem and Alvaro Sanchez, Hysteretic ac losses and susceptibility of thin
superconducting disks, Phys. Rev. B 50, 9355 (1993).
[7.16] Kazuo Funaki and Fumio Sumiyoshi, (Multilaments and Conduc-
tors in Japanese) (ISBN: 4782857527, 1995).
[7.17] W.J. Carr, Jr., AC Loss and Macroscopic Theory of Superconductors 2nd Ed. (Tay-
lor & Francis, London, 2001).
[7.18] W.J. Carr, Jr., Conductivity, permeability, and dielectric constant in a multila-
ment superconductor, J. Appl. Phys. 46, 4043 (1975).
[7.19] R.W. Fast, W.W. Craddock, M. Kobayashi, and M.T. Mruzek, Electrical and
mechanical properties of lead/tin solders and splices for superconducting cables,
Cryogenics 28, 7 (1988).
[7.20] J.W. Hafstrom, D.H. Killpatrick, R.C. Niemann, J.R. Purcell, and H.R. Thresh,
Joining NbTi superconductors by ultrasonic welding, IEEE Trans. Magn. MAG-
460 CHAPTER 7REFERENCES
13, 94 (1977).
[7.21] Hiroyuki Fujishiro (unpublished data, 2006).
[7.22] Jean-Marie Noterdaeme, Demountable resistive joint design for high current su-
perconductors, (M.S. Thesis, MIT Dept. Nuclear Engineering, May 1978).
[7.23] K. Sawa, M. Suzuki, M. Tomita and M. Murakami, A fundamental behavior on
mechanical contacts of YBCO bulks, Physica C: Superconduc. 378, 803 (2002).
[7.24] Hiroyuki Fujita, Katsuya Fukuda, Koichiro Sawa, Masaru Tomita, Masato Mu-
rakami, Naomichi Sakai, Izumi Hirabayashi, Contact resistance characteristics of
high temperature superconducting bulk Part V, Proc. 52nd IEEE Holm Conf.
Electrical Contacts, 124 (2006).
[7.25] Y. Kawate, R. Ogawa, and R. Hirose (personal communication, 1994).
[7.26] R.S. Kensley and Y. Iwasa, Frictional properties of metal insulator surfaces at
cryogenic temperatures, Cryogenics 20, 25 (1980).
[7.27] R.S. Kensley, H. Maeda, and Y. Iwasa, Transient slip behavior of metal/insulator
pairs at 4.2 K, Cryogenics 21, 479 (1981).
[7.28] H. Maeda, O. Tsukamoto, and Y. Iwasa, The mechanism of friction motion and
its eect at 4.2 K in superconducting magnet winding models, Cryogenics 22, 287
(1982).
[7.29] A. Iwabuchi and T. Honda, Temperature rise due to frictional sliding of SUS316
vs SUS316L and SUS316 vs polyimide at 4 K, Proc. 11 th Int. Conf. Magnet Tech.
(MT11) (Elsevier Applied Science, London), 686 (1990).
[7.30] P.C. Michael, E. Rabinowicz, and Y. Iwasa, Friction and wear of polymeric ma-
terials at 293, 77, and 4.2 K, Cryogenics 31, 695 (1991).
[7.31] T. Takao and O. Tsukamoto, Stability against the frictional motion of conductor
in superconducting windings, IEEE Trans. Magn. 27, 2147 (1991).
[7.32] Y. Iwasa, J.F. Maguire, and J.E.C. Williams, The eect on stability of frictional
decoupling for a composite superconductor, Proc. 8 th Symp. on Engr. Problems
of Fusion Research, (IEEE Publication 79CH1441-5, 1979), 1407.
[7.33] Y. Yasaka and Y. Iwasa, Stress-induced epoxy cracking energy release at 4.2 K in
epoxy-coated superconducting wires, Cryogenics 24, 423 (1984).
[7.34] S. Fuchino and Y. Iwasa, A cryomechanics technique to measure dissipative en-
ergies of 10 nJ, Exp. Mech. 30, 356 (1990).
[7.35] E.S. Bobrov and J.E.C. Williams, Direct optimization of the winding process for
superconducting solenoid magnets (linear programming approach), IEEE Trans.
Magn. MAG-17, 447 (1981).
[7.36] E.S. Bobrov, J.E.C. Williams, and Y. Iwasa, Experimental and theoretical investi-
gation of mechanical disturbances in epoxy-impregnated superconducting coils. 2.
Shear stress-induced epoxy fracture as the principal source of premature quenches
and trainingtheoretical analysis, Cryogenics 25, 307 (1985).
[7.37] H. Maeda, M. Urata, H. Ogiwara, S. Miyake, N. Aoki, M. Sugimoto, and J. Tani,
Stabilization for wind and react Nb3 Sn high eld insert coil, Proc. 11 th Int.
Conf. Magnet Tech. (MT-11) (Elsevier Applied Science, London, 1990), 1114.
[7.38] H. Nomura, K. Takahisa, K. Koyama, and T. Sakai, Acoustic emission from su-
perconducting magnets, Cryogenics 17, 471 (1977).
[7.39] Curt Schmidt and Gabriel Pasztor, Superconductors under dynamic mechanical
stress, IEEE Trans. Magn. MAG-13, 116 (1977).
AC AND OTHER LOSSESREFERENCES 461
922 (1984).
[7.57] Oluwasegun Olubunmi Ige, Mechanical disturbances in high-performance super-
conducting dipoles, (Ph.D. Thesis, MIT Dept. Mechanical Engineering, 1989).
[7.58] L. Wozny, P. Lubicki, and B. Mazurek, Acoustic emission from high Tc supercon-
ductors during current ow, Cryogenics 33, 825 (1993).
[7.59] Kazuaki Arai and Yukikazu Iwasa, Heating-induced acoustic emission in an adi-
abatic high-temperature superconducting winding, Cryogenics 37, 473 (1997).
[7.60] Haigun Lee, Ho Min Kim, Joseph Jankowski, Yukikazu Iwasa, Detection of hot
spots in HTS coils and test samples with acoustic emission signals, IEEE Trans.
Appl. Superconduc. 14, 1298 (2004).
[7.61] P.F. Dahl, G.H. Morgan, and W.B. Sampson, Loss measurements on twisted
multilamentary superconducting wires, J. Appl. Phys. 40, 2083 (1969).
[7.62] K. Kwasnitza and I. Horvath, Measurement of ac losses in multilament super-
conducting wires at frequencies between 1 and 100 Hz, Cryogenics 14, 71 (1974).
[7.63] W. David Lee, J. Thomas Broach, 60 Hz AC losses in superconducting solenoids,
IEEE Trans. Magn. MAG-13, 542 (1977).
[7.64] W.J. Carr, Jr., Ben Clawson, and Wayne Vogen, AC losses in superconducting
solenoids, IEEE Trans. Magn. MAG-14, 617 (1978).
[7.65] M.S. Walker, J.G. Declerq, B.A. Zeitlin, J.D. Scudiere, M.J. Ross, M.A. Janocko,
S.K. Singh, E.A. Ibrahim, P.W. Eckels, J.D. Rogers, and J.J. Wollan, Supercon-
ductor design and loss analysis for a 20 MJ induction heating coil, IEEE Trans.
Magn. MAG-17, 908 (1981).
[7.66] D. Ito, Y. Nakayama, and T. Ogasawara, Losses in superconducting magnets
under fast ramp rate operation, IEEE Trans. Magn. MAG-17, 971 (1981).
[7.67] K. Kwasnitza and I. Horvath, AC loss behaviour of the high-current NbTi super-
conductor for the Swiss LCT fusion coil, IEEE Trans. Magn. MAG-17, 2278
(1981).
[7.68] Kunishige Kuroda, ac losses of superconducting solenoidal coils, J. Appl. Phys.
53, 578 (1982).
[7.69] A. Lacaze, Y. Laumond, J.P. Tavergnier, A. Fevrier, T. Verhaege, B. Dalle, A.
Ansart, Coils performances of superconducting cables for 50/60 Hz applications,
IEEE Trans. Magn. 27, 2178 (1991).
[7.70] T. Ando, Y. Takahashi, K. Okuno, H. Tsuji, T. Hiyama, M. Nishi, E. Tada, K.
Yoshida, K. Koizumi, H. Nakajima, T. Kato, M. Sugimoto, T. Isono, K. Kawano,
M. Konno, J. Yoshida, H. Ishida, E. Kawagoe, Y. Kamiyauchi and S. Shimamoto,
AC loss results of the Nb3Sn Demo Poloidal Coil (DPC-EX), IEEE Trans. Magn.
28, 206 (1992).
[7.71] F. Sumiyoshi, S. Kawabata, Y. Kanai, T. Kawashima, T. Mito, K. Takahata and
J. Yamamoto, Losses in cable-in-conduit superconductors used for the poloidal
coil system of the Large Helical Device, IEEE Trans. Appl. Superconduc. 3, 476
(1993).
[7.72] J.P. Ozelis, S. Delchamps, S. Gourlay, T. Jaery, W. Kinney, W. Koska, M. Kuch-
nir, M.J. Lamm, P.O. Mazur, D. Orris, J. Strait, M. Wake, J. Dimarco, J. Kuzmin-
ski, and H. Zheng, AC loss measurements of model and full size 50 mm SSC col-
lider dipole magnets at Fermilab, IEEE Trans. Appl. Superconduc. 3, 678 (1993).
[7.73] Z. Ang, I. Bejar, L. Bottura, D. Richter, M. Sheahan, L. Walckiers, R. Wolf, Mea-
surement of AC loss and magnetic eld during ramps in the LHC model dipoles,
AC AND OTHER LOSSESREFERENCES 463
verse alternating magnetic eld by a pickup coil method, First edition, April
2003.
[7.104] Arend Nijhuis, Herman H.J. ten Kate, Pierluigi Bruzzone, Luca Bottura, Para-
metric study on coupling loss in subsize ITER Nb3 Sn cabled specimen, IEEE
Trans. Magn. 4. 2743 (1996).
[7.105] Arend Nijhuis, Herman H.J. ten Kate, Victor Pantsyrny, Alexander K. Shikov,
Marco Santini, Interstrand contact resistance and AC loss of a 48-strands Nb3
Sn CIC conductor with a Cr/Cr-oxide coating, IEEE Trans. Appl. Superconduc.
10, 1090 (2000).
[7.106] S. Egorov, I. Rodin, A. Lancetov, A. Bursikov, M. Astrov, S. Fedotova, Ch. Weber,
and J. Kaugerts, AC loss and interstrand resistance measurement for NbTi cable-
in-conduit conductor, IEEE Trans. Appl. Superconduc. 12, 1607 (2002).
[7.107] Takataro Hamajima, Naoyuki Harada, Takashi Satow, Hiroshi Shimamura, Kazu-
ya Takahata, and Makoto Tsuda, Long time constants of irregular AC coupling
losses in a large superconducting coil, IEEE Trans. Appl. Superconduc. 12, 1616
(2002).
[7.108] H. Eckelmann, M. Quilitz, C. Schmidt, W. Goldacker, M. Oomen, M. Leghissa,
AC losses in multilamentary low AC loss Bi(2223) tapes with novel interla-
mentary resistive carbonate barriers, IEEE Trans. Appl. Superconduc. 9, 762
(1999).
[7.109] Naoyuki Amemiya, Keiji Yoda, Satoshi Kasai, Zhenan Jiang, George A. Levin,
Paul N. Barnes, and Charles E. Oberly, AC loss characteristics of multila-
mentary YBCO coated conductors, IEEE Trans. Appl. Superconduc. 15, 1637
(2005).
[7.110] Mike D. Sumption, Paul N. Barnes, and Edward W. Collings, AC losses of coated
conductors in perpendicular elds and concepts for twisting, IEEE Trans. Appl.
Superconduc. 15, 2815 (2005).
[7.111] M. Majoros, B.A. Glowacki, A.M. Campbell, G.A. Levin, P.N. Barnes, and M. Po-
lak, AC losses in striated YBCO coated conductors, IEEE Trans. Appl. Super-
conduc. 15, 2819 (2005).
[7.112] Osami Tsukamoto, Naoki Sekine, Marian Ciszek, and Jun Ogawa, A method
to reduce magnetization losses in assembled conductors made of YBCO coated
conductors, IEEE Trans. Appl. Superconduc. 15, 2823 (2005).
[7.113] K. Osamura, N. Wada, T. Ogawa, and F. Nakao, Formation of submicron-thick
oxide barrier for reducing AC loss in multilamentary Bi2223 tape, IEEE Trans.
Appl. Superconduc. 15, 2875 (2005).
[7.114] Kunihiko Egawa (personal communication, 2003).
[7.115] Shunji Yamamoto, Tadatoshi Yamada and Masatoshi Iwamoto, Quench pro-
tection of persistent current switches using diodes in cryogenic temperature,
19th Annual IEEE Power Electronics Specialists Conf. 1, 321 (1988).
[7.116] Yukikazu Iwasa, Microampere ux pumps for superconducting NMR magnets
Part 1: Basic concept and microtesla ux measurement, Cryogenics 41, 384
(2001).
[7.117] S. Jeong, H. Lee, and Y. Iwasa, Superconducting ux pump for high temperature
superconducting insert coils of NMR magnets, Adv. Cryo. Engr. 47, 441 (2002).
[7.118] Haigun Lee, Homin Kim, and Yukikazu Iwasa, A ux pump for NMR magnets,
IEEE Trans. Appl. Superconduc. 13, 1640 (2003).
[7.119] Rocky Mai, Seung-yong Hahn, Haigun Lee, Juan Bascunan, and Yukikazu Iwasa,
466 CHAPTER 7REFERENCES
A digital ux injector for NMR Magnets, IEEE Trans. Appl. Superconduc. 15,
2348 (2005).
[7.120] M. Lakrimi, P. Bircher, G. Dunbar, and P. Noonan, Flux injector for NMR
magnets, IEEE Trans. Appl. Superconduc. 17, 1438 (2007).
[7.121] Weijun Yao, Woo-Seok Kim, Seungyong Hahn, J. Bascunan, Hai-Gun Lee, and
Yukikazu Iwasa, A digital ux injector operated with a 317-MHz NMR magnet,
IEEE Trans. Appl. Superconduc. 17, 1450 (2007).
[7.122] Yoondo Chung, Itsuya Muta, Tsutomu Hoshino, and Taketsune Nakamura, Per-
formance of a linear type magnetic ux pump for compensating a little decre-
mented persistent current of HTS magnets, IEEE Trans. Appl. Superconduc.
14, 1723 (2004).
[7.123] Marijin Oomen, Martino Leghissa, Guenter Ries, Norbert Proelss, Heinz-Werner
Neumueller, Florian Steinmeyer, Markus Vester, and Frank Daview, HTS ux
pump for cryogen-free HTS magnets, IEEE Trans. Appl. Superconduc. 15, 1465
(2005).
[7.124] Yoondo Chung, Tsutomu Hoshino, and Taketsune Nakamura, Current pumping
performance of linear-type magnetic ux pump with use of feedback control circuit
system, IEEE Trans. Appl. Superconduc. 16, 1638 (2006).
[7.125] W. Denis Markiewicz, Current injection for eld decay compensation in NMR
spectrometer magnets, IEEE Trans. Appl. Superconduc. 12, 1886 (2002).
CHAPTER 8
PROTECTION
8.1 Introductory Remarks
Protection is one of the ve key design and operation issuesthe others are stabil-
ity; mechanical integrity; cryogenics; conductor. As qualitatively shown in Fig. 1.6,
the diculty or cost of magnet protection increases with operating temperature,
while that of stability decreases. For HTS, protection can become a real chal-
lenge, while stability is benign, as discussed in CHAPTER 6. The question most
often asked concerning protection of HTS is this: If HTS is so stable, why is it
necessary to worry about protecting it? The answer comes down to the cost of
an HTS device vs. the cost of its protection and the probability of a system failure
mode that needs protection. To protect or not to protect an inherently stable
HTS magnet is a dilemma. This question is revisited in DISCUSSION 8.7.
Our focus is protection of the magnet winding; other parts of the magnet
mechanical, electrical, and cryogenicare not addressed, except in passing. The is-
sues included here are: 1) overheating; 2) overstraining (thermal and mechanical);
3) high internal voltage; and 4) protection techniques. PROBLEMS & DISCUSSIONS
also cover other relevant topics. Protection of course has been a key topic for super-
conducting magnets and has been addressed by many since the 1960s [1.27, 8.18.6],
updated periodically [8.78.11]; other references are cited later where appropriate.
That a magnet of even modest eld, e.g., 3 T, has been permanently damaged
by overheating demonstrates that a catastrophic energy concentration can occur
in real magnets. If only the local magnetic energy density were converted locally
into heat, eld energy density to enthalpy density computation similar to Eq. 8.1
gives temperatures below 200 K for elds up to 25 T; for elds above 25 T, see
Illustration in the next section.
Y. Iwasa, Case Studies in Superconducting Magnets: Design and Operational Issues, 467
DOI: 10.1007/b112047_8,
Springer Science + Business Media, LLC 2009
468 CHAPTER 8
Illustration Table 8.1 presents the hot-spot volume fractions, fr (%), of a sole-
noid ( = 1.5, = 2.0), in the range of B , 1.530 T, required to limit the nal
hot-spot temperature, Tf , to 200 K and 300 K for two initial winding temperatures,
Ti , 4 K and 80 K. Here, the entire winding is assumed copper (density: 8.96 g/cm3 ),
and Tf is computed from emr = hcu (Tf )hcu (Ti ), where emr is given by Eq. 8.7.
The table indicates that for this solenoid ( = 1.5, = 2.0) generating, for example,
1.5 T, a hot spot can be a fraction (<1%) of the winding volume for two limits
of Tf , 200 K and 300 K. This small fr to meet the Tf 200 K requirement has a
practical benet for detect-and-activate-the-heater active protection: for most
solenoids, a cumbersome protection heater, discussed in 8.8.4, planted within
the winding needs to convert (and expand) only a fraction of the winding to a
hot spot. However, as evident from Table 8.1, though fr 1 (100%) for solenoids
of B up to 25 T, for 30 T, fr exceeds 1 (100%), i.e., a portion of the stored
energy must be dissipated outside the solenoid, i.e., into a dump resistor (8.8.3).
level for Nb3 Sn and HTS. However, even if a uniform temperature is maintained
during heating, because a magnet winding consists of materials of dierent thermal
expansion coecients (Table 8.3), uniform heating can still induce overstraining
in a superconductor. It is generally safest to keep Tf below 200 K.
The temperature integral on the left-hand side of Eqs. 8.9c and 8.9d depends
monotonically on T ; we dene a Z(Tf , Ti ) function:
Tf
Cm (T )
Z(Tf , Ti ) dT (8.10a)
Ti m (T )
For alloy matrix metals with m (T ) suciently constant to be approximated by
a temperature-averaged resistivity, m , Eq. 8.10a may be simplied to:
Tf
1 Hm (Tf ) Hm (Ti )
Z(Tf , Ti ) Cm (Tf ) dT = (8.10b)
m Ti m
where Hm (T ) is the matrix metal volumetric enthalpy at T .
L
Iop
Iop r(T )
Figure 8.2 gives Z(T, 0) plots for silver (RRR = 1000, the ratio of electrical resis-
tivities at 0 C and 4.2 K; 100); copper (200; 100; 50); aluminum (grade: 1100);
and brass (70Cu-30Zn). The dashed line (barely discernable) is for brass with
m = 5.5108 m (Eq. 8.10b). Note that for any combination of Ti and Tf :
Z(Tf , Ti ) = Z(Tf , 0) Z(Ti , 0) = Z(Tf ) Z(Ti ) (8.11)
For any combination of Tf and Ti , there is a heating duration aih (Tf , Ti )the
superscript i denotes under constant-current (at Jm ) heatinggiven by:
i 1 + m/s Z(Tf , Ti )
ah (Tf , Ti ) = 2
(8.12a)
m/s Jm
20
Ag1000
15
Cu200
Z(T, 0) [1016 A2 s/m4 ]
Cu100 Cu50
Ag100
10
5
Al
Brass
0
0 50 100 150 200 250 300
Temperature [K]
Fig. 8.2 Z(T, 0) plots. Ag1000 (RRR:1000); Ag100; Cu200; Cu100; Cu50; Al (Grade
1100); Brass (70Cu-30Zn)also dashed line (Eq. 8.10b, with m = 5.5108 m).
474 CHAPTER 8
VD RD
+ Im (t) r(t)
The magnet inductance L and the dump resistance RD may be expressed in terms,
respectively, of the initial stored magnetic energy, Em , of the magnet and the initial
discharge voltage VD across RD . Thus:
2Em
L= 2
(8.17a)
Iop
VD
RD = (8.17b)
Iop
Combining Eqs. 8.16b, 8.17a, and 8.17b, we obtain:
2
Am Jm
Em
Z(Tf , Ti ) = (8.18a)
Acd VD Iop
Equation 8.18a shows that Tf is higher for combinations of greater Jm and/or
Em and smaller VD and/or Iop . We may consider the ratio Em /VD Iop to be an
eective discharge period, where VD Iop is the eective discharge power. Noting
that Jm = Iop /Am , we may reduce Eq. 8.18a to:
Jm Em
Z(Tf , Ti ) = (8.18b)
Acd VD
Under this discharge current mode, there is a maximum matrix current density,
D
Jm
, for any winding temperature limit Tf :
D Acd VD Z(Tf , Ti )
Jm
= (8.19)
Em
Equation 8.19 gives a criterion for Jm that is entirely dierent from the cryosta-
bility criterion of Eq. 6.22.
Im (t) r(T )
dIm (t)
L + r(T )Im (t) = 0 (8.20)
dt
r(T ), expressed as T -dependent, obviously increases with time for two reasons: 1)
magnetic energy is being converted to thermal energy to raise the normal zone
temperature; and 2) the normal zone itself is spreading in the winding. For the
purpose of our analysis here, the simplest assumption is used: a constant normal-
zone resistance, r(T ) = Rnz , given by:
m (Tf )nz
Rnz = (8.21)
4Am
where m (Tf ) is the matrix resistivity at the normal zones nal temperature; nz
is the total conductor length in the resistive state when the current has decayed
to zero. The factor of 4 in the denominator accounts for averaging, both in
time (factor of 2) and space (another factor of roughly 2), of the normal zone
temperature, from Ti to Tf . Note that Eq. 8.21 also assumes m (Tf ) m (Ti ),
which may not be valid for some combinations of Tf and Ti , but the uncertainty
introduced by this simplication is considered small compared with that arising
from the basic assumption of Eq. 8.21. For Rnz constant, Jm (t) = Im (t)/Am , with
Jm (t = 0) = Iop /Am = Jm , is given by:
where fr is the fraction of the winding volume in the resistive state, as in Eq. 8.6.
Combining Eqs. 8.21 and 8.24, we obtain:
2
Noting that Jm = Iop /Am and L = 2Em /Iop , where Em is the initial magnetic
energy stored by the magnet, we may express Eq. 8.28a as:
4 2 L(, )Em
dg = (8.28b)
fr ( + 1)m (Tf )Jm a1
sh 1 Acd a1
Jm = fr ( + 1)m (Tf )Z(Tf , Ti ) (8.30a)
2 Am 2 L(, )Em
sh
As expected Jm
increases with combinations of greater fr and Tf and smaller
Em . Because Em = a1 L(, )N 2 Iop
2 sh
/2, Jm
may also be given by:
sh 1 Acd fr ( + 1)m (Tf )Z(Tf , Ti )
Jm (Tf , Ti ) = (8.30b)
2 Am L(, )N Iop
Equation 8.30b states that a solenoid of large ampere-turns (N Iop ) must operate
sh
at smaller Jm
.
478 CHAPTER 8
which, for alloy matrix metals, may be simplied, as Eq. 8.10b, by:
Y (Tf , Ti ) m [Hm (Tf ) Hm (Ti )] (8.32b)
Similarly to Z(Tf , Ti ), Y (Tf , Ti ) is given by:
Y (Tf , Ti ) = Y (Tf ) Y (Ti ) (8.32c)
Figure 8.6 presents Y (T, 0) plots for the same metals as those for Z(T, 0) in Fig. 8.2.
Note that metal purity, which greatly inuences m for temperatures below 50 K
and thus has a large impact on Z(T, 0), has little impact on Y (T, 0) for T >100 K.
From Eq. 8.31b we may derive, similar to Eq. 8.9c:
2
Am Vop ah
Y (Tf , Ti ) = 2 (8.33)
Acd cd
+ L
Vop
Rm (T )
Fig. 8.5 Circuit for a superconducting magnet (L) with the entire winding
in the normal state, Rm (T ), under constant-voltage (Vop ) heating mode.
PROTECTION 479
Noting that cd for a solenoidal coil is given by Eq. 8.24 with fr = 1, and using
Eq. 8.26, we nd:
a1 L
cd = ( + 1) (8.34)
L(, )
10.0 40
Brass
Brass
7.5 30
Y (T, 0) [V 2 s/m2 ]
Al Cu
5.0 20
Ag
2.5 10
0 0
0 50 100 150 200 250 300
Temperature [K]
Fig. 8.6 Y (T, 0) plots. Left-hand vertical scale: Ag (100 to 1000); Cu (50 to 200); Al
(Grade: 1100). Right-hand vertical scale: brass (Cu70-Zn30)also dashed line (Eq.
8.32b with m = 5.5108 m). Especially for temperatures above 100 K, Y (T, 0)
is nearly independent of the purity of Ag and Cu.
480 CHAPTER 8
fr =0.1 fr =0.2
(a) (b)
fr =0.5 fr =1
(c) (d)
Fig. 8.7 Voltage distributions, at constant current, for dierent normal zone sizes
within a quenching superconducting magnet, both terminals grounded and unwound
from one end of the conductor to the other end. Within the resistive zone the conductor
resistivity is assumed constant. (a) 10% (fr = 0.1) of the winding is in the resistive
state; (b) 20% (fr = 0.2); (c) 50% (fr = 0.5); (d) 100% (fr = 1). The solid lines represent
the resistive voltage; the short dashed lines the inductive voltage; and the long dashed
lines the total internal voltage, given by the sum of the resistive and inductive voltages.
In each set of voltage distributions shown in Fig. 8.7, the solid line represents the
resistive voltage; the short dashed line the inductive voltage; and the long dashed
line the total internal voltage, given by the sum of the resistive and inductive
voltages. In each gure, the magnet current remains constant at Iop . In reality,
the current is decreasing with time; the eect of this decrease is included later.
The maximum resistive voltage, [Vr ]mx , occurs at one terminal of the magnet, and
because the terminal is grounded, it is matched exactly by an inductive voltage of
the same magnitude as that of [Vr ]mx = Rnz Iop , where Rnz is the total normal-zone
resistance, given by Eq. 8.25.
From Fig. 8.7, we note that the maximum internal voltage, [Vin ]mx is given by:
where fr , as noted rst with Eq. 8.6, is the fraction of the winding volume driven
to the resistive state. Note that the magnet current remains at Iop . Equation
8.38 shows that [Vin ]mx 0 as fr 0 or fr 1. The equation also shows that
[Vin ]mx peaks at fr = 0.5.
PROTECTION 483
V
Note that Jm
increases with the product of Vbk and Iop . Most importantly here,
V
Jm improves with the normal metals electrical conductivity. Also, a higher Vbk
V
implies a higher Jm
, but the winding pack density at Iop , Jop (6.3.3), could be
lower because of the need for more insulation.
U x
x=0
Similarly, a power density equation in the x-direction for the adiabatic case in the
superconducting region is given by:
Ts Ts
Cs (T ) = ks (T ) (8.41b)
t x x
Tn T z dT
= = U (8.42)
t z t dz
We may thus express Eq. 8.41 as:
dTn d dTn
Cn (T )U = kn (T ) + n (T )J 2 (8.43a)
dz dz dz
dTs d dTs
Cs (T )U = ks (T ) (8.43b)
dz dz dz
Rearranging Eqs. 8.43a and 8.43b, we have the following power density equations
for the superconductor in the normal (z < 0) and superconducting (z > 0) regions:
d dTn dTn
(z < 0) kn (T ) + Cn (T )U + n (T )J 2 = 0 (8.44a)
dz dz dz
d dTs dTs
(z > 0) ks (T ) + Cs (T )U =0 (8.44b)
dz dz dz
dTn
(z < 0) Cn U + n J 2 = 0 (8.45a)
dz
d2 Ts dTs
(z > 0) ks 2
+ Cs U =0 (8.45b)
dz dz
Ts (z) may be solved directly from Eq. 8.45b:
where Top is the operating temperature far away from z = 0, i.e., z 0, and
c = Cs U /ks . We also know that Ts = Tt at z = 0, where Tt is the transition
temperature for the superconductor carrying I. Thus:
Cs U
Ts (z) = (Tt Top ) exp z + Top (8.46b)
ks
486 CHAPTER 8
Another boundary condition is that the k(dT /dz) of each region should be equal
at z = 0heat ow must be continuous across the boundary:
dTn dTs
kn = ks (8.47a)
dz 0 dz 0
kn n J 2
= Cs U (Tt Top ) (8.47b)
Cn U
Important points to be noted from Eq. 8.48 are that U is directly proportional
to current density J and inversely proportional
to the geometric average of the
heat capacities in the two regions, Cn Cs . U given by Eq. 8.48 is valid for a
bare superconductor under adiabatic conditions. Although it is rarely necessary
to use an exact expression of U for which material properties are temperature
dependent, it is given below for the sake of completeness [8.47]:
n (Tt )kn (Tt )
U = J
Tt (8.49)
1 dkn Tt
C (T ) C (T )dT Cs (T ) dT
kn (Tt ) dT Tt Top
n t s
Top
Equation 8.49 has been found to agree well with U values measured for short
samples of coated YBCO tape in the temperature range 4577 K [8.66].
For constant material properties, we may note that Eq. 8.49 reduces to Eq. 8.48.
Also for the case Cn = Cs = C , Eq. 8.48 may be written as:
J n kn
U = (8.50a)
C (Tt Top )
For (Tt Top )/Top 1, a condition generally applicable to LTS but not HTS, we
may modify Eq. 8.50a to include the eects of T -dependent C , n , and kn :
J n (T )kn (T )
U = (8.50b)
C (T ) (Tt Top )
where T = (Tt +Top )/2. Equations 8.48 through 8.50 are valid for superconductors
having no matrix metal; in reality magnet-grade-conductors are composites, and
we may approximate the material properties by those of the matrix metal.
PROTECTION 487
Composite Superconductor
For a composite conductor with matrix metal of cross section Am , again for (Tt
Top )/Top 1 (thus valid only for LTS), with T -dependent properties, we may
generalize, with T = (Tt +Top )/2, Eq. 8.50b to:
Jm m (T )km (T )
U = (8.51a)
Ccd (T ) Tt Top
where Ccd (T ) is the conductors volumetric heat capacity averaged over the range
from Top to Tt , and Jm is the current density over the matrix metals cross section.
Because m (the matrix metals electrical resistivity) is much smaller than n , and
km (the matrix metals thermal conductivity) is much greater than kn , km (T ) and
m (T ) in Eq. 8.51a are quite appropriate. In Eq. 8.51a a transition temperature,
Tt (Tcs + Tc )/2, as suggested by Joshi [8.45], or simply Tt = Tcs , may be used.
This subtle dierence is dicult to verify with experimental data. Based on the
same approximation used in Eq. 8.9b, we may replace Ccd (T ) in Eq. 8.51a with
Ccd (T ) Cm (T ):
Jm m (T )km (T )
U (8.51b)
Cm (T ) Tt Top
The general validity of Eqs. 8.488.51 has been amply demonstrated by many
experiments, with LTS and more recently with HTS.
+ V1 + V2 + V3 V
I I V1 V2 V3
HEATER
(a) (b)
Fig. 8.9 (a) Schematic drawing of a setup to experimentally measure longitudinal NZP
velocity. (b) Oscillogram showing voltages traces recorded during an NZP event [8.55].
488 CHAPTER 8
Experiments with techniques similar to those used for LTS tests, one example of
which is described above, have been performed since the early 2000s. Here, without
any details of experimental setups, we present in Fig. 8.10 four sets of traces,
voltage (Figs. 8.10a8.10c) and temperature (Fig. 8.10d), recorded to determine
NZP velocities in YBCO tapes: a), b) and c) are V (t) traces for test samples of
lengths, respectively, 20 cm [8.65], 15 cm [8.70], and 18 cm [8.74]. In the 20-cm long
YBCO tape [8.65], creation of a local normal zone in the test sample, initially at
50 K, relied on the nonuniform distribution of critical current over the 20-cm long
tape. An over-current pulse of 72 A (shown in Figs. 8.10a and Fig. 8.10d) triggered
a quench in zone 5 (respectively, V5 and T5 ), inducing an NZP at a constant current
of 30 A (Fig. 8.10a); the time scales of V (t) traces in a) and of a companion set, the
T (t) traces in d), agree well. The V (t) traces shown in b), were recorded with the
15-cm long tape at 60 K [8.70], and in c), with the 18-cm long tape at 70 K [8.74].
The measured NZP velocities ranged 210 mm/s, as summarized in Table 8.5.
10
50
8
72 A (Over-Current)
40
6
V [mV]
30
V [mV]
30 A 4
20 V56 V67 V78
2
V5 V4
10 0
0
0 5 10 15 20 0 10 20 30 40 50
Time [s] Time [s]
(a) (b)
6 250
Heater Pulse
72 A (Over-Current)
200
4 V15
V [mV]
150
T [K]
30 A
V16
2 100 T5 T4
V17 V30 V31
50
0
0
0 2 4 6 8 10 12 0 5 10 15 20
Time [s] Time [s]
(c) (d)
Fig. 8.10 Longitudinal NZP signals from YBCO test samples: a), b), and c) are V (t)
traces recorded for test samples of lengths, respectively, 20 cm [8.67], 15 cm [8.70], and
18 cm [8.74]. T (t) traces shown in d) are a companion set to the V (t) traces shown in
a) [8.67]. Except the signal traces, the labeling style in each gure has been modied.
PROTECTION 489
Table 8.5: Selected Measured U for LTS and HTS, Bare and Composite
Superconductor Environment Top [K] Bex [T] J [A/mm2 ] U [mm/s]
Nb-Zr [8.26] Liquid helium 4.2 0 100* 933
(single strand; 1000* 9330
no matrix metal) 6 100* 5345
8.8 0 100* 1215
NbTi [8.36] Liquid helium 4.2 0 420 Recovery
(multilamentary 840 6800
composite) 4 420 4660
840 18600
Nb3 Sn [8.39] Adiabatic 4.2 0 630 1830
(multilamentary 6 315 1490
composite) 630 3720
Nb3 Sn [8.55] Quasi-adiabatic 12 0 700 510
(tape) 5.5 5 470 525
Bi2223-Ag [8.55] Quasi-adiabatic 40 0 230 2
YBCO [8.64] Adiabatic 46 0 1015 28
(coated) [8.69] Adiabatic 77 0 315 310
[8.73] Adiabatic 77 0 65 2.5
115 9
40 0 115 38
MgB2 [8.70] Quasi-adiabatic 4.2 4 26 No NZP
(single strand; 78 930
iron matrix) 212 6000
* I/(conductor cross section).
I/(matrix metal cross section).
Table 8.5 lists measured values of U for LTS and HTS. Although cooled by liquid
helium, U J for the Nb-Zr single strand [8.26] because in the absence of matrix
metal (pre-Stekly era superconductor) the normal-state Joule heating completely
overwhelms the cooling. Generally, these data show that NZP velocities for HTS
(Bi2223-Ag; coated YBCO) are about two to four orders of magnitude less than
those of LTS. Recovery for the NbTi [8.36] is discussed in 8.4.2 below. As for
MgB2 with iron matrix [8.70], the velocities are comparable with those of LTS,
chiey because of the absence of conductive matrix metal. Here at an overall
conductor current density (Jcd ) of 26 A/mm2 or below, no NZP takes place: the
normal-state superconductor is not generating full Joule heating, chiey be-
cause of its low index (n 15). Similar no-NZP behaviors were observed with
both Bi2223-Ag tape [8.55] and YBCO tapes [8.64, 8.69, 8.73].
Experimental Results
Despite the condition Ut U for LTS and HTS, in most LTS windings the dom-
inant direction of NZP is still transverse to the conductor axis, because in most
windings the conductor length, cd , is much greater than the winding dimensions,
e.g., a2 a1 for a solenoid: the condition (a2 a1 )/Ut cd /2U is generally met in
both LTS and HTS windings. However, as further discussed in 8.6, this condition
does not necessarily guarantee protection of a magnet, LTS or HTS.
Figure 8.11 shows four sets of transverse NZP signals for YBCO test assemblies
at 77 K: a) measured V (t) traces for two winding models, insulated with 38-m
thick Nomex spacersdry spacers (solid traces) and epoxied (dashed) [8.75]; b)
V (t) traces for the same epoxied winding model, experiment [solid; the same as
dashed in a)] and simulation (dashed), with Rthct1 = Rthct2 = 0; c) V (t) traces and
d) T (t) traces predicted for a model pancake coil of 100 mm i.d., 120 mm o.d., and
10 layers, epoxy-impregnatedthe transport current was shut o at t = 20 s [8.76].
The data give Ut in the range 0.11 mm/s, at least one order of magnitude less
than U . In the dry winding packs [8.75], measured Ut ranges from 0.1 mm/s at
a contact pressure of 10 MPa to 0.2 mm/s at a contact pressure of 25 MPa, while
with the epoxied winding packs, the transverse velocity is 1 mm/s.
100 100
80 80
V [mV]
60 60
V [mV]
40 40
Heater Pulse
20 20
0 0
0 1 2 3 4 5 0 0.5 1
Time [s] Time [s]
(a) (b)
2.0 170
V1 160 T1
1.5
V3 140 Heater
T [K]
Pulse T2
V [V]
1.0 120
Heater V2
0.5 Pulse 100 T3
0 80
0 10 20 30 0 10 20 30
Time [s] Time [s]
(c) (d)
Fig. 8.11 Transverse NZP signals from YBCO test assemblies at 77 K, each event trig-
gered by a heater pulse: a) V (t) traces for winding models insulated with 38-m thick
Nomex spacersdry spacers (solid traces) and epoxied spacers (dashed); b) experi-
mental, epoxied (solid) and simulation (dashed), for the same winding model [8.75]; c)
V (t) traces and d) T (t) traces for an epoxy-impregnated model pancake coil [8.76].
492 CHAPTER 8
2a1
Ut Ut
Ut Ut
Ut Ut
Ut Ut
Combining Eqs. 8.51b, 8.52, and 8.58 with Jm = Jm , we obtain a size limit,
[a1 ( 1)]iah , for the winding build of a self-protecting magnet under adiabatic,
constant-current heating condition:
Z(T f , T i ) m (T )ki (T )cd
[a1 ( 1)]iah = (8.59)
Jm Cm (T ) 2i (Tt Top )
Equation 8.59 states that the permissible magnet size, as expected, decreases with
Jm and Cm (T ) and increases with Z(Tf , Ti ). It also shows that the size limit
expands with m (T ) and ki (T ), and shrinks with (Tt Top ). The Cm (T ) and
(Tt Top ) dependences imply that for the same Jm and Tf , a self-protecting HTS
magnet, if such a magnet of practical utility can indeed exist, would have to be
more compact than its LTS counterpart.
PROTECTION 495
or in terms of Em :
1 m (T )ki (T )cd
[a1 ( 1)]sh =
ah
Cm (T ) 2i (Tt Top )
4 2 L(, )Em
(8.60d)
fr ( + 1)m (Tf ) a1
Again, as with the constant-current heating case treated above, because both
Cm (T ) and Tt Top appear in the denominator of Eqs. 8.60c and 8.60d, for the
same operating parameters a self-protecting HTS magnet must be considerably
smaller than its LTS counterpart operating at liquid helium temperature.
Answer to TRIVIA 8.1 iii). The German physicist Gustav R. Kirchho (1824
1887) taught the young Kamerlingh Onnes (18531926) in Heidelberg (c. 1870);
most noted as the developer, with R.W. von Bunsen (18111899), of the rst spec-
troscope; identied a half dozen elements, including gold, in the sun. Once asked
by his banker, unimpressed by his claim of a spectroscopes ability to nd gold in
the sun, Of what use is gold in the sun if I cannot bring it down to earth?
496 CHAPTER 8
Er 0.5(1k) + (1+k)
= (8.61)
Em + (1+k)
COLD ENVIRONMENT
I1
R1 L1
HEATER
PCS M
I2
R2 L2
Two-Coil Magnet
A 2-coil magnet, with its circuit in Fig. 8.14 [8.45], is studied here as an illustrative
case. The two coils, each wound (close-packed hexagonal) with an insulated NbTi
composite wire and shunted with a 0.5- resistor, are connected in series. The
power supply may be modeled as a constant current source for voltages up to 10 V.
In the analysis, it is assumed that the normal-zone propagation is dominated by
transverse heat conduction. The normal-zone growth is three dimensionalaxial
and radial over the entire circumference. Although U Ut , because 2a11 U
d1 Ut , where a11 and d1 are, respectively, the inner winding radius and conductor
diameter of Coil 1, transverse propagation predominates in both sections.
Table 8.6 gives the coils appropriate parameters. The total inductance is 1.52 H.
Because Coil 2 is wound directly over Coil 1, the two coils are in good thermal
contact at the interface: the entire coil may be considered as one homogeneous
thermal unit. (Note that, as indicated in Table 8.6, the two coils are wound with
conductors of dierent diameters, and thus the NZP velocities are dierent.)
A heater, placed at the midplane of the innermost radius of Coil 1, is used to
initiate a quench. We may thus assume that the normal zone starts as a ring at
the midplane of the innermost radius of Coil 1 and spreads as depicted in Fig. 8.12.
Figure 8.15 shows current and voltage vs. time traces for a heater-driven quench
in which the magnet is initially at 100 A; the power supply voltage limit is 10 V.
Both current and voltage plots consist of four traces: the solid traces (experiment)
and dashed traces (simulation). For both current and voltage traces, the curves
labeled 1 are for Section 1 and the curves labeled 2 are for Section 2. In answering
the following questions, you may ignore the analytical curves.
The following observations may be made from the traces shown in Fig. 8.15:
1. I1 decreases initially because Section 1 is where the quench is initiated.
2. I2 increases initially to keep the ux constant.
3. The behaviors of I1 and I2 are also reected in V1 and V2 . V1 increases
because I1 , not owing through Section 1, is now owing through R1 .
125
100
Current [A]
75 1
50
2
25
0
0 0.5 1.0 1.5 2.0
15
2
10
Voltage [V]
0
1
5
10
0 0.5 1.0 1.5 2.0
Time [s]
Fig. 8.15 Current and voltage vs. time traces of Coil 1 (labeled 1): experiment
(solid lines) and simulation (dashed); and Coil 2 (labeled 2) for a quench with
the magnet initially at 100 A [8.45].
4. To keep the terminal voltage zero (at least initially), V2 swings negative.
These initial responses are consistent with results discussed in PROBLEM
8.9. Eventually V1 + V2 climbs up to 10 V, the power supply limit.
5. At t 0.2 s, V2 starts climbing up, a denite indication that a normal zone
has been induced in or has reached Coil 2.
6. I2 thus begins to drop, and I1 increases, trying to keep the ux constant.
7. At t 0.4 s, V1 + V2 reaches 10 V, and I1 must start decreasing.
8. V1 + V2 = 10 V for t > 0.4 s.
The total energy dissipated in the magnet, Ed , may be given by:
where Em is the total energy stored in the magnet initially, Es is the energy
supplied by the power supply between t = 0 and t = 2 s, and ER1 and ER2
are, respectively, energies dissipated in resistor R1 and R2 . Em is 7600 J [=
(0.5)(1.52 H)(100 A)2 ]. Es is given by Vs (t)Is (t) integrated for 0 t 2 s. Vs (t)
and Is (t) are, respectively, the power supply voltage and current. The power
supply may be modeled as a constant current supply (100 A) for 0 t 0.4 s
and a constant voltage supply (10 V) for t 0.4 s. We have, for 0 t 0.4 s,
Vs (t) = V1 (t) + V2 (t) and, for t 0.4 s, Is (t) = I1 (t) + V1 (t)/R1 . (A proof of a
relationship similar to this involving more coils is a question in PROBLEM 8.11.)
500 CHAPTER 8
Using traces shown in Fig. 8.15, we can compute Es , ER1 , and ER2 :
0.4 s 2s
V1 (t)
Es = (100 A) [V1 (t) + V2 (t)] dt + (10 V) I1 (t) + dt
0 0.4 s R1
200 J + 650 J 850 J
1 2s 1 2s
ER1 = V1 (t)2 dt 50 J ER2 = V2 (t)2 dt 300 J
R1 0 R2 0
Vcd [hcu (Tf ) hcu (Top ] (694 cm3 )[hcu (Tf )] = 5500 J
where Vcd is winding volume and hcu is coppers volumetric enthalpy. For
Tf > Top = 4.2 K, hcu (Tf ) hcu (Top ). From Fig. A3.3, we nd Tf 50 K,
which roughly agrees with a simulation value of 47 K (see Fig. 8.16).
When aluminum is substituted for copper, Fig. A3.3 gives Tf = 75 K. A simu-
lation gives a temperature of 57 K.
Figure 8.16 shows spatially averaged temperature plots for Section 1 and Section
2 of this magnet. The solid curves correspond to Nb-Ti/copper wires, while the
dotted curves correspond to Nb-Ti/aluminum wires [8.46].
60 2
1
2
1
40
T [K]
20
0
0 0.5 1.0 1.5 2.0
Time [s]
Fig. 8.16 Spatially averaged temperature vs. time plots for Coils 1 and 2. Solid
curves: NbTi/copper wires; dotted curves: NbTi/aluminum wires [8.45].
PROTECTION 501
Thus, in approach 1), the eective Em converted into heating in the winding is
decreased even when fr remains small; in approach 2), the eective Em remains
the same as the original stored energy, but fr is increased. Each thus lowers emr .
COLD ENVIRONMENT
S
Iop
r(t)
RD VD L
What all this leads to is an overheating criterion for matrix metal current density
D
at operating current, Jm
, already derived and given by Eq. 8.19:
Discharge Voltage: VD
For a given Z(Tf , Ti ), as may be inferred, for example, from Eq. 8.18, there are
ve design parameters: Jm ; Em ; m/s ; VD ; and Iop . VD is thus given by:
Jm Em
VD = (8.65)
Acd Z(Tf , Ti )
Equation 8.65 states that VD increases linearly with Em and Jm and decreases
inversely with Acd (and hence Iop ) and Z(Tf , Ti ).
Answer to TRIVIA 8.2 Tsunami (200, in m/s); fastball (42.2); car (2); NZP (0.01).
504 CHAPTER 8
Switching Delay
Equations 8.18 and 8.19 are both based on the assumption that the current dis-
charge begins at the instant a non-recovering normal zone is created. In realty,
there is a delay, dl , between the creation of this normal zone and the start of the
current discharge: dl is the sum of the delay in the normal zone detection and the
circuit delay for the switch to actually open. During this period the current will
remain at its initial value, Iop . Thus, to compute Tf , through Z(Tf , Ti ), we shall
i
combine Eq. 8.12a, with dl substituted for ah (Tf , Ti )), and Eq. 8.16a:
Am 2 2
Z(Tf , Ti ) = (Jm + 12 Jm
dl
)
dg
(8.66a)
Acd
Am 2
= (dl + 12 dg )Jm
(8.66b)
Acd
COLD ENVIRONMENT
COIL 1
Iop Rs Iop
r(t)
MAIN
PCS
Ih
COIL 2
COIL 2 HEATER
Vh
PCS-H HEATER
I2 Ih
A
PCS-H Ih
r(t)
+ Vout (t)
Vcl
Coil 2 (L2 ) R2
In the following analysis, we assume that all circuit elements, including r, are
constant; also we assume that R1 and R2 are suciently large so that they do not
load the bridge circuit.
For the case when R1 and R2 are large, the total voltage across the two coils,
Vcl (t), is given by:
dI(t) dI(t)
Vcl (t) = L1 + rI(t) + L2 (8.67a)
dt dt
For the same condition, the current through the resistors R1 and R2 , iR (t), is
given by:
Vcl (t)
iR (t) = (8.67b)
R1 + R2
From the circuit shown in Fig. 8.19, we have:
dI(t)
Vout (t) = L1 + rI(t) R1 iR (t) (8.67c)
dt
Combining Eqs. 8.67a8.67c, we obtain:
dI(t)
Vout (t) = L1 + rI(t)
dt
R1 dI(t) dI(t)
L1 + rI(t) + L2
R 1 + R2 dt dt
R2 dI(t)
= L1
R1 + R2 dt
R1 dI(t) R2
L2 + rI(t) (8.68)
R1 + R2 dt R1 + R2
To make Vout (t) proportional only to rI(t), the rst two terms in the right-hand
side of Eq. 8.68 must sum to zero:
R2 dI(t) R1 dI(t)
L1 L2 =0 (8.69)
R 1 + R2 dt R1 + R 2 dt
For copper RRR = 100, Z(Tf , Ti ) is given in Fig. 8.2: Z(Tf = 300 K, Ti = 10 K) =
15.11016 A2 s/m4 . Solving for tIw |300K
10K , we obtain:
With a constant-current supply feeding 25 A into the magnet, the magnet warms
up from 10 K to 300 K in slightly less than two days.
b) We may use Eq. 8.71b to solve for tVw |300K
10K , the warm-up time from 10 K to
300 K under constant-voltage heating:
Vcd cd 300 K
tVw |300K
10K = Ccu (T )cu (T ) dT
Am V2 10 K
Vcd cd
= Y (Tf = 300 K, Ti = 10 K) (S1.3)
Am V2
For copper RRR = 100, Y (Tf , Ti ) is given in Fig. 8.6: Y (Tf = 300 K, Ti = 10 K) =
7.25 V2 s/m2 . Solving for tVw |300K
10K , we obtain:
A. Technique I
In this technique, the magnet is divided into A +A and B +B. Although inductive
voltage canceling achieved by this technique is slightly better than that achieved
by a more conventional technique that divides the magnet into B +A and A+B,
it is still not entirely satisfactory. The technique cannot completely eliminate all
inductive voltages that are developed whenever either the water-cooled insert or
the NbTi coil is energized.
B. Technique II
The second technique, developed by Ishigohka [8.114], employs voltage taps on
all 22 double pancakes, and combines them into two major components: one that
contains odd-numbered double pancakes, V2n1 (t), and the other that contains
RADIATION SHIELDS (TOP)
SCM
DP 22
B
BITTER INSERT
DP 16
DP 15
A
DP 12
DP 11
A
DP 8
DP 7
B
DP 1
1s
VAA
VBB
5V
100 mV
V
Fig. 8.21 Measured unbalanced voltages from an insert trip in Hybrid II [8.114].
PROTECTIONPROBLEMS & DISCUSSIONS 517
RD = (S5.1a)
w
or
w = (S5.1b)
RD
Em = wCp T (S5.2a)
or
Em
w = (S5.2b)
Cp T
= 86.6 m
(106 m)(2107 J)
w
(0.3 )(4106 J/m2 K)(200 K)
The dump resistor constructed for Hybrid III consisted of nearly 90 steel bars,
each 1-m long, 5-cm wide, and 6-mm thick, electrically in series. Note that if
T had been limited to only 100 K, both and cross section w would have had
to have been increased by 40%, i.e., 120 m and w 410 mm2 .
PROTECTIONPROBLEMS & DISCUSSIONS 519
A. Low-Resistance Resistor
When rd is activated in the circuit, Vm (t) = (rd + r )Im (t): Im (t) decreases ex-
ponentially with m = L/(r +rd ). Obviously, however, the rate of current decay,
dIm (t)/dt, is not constant; it decreases with Im (t):
dIm (t) Im (t)
= (8.77a)
dt m
For a magnet operating current of, for example, 250 A, rd as high as 10 m may
be used. Thus a magnet inductance of 100 H will result in m 104 s; the magnet
will discharge in about half a day.
Design criteria for rd is studied below in DISCUSSION 8.3.
r r
+ + + +
Vp 0 Vm (t) L Vp 0 Vm (t) L
Im (t) Im (t)
S S
rd Vdd +
(a) (b)
Fig. 8.23 Circuits for slow discharge modes: (a) low-resistance discharge
resistor, rd , activated by switch S; (b) a set of series-connected diodes acti-
vated. In each circuit r represents the total line resistance and Vp 0.
520 CHAPTER 8PROBLEMS & DISCUSSIONS
where Im (0) is the magnet current when the resistor is activated and gcv , assumed
constant here, is the natural convective heat transfer ux [W/m2 ] on the resistor
surface, generally (as applied here) of still air or, if the resistor is immersed in a
bucket of water, of still waterthe bucket must be large enough for the water to
absorb the total dissipation energy without too much temperature rise.
Equation 8.78b expresses that the total Joule heating generated in the resistor,
rd Im (0)2 , is balanced by the total convective heat emanating from the resistors
surface, given approximately by 2wgcv for w . Note also that Im (t), which
actually is decaying slowly, is assumed constant and equal to Im (0) in Eq. 8.78b.
Solving for from both equations, we obtain:
rd Im (0) (8.79a)
2gcv
w Im (0) (8.79b)
2gcv
6.3 m
(106 m)
w (250 A)
2(250106 m)(20 W/m2 )
= 2.5 m
Note that this 10-m resistor must handle, initially, a total power of 625 W. A
10-m air-cooled resistor satisfying the above two conditions may be constructed,
for example, with 50 steel strips, each strip 250-m thick, 6.3-m long and 5-cm
wide, connected electrically in parallel.
Water-Cooled
(250106 m)
(0.01 )(250 A)
2(106 m)(20103 W/m2 )
20 cm
(106 m)
w (250 A)
2(250106 m)(20103 W/m2 )
8 cm
As may be seen from Eqs. 8.79a and 8.79b, both and w approach zero with
increasing gcv ; however, the ratio /w is independent of gcv (and rd ). Thus, if a
combination of Im (0), gcv , and rd results, for example, in = 5 cm and w = 2 cm,
dimensions perhaps too uncomfortably small, a practical water-cooled version
of such a resistor may have two 250-m thick steel strips, each 25-cm long and
5-cm wide, again, connected in parallel.
sh
Jm op
805 A/mm2 is greater than most matrix current densities used even in
high-performance magnets. What this implies is that because Jm sh
op
fr , at
Jmop = 300 A/mm , for example, fr 0.2 suces to limit Tf to 200 K.
sh 2
V
Let us now apply the internal voltage criterion (Eq. 8.40b) to solve Jm op
, the
matrix current density meeting the voltage criterion, for the same magnet. Here
let us take Vbk = 10 kV.
= 8(0.102)(2.701014 A2 /m)(2.36106 A1 m1 )
= 5.2108 A/m2 = 520 A/mm2
m It2 2
Tpk 12 (Tcl + Twm ) +
8a2m km
(1106 cm)(50 A)2 (15 cm)2
= 12 (10 K + 70 K) + 590 K
8(8103 cm2 )2 (2 W/cm K)
B. Meltdown Time
An energy density of 2900 J/cm3 is required to raise the temperature of silver
from 70 K to 1223 K, its melting temperature. (For estimating the meltdown
time for this tape, we assume the enthalpy of the silver-gold alloy to be that of
silver.) With an average resistivity of this matrix between 70 K and 1200 K of 5
106 cm and a constant matrix current density of 6000 A/cm2 (= It /am with
It = 50 A and am = 0.008 cm2 ), the matrix is heated at an average Joule heating
of 200 W/cm3 . Thus under an adiabatic condition in an evacuated environment
of a cryocooled magnet where these leads are placed, it will take less than 15 s
[=(2900 J/cm3 )/(200 W/cm3 )] to melt the tape; the tapes need to be continuously
monitored and, as discussed below, protected.
C. Additional Remarks
As stated often in this book HTS is very stable, and an event such as that described
above should not occur. However, although a short length of Bi2223/Ag-Au tape
can readily, and quite inexpensively, be replaced, the consequence of a suddenly
opened circuit can be highly dangerous and damaging to the system if the magnet
to which the current leads are connected happens to have a large inductancenote
that V L(dI/dt): the leads should never be open-circuited, unless the magnet
terminals are shunted with a resistor robust enough to absorb the magnet energy
and remain intact.
As already discussed in DISCUSSION 4.15 and further studied in PROBLEMS 4.6B
and 4.6C, these tapes need to be protected with an additional strip of normal
metal having a low thermal conductivity (to keep conduction heat input low)
and low electrical resistivity.
Answer to TRIVIA 8.3 Quake (1018 , in J); LHD (109 ); turkey (108 ); sleep (106 ).
PROTECTIONPROBLEMS & DISCUSSIONS 525
C1 C2
M
LEC1 (z) LEC2 (z)
z
LEM (z) (0, 0)
Joint #4
st
Coil M Start of 1 Layer (Coil M)
Fig. 8.25 Schematic drawing of a 3-coil arrangement, shunted by PCS for persistent-
mode operation. The dots represent 4 superconducting MgB2 /MgB2 splices.
D. Protection Heater
We discuss key design issues for the Protection Heater for this 3-coil magnet:
1) location; 2) power requirement; 3) heater wire & placement; and 4) heater
resistance & power supply.
Location As stated above, the location of a protection heater is not restricted:
the designer can place it where it is easiest to install under real magnet assembly
conditions. Most likely locations for this magnet are over the entire or a portion
of either the 5th layer of Coil M or the 3rd layers of Coils C1 and C2. For this
discussion, let us place the protection heater over Coils C1 and C2. A heater wire
will be wound over the 3rd layers of Coils C1 and C2. Each 3rd layer has 88 turns
(Table 8.10), containing 78-m long MgB2 wire. The two 3rd layers thus have a
total volume of 86 cm3 . An energy input of 3.9-kJ to the two 3rd layers translates
to an energy density of 45 J/cm3 (or 5 J/g): each 3rd layer will be heated to 75 K.
8 9
7
6
5
34
12 12
10
11
8 9a 9b 10 11 12
Figure 8.27 includes neither a persistent switch nor diodes shunting the entire
system because discussion here chiey concerns premature quenching that occurs
during system charge up when both the switch and the diodes are open. Because
of the presence of the power supply, the entire system is shorted, as is the case
when the system is in persistent mode.
The values of the inductance matrix and shunt resistors are given, respectively, in
Tables 8.12 and 8.13. As is evident from Table 8.12, although ideally identical,
Coils 11 and 12 actually have slightly dierent values of inductance.
Our rule of thumb in determining values of shunt resistors in such a multi-coil
system is to rst choose a total value of shunt resistance from voltage consider-
ations. In this particular example, a value of 1.5 was selected because at an
operating current of 310 A, it would translate to a voltage level of 500 Va safe
level. Each shunt resistor is then selected to be roughly proportional to each coils
total stored energy.
40
9b
20
8
9a
Voltage [V]
9a 9b
10
0
10 8
11 & 12
20
40
0 0.5 1 1.5 2 2.5 3 3.5 4
Time [s]
Fig. 8.28 Voltage traces recorded across Coils 8, 9a, 9b, 10, and 11/12
following a quench at 227 A [8.80].
PROTECTIONPROBLEMS & DISCUSSIONS 533
300
11 & 12
10
10
Current [A]
200
9a
8
9a
100
9b
9b
0
0 0.5 1 1.5 2 2.5 3 3.5 4
Time [s]
Fig. 8.29 Current traces through Coils 8, 9a, 9b, 10, and 11/12 corresponding
to the voltage traces of Fig. 8.28. The dotted curve is the supply current, I0 .
V8
I8 = I0 (8.84a)
S8
V9a
I9a = I0 (8.84b)
S9a
V9b
I9b = I0 (8.84c)
S9b
V10
I10 = I0 (8.84d)
S10
V11/12
I11/12 = I0 (8.84e)
S11/12
12
Pmg = Vr |n In (S6.4)
n=8
Note that the term within the parentheses in Eq. S6.5 is equal to the approximate
resistive voltage across the entire magnet. From Table 8.12, we obtain the sum of
the inductances to be 60.25 H. With dI/dt 50 A/s and I 90 A, we have:
Because V 0, the inductive and resistive voltages are nearly balanced, and at this
time the total resistive voltage has an amplitude of 3000 V (270,000 W/90 A)
this is an example of an internal voltage generated within a quenching magnet,
discussed in 8.3.3. From this voltage, we estimate that the total magnet resistance
has grown to 33 (3000 V/90 A).
PROTECTIONPROBLEMS & DISCUSSIONS 537
REFERENCES
[8.1] P.F. Smith, Protection of superconducting coils, Rev. Sci. Instr. 34, 368 (1963).
[8.2] Z.J.J. Stekly, Theoretical and experimental study of an unprotected supercon-
ducting coil going normal, Adv. Cryogenic Eng. 8, 585 (1963).
[8.3] B.J. Maddock and G.B. James, Protection and stabilization of large supercon-
ducting coils, Rev. Sci. Instrum. 34, 368 (1963).
[8.4] D.L. Atherton, Theoretical treatment of internal shunt protection for supercon-
ducting magnets, J. Phys. E: Sci. Instr. 4, 653 (1971)
[8.5] M.A. Green, Quench protection and design of large high current density super-
conducting magnets, IEEE Trans. Magn. MAG-17, 1793 (1981).
[8.6] D. Ciazynski, Protection of high current density superconducting magnets, IEEE
Trans. Magn. MAG-19, 700 (1983).
[8.7] M.A. Hilal and Y.M. Eyssa, Self protection of high current density superconduct-
ing magnets, IEEE Trans. Magn. 25, 1604 (1989).
[8.8] Lembit Salasoo, Superconducting magnet quench protection analysis and design,
IEEE Trans. Magn. 27, 1908 (1991).
[8.9] Alexei V. Dudarev, Victor E. Keilin, Yurii D. Kuroedov, Alexei A. Konjukhov and
Vytaly S. Vysotsky, Quench protection of very large superconducting magnets,
IEEE Trans. Appl. Superconduc. 5, 226 (1995).
[8.10] L. Bottura, Stability and protection of CICCs: an updated designers view, Cryo-
genics 38, 491 (1998).
[8.11] Joel H. Schultz, Protection of superconducting magnets, IEEE Trans. Appl.
Superconduc. 12, 1390 (2002).
[8.12] J.W. Lue, R. C. Duckworth, M.J. Gouge, Short-circuit over-current limitation of
HTS tapes, IEEE Trans. Appl. Superconduc. 15 (2005).
PROTECTIONREFERENCES 539
[8.13] M. Okaji, K. Nara, H. Kato, K. Michishita, and Y. Kubo, The thermal expansion
of some advanced ceramics applicable as specimen holders of high Tc superconduc-
tors, Cryogenics 34, 163 (1994).
[8.14] Cees Thieme and Garry Ferguson of American Superconductor Corp., (personal
communication, 2004).
[8.15] J.P. Voccio, O.O. Ige, S.J. Young and C.C. Duchaine, The eect of longitudinal
compressive strain on critical current in HTS tapes, IEEE Trans. Appl. Super-
conduc. 11, 3070 (2001).
[8.16] NIST website: www.ceramics.nist.gov/srd/summary/htsy123.htm.
[8.17] Volker Pasler, Gunter Bonisch, Rainer Meyder, and Gernot Schmitz, Deployment
of MAGS, a comprehensive tool for magnet design and safety analysis, for quench
and arc simulation of an ITER-FEAT coil, IEEE Trans. Appl. Superconduc. 12,
1574 (2002).
[8.18] N.G. Anishchenko, R. Heller, Yu.A. Shishov, G.P. Tsvineva, and V.Ya.Vokov, High
voltage heavy current leads for liquid helium cryostats, Cryogenics 22, 609 (1982).
[8.19] J. Gerhold, Design criteria for high voltage leads for superconducting power sys-
tems, Cryogenics 24, 73 (1984).
[8.20] B. Fallou, J. Galand, and B. Bouvier, Dielectric breakdown of gaseous helium at
very low temperatures, Cryogenics 10, 142 (1970).
[8.21] J. Thoris, B. Leon, A. Dubois, and J.C. Bobo, Dielectric breakdown of cold gaseous
helium, Cryogenics 19, 147 (1970).
[8.22] J. Gerhold, Dielectric breakdown of cryogenic gases and liquids, Cryogenics 19,
571 (1979).
[8.23] M.S. Naidu and V. Kamaraju, High Voltage Engineering 2nd Ed., (McGraw Hill,
New York, 1995).
[8.24] T. Tominaka, N. Hara, and K. Kuroda, Estimation of maximum voltage of su-
perconducting magnet systems during a quench, Cryogenics 31, 566 (1991).
[8.25] W.H. Cherry and J.I. Gittleman, Thermal and electrodynamic aspects of the
superconductive transition process, Solid State Electronics 1, 287 (1960).
[8.26] C.N. Whetstone and C. Roos, Thermal transitions in superconducting NbZr al-
loys, J. Appl. Phys. 36, 783 (1965).
[8.27] V.V. Altov, M.G. Kremlev, V.V. Sytchev and V.B. Zenkevitch, Calculation of
propagation velocity of normal and superconducting regions in composite conduc-
tors, Cryogenics 13, 420 (1973).
[8.28] D. Hagendorn and P. Dullenkopf, The propagation of the resistive region in high
current density coils, Cryogenics 14, 264 (1974).
[8.29] L. Dresner, Propagation of normal zones in composite superconductors, Cryo-
genics 16, 675 (1976).
[8.30] K. Ishibashi, M. Wake, M. Kobayashi and A. Katase, Propagation velocity of
normal zones in a sc braid, Cryogenics 19, 467 (1979).
[8.31] Osami Tsukamoto, Propagation velocities of normal zones in a forced-ow cooled
superconductor, IEEE Trans. Magn. MAG-15, 1158 (1979).
[8.32] B. Turck, About the propagation velocity in superconducting composites, Cryo-
genics 20, 146 (1980).
[8.33] A.Vll. Gurevich, R.G. Mints, On the theory of normal zone propagation in su-
perconductors, IEEE Trans. Magn. MAG-17, 220 (1981).
540 CHAPTER 8REFERENCES
[8.34] P.H. Eberhard, G.H. Gibson, M.A. Green, E. Grossman, R.R. Ross, and J.D. Tay-
lor, The measurement and theoretical calculation of quench velocities within large
fully epoxy impregnated superconducting coils, IEEE Trans. Magn. MAG-17,
1803 (1981).
[8.35] M. Kuchnir, J.A. Carson, R.W. Hanft, P.O. Mazur, A.D. McIntur and J.B. Strait,
Transverse quench propagation measurement, IEEE Trans. Magn. MAG-23,
503 (1987).
[8.36] H.H.J. ten Kate, H. Boschamn, L.J.M. van de Klundert, Longitudinal propagation
velocity of the normal zone in superconducting wires, IEEE Trans. Magn. MAG-
23, 1557 (1987).
[8.37] K. Funaki, K. Ikeda, M. Takeo, K. Yamafuji, J. Chikaba and F. Irie, Normal-zone
propagation inside a layer and between layers in a superconducting coil, IEEE
Trans. Magn. MAG-23, 1561 (1987).
[8.38] D.A. Gross, Quench propagation analysis in large solenoidal magnets, IEEE
Trans. Magn. 24, 1190 (1988).
[8.39] A. Ishiyama and Y. Iwasa, Quench propagation velocities in an epoxy-impregnated
Nb3 Sn superconducting winding model, IEEE Trans. Magn. 24, 1194 (1988).
[8.40] Y.Z. Lei, S. Han, Quenching dynamic process and protection by shunt resistors
of solenoid superconducting magnets with graded current density, IEEE Trans.
Magn. 24, 1197 (1988).
[8.41] Yan Luguang, Li Yiping and Liu Decheng, Experimental investigation on normal
zone propagation in a close-packed superconducting solenoid, IEEE Trans. Magn.
24, 1201 (1988).
[8.42] D. Ciazynski, C. Cure, J.L. Duchateau, J. Parain, P. Riband, B. Turck, Quench
and safety tests on a toroidal eld coil of Tore Supra, IEEE Trans. Magn. 24,
1567 (1988).
[8.43] A.A. Konjukhov, V.A. Malginow, V.V. Matokhin, V.R. Karasik, Quenching of
multisection superconducting magnets with internal and external shunt resistors,
IEEE Trans. Magn. 25, 1538 (1989).
[8.44] Arnaud Devred, General formulas for the adiabatic propagation velocity of the
normal zone, IEEE Trans. Magn. 25, 1698 (1989).
[8.45] C.H. Joshi and Y. Iwasa, Prediction of current decay and terminal voltages in
adiabatic superconducting magnets, Cryogenics 29, 157 (1989).
[8.46] Yu.M. Lvovsky, Thermal propagation of normal zone with increasing temperature
level in helium-cooled and high temperature superconductors, Cryogenics 30, 754
(1990).
[8.47] Z.P. Zhao and Y. Iwasa, Normal-zone propagation in adiabatic superconduct-
ing magnets: I. Normal-zone propagation velocity in superconducting composites,
Cryogenics 31, 817 (1991).
[8.48] G. Lopez and G. Snitchler, Quench propagation in the SSC dipole magnets IEEE
Trans. Magn. 27, 1973 (1991).
[8.49] A. Ishiyama, Y. Sato, and M. Tsuda, Normal-zone propagation velocity in super-
conducting wires having a CuNi matrix, IEEE Trans. Magn. 27, 2076 (1991).
[8.50] M. Iwakuma, K. Funaki, M. Takeo and K. Yamafuji, Quench protection of super-
conducting transformers, IEEE Trans. Magn. 27, 2080 (1991).
[8.51] S. Fujimura and M. Morita, Quench simulation of 4.7 tesla superconducting mag-
net for magnetic resonance spectroscopy, IEEE Trans. Magn. 27, 2084 (1991).
PROTECTIONREFERENCES 541
[8.86] V. Kadambi and B. Dorri, Current decay and temperatures during supercon-
ducting magnet coil quench, Cryogenics 26, 157 (1986).
[8.87] K. Kuroda, S. Uchikawa, N. Hara, R. Saito, R. Takeda, K. Murai, T. Kobayashi,
S. Suzuki, and T. Nakayama, Quench simulation analysis of a superconducting
coil, Cryogenics 29, 814 (1989).
[8.88] O. Ozaki, Y. Fukumoto, R. Hirose, Y. Inoue, T. Kamikado, Y. Murakami,
R. Ogawa and M. Yoshikawa, Quench analysis of multisection superconducting
magnet, IEEE Trans. Appl. Superconduc. 5, 483 (1995).
[8.89] Yehia M. Eyssa and W. Denis Markiewicz, Quench simulation and thermal dif-
fusion in epoxy-impregnated magnet system, IEEE Trans. Appl. Superconduc.
5, 487 (1995).
[8.90] Tomoyuki Murakami, Satoru Murase, Susumu Shimamoto, Satoshi Awaji, Kazuo
Watanabe, Two-dimensional quench simulation of composite CuNb/Nb3 Sn con-
ductors, Cryogenics 40, 393 (2000).
[8.91] V.S. Vysotsky, Yu. A. Ilyin, A.L. Rakhmanov and M. Takeo, Quench develop-
ment analysis in HTSC coils by use of the universal scaling theory, IEEE Trans.
Appl. Superconduc. 11, 1824 (2001).
[8.92] A. Korpela, T. Kalliohaka, J. Lehtonen and R. Mikkonen, Protection of conduc-
tion cooled Nb3 Sn SMES coil, IEEE Trans. Appl. Superconduc. 11, 2591 (2001).
[8.93] L. Imbasciati, P. Bauer, G. Ambrosio, V. Kashikin, M. Lamm, A.V. Zlobin,
Quench protection of high eld Nb3Sn magnets for VLHC, Proc. 2001 Particle
Accelerator Conf., 3454 (2001).
[8.94] Yuri Lvovsky, Conduction crisis and quench dynamics in cryocooler-cooled HTS
magnets, IEEE Trans. Appl. Superconduc. 12, 1565 (2002).
[8.95] Ryuji Yamada, Eric Marscin, Ang Lee, Masayoshi Wake, and Jean-Michel Rey,
2-D/3-D quench simulation using ANSYS for epoxy impregnated Nb3 Sn high
eld magnets, IEEE Trans. Appl. Superconduc. 13, 159 (2003).
[8.96] Qiuliang Wang, Peide Weng, Moyan He, Simulation of quench for the cable-
in-conduit-conductor in HT-7U superconducting Tokamak magnets using porous
medium model, Cryogenics 44, 81 (2004).
[8.97] C. Berriaud, F.P. Juster, M. Arnaud, Ph. Benoit, F. Broggi, L. Deront, A. Du-
darev, A. Foussat, M. Humeau, S. Junker, N. Kopeykin, R. Leboeuf, C. Mayri,
G. Olesen, R. Gengo, S. Ravat, J-M. Rey, E. Sbrissa, V. Stepanov, H.H.J. ten
Kate, P. Vedrine, and G. Volpini, Hot spot in ATLAS barrel toroid quenches,
IEEE Trans. Appl. Superconduc. 18, 1313 (2008).
[8.98] Taotao Huang, Elena Martnez, Chris Friend, and Yifeng Yang, Quench charac-
teristics of HTS conductors at low temperatures, IEEE Trans. Appl. Supercon-
duc. 18, 1317 (2008).
[8.99] Philippe J. Masson, Vincent R. Rouault, Guillaume Homan, and Cesar A. Lu-
ongo, Development of quench propagation models for coated conductors, IEEE
Trans. Appl. Superconduc. 18, 1321 (2008).
[8.100] I. Terechkine and V. Veretennikov, Normal zone propagation in superconduct-
ing focusing solenoids and related quench protection issues, IEEE Trans. Appl.
Superconduc. 18, 1325 (2008).
[8.101] A. Zhukovsky, Y. Iwasa, E.S. Bobrov, J. Ludlam, J.E.C. Williams, R. Hirose,
Z. Ping Zhao, 750 MHz NMR Magnet Development, IEEE Trans. Magn. 28,
644 (1992).
[8.102] R. Stiening, R. Flora, R. Lauckner, G. Tool, A superconducting synchrotron
544 CHAPTER 8REFERENCES
power supply and quench protection scheme, IEEE Trans. Magn. MAG-15,
670 (1979).
[8.103] D. Bonmann, K.-H. Me, P. Schmuser, M. Schweiger, Heater-induced quenches in
a superconducting HERA test dipole, IEEE Trans. Magn. MAG-15, 670 (1979).
[8.104] A. Devred, M. Chapman, J. Cortella, A. Desportes, J. DiMarco, J. Kaugerts,
R. Schermer, J.C. Tomkins, J. Turner, J.G. Cottingham, P. Dahl, G. Ganetis,
M. Garber, A. Ghosh, C. Goodzeit, A. Greene, J. Herrera, S. Kahn, E. Kelly,
G. Morgan, A. Prodell, E.P. Rohrer, W. Sampson, R. Shutt, P. Thompson,
P. Wanderer, E. Willen, M. Bleadon, B.C. Brown, R. Hanft, M. Kuchnir, M. Lamm,
P. Mantsch, D. Orris, J. Peoples, J. Strait, G. Tool, S. Caspi, W. Gilbert, C. Pe-
ters, J. Rechen, J. Royer, R. Scanlan, C. Taylor, and J. Zbasnik, Investigation
of heater-induced quenches in a full-length SSC R&D dipole, Proc. Intnl Conf.
Magnet Tech. (MT11), (1990).
[8.105] H. Yoshimura, A. Ueda, M. Morita, S. Maeda, M. Nagao, K. Shimohata, Y. Mat-
suo, Y. Nagata, T. Yamada, M. Tanaka, Heater-induced quenching in a model
eld winding for the 70MW class superconducting generator, IEEE Trans. Magn.
27, 2088 (1991).
[8.106] E. Acerbi, M. Sorbi, G. Volpini, A. Dael, C. Lesmond, The protection system
of the superconducting coils in the Barrel toroid of ATLAS, IEEE Trans. Appl.
Superconduc. 9, 1101 (1999).
[8.107] Yehiha M. Eyssa, W. Denis Markiewicz, and Charles A. Swenson, Quench heater
simulation for protection of superconducting coils, IEEE Trans. Appl. Supercon-
duc. 9, 1117 (1999).
[8.108] V. Maroussov, S. Sanlippo, A. Siemko, Temperature proles during quenches in
LHC superconducting dipole magnets protected by quench heaters, IEEE Trans.
Appl. Superconduc. 10, 661 (2000).
[8.109] E.E. Burkhardt, A. Yamamoto, T. Nakamoto, T. Shintomi and K. Tsuchiya,
Quench protection heater studies for the 1-m model magnets for the LHC low-
quadrupoles, IEEE Trans. Appl. Superconduc. 10, 681 (2000).
[8.110] Iain R. Dixon and W. Denis Markiewicz, Protection heater performance of Nb3 Sn
epoxy impregnated superconducting solenoids, IEEE Trans. Appl. Superconduc.
11, 2583 (2001).
[8.111] W. Denis Markiewicz, Protection of HTS coils in the limit of zero quench prop-
agation velocity, IEEE Trans. Appl. Superconduc. 18, 1333 (2008).
[8.112] J.M. Pfotenhauer, F. Kessler, and M.A. Hilal, Voltage detection and magnet
protection, IEEE Trans. Appl. Superconduc. 3, 273 (1993).
[8.113] I.R. Dixon, W.D. Markiewicz, P. Murphy, T.A. Painter, and A. Powell, Quench
detection and protection of the wide bore 900 MHz NMR magnet at the Na-
tional High Magnetic Field Laboratory, IEEE Trans. Appl. Superconduc. 14,
1260 (2004).
[8.114] T. Ishigohka and Y. Iwasa, Protection of large superconducting magnets: a
normal-zone voltage detection method, Proc. 10th Sympo. Fusion Eng. (IEEE
CH1916-6/83/0000-2050, 1983), 2050.
[8.115] Charles A. Swenson, Yehia M. Eyssa, and W. Denis Markiewicz, Quench protec-
tion heater design for superconducting solenoids, IEEE Trans. Magn. 32, 2659
(1996).
[8.116] Thomas Rummel, Osvin Gaupp, Georg Lochner, and Joerg Sapper,Quench pro-
tection for the superconducting magnet system of Wendelstein 7-X, IEEE Trans.
Appl. Superconduc. 12, 1382 (2002).
CHAPTER 9
SOLENOID EXAMPLES, HTS MAGNETS &
CONCLUDING REMARKS
9.1 Introduction
This chapter consists of three segments. The rst segment presents solenoid
magnet EXAMPLES, each accompanied by a study section, Questions/Answers
(Q/A). The next segment discusses HTS, magnet applications and outlook. The
chapter ends with brief concluding remarks.
The four solenoid magnet EXAMPLES described and studied here are selected not
because of their especial importance nor uniquenessno magnet system is unique
or, perhaps to some, every magnet is unique. The selection stems chiey from
the authors familiarity with these magnets. In the Q/A section that follows each
description, some of the design and operation issues, covered in the previous seven
chapters, CHAPTERS 28, are studied and revisited.
Here, we emphasize once again our basic philosophy, rst stated in CHAPTER 1,
that in any problem solving that requires numerical solution, the very rst step
is to get ballpark gures, computed on a simple model amenable to numerical so-
lution. The ballpark gures quickly tell the magnet designer if the magnet is on
the right track. This exercise is important with any magnet, simple or complex.
An innovative magnet idea often begins with an individual. To assess whether
the idea is realistic and worth pursuing further with colleagues or even forming
a design team, the initiator must rst compute ballpark gures of key design and
operation parameters, those covered in CHAPTERS 28, e.g., from simple param-
eters like total ampere turns, overall operating current density, size and weight
of the magnet, total length of conductor, to more intricate ones like stability and
protection, forces, and cryogenic requirements. The key word here is ballpark: in
the later stages of a magnet project, specialists in the design team, armed with so-
phisticated codes, will compute accurate parameter values. Leave the exact values
to the specialists, but be prepared to verify that theirs indeed fall within the range
of independently computed ballpark gures. The author hopes that after having
studied CHAPTERS 28, the readera specialist in whatever area, electromagnetic
elds, stresses, cryogenics, or even materialswill be capable of handling most of
the questions included in the four magnet EXAMPLES presented below.
Y. Iwasa, Case Studies in Superconducting Magnets: Design and Operational Issues, 545
DOI: 10.1007/b112047_9,
Springer Science + Business Media, LLC 2009
546 CHAPTER 9
305 305
942
(0,0)
15114
12
3
4
5
5-Coil Resistive Insert
Superconducting Magnet
Fig. 9.1 Cross sectional view of one half of the SCH magnet at the NHMFL [9.1]
the superconducting shield coil (of radius 1 m) is not shown here. The winding
dimensions, in mm, are approximate; (in Coil 5) marks its magnetic center after a
fault mode, discussed in the Q/A section.
SOLENOID EXAMPLES, HTS MAGNETS & CONCLUDING REMARKS 547
20-kA/2-kV Breaker
Dump Superconducting
Resistor Magnet
RD = 0.1 Ls = 260 mH
20-kA/600-V
DC Power Msr = 17 mH
Supply
Resistive
Magnet
Lr = 10 mH
Rr = 30 m
Fig. 9.2 Circuit diagram of the SCH magnet. [Courtesy: NHMFL, 2005]
Figure 9.2 shows a circuit diagram of the SCH, in which a superconducting magnet
(of self inductance, Ls = 260 mH) is connected in series with a resistive magnet
(of Lr = 10 mH, and resistance, Rr = 30 m). The magnets are powered by a
20-kA/600-V supply. For protection, the superconducting magnet is shunted by a
dump resistor, RD = 0.1 , with a diode connected in series; the resistive magnet
is shunted by a diode. (Assume that both diodes are ideal, i.e., zero forward
resistance and innite backward resistance.) As indicated in Fig. 9.2 the mutual
inductance, Msr , between the two magnets is 17 mH. In case of a malfunction of
either magnet, the 20-kA/2-kV breaker is opened.
Table 9.1 lists key parameter values of the superconducting magnet and the CIC
conductor used in the innermost layer of the magnet. Note that Asc , Am , Am ,
and Acl are cross sectional areas in the CIC conductor, respectively, of Nb3 Sn;
non-matrix metal; matrix metal (copper); and supercritical helium at 4.5 K.
b) Central Field What is the eld (magnetic induction) at the center, Bz (0, 0),
generated by the superconducting magnet at Iop = 20 kA?
S imilarly from Table 9.1, we have: = (1220.2 mm)/(610.0 mm) = 2.00; =
(942.0 mm)/(610.0 mm) = 1.544. Applying Eq. 3.110, we have:
N I + 2 + 2
Bz (0, 0) = ln (3.110)
2a1 ( 1) 1 + 1 + 2
Thus:
(4107 H/m)(756)(20103 A) 2.00+ (2.00)2 +(1.544)2
Bz (0, 0) = ln
(0.610 m)(2.001) 1+ 1+(1.544)2
= 14.52 T
c) Midplane Radial Field What is the radial component of the eld at the
midplane (z = 0) at r = a2 , Br (z = 0, r = 2a2 ), generated by the superconducting
magnet? Assume the magnet winding to be an ideal solenoid.
T he radial component of eld along the midplane of an ideal solenoidal magnet
or nested-coil magnet symmetric about its midplane is always zero: Br (0, a2 ) = 0.
d) Inductance Using Eq. 3.81 and Fig. 3.14, compute an approximate value of
the magnets self inductance, Ls . As noted above the exact value is 260 mH.
Applying Eq. 3.81 and from Fig. 3.14, L( = 2.00, = 1.544) 1.2, we obtain:
L = a1 N 2 L(, ) (3.81)
7 2
Ls = (410 H/m)(0.305 m)(756) (1.2) = 263 mH
e) Stored Magnet Energy What is the total magnetic energy, Ems , of the
superconducting magnet at the operating current, Iop , of 20 kA?
Here we must include the eect of the mutual inductance. Thus:
2
Ems = 12 (Ls + Msr )Iop
= 12 (260 mH + 17 mH)(20 kA)2 = 55.4 MJ
SOLENOID EXAMPLES, HTS MAGNETS & CONCLUDING REMARKS 549
g) Charging Voltage For a constant charging rate of 400 A/s, dIs /dt = dIr /dt
= dIS /dt = 400 A/s, where Is is the current through the superconducting magnet,
Ir is the current through the resistive magnet, and IS is the power supply current,
what is the required supply voltage, VS , at Is = Ir = IS = 10 kA?
T he supply voltage VS is given by:
dIs dIr dIs dIr
VS = Ls + Msr + Msr + Lr + Rr Ir (g.1a)
dt dt dt dt
dIS
= (Ls + 2Msr + Lr ) + Rr IS (g.1b)
dt
Inserting appropriate values in Eq. g.1b, we have:
VS = (260 mH + 217 mH + 10 mH)(400 A/s) + (30 m)(10 kA) = 421.6 V
TRIVIA 9.1 List the items below in descending order of stress (tension or compression).
i) Maximum tension in the winding of a 35-T magnet;
ii) On the Titanic, at the bottom of the Atlantic;
iii) Sound of rock music;
iv) Synthesizing diamond.
550 CHAPTER 9
k) CICHelium Flow The CIC conductor has a helium cross sectional area
of Acl = 76.0 mm2 (Table 9.1). Supercritical helium of 3.5 atm and 4.5 K ows at
a mass rate of mhe = 5 g/s through the conduit. Show that the ow is turbulent
with a Reynolds number, Re 105 . Use the following parameter valueshelium
density: he = 0.132 g/cm3 ; helium viscosity: he = 35.9106 g/cm s; hydraulic
diameter: Dhe = 1 cm.
F luid velocity, vhe , is given by: mhe = he Acl vhe . Thus:
mhe
vhe =
he Acl
(5 g/s)
= 50 cm/s
(0.132 g/cm3 )(0.760 cm2 )
Reynolds number Re is given by:
he vhe Dhe
Re =
he
(0.132 g/cm3 )(50 cm/s)(1 cm)
1.8105
(35.9106 g/cm s)
Flow is turbulent when its Reynolds number exceeds 2300.
SOLENOID EXAMPLES, HTS MAGNETS & CONCLUDING REMARKS 551
Figure 6.3 (p. 359) shows liquid helium heat transfer coecients. From this gure,
we nd that for P = 3.5 atm, Top = 4.5 K, and T = 5.8 K, hq 0.26 W/cm2 K at
Re = 105 . Noting that hhe Re0.8 (Eq. 6.29), hq 0.42 W/cm2 K at Re = 1.8105 .
Thus:
(57.4102 cm2 )(3 cm)(0.42 W/cm2 K)(5.8 K)
Ilim = 14.4 kA < 20 kA
(2108 cm)
That is, this superconducting magnet operates at 40% more than the Stekly
current of 14.4 kA: this CIC conductor is not cryostable at 20 kA.
m) Current Dump Suppose that both magnets are at 20 kA, and the pro-
tection system detects a fault in either one of the magnets at t = 0 and opens
the 20-kA/2-kV breaker without a delay, i.e., at t = 0. For a quick estimate
of Is (t) and Ir (t), solve for Is (t) and Ir (t) assuming that Msr = 0, i.e., the two
magnet circuits
are uncoupled. Actually, the coupling is not quite negligible, i.e.,
ksr = Msr / Ls Lr = (0.017 H)/ (0.260 H)(0.010 H) = 0.333; nevertheless, the re-
sults with Msr = 0 are informative and give a feel for the time scale.
With Msr = 0 the circuit equation for each magnet is given by:
dIs (t) dIr (t)
Ls + Msr + RD Is (t) = 0 (m.1a)
dt dt
dIs (t) dIr (t)
Msr + Lr + Rr Ir (t) = 0 (m.1b)
dt dt
With Msr = 0 Eqs. m.1a and m.1b are then simplied to:
dIs (t)
Ls + RD Is (t) = 0 (m.2a)
dt
dIr (t)
Lr + Rr Ir (t) = 0 (m.2b)
dt
Equations m.2a and m.2b may be solved independently of each other:
Is (t) = I etRD /Ls (m.3a)
Ir (t) = I etRr /Lr (m.3b)
where I = 20 kA, Ls /RD = 2.6 s (= 260 mH/0.1 ), and Lr /Rr = 0.33 s (= 10 mH/
30 m). Equations m.3a (solid circles) and m.3b (open) are shown in Fig. 9.3.
552 CHAPTER 9
Ir (t)
5
0
0 2 4 6 8 10
Time [s]
Fig. 9.3 Is (t) by Eq. m.3a (solid circles) and Ir (t) by Eq. m.3b (open circles),
both with Msr = 0. The curve nearby each gives the solution with Msr = 0.017 H.
o) Eective Decay Time Constant As stated above in n), when the induc-
tive coupling of the two magnets is included, Is (t) and Ir (t) decay more slowly
than in the uncoupled system of m). Assuming that the total magnetic energy
stored in the superconducting magnet at 20 kA of Es = 55.4 MJ, as computed in
e), is entirely dissipated in the dump resistor (RD ) and that Is (t) in the coupled
system is represented by a single eective time constant of eff , compute eff .
T he magnetic energy of the superconducting magnet is dissipated all in RD :
Es = RD Is2 (t) dt (o.1a)
0
And because Is (t) = I et/eff , Eq. o.1a becomes:
RD I2 eff
Es = RD I2 e2t/eff dt = (o.1b)
0 2
Solving Eq. o.1b for eff , we have:
2Es 2(55.4106 J)
= eff = = 2.77 s
RD I2 (0.1 )(2104 A)2
Note that this is 6% greater than the 2.6 s computed for the uncoupled system.
SOLENOID EXAMPLES, HTS MAGNETS & CONCLUDING REMARKS 553
2df Bm
ehy1 Jc (B, T,
)dB (p.1)
3 0
Here, Jc (B, T,
), the critical current density of Nb3 Sn, includes dependences of Jc
not only on B and T but also on
, strain, because as the magnet is energized or
discharged, strain acting on each composite Nb3 Sn strand changes. The depen-
dence of T can also be important, because the strand temperature can rise unless
the magnet is energized or discharged slowly enough for the composite to transfer
heat to the coolant with negligible temperature rise. Generally the integration of
Eq. p.1 is too complicated to be performed in a closed form.
However, we shall here perform a closed-form integration by: 1) changing the eld
at a rate slow enough to keep the composite at 4.5 K; and 2) neglecting the strain
eects on Jc . For the composite Nb3 Sn strands in the magnet midplane and the
innermost layer, we may use the average current density, Jc (B, 4.5 K), given below:
b
Jc (B, 4.5 K,
= 0) = Jc (0, 4.5 K) (p.2)
b + B
where Jc (0, 4.5 K) = 42 109 A/m2 and b = 1 T. Equation p.2 is a rough approxi-
mation of Jc (B, T,
) for Nb3 Sn at 4.5 K and for
= 0.
U sing Eq. p.1, we rst perform the integral within Eq. p.1:
Bm
Bm
dB b +Bm
Jc (B, 4.5 K) dB = Jc (0, 4.5 K)b
= Jc (0, 4.5 K)b ln
0 0 b +B b
= (42109 J/m2 )(1 T) ln 15 = 113.7109 J/m4
Inserting this into Eq. p.1, we obtain:
2(42106 m)
ehy1 (113.7109 J/m4 ) 1014 kJ/m3
3
Using GANDALF, Gavrilin obtains ehy1 = 1039 kJ/m3 for Eq. p.1; for the case
T = 4.5 K and
= 0, as in our simple model above, he obtains 1122 kJ/m3 greater
than 1039 kJ/m3 because Jc (B, 4.5 K,
= 0) > Jc (B, T > 4.5 K,
> 0) [9.2].
554 CHAPTER 9
(Dmf /p )2
1 and Hm = Bm / in Eq. 7.36; the charging time period covers
only 1/2 of the complete triangular excitation (Fig. 7.18), thus a factor of 1/2
must also be introduced into Eq. 7.36. Thus, ecp becomes:
2
B
ecp 2 2 m
1
2
4Bm cp
= (r.1)
m
Inserting Bm = 14 T, cp = 30 ms, and m = 50 s, we obtain:
With parameter values from Table 9.1, Am = 57.4 mm2 ; Acd = Am +Asc +Am =
97.6 mm2 ; and Jmop = Iop /Am = (20, 000 A)/(57.4 106 m2 ) = 3.48 108 A/m2 ,
inserted to Eq. 8.16a and dg = eff = 2.77 s in Eq. s.1, and solving Zcu (Tf , Ti ),
Z(Tf , Ti ) for copper, we have:
57.4106 m2 8
2 2 2.77 s
Zcu (Tf , Ti ) = 3.4810 A/m
97.6106 m2 2
9.91016 A2 s/m4
From Fig. 8.2, we nd that Zcu (Tf , Ti = 4.5 K) = 9.91016 A2 s/m4 corresponds,
for copper (RRR = 100), to Tf 125 K.
13.51016 A2 s/m4
From Fig. 8.2, we nd that Zcu (Tf , Ti = 4.5 K) = 13.51016 A2 s/m4 corresponds,
for copper (RRR = 100), to Tf 225 K, which is nearly the limit of acceptable hot
spot temperature. Although, in reality, because of cooling by the coolant, Tf will
likely be < 225 K, it is prudent to make sure that for this magnet the breaker switch
is opened no later than 0.5 s after the initiation of a hot spot. This may imply
that the hot spot must be detected within much less than 0.5 s of its initiation.
Figure 9.4 shows temperature (solid) and pressure (dashed) vs. time plots, com-
puted by Gavrilin [9.2], who applied GANDALF for the case similar to that con-
sidered in s), except in this analysis not only is the exact solution of Is (t) used
but also eects of helium cooling and AC losses are included. Chiey because of
helium cooling, the hot spot temperature, Tf , reaches only 95 K instead of the
125 K computed in s) under adiabatic conditions and with dl = 0.. The analysis
also shows that the peak helium pressure reached in the conduit is 23.5 atm.
100 25
Hot-Spot Temperature [K]
80 20
60 15
40 10
20 5
0 0
0 2 4 6 8 10
Time [s]
Fig. 9.4 Temperature (solid, the left y-axis) and pressure (dashed, the right
y-axis) vs. time plots after a dump from 20 kA, computed by Gavrilin [9.2].
558 CHAPTER 9
u) Fault-Mode Axial Forces One likely fault mode in the SCH magnet is a
failure, at the nominal operating current of 20 kA, of the 5-coil resistive insert
(Fig. 9.1), in which, under the worst scenario, an entire half of each coil is de-
stroyed, and electrically shorted, with the remaining half still alive and generating
eld. With the top half of each coil shorted, the axial eld center of each coil moves
downward by b/2; in the case of Coil 5 (outermost) this is 151 14 mm, as indicated
in Fig. 9.1. Table 9.2 lists key parameters of the 5-coil resistive insert [9.3].
When this happens the superconducting magnet will experience a downward axial
force, which is matched by an upward force on the resistive magnet. Looking at
Fig. 9.1, we may conclude that the largest force on the superconducting magnet
will come from Coil 5, despite its central eld being smallest, because: 1) Coil 5
is the largest; and 2) it is coupled most closely with the superconducting magnet.
Following a step-by-step simple but well-reasoned procedure described below,
compute ballpark forces on the superconducting magnet exerted by Coils 5 and 4.
The aim here is to enable a non-specialist on the design team to obtain ballpark
gures, realizing full well that the teams specialist will use a code to compute
accurate values.
In 3.5 analytical methods are discussed to compute axial forces between combina-
tions of simple solenoids, specically, ring coils, thin-walled coils, and other
cases. Of these combinations, the simplest is the force computation between two
ring coils (Fig. 3.5), Coil A of diameter aA , total turns NA , and current IA and Coil
B of diameter aB , total turns NB , and current IB .
Step-by-Step Procedure
The procedure consists of the following two steps.
Step 1 Model each coil as a ring coil that generates the same central eld as the
original coil. Compute the total number of turns, N , for the ring coil at
20 kA, the operating current of the SCH. Although the superconducting
magnet and resistive coils both are more accurately modeled as thin-
walled solenoids (Fig. 3.7) than as ring coils, the corresponding axial force
expression is more complex than necessary for estimating these gures.
Step 2 Use Eq. 3.34 to compute the axial force between two ring coils: rst between
the superconducting ring coil (Coil A) and the resistive ring Coil 5 (Coil
B); then between the superconducting ring coil (Coil A) and the resistive
ring Coil 4 (Coil B).
SOLENOID EXAMPLES, HTS MAGNETS & CONCLUDING REMARKS 559
K(k) and E(k) are the complete elliptic integrals, respectively, of the rst and
second kinds. The modulus, k, of the elliptic integrals for this system is given by:
4aA aB
k2 = (3.36)
(aA + aB )2 + 2
We may compute NA for the superconducting ring Coil A by using the eld
expression given by Eq. 3.111a:
N I
Bz (0, 0) = (3.111a)
2a1
Here, N = NA , I = IA , and a1 = aA . An appropriate value of aA for the super-
conducting coil is the mean winding radius given by its i.d. and o.d. (Table 9.1):
aA 458 mm. Thus, with Bz (0, 0) = 14.52 T and IA = 20 kA, we obtain:
2aA Bz (0, 0) 2(0.458 m)(14.52 T)
NA = = = 529
IA (4107 H/m)(2104 A)
Note that NA is less than the magnets real number of turns, 756, because to
generate the same central eld, a ring coil is much more ecient than the real
magnet in which turns are distributed over a large winding cross section.
We must be a bit more careful modeling Coil 5 as ring coil 5 (Coil B): the central
eld of 3.72 T cannot be used, because it is the central eld generated by Coil 5
before the fault. Now the axial length (2b) is halved, and thus we must rst
compute the central eld generated by a Coil 5 of = /2. For a Bitter magnet,
the eld factor, [F (, )]B , is given by Eq. 3.115b:
+ 1+ 2
[F (, )]B = ln (3.115b)
+ 2 + 2
Because the rest of the coil parameters remain the same in the healthy half of the
half-burned coil, the new central eld, [Bz (0, 0)]B may be given in terms of the
original central eld [Bz (0, 0)]B by:
+ 1+ 2
ln
[Bz (0, 0)]B [F (, ]B + 2 + 2
= = (u.1)
[Bz (0, 0)]B [F (, )]B + 1+ 2
ln
+ 2 + 2
560 CHAPTER 9
a2ps a2as
rth a2 (610.1 mm)
ab
a1 (305.0 mm)
a1ps a1as
Steel Shell Main Superconducting Magnet Shield Coil
(Passive Shielding) (N =756; Iop = 20 kA) (Active Shielding: Nas ; Ias )
Fig. 9.5 Cross sectional view of the main superconducting magnet with either a shield
coil (active shielding) or a steel shell (passive shielding). Note that the active shield coil
consists of two subcoils, one above and the other below the midplane.
562 CHAPTER 9
3
Re 1
Hf = H0 (cos r + 2 sin ) (3.163)
r
where a1 and a2 are, respectively, the inside and outside winding radii of the
magnet. Because of H f (Re /r)3 (Eq. 3.163) and H0 R3 a2 (Eq. 3.165), we may
e 1
indeed ignore the resistive insert herea1 = 173.9 mm for Coil 5 and a1 = 305.0 mm
for the superconducting magnet. Applying Eqs. 3.163 and 3.165 to the main
superconducting magnet and shield coil (subscript as), we have:
where as = a2as /a1as . Solving Eq. v.1 for Nas , with Ias = Iop , we obtain:
2
a1 1+2 +
Nas = 2 +
N (v.2)
a1as 1+as as
2
0.305 m 1+4+2
Nas = (756)
1.203 m 1+1.08+1.04
109
The actual value is 78 for the shield coil [9.4], which actually consists of two sub-
coils, as shown in Fig. 9.5. The magnetic moment is undiminished by axial dis-
placement of loops; only the eld, not the magnetic moment, decreases if turns
are spread out axially.
SOLENOID EXAMPLES, HTS MAGNETS & CONCLUDING REMARKS 563
x1) Active Shield CoilInteraction Force: 1 If the shield coil were a single
coil as modeled, then there would be no net interaction force between it and the
main magnet. In the original system, the shield coil is split into two subcoils. Still,
there will be no net interaction force between the shield coil as a whole and the
main magnet, but there is between each subcoil and the the main magnet. We shall
here compute its ballpark magnitude by modeling both the main magnet and one
of the subcoils as ring coils. Take = 0.277 m (center-to-center displacement).
Note that because ab in Fig. 9.5 is equal to 2, ab = 0.554 m.
The main magnet was already modeled as a ring coil in t) above: NA = 529; aA
458 mm. Similarly, we may model one of the subcoils as a ring coil generating
a central eld of 0.375 T (half of 0.75 T, though each is displaced vertically): aB =
1.227 m (average of a1as and a2as ), and NB is given by:
(0.375 T)2aB (0.375 T)2(1.227 m)
NB = 37
IB (4107 H/m)(20103 A)
It is not surprising that the eective number of turns of 37 is very close to the
actual number of turns of 39, because each subcoil, with = 1.04 and small
clearly indicates that it is essentially a ring coil ( = 1, = 0). First, we compute
the modulus constant, k, given by Eq. 3.36 used in t) above.
4aA aB
k2 = (3.36)
(aA + aB )2 + 2
4(0.458 m)(1.227 m)
= 0.7709 (k 0.8780)
(0.458 m + 1.227 m)2 + (0.277 m)2
564 CHAPTER 9
(4107 H/m)
FzA (0.277 m) = (529)(2104 A)(37)(2104 A)
2
(0.277 m) (1.685 m)2 + (0.277 m)2
(0.769 m)2 + (0.277 m)2
0.7709(2.1957) + (0.77092)(2.19571.1977)
The interaction force on the main magnet is negative (downward); the same force
acts upward on the shield subcoil above the midplane; i.e., if the coil werent
restrained, it would y away from the midplane. Of course combined with the
second subcoil below, the net interaction force from the main magnet is zero.
This agrees quite well with 1.6 MN computed with a code [9.3].
SOLENOID EXAMPLES, HTS MAGNETS & CONCLUDING REMARKS 565
The steel cylinder thickness, a2ps a1ps thus becomes 72 cm. The total weight of
this steel cylinder, 2.4-m high, is 133,000 kg or over 100 tonnes.
Because the eld within the steel cylinder is chiey axially (z)-directed and because
the tangential eld (here z-directed) must be continuous at the steel-air boundary,
the magnetic eld is given by Ms /(/ )dif , i.e., 1.25 T/180 = 0.0069 T or
69 gauss at r = a1ps = 1.203 m and r = a2ps = 1.924 m; by r 20 m, the eld
falls to less than 1 gauss. From the law of ux continuity in the normal (axial)
direction, the ux in the steel cylinder, 1.25 T and directed downward, becomes
upward as it returns towards the magnet midplane, enhancing the central eld
of 14.52 T generated by the main superconducting magnet. In this respect this
passive shielding is superior to active shielding.
y2) Passive Shield Shell: 2 Clearly, the further away from the main magnet
the steel shell is placed, the thinner will be the shell thickness required, as indicated
explicitly by Eq. y1.3. However, according to a basic principle of passive shielding,
to cancel a given dipole moment the amount of steel is independent of the location
of steel shielding, if the steel is magnetized to the same level. Here, consider Shell 2
of a1ps = 1.805 m (3.6-m high), 50% larger than Shell 1, and compute new a2ps and
check to see if the mass of Shell 2 is still 133,000 kg, computed above for Shell 1.
Using the same argument as above, we obtain a new threshold radius for ux
lines: rth = a1ps b = 1.805 m0.471 m = 1.334 m. Inserting appropriate parameter
values in Eq. w.3, we compute
(4107 H/m)(0.305 m)2 (7)(756)(20103 A)
a2ps = (1.805 m)2 +
6(1.334 m)(1.25 T)
= 3.258 m2 + 1.237 m2 = 2.120 m
The cylinder thickness, a2ps a1ps , thus becomes 32 cm, less than half that of
Shell 1, resulting in a total mass of 109,000 kg for Shell 2 cylinder (3.6-m high),
somewhat less than that for Shell 1. This discrepancy implies that this simple
analytic approach is valid only for a ballpark estimate of the mass of a steel shield.
Answer to TRIVIA 9.2 iv). The French mathematician Jean Baptise Joseph Fourier
(17681830) made it clear, in his work on heat conduction, that a scientic equation
must involve a consistent set of units.
SOLENOID EXAMPLES, HTS MAGNETS & CONCLUDING REMARKS 567
Parameters Value
Winding i.d. (2a1 ) [mm] 40
Winding o.d. (2a2 ) [mm] 70
Winding height (2b) [mm] 10
Total turns (N ) 150
70
40
z
10 r
(0, 0)
st Steel Plate
0.15
z=0
5
0.10
7.5
10
Bz (z, r) [T]
12.5
0.05 15
17.5
20
0.05
0 5 10 15 20 25 30 35 40
r [mm]
Fig. 9.7 Bz (z, r) generated by the coil at z = 0, 5, 7.5 10, 12.5, 15, 17.5, and 20 mm
at 25 A without the steel plate. Bz (0, 20 mm) = 0.158 T and Bz (0, 35 mm) = 0.068 T.
Review of Inductance
As discussed in 3.7 the coils self inductance, L, may be given from an equation
relating and I:
= LI (3.78)
is the total magnetic ux linkages at current I:
N Rj
= 2 r Bz (zj , r) dr (9.1)
j=1 0
c) Field Enhancement by Steel Plate Using the model shown in Fig. 9.9,
compute the coils eld at the center, Bz (0, 0), for the system shown in Fig. 9.6
when the coil is energized at 25 A.
Without the steel plate, the coil at 25 A, from Fig. 9.7, generates [Bz (0, 0)]w/o
0.0863 T. With the steel plate, we must include the eld contribution of the image
coil, centered at r = 0 and z = 10 mm.
[Bz (0, 0)]with = [Bz (0, 0) + B(z = 10 mm, 0)]w/o
0.0863 + 0.0711 0.1574 T
The second term in the right-hand side is the contribution at (z = 0, 0) of the
image coil, whose center is at (z = 10 mm, 0). Note that the steel plate does not
quite double the eld in the upper regionit does so only at z = 5 mm. It is as if
the steel plate ipped the eld in the lower region over to the upper region about
the x-y plane at z = 5 mm. Below an ideal steel plate, the eld is zero.
r
(0,0)
Fig. 9.9 Coil arrangement to model the eect of the steel plate. Dimensions are
the same as in Fig. 9.6. An image coil identical to the original and directly beneath
it replaces the steel plate, here idealized to be of of / = .
SOLENOID EXAMPLES, HTS MAGNETS & CONCLUDING REMARKS 571
The total ux leaving the steel disk in the range 0 r r = 31.56 mm (Fig. 9.10),
r , is thus:
r = 2(7.22105 T m2 ) 45.4105 T m2
This ux is owing inward at r = r and equal to the total ux owing through
the steel, st :
st = 2 r st ( Mst ) = r
That is, the steel plate should extend to capture at least 80% of the ux leaving
the plate from the inner region (0 r r ). To satisfy the above condition, it
turns out that rod close to 60 mm is adequate.
z
Flat HTS Plate
d
2Rd
z
r
(0, 0)
Steel Plate
Coil 2
Coil 1
0.15
z=0
5
0.10
7.5
10
Bz (z, r) [T]
12.5
0.05 15
17.5
20
0.05
0 5 10 15 20 25 30 35 40
r [mm]
Fig. 9.12 Bz (z, r) vs. r plots, at z = 0, 5, 7.5, 10, 12.5, 15, 17.5, and
20 mm, with Coil 1 at 25 A and Coil 2 at 15 A without the steel plate.
0.030
17.5
0.025
0.020 15
Br (z, r) [T]
0.015 12.5
0.010 10
0.005 r =5
0
0.005
0 5 10 15 20
z [mm]
Fig. 9.13 Br (z, r) vs. z plots, at r = 5, 10, 12.5, 15, and 17.5 mm, with
Coil 1 at 25 A and Coil 2 at 15 A without the steel plate.
574 CHAPTER 9
mp = Rd2 d
= (1.5 cm)2 (0.5 cm)(6.4 g/cm3 ) 23 g
Thus even in the absence of the steel plate, the levitation force on the plate at
z = 10 mm is double the plate weight.
c) Induced Supercurrent Again, assuming the absence of the steel plate, com-
pute the supercurrent layer, s , for the same YBCO plate at z = 10 mm and show
s
Rd . Use Jc = 108 A/m2 . Also compute Is .
Applying Eq. a.1 and noting Bz (z , Rd ) 0.070 T, we have:
Hz (z , Rd ) 1 Bz (z , Rd )
s = = (a.1)
Jc Jc
(0.070 T)
0.56 mm
(108 A/m2 )(4107 H/m)
Is = d s Jc (a.2)
(5103 m)(0.56103 m)(108 A/m2 ) 280 A
This supercurrent of 280 A is induced in the YBCO plate when the plate is exposed
to the axial eld, Bz (Rd , z ), of 0.07 T.
576 CHAPTER 9
The minus sign indicates that Fz (z, Rd ) increases with decreasing z. Combining
Eqs. 9.3b and d.1, we have:
Br (z , Rd ) Bz (z, Rd ) Bz (z , Rd ) Br (z, Rd )
kz = 2Rd d + (9.4)
z z z z
kz 2(15103 m)(5103 m)
0.0181 T 0.070 T
(5.2 T/m)+ (1.1 T/m)
4107 H/m 4107 H/m
7
6 10
= 2(7510 m ) 2
m/H (0.0943 T2 /m + 0.0758 T2 /m)
4
6.9 N/m
Thus:
1 (6.9 N/m)
z
2 (23103 kg)
3 Hz
When the plate is displaced vertically, it will oscillate with a frequency of 3 Hz.
SOLENOID EXAMPLES, HTS MAGNETS & CONCLUDING REMARKS 577
x
Rd +x 0 Rd +x
These values are in the order of magnitude of measured values [9.5, 9.6]
For Coil 1 with a1 = 20 103 m; = 2a2 /2a1 = (70 mm)/(40 mm) = 1.75 =
2b/2a1 = (10 mm)/(40 mm) = 0.25, Eq. g.1 gives:
(55 W)
Q = = 0.34 cm3 /s = 1224 cm3 /h
(161 J/cm3 )
1 liter/h
For long operation of this system the cryostat holding the coils may require liquid
nitrogen replenishment at a rate of 1 liter/h.
580 CHAPTER 9
r
(0, 0)
Fig. 9.15 Coil arrangement to model the presence of the steel plate in Fig. 9.11.
We shall now study the eects of the steel plate in Fig. 9.11. To simplify the
problem, we rst assume the plate to be of innite permeability, i.e., / = .
Next, to take account of the plates eect on the magnetic eld above the plate,
i.e., z 5 mm, we replace the steel plate with a two-coil unit identical to the
original two-coil unit, placing it directly beneath the original unit, as illustrated
in Fig. 9.15. (Note also here that each coil form is assumed to be formless.)
i) With Steel Plate Now compute the levitation force Fzsteel (z ) on the plate
at z = 10 mm with an ideal steel plate (/ = ). Also, discuss stability in the
axial and radial directions.
W ithout the steel plate, we have, from Figs. 9.12 and 9.13, [Bz (z , Rd )]w/o
0.070 T and [Br (z , Rd )]w/o 0.019 T. With an ideal steel plate, the contribution
of each of the image coils located at z = 10 mm must be included.
Thus, with an ideal steel plate, the system generates a levitation force that is six
times the plate weight (0.23 N), i.e., the plate can support a load mass of more
than 110 g. The system is stable in the axial direction (kz 6.9 N/m) as well as in
the lateral direction (kx 9.1 N/m).
SOLENOID EXAMPLES, HTS MAGNETS & CONCLUDING REMARKS 581
Bz
2b
b
2a1
2a2
(a)
p f
2a1
2a2
(b) (c)
Fig. 9.16 Schematic cross sectional views, each of 2a1 i.d. and 2a2 o.d.:
a) bulk annulus of thickness b ; b) thin-plate annulus of overall thickness
p and superconducting lm thickness f ; and c) a magnet of overall
height 2b assembled from a stack of bulk (or plate) annuli.
582 CHAPTER 9
Note that Bz (0, 0) is independent of the magnet length and directly proportional
to the winding build. Solving Eq. 3.111e for J, we obtain:
Bz (0, 0) (11.74 T)
J = = = 7.5108 A/m2
a1 (1) (4107 H/m)(0.025 m)(1.51)
Fig. 9.18 Schematic drawing of an annulus magnet system after Step 5 of the energizing
sequence. The annulus magnet, now cold and superconducting, is generating a persistent-
mode eld, which in turn is generated by the annuli, each with a supercurrent distribution
induced when the external magnet was discharged to zero. A eld-shielding assembly
made of steel annuli surrounds the magnet/cryostat unit.
REFERENCES
[9.1] John R. Miller and Tom Painter (SCH presentations, April and August 2005).
[9.2] Andrey V. Gavrilin (SCH presentation, April 2005).
[9.3] Mark D. Bird (SCH presentation, August 2005).
[9.4] Iain Dixson (SCH presentation, August 2005).
[9.5] Yukikazu Iwasa and Haigun Lee, Electromaglevmagnetic levitation of a super-
conducting disk with a DC eld generated by electromagnets: Part 1 Theoretical
and experimental results on operating modes, lift-to-weight ratio, and suspension
stiness, Cryogenics 37, 807 (1997).
[9.6] Haigun Lee, Makoto Tsuda, and Yukikazu Iwasa, Electromaglev active-maglev
magnetic levitation of a superconducting disk with a DC eld generated by
electromagnets: Part 2 Theoretical and experimental results on lift-to-weight ratio
and stiness, Cryogenics 38, 419 (1998).
[9.7] Makoto Tsuda, Haigun Lee, So Noguchi, and Yukikazu Iwasa, Electromaglev
(active-maglev) magnetic levitation of a superconducting disk with a DC eld
generated by electromagnets: Part 4 Theoretical and experimental results on su-
percurrent distributions in eld-cooled YBCO disks, Cryogenics 39, 893 (1999).
[9.8] Y. Iwasa, H. Lee, M. Tsuda, M. Murakami, T. Miyamoto, K. Sawa, K. Nishi,
H. Fujimoto, and K. Nagashima, Electromaglevlevitation data for single and
multiple bulk YBCO disks, IEEE Trans. Appl. Supercond. 9, 984 (1999).
[9.9] Stephen Hawking, A Brief History of Time (Bantam Books, New York, 1988).
[9.10] S. Jin, T.H. Tiefel, R.C. Sherwood, M.E. Davis, P.B. van Dover, G.W. Kammlott,
R.A. Fastnacht, and H.D. Keith, High critical currents in Y-Ba-Cu-O supercon-
ductors, Appl. Phys. Lett. 52, 2074 (1988).
[9.11] K. Salama, V. Selvamanickam, L. Gao, and K. Sun, High current density in bulk
YBa2 Cu3 Ox superconductor, Appl. Phys. Lett. 54, 2352 (1989).
[9.12] S. Gotoh, M. Murakami, H. Fujimoto, and N. Koshizuka, Magnetic properties of
superconducting YBa2 Cu3 Ox permanent magnets prepared by the melt process,
J. Appl. Phys. 6, 2404 (1992).
[9.13] N. Sakai, K. Ogasawara, K. Inoue, D. Ishihara, and M. Murakami, Fabrication
of melt-processed RE-Ba-Cu-O bulk superconductors with high densities, IEEE
Trans. Appl. Supercond. 11, 3509 (2001).
[9.14] Masaru Tomita and Masato Murakami, High-temperature superconductor bulk
magnets that can trap magnetic elds of over 17 tesla at 29 K, Nature 421, 517
(2003).
[9.15] Yukikazu Iwasa, Seung-yong Hahn, Masaru Tomita, Haigun Lee, and Juan
Bascunan, A persistent-mode magnet comprised of YBCO annuli, IEEE Trans.
Appl. Superconduc. 15, 2352 (2005).
[9.16] M. Konishi, S. Hahakura, K. Ohmatsu, K. Hayashi, K. Yasuda, HoBCO thin lms
for SN transition type fault current limiter, Physica C 412414, 1056 (2004).
[9.17] Marty Rupitch (personal communication, 2007).
[9.18] Y. Sakai, K. Inoue, T. Asano, and H. Maeda, Development of a high strength,
high conductivity copper-silver alloy for pulsed magnets, IEEE Trans. Magn. 28,
888 (1992).
[9.19] H. Fujishiro and S. Kobayashi, Thermal conductivity, thermal diusivity and
thermoelectric power in Sam-based bulk superconductors, IEEE Trans. Appl.
Supercond. 12, 1124 (2002).
SOLENOID EXAMPLES, HTS MAGNETS & CONCLUDING REMARKS 591
[9.20] Hiroyuki Fujishiro, Tetsuo Oka, Kazuya Yokoyama, Masahiko Kaneyama, and Ko-
shichi Noto, Flux motion studies by means of temperature measurement in magne-
tizing processes for HTSC bulks, IEEE Trans. Appl. Supercond. 14, 1054 (2004).
[9.21] K. Okuno, K. Sawa and Y. Iwasa, Performance of the HTS bulk magnet in cry-
ocooler system with cyclic temperature variation, Physica C: Supercond. 426
431, 809 (2005).
[9.22] N. Koshizuka, R&D of superconducting bearing technologies for ywheel energy
storage systems, Advances in Superconductivity XVIII (Elsevier, 2006), 1103.
[9.23] Christopher Rey (private communication, 2007).
Papers Cited in Table 9.7: Electric Power
General Overview
[9.24] Mario Rabinowitz, The Electric Power Research Institutes role in applying super-
conductivity to future utility systems, IEEE Trans. Magn. MAG-11, 105 (1975).
[9.25] Paul M. Grant Superconductivity and electric power: promises, promises ... past,
present and future, IEEE Trans. Appl. Superconduc. 7, 112 (1997).
[9.26] William V. Hassenzahl, Superconductivity, an enabling technology for 21st cen-
tury power systems?, IEEE Trans. Appl. Superconduc. 11, 1447 (2001).
[9.27] Donald U. Gubser, Superconductivity: an emerging power-dense energy-ecient
technology, IEEE Trans. Appl. Superconduc. 14, 2037 (2004).
[9.28] Osami Tsukamoto, Roads for HTS power applications to go into the real world:
cost issues and technical issues, Cryogenics 45, 3 (2005).
[9.29] Alex P. Malozemo, The new generation of superconductor equipment for the
electric power grid, IEEE Trans. Appl. Superconduc. 16, 54 (2006).
FusionTore Supra
[9.30] B. Turck, Six years of operating experience with Tore Supra, the largest Tokamak
with superconducting coils, IEEE Trans. Magn. 32, 2264 (1996).
[9.31] J.L. Duchateau and B. Turck, Application of superuid helium cooling techniques
to the toroidal eld systems of tokamaks, IEEE Trans. Appl. Superconduc. 9, 157
(1999).
FusionLarge Helical Device (LHD)
[9.32] T. Satow, N. Yanagi, S. Imagawa, H. Tamura, K. Takahata, T. Mito, H. Chikara-
ishi, S. Yamada, A. Nishimura, R. Maekawa, A. Iwamoto, N. Inoue, Y. Nakamura,
K. Watanabe, H. Yamada, A. Komori, I. Ohtake, M. Iima, S. Satoh, O. Motojima,
and LHD Group, Completion and trial operation of the superconducting magnets
for the Large Helical Device, IEEE Trans. Appl. Superconduc. 9, 1008 (1999).
[9.33] S. Imagawa, N. Yanagi, H. Sekiguchi, T. Mito, and O. Motojima, Performance of
the helical coil for the Large Helical Device in six years operation, IEEE Trans.
Appl. Superconduc. 14, 629 (2004).
[9.34] S. Imagawa, T. Obana, S. Hamaguchi, N. Yanagi, T. Mito, S. Moriuchi, H. Seki-
guchi, K. Ooba, T. Okamura, A. Komori, and O. Motojima, Results of the ex-
citation test of the LHD helical coils cooled by subcooled helium, IEEE Trans.
Appl. Superconduc. 18, 455 (2008).
FusionEAST
[9.35] Peide Weng, Qiuliang Wang, Ping Yuan, Qiaogen Zhou, and Zian Zhu, Recent
development of magnet technology in China: Large devices for fusion and other
applications, IEEE Trans. Appl. Superconduc. 16, 731 (2006).
592 CHAPTER 9REFERENCES
Generator [LTS]
[9.59] J.L. Smith, Jr., J.L. Kirtley, Jr., P. Thullen, Superconducting rotating machines,
IEEE Trans. Magn. MAG-11, 128 (1975).
[9.60] C.E. Oberly, Air Force application of lightweight superconducting machinery,
IEEE Trans. Magn. MAG-13, 260 (1977).
594 CHAPTER 9REFERENCES
SMES/Flywheel [HTS]
[9.81] P. Stoye, G. Fuchs, W. Gawalek, P. Gornert, A. Gladun, Static forces in a super-
conducting magnet bearing, IEEE Trans. Magn. 31, 4220 (1995).
[9.82] P. Tixador, P. Hiebel, Y. Brunet, X. Chaud, P. Gautier-Picard, Hybrid supercon-
ducting magnetic suspensions, IEEE Trans. Magn. 32, 2578 (1996).
[9.83] S.S. Kalsi, D. Aized, B. Conner, G. Snitchler, J. Campbell, R.E. Schwall, J. Kellers,
Th. Stephanblome, A. Tromm, P. Winn, HTS SMES magnet design and test
results, IEEE Trans. Appl. Superconduc. 7, 971 (1997).
[9.84] S. Ohashi, S. Tamura, and K. Hirane, Levitation characteristics of the HTSC-
permanent magnet hybrid ywheel system, IEEE Trans. Appl. Superconduc. 9,
988 (1999).
[9.85] Y. Miyagawa, H. Kameno, R. Takahata and H. Ueyama, A 0.5 kWh ywheel
energy storage system using a high-Tc superconducting magnetic bearing, IEEE
Trans. Appl. Superconduc. 9, 996 (1999).
[9.86] Shigeo Nagaya, Naoji Kashima, Masaharu Minami, Hiroshi Kawashima and
Shigeru Unisuga, Study on high temperature superconducting magnetic bear-
ing for 10 kWh ywheel energy storage system, IEEE Trans. Appl. Superconduc.
11, 1649 (2001).
[9.87] J.R. Fang, L.Z. Lin, L.G. Yan, and L.Y. Xiao, A new ywheel energy storage
system using hybrid superconducting magnetic bearings, IEEE Trans. Appl. Su-
perconduc. 11, 1657 (2001).
[9.88] Yevgeniy Postrekhin, Ki Bui Ma and Wei-Kan Chu, Drag torque in high Tc
superconducting magnetic bearings with multi-piece superconductors in low speed
high load applications, IEEE Trans. Appl. Superconduc. 11, 1661 (2001).
[9.89] Thomas M. Mulcahy, John R. Hull, Kenneth L. Uherka, Robert G. Abboud, John
J. Juna, Test results of 2-kWh ywheel using passive PM and HTS bearings,
IEEE Trans. Appl. Superconduc. 11, 1729 (2001).
[9.90] Amit Rastogi, David Ruiz Alonso, T.A. Coombs, and A.M. Campbell, Axial and
journal bearings for superconducting ywheel systems, IEEE Trans. Appl. Super-
596 CHAPTER 9REFERENCES
Transmission [LTS]
[9.172] R.L. Garwin and J. Matisoo, Superconducting lines for the transmission of large
amounts of electrical power over great distances, Proc. IEEE 55, 538 (1967).
602 CHAPTER 9REFERENCES
Transmission [HTS]
[9.180] J.W. Lue, M.S. Lubell, E.C. Jones, J.A. Demko, D.M. Kroeger, P.M. Martin, U.
Sinha, and R.L. Hughey, Test of two prototype high-temperature superconduct-
ing transmission cables, IEEE Trans. Appl. Superconduc. 7, 302 (1997).
[9.181] M. Leghissa, J. Rieger, H.-W. Neumuller, J. Wiezoreck, F. Schmidt, W. Nick, P.
van Hasselt, R. Schroth, Development of HTS power transmission cables, IEEE
Trans. Appl. Superconduc. 9, 406 (1999).
[9.182] Y.B. Lin, L.Z. Lin, Z.Y. Gao, H.M. Wen, L. Xu, L. Shu, J. Li, L.Y. Xiao, L. Zhou,
and G.S. Yuan, Development of HTS transmission power cable, IEEE Trans.
Appl. Superconduc. 11, 2371 (2001).
[9.183] J.P. Stovall, J.A. Demko, P.W. Fisher, M.J. Gouge, J.W. Lue, U.K. Sinha, J.W.
Armstrong, R.L. Hughey, D. Lindsay, and J.C. Tolbert, Installation and opera-
tion of the Southwire 30-meter high-temperature superconducting power cable,
IEEE Trans. Appl. Superconduc. 11, 2467 (2001).
[9.184] D.W.A. Willen, F. Hansen, C.N. Rasmussen, M. Daumling, O.E. Schuppach, E.
Hansen, J. Baerentzen, B. Svarrer-Hansen, Chresten Traeholt, S.K. Olsen, C. Ra-
mussen, E. Veje, K.H. Jensen, J. stergaard, S.D. Mikkelsen, J. Mortensen, M.
Dam-Andersen, Test results of full-scale HTS cable models and plans for a 36 kV,
2 kArms utility demonstration, IEEE Trans. Appl. Superconduc. 11, 2473 (2001).
[9.185] Jeonwook Cho, Joon-Han Bae, Hae-Jong Kim, Ki-Deok Sim, Ki-Chul Seong,
Hyun-Man Jang, and Dong-Wook Kim, Development and testing of 30 m HTS
power transmission cable, IEEE Trans. Appl. Superconduc. 15, 1719 (2005).
[9.186] Shinichi Mukoyama, Noboru Ishii, Masashi Yagi, Satoru Tanaka, Satoru Maru-
yama, Osamu Sato, and Akio Kimura, Manufacturing and installation of the
worlds longest HTS cable in the Super-ACE project, IEEE Trans. Appl. Super-
conduc. 15, 1763 (2005).
[9.187] Ying Xin, Bo Hou, Yanfang Bi, Haixia Xi, Yong Zhang, Anlin Ren, Xicheng Yang,
Zhenghe Han, Songtao Wu, and Huaikuang Ding, Introduction of Chinas rst
live grid installed HTS power cable system, IEEE Trans. Appl. Superconduc. 15,
SOLENOID EXAMPLES, HTS MAGNETS & CONCLUDING REMARKS 603
Detector Magnets
[9.285] P.H. Eberhard, M.A. Green, W.B. Michael, J.D. Taylor and W.A. Wenzel, Tests
on large diameter superconducting solenoids designed for colliding beam acceler-
ators, IEEE Trans. Magn. MAG-13, 78 (1977).
[9.286] H. Desportes, J. Le Bars, and G. Mayayx, Construction and test of the CELLO
thin-walled solenoid, Adv. Cryo. Engr. 25, 175 (1980).
[9.287] W.V. Hassenzahl, Quenches in the superconducting magnet CELLO, Adv.
Cryo. Engr. 25, 185 (1980).
[9.288] Stefan Wipf, Superconducting magnet system for a 750 GeV MUON spectrom-
eter, IEEE Trans. Magn. MAG-17, 192 (1981).
[9.289] R.W. Fast, E.W. Bosworth, C.N. Brown, D.A. Finley, A.M. Glowacki, J.M. Jag-
ger and S.P. Sobczynski, 14.4 m large aperture analysis magnet with aluminum
coils, IEEE Trans. Magn. MAG-17, 1903 (1981).
[9.290] R. Bruzzese, S. Ceresara, G. Donati, S. Rossi, N. Sacchetti, M. Spadoni, The
aluminum stabilized Nb-Ti conductor for the ZEUS thin solenoid, IEEE Trans.
Magn. 25, 1827 (1989).
[9.291] A. Bonito Oliva, O. Dormicchi, M. Losasso, and Q. Lin, Zeus thin solenoid: test
results analysis, IEEE Trans. Magn. 27, 1954 (1991).
[9.292] F. Kircher, P. Bredy, A. Calvo, B. Cure, D. Campi, A. Desirelli, P. Fabbricatore,
S. Farinon, A. Herve, I. Horvath, V. Klioukhine, B. Levesy, M. Losasso, J.P. Lot-
tin, R. Musenich, Y. Pabot, A. Payn, C. Pes, C. Priano, F. Rondeaux, S. Sgobba,
Final design of the CMS solenoid cold mass, IEEE Trans. Appl. Superconduc.
10, 407 (2000).
[9.293] A. Dael, B. Gastineau, J.E. Ducret, and V.S. Vysotsky, Design study of the
superconducting magnet for a large acceptance spectrometer, IEEE Trans. Appl.
Superconduc. 12, 353 (2002).
[9.294] S. Mizumaki, Y. Makida, T. Kobayashi, H. Yamaoka, Y. Kondo, M. Kawai,
Y. Doi, T. Haruyama, S. Mine, H. Takano, A. Yamamoto, T. Kondo, and H. ten
Kate, Fabrication and mechanical performance of the ATLAS central solenoid,
IEEE Trans. Appl. Superconduc. 12, 416 (2002).
610 CHAPTER 9REFERENCES
617
618 APPENDIX 1A
Electromagnetic
1 gauss 104 T
1 oersted 250/ A/m
3
1 emu/cm 1000 A/m
Pressure
1 mmHg (1 torr) 133 Pa
1 atm (760 torr) 101 kPa
1 bar (750 torr) 0.1 MPa
1 psi (52 torr) 6.9 kPa
Viscosity
1 poise 0.1 Pa s (0.1 kg/m s)
Temperature
0 C 273 K
1 eV 11600 K
Mass
1 lb 0.45 kg
1 metric ton 1000 kg
Dimension
1 in 25.4 mm
1 French league 4 km
1 liter (1000 cm3 ) 0.001 m3
1 ft3 (28.3 liter) 0.0283 m3
* Values in italics are approximate.
APPENDIX 1B
UNIFORM-CURRENT-DENSITY SOLENOIDAL COIL
FIELD ERROR COEFFICIENTS
n = 2 : M2 = 2; f2 (, ) = 1
n = 3 : M3 = 6; f3 (, ) = 2 +4 2
619
620 APPENDIX IB
Most thermodynamic property values given here (and generally given only up to
three signicant gures) are for ballpark computation and rst-cut design.
For hydrogen (H2 ), properties of normal hydrogen (n-H2 ) are listed. As for
enthalpy, it is its dierence between two temperatures or at a phase transition
that matters; i.e., an enthalpy is not always set to 0 at 0 K as is the case with
the specic heat from which the enthalpy vs. temperature data are computed.
Therefore, enthalpy values given here may numerically dier from those given in
other sources, but the dierence agrees reasonably well.
Symbols: Tb , Tct , Tm , Ts , and T , respectively, boiling, critical, melting, satu-
ration, and triple temperatures [K]; Ps : saturation pressure [torr/atm]; : den-
sity [kg/m3 ]; h: enthalpy:[kJ/kg]; hL : heat of fusion (at Tm or T ) or vapor-
ization (at Tb ) [kJ/kg]; h : volumetric heat of vaporization [J/cm3 ]; k: thermal
conductivity [W/m K]; : viscosity [Pa s (pascal second)]; Pr: Prandtl number.
621
622 APPENDIX II
Table A2.2: Density and Enthalpy Data of Solid Cryogens at 1 bar*/1 atm
Neon (1 bar) Nitrogen (1 atm) Argon (1 bar)
T [K] [kg/m3 ] h [kJ/kg] [kg/m3 ] h [kJ/kg] [kg/m3 ] h [kJ/kg]
4 1507 0.02 1032 0.008 1771 0.01
5 1507 0.09 1032 0.021 1771 0.015
6 1506 0.10 1032 0.044 1771 0.024
8 1505 0.31 1032 0.15 1771 0.08
10 1503 0.75 1031 0.38 1771 0.20
12 1499 1.43 1031 0.80 1771 0.42
15 1490 2.76 1029 1.85 1769 0.96
20 1466 6.78 1027 4.74 1765 2.20
22 1452 8.84 1025 6.29 1763 2.80
24.56 1429 12.0 1022 8.51 1760 3.90
1240 24.2
30 1016 14.5 1752 6.12
35.61 1002 22.7 1742 8.90
994 30.8
40 989 37.1 1731 11.3
45 981 44.3 1725 14.2
50 970 52.1 1715 17.3
60 949 68.9 1691 23.9
946 74.3
63.16 877 100.0 1683 25.5
70 1664 31.1
80 1633 39.1
83.81 1619 42.3
71.6
* 1 bar = 750 torr = 0.987 atm.
Liquid phase.
Solid-to-solid phase transition; nitrogen expands and absorbs an energy density of 8.17
103 [kJ/kg].
624 APPENDIX II
T p* h
[K] [torr/atm] [kg/m3 ] [kJ/kg]
1.50 3.53 145 0.15 0.26 22.61
1.55 4.47 145 0.19 0.32 22.85
1.60 5.59 145 0.23 0.39 23.09
1.65 6.90 145 0.27 0.48 23.33
1.70 8.45 145 0.33 0.58 23.56
1.75 10.2 145 0.38 0.70 23.79
1.80 12.3 145 0.45 0.84 24.02
1.85 14.6 145 0.52 1.00 24.25
1.90 17.2 145 0.60 1.18 24.47
1.95 20.2 145 0.69 1.38 24.69
2.00 23.4 146 0.78 1.63 24.91
2.05 27.0 146 0.89 1.92 25.13
2.10 31.0 146 0.99 2.23 25.33
T : 2.18 38.0 146 1.18 25.41
2.20 39.9 146 1.24 3.28 25.51
2.30 50.4 146 1.50 3.57 25.91
2.40 62.5 145 1.81 3.82 26.30
2.50 76.6 145 2.14 4.05 26.68
2.60 92.6 144 2.52 4.27 27.04
2.80 133 142 3.40 4.73 27.71
3.00 182 141 4.46 5.23 28.33
3.25 257 139 6.08 5.93 28.99
3.50 352 136 8.09 6.72 29.52
3.75 470 133 10.5 7.62 29.91
4.00 615 129 13.6 8.65 30.12
4.22 760/1.00 125 16.9 9.71 30.13
4.30 1.07 124 18.2 10.1 30.1
4.40 1.17 121 20.2 10.7 30.0
4.50 1.28 119 22.1 11.3 29.8
4.75 1.58 112 28.7 13.0 29.0
5.00 1.93 101 39.3 15.4 27.3
Tct : 5.20 2.24 69.6 21.4
* In torr below 760 torr (1 atm) and in atm above 760 torr.
Italics are for the vapor phase.
THERMODYNAMIC PROPERTIES OF CRYOGENS 625
Table A2.4b: Density and Enthalpy Data of Nitrogen at 1, 15, and 20 atm
1 atm 15 atm 20 atm
3
T [K] [kg/m ] h [kJ/kg] [kg/m3 ] h [kJ/kg] [kg/m3 ] h [kJ/kg]
64 865 1.82 867 2.97 897 3.39
65 861 3.87 863 5.02 864 5.43
66 857 5.93 859 7.06 860 7.47
67 853 7.99 855 9.12 856 9.52
68 849 10.1 851 11.2 852 11.6
69 844 12.1 847 13.2 848 13.6
70 840 14.2 843 15.3 844 15.7
71 836 16.3 839 17.3 839 17.7
72 832 18.3 834 19.4 835 19.8
73 827 20.4 830 21.4 831 21.8
74 823 22.5 826 23.5 827 23.9
75 818 24.5 821 25.5 822 25.9
76 814 26.6 817 27.6 818 28.0
77 809 28.6 812 29.6 813 30.0
807 29.4
77.364 4.60 228.7 811 30.4 812 30.7
79 4.50 230 803 33.7 804 34.1
80 4.44 232 799 35.8 800 36.1
85 4.15 237 775 46.0 776 46.3
90 3.90 242 750 56.2 752 56.5
95 3.68 248 723 66.8 725 67.0
100 3.48 253 694 77.8 697 77.9
105 3.31 258 661 89.6 665 89.5
108 3.21 262 639 97.1 644 96.8
617 104.0
110.60 3.13 264 6.51 237.0 623 104
112 3.09 266 6.29 240 611 107
114 3.04 268 6.01 243 592 113
571 118.6
115.823 2.99 270 5.79 246 9.22 232.6
118 2.93 272 5.56 250 8.59 238
120 2.88 274 5.37 253 8.13 242
125 2.76 280 4.97 260 7.28 252
130 2.65 285 4.65 267 6.67 260
135 2.55 290 4.37 274 6.19 268
140 2.46 295 4.14 281 5.81 275
150 2.29 306 3.76 293 5.20 288
200 1.71 358 2.64 351 3.56 349
250 1.37 410 2.07 406 2.77 404
275 1.24 436 1.87 433 2.50 431
300 1.14 462 1.71 459 2.28 458
628 APPENDIX II
Table A2.6: Density and Enthalpy Data of Neon at 1, 10, and 20 atm
1 atm 10 atm 20 atm
3
T [K] [kg/m ] h [kJ/kg] [kg/m3 ] h [kJ/kg] [kg/m3 ] h [kJ/kg]
1207 0
27.10 9.58 85.8 1213 0.43 1218 0.97
28 9.22 86.9 1197 2.12 1203 2.63
29 8.85 88.1 1179 4.02 1185 4.50
30 8.52 89.2 1160 5.97 1167 6.42
32 7.94 91.5 1119 10.06 1128 10.4
33 7.68 92.5 1097 12.21 1108 12.5
34 7.43 93.6 1074 14.43 1086 14.7
36 6.99 95.8 1023 19.13 1038 19.2
37.62 6.67 97.5 974.5 23.31 994.6 23.1
88.2 86.80
40 6.25 100 77.0 91.3 916.9 29.7
787.3 39.1
42.34 5.89 103 69.5 95.0 212.9 78.6
44 5.66 104 65.2 97.4 172.1 85.5
46 5.41 106 61.0 100 148.9 90.7
48 5.18 109 57.4 103 134.0 94.8
50 4.96 111 54.3 105 123.1 98.3
55 4.50 116 48.0 111 104.1 106
60 4.12 121 43.2 117 91.5 113
65 3.80 126 39.4 123 82.1 119
70 3.52 131 36.2 129 74.7 125
75 3.29 137 33.6 134 68.7 131
80 3.08 142 31.3 140 63.7 137
85 2.90 147 29.3 145 59.4 143
90 2.74 152 27.6 150 55.7 148
95 2.59 157 26.1 156 52.5 154
100 2.46 163 24.7 161 49.7 159
110 2.24 173 22.4 172 44.9 170
120 2.05 183 20.5 182 40.9 181
130 1.89 193 18.9 193 37.7 192
140 1.76 204 17.5 203 34.9 202
150 1.64 214 16.3 214 32.5 213
160 1.54 224 15.3 224 30.5 223
180 1.37 245 13.6 245 27.1 244
200 1.23 266 12.2 266 24.3 265
220 1.12 286 11.1 286 22.1 286
240 1.02 307 10.2 307 20.3 307
260 0.945 328 9.4 328 18.7 328
280 0.878 348 8.7 348 17.4 349
300 0.819 369 8.2 369 16.2 369
630 APPENDIX II
In using property data presented in this Appendix, we should be well aware that
although most data are given to at least three signicant gures, implying these
particular data are quite accurate, they do not necessarily accurately represent
the property value of the specic material for which we seek information. Among
property data presented here, those that are subject to considerable degrees of
variation from one material batch to another include: thermal conductivity data
(Fig. A3.1); mechanical property data (Table A3.1); thermal expansion data (Table
A3.2), particularly of non-metals.
631
632 APPENDIX III
105
104
Sapphire
Diamond
Copper (100)
Cu (50)
3
10
In
Thermal Conductivity [W/m K]
Aluminum 1100F
Cu (100)
Aluminum 6063
Copper (50)
Indium Bronze Sn2PO
100
Sn60-Pb40
Cu90-Ni10
Sapphire
Silicon Bronze
Al 6063
Diamond Constantan &
Monel
Sn60-Pb40
10
Bronze Sn2PO Hastelloy C276
1
Brass (Cu70-Zn30)
Cu90-Ni10
Manganin
Neon (solid)
Nitrogen (solid)
Aramid
Kevlar 49
1 Stycast 2850
Resin
G-10
Thermal Conductivity [W/m K]
G-10 Teflon
Apiezon N
Mylar
Resin (Epitoke 828) Nylon
0.1
Apiezon N
Teflon
Mylar
Kevlar 49
He (gas)
Nylon
Styrofoam
102
Nitrogen (gas)
Kapton
103
1 2 5 10 20 50 100 200 300
Temperature [K]
Fig. A3.1b Thermal conductivity vs. temperature plots for nonconductive materials;
G-10 in the direction parallel to warp fibers and G-10 in the direction normal to cloth.
(For clarity, selected plots are in dashed lines.)
634 APPENDIX III
104
Stainless steel
Copper
Silver Aluminum
103
100
Heat Capacity [kJ/m3 K]
Silver
Stainless steel Copper
Aluminum
10
Al
Cu
Ag
0.1
0.01
1 2 5 10 20 50 100 200 300
Temperature [K]
Fig. A3.2a Heat capacity vs. temperature plots for aluminum, copper, silver,
and stainless steel. Converted from specific heat [J/kg K] data with constant
densities: Aluminum (2700 kg/m3 ); Copper (8960 kg/m3 ); Silver (10490 kg/m3 );
Stainless steel (7900 kg/m3 ).
PHYSICAL PROPERTIES OF MATERIALS 635
Copper
Aluminum
105
Solid N2
Solid Ne Silver
104
103
Enthalpy [kJ/m3 ]
Copper
Solid Ne
100
Stainless steel
10
Solid N2
Al
0.1 Cu
Ag
0.01
1 2 5 10 20 50 100 200 300
Temperature [K]
Fig. A3.2b Enthalpy vs. temperature plots for aluminum, copper, silver, stainless
steel. Converted from specific heat [J/kg K] data with constant densities: Alu-
minum (2700 kg/m3 ); Copper (8960 kg/m3 ); Silver (10490 kg/m3 ); Stainless steel
(7900 kg/m3 ). For densities of solid neon and nitrogen, at 1 bar, see Table A2.2.
636 APPENDIX III
Material T [K]
20 80 140 200 973
Aluminum 0.415* 0.391 0.312 0.201
Brass (70Cu-30Zn) 0.369 0.337 0.260 0.163 +1.30
Bronze 0.330 0.304 0.237 0.150 +1.33
Copper 0.324 0.300 0.234 0.148 +1.3
Nickel 0.224 0.211 0.171 0.111
Silver 0.409 0.360 0.270 0.171 +1.50
Stainless steel 304 0.306 0.281 0.222 1.40 +1.32
Epoxy 1.15 1.02 0.899 0.550
G-10 (warp) 0.241 0.211 0.165 0.108
G-10 (normal) 0.706 0.638 0.517 0.346
Phenolic (normal) 0.730 0.643 0.513 0.341
Teon (TFE) 2.11 1.93 1.66 1.24
Nb3 Sn 0.171 0.141 0.102 0.067 +0.55
Cu/NbTi wire 0.265 0.245 0.190 0.117
Solder (50Sn-50Pb) 0.510 0.365 0.229
* An aluminum bar 1-m long at 293 K shrinks by 4.15 mm when cooled to 20 K.
A common Nb3 Sn reaction temperature.
crosswise.
APPENDIX IV
ELECTRICAL PROPERTIES OF NORMAL METALS
A.F. Clark, G.E. Childs, and G.H. Wallace, Electrical resistivity of some engineering
alloys at low temperatures, Cryogenics 10, 295 (1970).
F.R. Fickett, Aluminum 1. A review of resistive mechanisms in aluminum, Cryo-
genics 11, 349 (1971).
F.R. Fickett, Oxygen-free copper at 4 K: resistance and magnetoresistance, IEEE
Trans. Magn. MAG-19, 228 (1983).
A.M. Hatch, R.C. Beals, and A.J. Soa, Bench-scale experiments on superconducting
material joints (Final Report, Avco Everett Research Laboratory, November 1976;
unpublished).
Y. Iwasa, E.J. McNi, R.H. Bellis, and K. Sato, Magnetoresistivity of silver over the
range 4.2159 K, Cryogenics 33, 836 (1993).
G.T. Meaden, Electrical Resistance of Metals (Plenum Press, New York 1965).
Materials at low temperatures, Eds. Richard P. Reed and Alan F. Clark (American
Society For Metals, 1983).
Jean-Marie Noterdaeme, Demountable resistive joint design for high current supercon-
ductors (Master of Science Thesis, Department of Nuclear Engineering, Massachusetts
Institute of Technology, 1978).
M.T. Taylor, A. Woolcock, and A.C. Barber, Strengthening superconducting com-
posite conductors for large magnet construction, Cryogenics 8, 317 (1968).
Guy Kendall White, Experimental Techniques in Low-Temperature Physics (Clarendon
Press, Oxford, 1959).
J.L. Zar, Electrical switch contact resistance at 4.2 K, Adv. Cryo. Engr. 13, 95
(1968).
639
640 APPENDIX IV
Stainless steel
0.5
Silver Aluminum
0.1
0.05
Normalized Resistivity [ (T )/(293 K)]
Copper
(50)
Aluminum
0.01
Copper (100)
0.005
Copper (200)
Silver
0.001
0.0005
0.0001
0 50 100 150 200 250 300
Temperature [K]
Fig. A4.1 Normalized zero-eld (T ) plots for conductive metals and stainless steel:
aluminum (99.99%, 26.44 n m); copper (RRR= 50, 17.14 n m); copper (100,
17.03 n m); copper (200, 16.93 n m); silver (99.99%, 16.00 n m). Stainless
steel (304L 725 n m) represents a typical alloy. For each metal, (T ) is normalized
to its zero-eld resistivity at 293 K.
ELECTRICAL PROPERTIES OF NORMAL METALS 641
400
4.2 K
300
RRR
19 K
200
100
42 K
77 K
0
0 2 4 6 8 10
B [T]
Fig. A4.2 RRR vs. magnetic eld plots: copper (solid) and aluminum (dotted),
both at 4.2 K, and silver (dash-dotted, with temperatures indicatedRRR = 735 at
0 T). At 77 K, eld-dependence of RRR for copper is similar to that of silver.
Metal Resistivity [n m]
T [K] 4 20 40 76 273 295*
Aluminum 1100 0.82 0.84 1.14 3.10 26.7 29.3 0.120
Aluminum 6061 13.8 13.9 14.1 16.6 39.4 41.9 0.116
Hastelloy C [a] 1230 1230 1236 1240 1268 1271 0.142
Copper OFHC 0.16 0.17 0.28 2.00 15.6 17.1 0.069
Brass (Cartridge) [b] 42 42 43 47 67 69 0.102
(Naval) [c] 33 33 35 39 63 66 0.122
(Red) [d] 27 27 28 30 47 49 0.086
Bronze (Aluminum) [e] 139 138 140 140 162 164 0.112
(Commercial) [f] 21 21 22 24 39 41 0.076
(Ni-Al) [g] 157 157 158 160 184 187 0.122
(Phosphor) [h] 86 86 88 89 105 107 0.081
90Cu-10Ni 167 167 168 169 184 186 0.076
70Cu-30Ni 364 365 366 367 384 386 0.086
Stainless steel 304L 496 494 503 514 704 725 0.964
310 685 688 694 724 873 890 0.756
316 539 538 546 566 750 771 0.934
* Based on Eq. A2.1 given at the top of this table.
[a]: 58Ni-16Mo-15Cr-5Fe-3W; [b]: 70Cu-30Zn; [c]: 59Cu-40Zn; [d]: 84Cu-15Zn
[e]: 91Cu-7Al; [f]: really brass, 89Cu-10Zn; [g]: 81Cu-10Al-5Ni; [h]: 94Cu-5Sn-0.2P
Heater Metal [n m]
20 K 76 K 295 K
Evanohm (75Ni-20Cr-2.5Al-2.5Cu) 1326 1328 1337
Chromel A (80Ni-20Cr) 1020 1027 1055
Constantan (60Cu-40Ni) 460 466 490
Cupron (55Cu-45Ni) 431 458 476
Manganin (86Cu-12Mn-2Ni) 425 444 475
Tophet A (80Ni-20Cr) 1087 1090 1120
NbTi 600
* Data for brass, bronze, and stainless steel are given in Table A4.2.
In the normal state.
APPENDIX V
PROPERTIES OF SUPERCONDUCTORS
643
644 APPENDIX V
Nb3 Sn
Critical Current Density, Jc (b, t,) The critical current density as a function
of reduced eld, b B/Bc2 (T ), t, and strain , Jc (b, t, ) [A/m2 ], is given by:
C
s()(1 t1.52 )(1 t2 )bp (1 b)q
Jc (b, t, ) = (A5.3)
B
where C [T], p, and q are parameters. s() is a very complicated function of that
contains several parameters. The important point here is that Jc , as with that for
NbTi, decreases with magnetic eld as 1/B and its temperature dependence is
steeper than that for NbTi.
PROPERTIES OF SUPERCONDUCTORS 645
Table A5.1: Jc or Je Data: NbTi, Nb3 Sn, MgB2 (Wires); YBCO, Bi2223 (Tapes)
YBCO & Bi2223 /:* Field Parallel/Perpendicular to Tapes Broad Surface
Jc (NbTi, MgB2 ), Non-Copper Jc (Nb3 Sn), or Je (YBCO & Bi2223) [A/mm2 ]
T [K] B [T] 3 5 8 12 15 20 25 30
1.8 NbTi >5000 >4000 >3000 900 NA NA NA NA
Nb3 Sna) >2000 >2000 >2000 >2000 >2000 800 155 NA
4.2 NbTi >3000 2500 1000 NA NA NA NA NA
Nb3 Sna) >2000 >2000 >2000 >2000 1700 475 NA
Nb3 Snb) >2000 >2000 1400 840 475 130 NA
Nb3 Snc) 3350 2100 1250 660 375 100 NA
MgB2 1930 900 165 NA NA NA NA
YBCO 1350 1270 1155 980 905 810 NA NA
425 310 230 175 155 NA NA NA
Bi2223 900 900 900 800 700 590 490 400
560 530 500 485 465 455 440 435
10 Nb3 Snc) 1400 850 375 70 10
MgB2 1595 610 75 NA NA NA NA
Bi2223 725 725 725 630 560 450 350 300
505 455 390 350 325 285 265 245
15 MgB2 1160 280 20 NA NA NA NA NA
20 MgB2 585 80 NA NA NA NA NA
YBCO 130 90 65 NA NA NA NA
Bi2223 730 650 575 510 460 370 280 200
385 325 260 200 170 130 100 75
30 Bi2223 640 540 480 365 300 200 150 100
230 150 80 45
35 YBCO 80 55 35 NA NA NA NA
T [K] B [T] s.f. 0.1 0.3 0.5 0.8 1 3 5
50 Bi2223 760 760 590 525 455 420 250 165
760 485 305 215 140 105
65 YBCO 205 195 180 160 145 100 NA NA
205 115 80 55 45 20 NA NA
66 Bi2223 405 405 295 250 200 185 75 25
405 225 105 50
70 Bi2223 330 330 240 175 160 135 40
330 175 70 25
77 YBCO 100 95 85 75 60 55 NA NA
100 80 45 30 15
Bi2223 220 220 150 120 90 75
220 100 30
* Je (B, T )/Je (77 K, s.f.) and Je (B, T )/Je (77 K, s.f.) ratios for YBCO and Bi2223 may be
assumed constant for dierent values of Je (77 K, s.f.) = Je (77 K, s.f.) = Je (77 K, s.f.).
Data not available
Jc or Je too low for magnet applications.
Self eld: 0.1 T for parallel eld; 0.01 T for perpendicular eld.
a) Internal tin; b) ITER; c) ITER barrel.
646 APPENDIX V
A-15: The cubic crystalline structure of most intermetallic compound LTS, e.g., Nb3 Al,
Nb3 Sn, V3 Ga. Also known as the beta-tungsten (-W) structure.
* Includes terms discussed and mentioned only briey or not at all in the main text but which
are of general interest in superconducting magnet technology and its applications. Also many
oft-cited acronyms in the superconducting magnet community are briey described.
649
650 APPENDIX VI
Carnot cycle: A reversible thermodynamic cycle, composed of two adiabatic and two
isothermal processes, in which a working uid operates between two thermal reservoirs to
produce work or refrigeration most eciently. The French physicist and military engineer
N.L.S. Carnot (17961832) published the theory in 1824: Reexions sur la Puissance
Motrice du Feu (Reections on the Motive Power of Fire). See STIRLING CYCLE.
CEA (Commissariat a lEnergie Atomique): A French government research organization
with centers throughout France; its superconducting magnet and cryogenics activities are
at Saclay (30 km southwest of Paris) and Cadarache (40 km north of Aix-en-Provence).
CERN (Conseil Europeen pour la Recherche Nucleaire): Located outside Geneva on the
Swiss/French border, an international center for high-energy physics research. See LHC.
CIC (Cable-In-Conduit) conductor: A cable of transposed MULTIFILAMENTARY CON-
DUCTORs (usually NbTi or Nb3 Sn) encased in a high-strength conduit, through which
cooling uid, generally SUPERCRITICAL HELIUM, is forced.
CMS (Compact Muon Solenoid): A superconducting detector magnet (together with
ATLAS and two others) for LHC; with overall dimensions of 15-m diameter, 21.5-m long
and a mass of over 1.25106 kg, it generates a center eld of 4 T.
Coercive force: The magnetic eld required to demagnetize a magnetized material.
Coherence length: The distance over which the superconducting-normal transition
takes place; introduced by A.B. Pippard (Cambridge University) in the early 1950s.
Composite superconductor: A conductor with one or more strands or tapes of super-
conductor in a matrix of normal metal; may be a MULTIFILAMENTARY CONDUCTOR.
Cooper pair: The paired electrons responsible for superconductivity (BCS THEORY).
Copper-to-superconductor ratio: The volumetric ratio of copper to superconductor
in a COMPOSITE SUPERCONDUCTOR (like NbTi), or the ratio of copper to non-copper in
a composite superconductor (like Nb3 Sn) which includes other metals.
Coupling loss: AC LOSS generated by eld-induced currents between laments and/or
strands in a MULTIFILAMENTARY SUPERCONDUCTOR or CIC CONDUCTOR.
Coupling time constant: The predominant decay time constant of eld-induced cur-
rents in a MULTIFILAMENTARY SUPERCONDUCTOR or CIC CONDUCTOR.
Critical current, Ic : The maximum current a conductor can carry and still remain
superconducting at a given temperature and magnetic eld.
Critical current density, Jc : One of the three material-specic parameters that denes
the critical surface for superconductivity. In a TYPE II SUPERCONDUCTOR, Jc is sensitive
to metallurgical processing.
Critical eld, Hc : One of the three material-specic parameters that denes supercon-
ductivity; insensitive to metallurgical processing. In a TYPE II SUPERCONDUCTOR, there
are two critical elds: LOWER (Hc1 ) and UPPER (Hc2 ).
Critical state model: For magnetic behavior of a BEAN SLAB; in the critical state,
every part of the slab operates at its CRITICAL CURRENT DENSITY.
Critical temperature, Tc : The temperature that denes the transition from the super-
conducting to the normal state; distinguishes HTS from LTS.
Cryocirculator: A CRYOCOOLER equipped with two helium circulators that provide
high-pressure streams of helium gas, typically one at 80 K and the other at 20 K.
Cryocooled magnet: A magnet cooled by a CRYOCOOLER or CRYOGENERATOR.
Gradient coil: A coil that generates an axial H-eld that varies linearly with position.
Hall probe: A sensor for measuring magnetic eld. Based on the principle of the Hall
eect: its output voltage, for a given supply current, is proportional to eld strength.
He I: Another name for the non-superuid liquid of ordinary helium (He4 ).
He II: Another name for SUPERFLUID HELIUM.
He : A helium isotope with atomic weight 3; boils at 3.19 K at 1 atm. Naturally, 1/106
3
Hysteresis loss: An energy loss due to the hysteretic eect of a material property, e.g.,
magnetization of a TYPE II SUPERCONDUCTOR or ferromagnetic material.
Index number: The exponent n in the voltage (electric eld) vs. current (current den-
sity) relationship for a superconductor; typically, n is between 10 and 100.
Induction heating: Heating generated in conductive metal by a time-varying magnetic
eldcalled induction heating when desired and EDDY-CURRENT LOSS when it is not.
Internal diusion process: A modied BRONZE PROCESS for Nb3 Sn developed in 1974
by Y. Hashimoto of the Mitsubishi Electric Corp.
Irreversible eld: A magnetic eld above which a superconductor carries insucient
transport current to be useful for magnet operation.
Iseult: A French-German project at CEA Saclay to construct a 11.74-T (500-MHz)/900-
mm bore MRI superconducting magnet for cerebral imagingall NbTi, operated at 1.8 K
with a magnetic energy of 330 MJ; in Germany, INUMAC (Isolde, initially).
ITER (International Thermonuclear Experimental Reactor): A multi-nation project to
construct a break-even TOKAMAK, in the CEA Cadarache facilities, France.
Jelly-roll process: A process for making NbTi, Nb3 Sn, and other LTS, in which foiled
conductor ingredients are rolled to form the basic ingot. Developed in 1976 by W.K. Mc-
Donald of Teledyne Wah Chang; also used to manufacture some HTS.
Josephson eect: A quantum eect characterized by the tunnelling of superelectrons
through the insulator of a JOSEPHSON JUNCTION, and observable as current ow without
a driving potential; based on the theoretical work (1964) of B. Josephson.
Josephson junction: A device with two superconducting plates separated by an oxide
layer. Josephson junctions are used in SQUID and other micro-scale electronic devices.
J-T (Joule-Thomson) valve: A needle valve across which the working uid expands
adiabatically and isenthalpically; the expansion process is irreversible.
Kaiser eect: A mechanical behavior observed in a structure under cyclic loading in
which events, e.g., conductor motion and epoxy fracture in a superconducting magnet,
appear only when the loading exceeds the maximum level achieved in the previous cycle.
Discovered by J. Kaiser in the early 1950s. See AE (ACOUSTIC EMISSION).
Kapitza resistance: The thermal boundary resistance which occurs at the interface
when heat ows from a solid to liquid helium; discovered by P. Kapitza in 1941.
Kapton: Trade name for a polyimide lm; useful as an insulating material.
Kevlar: Trade name for a strong and lightweight para-aramid synthetic ber. Of three
varieties of Kevlar, 29, 49, and 149, 49 is used in most cryogenic applications.
Kohler plot: A plot that combines the eects of magnetic eld and temperature on the
electrical resistivity of a conductive metal at CRYOGENIC TEMPERATUREs.
KSTAR (Korean Superconducting TokAmak Reactor): A superconducting TOKAMAK
for fusion science and technology R&D, at the Korea Basic Science Institute in Taejon.
Lambda point: The temperature below which ordinary liquid helium (He4 ) becomes
SUPERFLUID HELIUM; it is 2.172 K at a pressure of 38 torr (0.050 atm).
Laser cooling: A technique that uses photons, an atom laser, to remove heat from
atoms to cool them to a very low temperaturenanokelvins and picokelvins.
Layer winding: A technique to fabricate a solenoidal coil by winding conductor helically
around a mandrel, one layer at a time.
654 APPENDIX VI
LHC (Large Hadron Collider): The largest particle accelerator and collider built to date,
situated in CERN with a main ring (nearly circular) of 8.5-km diameter. See ATLAS and
CMS; SSC and TEVATRON.
MPZ (Minimum Propagating Zone): The largest volume within a magnet winding that
can support Joule heating without growing.
MRI (Magnetic Resonance Imaging): Magnetic elds to create, through NMR, visual
images of the brain and other body parts for diagnostic purposes. See ISEULT.
GLOSSARY 655
Perovskite: The cubic structure of most HTS, ferroelectric, ferro- and antiferri-magnetic
materials. Named for the Russian minister Count L.A. von Perovski (17921856).
Persistent-mode: A mode for an energized superconducting magnet, decoupled from its
power source and shunted by a PCS, to maintain a constant eld. See DRIVEN-MODE.
Phenolic: A thermosetting resin, usually reinforced with laminae of linen, cotton, or
even paper; weaker but easier to machine than G-10.
Piezoelectric eect: The coupling of mechanical and electric eects in which a strain
in some crystals, e.g., quartz, induces an electric potential and vice versa; an AE sensor
utilizes this eect. Discovered in 1880 by P. Curie (18591906).
Poloidal magnet: A PULSE MAGNET that generates a eld in the axial (vertical) direc-
tion for plasma stabilization in a MAGNETIC CONFINEMENT-based fusion machine.
656 APPENDIX VI
ppm (parts per million): A dimensionless number to express a trace quantity of a sub-
stance or a deviation from a norm, e.g., a small change in eld strength.
Prandtl number: A dimensionless coecient equal to the ratio of kinematic viscosity
to thermal diusivity, a measure of the relative diusion rates of momentum and heat in
convective heat transfer. Named for L. Prandtl (18751953).
Premature quench: A QUENCH of a superconducting magnet below the designed op-
erating current; still occasionally occurs in some LTS-based ADIABATIC MAGNETs.
Pulse magnet: A magnet that generates a eld over a short duration, ranging from
microseconds to a fraction of a second.
Quadrupole magnet: A magnet that generates a linear gradient eld transverse to its
axis over the central region of its bore; focuses particles in particle accelerators.
Quench: The superconducting-to-normal transition; specically, the rapid irreversible
process in which a magnet or a part of a magnet is driven fully normal.
R3B-GLAD (Reactions with Relativistic Radioactive Beams of Exotic Nuclei-GSI Large
Acceptance Dipole): A CEA Saclay designed superconducting magnet system (six RACE-
TRACK MAGNETs), to be installed at GSI (Gesellschaft fur Schwerionenforschung, Darm-
stadt, Germany), to study the structure of exotic nuclei or the reaction mechanisms.
Racetrack magnet: A magnet resembling a racetrack, wound in a plane, each turn
having two parallel sides, joined by a semi-circle at each end.
React-and-wind: A coil-winding technique used for a superconductor, e.g., Nb3 Sn and
HTS, that has already been reacted; applicable only when winding-induced strains in the
superconductor can be kept below the superconductors strain tolerance limit.
Regenerator: A component in STIRLING and G-M CYCLEs, storing and releasing heat
during isochoric processes; for operation below 10 K, rare-earth materials proven useful.
Reynolds number: The dimensionless ratio of inertial forces to viscous forces; charac-
terizes the viscous ow of uids. Named for O. Reynolds (18421912).
RHIC (Relativistic Heavy Ion Collider): A superconducting particle accelerator and col-
lider in operation since 2000 at the Brookhaven National Laboratory, Long Island, NY.
RRR (Residual Resistivity Ratio): The ratio of a metals electrical resistivity at 273 K
to that below 10 K (often 4.2 K); indicative of the metals purity.
Rutherford cable: A at two-layer cable suitable for low AC LOSS applications, devel-
oped in the 1970s at the Rutherford Appleton Laboratory, England. Laid at an angle
to the conductor axis, MULTIFILAMENTARY CONDUCTOR strands in each layer alternate
on two sides of the cable, with a TWIST PITCH LENGTH dened by the tilt angle and the
cable width; some have a high-strength strip between the strand layers for reinforcement.
Saddle magnet: A magnet with windings that resemble a saddle; saddles may be
congured to produce the elds of a dipole (2); a quadrupole (4), a sextuple (6), . . .
Saturation magnetization: The maximum magnetization of a ferromagnetic material.
Search coil: A coil for measuring a magnetic eld; requires a very stable integrator to
convert the search coil output voltage to a voltage proportional to the eld strength.
Self-eld loss: A hysteresis AC LOSS in a superconductor due to the magnetic eld
generated by a transport current.
SFCL (Superconducting Fault Current Limiter): A superconducting version of an FCL,
a device to limit the current surge during a fault in a utility power line.
Shim coils: A set of corrective coils, superconducting or copper (operated at room-
temperature) or both, added to the main eld to make the resultant eld meet spatial
eld requirements. Sometimes ferromagnetic materials are used to achieve the same goal.
GLOSSARY 657
Skin depth: The distance from a metal surface at which the amplitude of a sinusoidally
time-varying magnetic eld, due to the SKIN EFFECT, is 0.37 (= 1/e) the surface eld.
Skin eect: A phenomenon of an induced surface current in a metal shielding most of
the interior of the metal from a time-varying magnetic eld. See SKIN DEPTH.
SMES (Superconducting Magnetic Energy Storage): The magnetic storage of electrical
energy in a superconducting magnet for power conditioning by an electric utility.
SOR (Synchrotron Orbital Radiation): A device to produce x-rays by means of acceler-
ating electrons in a magnetic eld, generated usually by superconducting magnets.
SQUID (Superconducting QUantum Interference Device): A device using the JOSEPH-
SON EFFECTto measure the smallest possible change in magnetic ux.
SSC (Superconducting Super Collider): Terminated in 1993, would have been the largest
(larger than LHC) particle accelerator and collider for high-energy physics research.
Stekly criterion: A design criterion for a CRYOSTABLE MAGNET, in which the cooling
ux matches the conductors full normal-state Joule heating ux.
Stellarator: A toroidal plasma fusion MAGNETIC CONFINEMENT machine, invented by
Lyman Spitzer (1914-1997) of the Princeton University; the rst Stellarator built in 1951
at the Princeton Plasma Physics Laboratory. See LHD and WENDELSTEIN 7-X (W7-X).
Stirling cycle: A cycle incorporating a REGENERATOR. Invented, in the late 1810s, by
a Scottish minister, R. Stirling (17901878); has the same eciency as CARNOT CYCLE.
Storage dewar: An insulated container of liquid CRYOGEN. See CRYOSTAT and DEWAR.
Stycast: Trade name for an epoxy resin. Stycast 2850 is used in cryogenic experiments
as a sealant and as a thermally conductive (among organic materials) adhesive.
Styrofoam: Trade name for polysterene foam; usable for a liquid-nitrogen CRYOSTAT.
Toroidal magnet: A magnet that generates a eld in the toroidal (azimuthal) direction
to conne a hot plasma; the superconducting version is DC.
Training: Behavior of a HIGH-PERFORMANCE MAGNET whose successive PREMATURE
QUENCHes gradually approach the design operating current.
Transfer line: A double-walled vacuum insulated line for transferring liquid helium
from a STORAGE DEWAR to a CRYOSTAT.
Transposition: In a transposed cable, every strand occupies every radial location in the
cables overall diameter over its repeat length.
Triple point: The equilibrium state for solid, liquid, and vapor phases. Triple points of
hydrogen (13.8033 K), neon (24.5561 K), oxygen (54.3584 K), argon (83.8058 K), mercury
(234.3156 K), and water (273.16 K) are xed points in the International Temperature
Scale of 1990 (ITS-90). Helium does not have a triple point.
Twist pitch length: The linear distance over which 1) a lament of a twisted MUL-
TIFILAMENTARY CONDUCTOR makes one complete spiral, or 2) a strand of transposed
cable completes its repeat pattern. See TRANSPOSITION.
Type I superconductor: A superconductor that exhibits the MEISSNER EFFECT up to
its critical eld; called a soft superconductor because of its low mechanical strength.
Type II superconductor: A superconductor exhibiting the MIXED STATE; also called
a hard superconductor; an alloy of lead-bismuth was the rst Type II superconductor,
discovered in 1930 by W. J. de Haas and J. Voogd of the Leiden University.
Upper critical eld, Hc2 : The magnetic eld at which a TYPE II SUPERCONDUCTOR
loses its superconductivity completely.
Vapor-cooled lead: See GAS (OR VAPOR)-COOLED LEAD.
von Mises stress: A measure of stress particularly appropriate for ductile materials.
principal stresses, 1 , 2 , and 3 , will begin to yield when its von Mises
A material with
stress, given by [(1 2 )2 +(2 3 )2 +(3 1 )2 ]/2, exceeds the YIELD STRESS.
Water-cooled magnet: A magnet, usually made of copper or copper alloys, cooled by
water, generally forced through the winding or through tubes in a cooling plate.
Wendelstein 7-X (W7-X): An R&D superconducting STELLARATOR, to be installed
in Greifswald, Germany, for the Max Planck Institute for Plasma Physics; named for the
Wendelstein peak, south of Munich, bordering Austria. See LHD.
Wind-and-react: A coil preparation technique consisting of two stages: winding of a
coil with an unreacted superconductor, followed by heat treatment of the wound coil;
used for a conductor such as Nb3 Sn when winding-induced strains likely would degrade
the superconductor. The process is more dicult than the REACT-AND-WIND process,
primarily because the high reaction temperature precludes the use of many materials,
e.g., organic insulators, in the winding before the heat treatment.
YBCO: An yttrium-based HTS (YBa2 Cu3 O7 ) with Tc = 93 K, discovered in 1987 by
groups, at the University of Alabama and the University of Houston, led by P.W. Chu.
Initially fabricated only in bulk; now also available as coated conductor (tape).
Yield stress: The stress at which a material begins to deform plastically, deviating from
the materials initial elastic, i.e., linear, stress-strain behavior. See VON MISES STRESS.
Youngs modulus: The materials stiness, equal to the ratio of stress to strain in its
elastic stress-strain range. Known by various terms, including modulus of elasticity.
APPENDIX VII
QUOTATION SOURCES AND CHARACTER IDENTIFICATION
Jonathan Livingston (Preface, p. vii)
A seagull in Richard Bachs Jonathan Livingston Seagull, A Story (Avon Books, 1970);
devotes his life perfecting, and teaching others, ying skills.
659
660 APPENDIX VII
661
662 INDEX
Davis, Helen 286 Discharge 249, 286, 449, 450, 453, 475,
dB/dt (-induced) heating 492, 512 504505, 511, 520, 549, 553, 584
DC response (of superconductors) 6 power 475
Decay rate 455, 492, 502, 520 slow 519520
Delay time 510, 554 low-resistance resistor 519
constant 406, 455, 552, 556 series of diodes 520
Detect-and-activate-the-heater 469, 504, time constant 476, 477, 510, 513
525, 526, 528 eective 476
Detect-and-dump 502, 510512, 526 voltage(s) 475, 480, 481, 502503, 511,
Detection of mechanical events 413 649, 651
AE/voltage technique 413 voltages across current leads 480
Detector (superconducting) magnet 176, Disconnectable leads See Current lead(s)
649, 650 Discovery of superconductivity 1
Dewar, Sir James 244 Dislocations (to create pinning centers ) 9
ask/storage 244, 651, 657, 658 Displacements and strains 99
Diamagnet(s; ic; ism) 3, 7, 27, 39, 291, 314, Dissipation density in adiabatic winding
320, 654 392
Diamond 490 dissipation-density limit vs. 394
thermal conductivity vs. T plot 632 Disturbance(s) 14, 58, 256, 258, 353358,
Dielectric breakdown of cryogens 480 390, 391, 399, 408, 410412, 471, 484,
Dierential, / 49, 50 586 See also Mechanical disturbance
annealed ingot iron, as-cast steel, va- spectra 356, 399
nadium permendur 50 Dittus-Boelter-Giarratano-Yaskin correla-
resistance 371, 372 tion (heat transfer) 379
thermal expansion 12, 526 Double-pancake (coils/magnets) 129, 134,
Diusion equations, magnetic and thermal 136137, 150, 651, 655
336 dissipation (vs. He II cooling ) 452
Diusion pump/cold trap 252 HTS magnet 193
Diusivities, magnetic & thermal, of stain- stability analysis 386
less steel and copper at 4 K and 80 K vs. layer-wound 136137
337 advantages & disadvantages 137
thermal (460 K ) of Cu, SN2 , SNe 255 Dresner, L. 381
Digital ux injector 458 Driven-mode magnet 21
Diode(s) 455457, 495, 496, 519, 520, 531, Dry (current ) lead(s) See Current lead(s)
547, 549 Dry (cryogen-free) magnet(s) 1, 2, 219,
thermometers 264, 266, 267 220, 224, 247, 651
Dipole(s) 71, 160, 171, 208, 410, 413, 588, Dual-stability regime (CIC conductors)
655 381
eld 31, 32, 44, 52, 54, 200 Dump(ed) 449, 484, 502, 551552, 651
far from cluster 5253 initiation delay 557
(inside) sphere 32 resistor 474, 502, 503, 510, 511, 513,
(outside) sphere 32 526, 547, 552, 651
spherical 52 design of 517518
iron magnet 54 (for) Hybrid III 518
(ideal) magnet(s)/winding(s) 42, 109, voltage 474
154158, 162, 175, 176, 567, 651 Dunk mode of cooling 229, 230
(in) atom smashers 42 Dynamic stability 337, 352, 353 See also
cumulative force 157 Stability
eld and force vectors 156
force density 157 E(k) See Complete elliptic integral(s)
inductance 158 Earth eld 22, 80, 242
(with) iron yoke 155 EAST 587, 651, 652, 657
magnetic energy 157 Eddy-current (energy) density
surface current density 156 loss 58, 65, 66, 407, 446, 649, 651, 653,
moment 54, 562, 565566 654
shell (main coil) 565 wire and tape (table ) 443
CASE STUDIES IN SUPERCONDUCTING MAGNETS 667
Eect of transport current on magnetiza- vs. magnetic energy density 411, 467
tion 316, 317 vs. temperature plots (Ag, Al, Cu, SN2 ,
Eective matrix resistivity 406, 407, 446 SNe, stainless steel) 635
Eciency Epoxy (resin)
generation of magnetic eld (as bonding agent ) 386
Bitter, Gaume, Kelvin, polyhelix, enthalpy data 637
uniform-current-density 125, 137 lling material (ller) 360, 500
refrigeration fracture (cracking)
Carnot 225 detection by AE 411, 413, 649
cryogenic 589, 650, 655 (source of dissipation) 391, 399, 413
liquid production 241 heat capacity data 354, 636
(superconducting magnet) 366, 399 -impregnated (impregnation)
Electric, energy 29, 521 solenoid/winding 191, 392, 408, 411,
eld 25, 26, 2832, 56, 57, 59, 65, 175, 491, 493, 651
331, 338, 339, 345, 370, 457, 653 NMR magnet 530
generators 2, 154 mechanical properties 638
permittivity, free space 25 (against ) positive radial stress 105
Electrical resistivity(ies) (compared with ) solid nitrogen 224
Ag-Au alloys, vs. T plots 284 thermal conductivity data 355 See also
copper and stainless steel 337, 355 Stycast, vs. temperature plot
dump resistor 517 thermal expansion data 471, 638
selected heater metals 642 Equal-area criterion 353, 369 See also
selected metals 642 Stability
solder alloys 410 Er3 Ni 224
(normalized) vs. T plots (Ag, Al, Cu, Essmann, U. 5
stainless steel) 640 Ethane (C2 H6 ) 8
Electromagnet(s) 7, 80, 84, 132, 561, 585 Evanohm, electrical resistivity data 642
Bitter 1, 2, 122
iron, eld limit 54 Fahrenheit, Gabriel Daniel 267
pole shape 54, 55 Faradays law 25, 26, 28
Electromagnetic quantities (SI units) 25 Fault current limiter(s) 492, 587, 588
Elliptic integral(s), complete See Complete Fault forces 136, 204
elliptic integral(s) Fault-mode axial forces 558560
Embossed radiation shields 247 Feynman, Richard P. 534, 660
Emissivity 248 Field(s) (chiey magnetic )
practical considerations 250 analysis 14, 34, 71, 73
Enabling technology 587 Maxwells equations (0 th and 1st or-
End eld of solenoid See Field(s) der) 28
End user (of electric power ) 587, 588 solenoidal coil 74
Energy margin 357, 381, 586, 651 See also error coecients 150, 619620
Stability nested-coil magnet 77
selected values for LTS and HTS 358 pancake-coil magnet 150153
Engineering critical current density 360 simple coils 78
Enthalpy superposition technique 130
helium at 1, 6, 10 atm 625 end 130, 131
Joule-Thomson process 241 o-center axial 130, 131
neon at 1, 10, 20 atm 629 -cooling (annulus magnet, HTS ) 584
nitrogen at 1, 15, 20 atm 627 decay rate, in NMR & MRI magnets 455
(at) saturation, argon 628 expansion in Cartesian coordinates 145
helium 624 factor, F (, ) 76, 116, 559
(normal) hydrogen 628 Bitter magnet, FB (, ) 123
neon 628 derivation of 116
nitrogen 626 far from four dipoles 5253
selected materialsCu, epoxy, Kapton, homogeneity 22, 74, 78, 124, 139, 142,
Ni, Teon, W 230 (only Cu ), 637 173, 258, 263, 530, 585
solid cryogens (SAr, SN2 , SNe) 623 line(s) 40, 45, 50, 74, 156, 160, 322, 583
668 INDEX
Field(s) (continuation) Fracture 411413, 649, 653 See also Epoxy
(and ) power (Bitter magnet ) 124 fracture (cracking)
quasi-static 28 (of) ller material and remedy 412
saturation 39 (of) impregnated lling material 410
-shielding of annulus magnet, HTS 584 Francis Bitter (National) Magnet Labora-
solutions from scalar potentials 30 tory/FB(N)ML 124, 191, 254, 387,
upper limit, (iron magnet ) 54 390, 493, 515, 530, 537
(water magnet ) 135 Freon-14 (CF4 ) 8
vs. power 118 Friction and Wear of Materials 412
Filament, decoupling 344 Fringing eld 171, 172, 253, 449, 450, 649,
fatigue (hot-cathode vacuum gauge) 253 650, 654
Film boiling 222, 358, 359, 367, 368, 377 (of) magnet 37, 45
Floating winding 412 Fujishiro, Hiroyuki 410
Florida-Bitter plate 122 Fusion (machine/project/reactor) 70, 170,
Florida State University 132 265, 447, 587, 588, 653655, 657
Flux (Type II superconductor ), bundles 6 magnets 2, 13, 381, 399, 587
ow 6, 48, 355 CIC conductors 378, 379
quanta 328 protection 511
Flux jump(s)/jumping 2, 313, 332, 342,
344, 347, 352, 353, 356, 358, 448, 652 G-10 652
criterion 336, 338 mechanical properties 638
(in) HTS 349 thermal conductivity
Fluxoids 6 T -averaged/ultimate strength 307
Flywheel 587, 588 vs. temperature plots 633
Force(s)/density 71 See also Axial force thermal contraction data 638
axial See Axial force GaA1As (thermometer) 264
body force 105, 191 signal level & sensitivity 267
hybrid magnet 204, 558560 GANDALF (code ) 13, 553, 557
active shield coil 563564 Garrett, M.W. 83, 88
fault/misalignment 204 Gaume coil 125
interaction force 134, 136 Gauss, Karl Friedrich 33, 80
Lorentz/magnetic 9, 59, 71, 7273, 98, law 25, 26
366, 399, 545, 654 Gavrilin, Andrey V. 553, 557
annulus magnet, HTS 583, 585 Generation & storage (electric power ) 587,
circulating proton 175, 176 588
(on) current leads 503 Generator(s) 2, 154, 162, 214, 587, 588,
dipole 154, 155, 156, 157 657
entire winding body 412, 413 Gerhold, J. 480
friction(al) 411 Ginzburg, Vitaly L. 6, 652
levitation 574576, 580 GLAG theory 2, 6, 652
lateral force 577578 Goodman, B.B. 5
quadrupole 159, 160, 161 Gorkov, Peter L. 6, 652
toroidal magnet 168, 169 Gradient 22, 30, 587
two-coil magnet 180 coil (magnet) 22, 138, 529, 652
radial 203 eld 138, 139, 144, 656
two-racetrack coil magnet 166167 slew rate (MRI ) 22
interaction forces 167 temperature (thermal) 233, 272, 337,
midplane See Axial force 471
(on) sphere, iron 200202 Grenoble Magnet Laboratory 132
magnetic 37
vs. mutual inductance 112, 114, 213 Hand-shake lap splice 408
Force/forced (ow of uid ) 122, 220, 378, Harmonic errors 81
492, 658 Hastelloy, electrical resistivity vs. temper-
-cooled 220, 221, 254 ature data 642
supercritical helium 358, 359, 378380, thermal conductivity vs. T plot 632
650 Heller, R. 480
CASE STUDIES IN SUPERCONDUCTING MAGNETS 669