Materiais Temperaturaws Criogênicas
Materiais Temperaturaws Criogênicas
Materiais Temperaturaws Criogênicas
3rd Law: C 0 as T 0
These two forms of the heat capacity are
related through the following thermodynamic
relation, 2
∂v ⎞ ∂p ⎞ Tvβ 2 1 ∂v ⎞ 1 ∂v ⎞
C p − Cv = −T ⎟ ⎟ = κ = − ⎟⎟ β =− ⎟
∂T ⎠ p ∂v ⎠T κ v ∂p ⎠T v ∂T ⎠ p
Isothermal Volume
Note: Cp – Cv is small except for expansivity
compressibility
gases, where ~ R = 8.31 J/mole K.
USPAS Cryogenics Short Course Boston, MA 6/14 to 6/18/2010 3
Heat Capacity of Solids (Lattice Contribution)
1
n(ω ) = hω
h = Planck’s constant = 6.63 x 10-34 J.s
2πk BT kB = Boltzmann’s constant = 1.38 x 10-23 J/K
e −1
K-l
3x
⎛T ⎞ D
x3
E ph = 9 RT ⎜⎜ ⎟⎟
⎝ θD ⎠
∫0 dx e x − 1
3x
⎛T ⎞ D
x 4e x
C ph = 9 R⎜⎜ ⎟⎟
⎝ θD ⎠
∫ dx (e
0
x
− 1)
2
and V ⎛ 8π m ⎞ 2 12 2
D(ε ) = ⎜ 2 ⎟⎟ ε
2 ⎜
2π ⎝ h ⎠
At low temperatures, T << εf/kB ~ 104 K
Ce ≈ π D (ε f )k B2T = γT
1 2
3
The electronic and phonon contributions to the heat
capacity of copper are approximately equal at 4 K
USPAS Cryogenics Short Course Boston, MA 6/14 to 6/18/2010 7
Summary: Specific Heat of Materials
General characteristics:
Specific heat decreases by
~ 10x between 300 K and
LN2 temperature (77 K)
Decreases by factor of ~
1000x between RT and 4 K
Temperature dependence
C ~ constant near RT
C ~ Tn, n ~ 3 for T < 100 K
C ~ T for metals at T < 1 K
T1 T~2 <
300
T1K D D-ΔD
ΔL
Linear expansion coefficient
1 ∂L ⎞ 1 For isotropic
α= ⎟ = β
L ∂T ⎠ p 3 materials
∂v ⎞ ∂s ⎞
⎟ = − ⎟⎟ → 0
∂T ⎠ p ∂p ⎠T T →0
(Third law: s 0)
USPAS Cryogenics Short Course Boston, MA 6/14 to 6/18/2010 10
Expansion coefficient for materials
A1
L
Assumptions:
Force balance
Composite is stress free at T0 F1 = σ 1 A1 = F2 = σ 2 A2
Materials remain elastic: σ = E yε ΔL ΔL
ε1 = −
No slippage at boundary L 1 L composite
Ends are free ΔL ΔL
ε2 = −
L composite L 2
⎡ ⎤
⎢ ΔL − ΔL ⎥
L L
σ 1 = E y1ε1 = E y1 ⎢ 1 2
⎥
⎢ ⎛1 + E y1 A1 ⎞⎥
⎜
⎢⎣ ⎝ E y 2 A2 ⎟⎠ ⎥⎦
V A
I
Heat generation by electrical conduction Q = I2R
V
R(T ) =
I
dR/dT ~ constant
Residual
Resistivity
R ~ constant
Example:
RRR = 100 ( ) = 1000
B = 10 T
ΔR/R = 3
R(10 T)/R(0 T) = 4
RRR (equival.) = 25
Carrier concentration
Energy gap
Impurity concentration
ρ ~ 10-4 to 107 Ω−m
Basis for low temperature
thermometry
Log R
T(K)
Q& = −k (T )A
dT
dx A
π 2 k B2
= f (T ) =
k L0 = Lorentz number = 2.443 x 10-8 WΩ/K2
T ≡ L0T
σ 3e 2
Copper
k~T
k ~ constant
Alloys
Non-crystalline
Non-metallics
Q Q
Heat transfer coefficient
ΔTc Q = hcAΔT
ideal real
USPAS Cryogenics Short Course Boston, MA 6/14 to 6/18/2010 30
Thermal contact conductance (conductive)
Contact point between two materials can produce significant thermal resistance
onds
b
t allic
M e
l)
(A
cts
a
nt
co
y
Dr
h~T
Example:
USPAS Cryogenics Short Course Boston, MA 6/14 to 6/18/2010 31
Contact conductance (Insulating)
ed
nd
Bo
cts
a
nt
co
y
Dr
h ~ T3
Resistive Material
V = IR; P = I2R
Copper and copper alloys V Rm
Aluminum and Al alloys
Stainless steel
I
Superconducting Materials
V = 0 below Ic
V ~ In above Ic
Transport properties depend on V (I/Ic)n
Temperature n ≈ 10-40
Ic
Magnetic field
Metallurgical processes
Strain in conductor I
“Critical Surface”:
Bi2Sr2Ca2Cu3Ox/Ag (A)
( C)
(F) ~0.5 mm
SuperPower
RRP Nb3Sn tape used in
Nb-Ti record breaking
Complied from
NHMFL insert
ASC'02 and
1000 ICMC'03
coil 2007
JE (A/mm²)
Engineering
USPAScritical current
Cryogenics density
Short in HTS and LTS
Course conductors.
Boston, Image
MA 6/14 to courtesy of Peter Lee,
6/18/2010
http://magnet.fsu.edu/~lee/plot/plot.htm (October 2008) 47
Stress effects in SC?
I
ε > 0; ε = 0
V
V
F I
Ey (NbTi) ≈ 82 GPa
σU (ε= 2%) ≈ 2200 MPa
Ey (NbTi) ≈ 82 GPa
σU (ε= 2%) ≈ 130 MPa
Composite is average between
Copper and NbTi
Nb3Sn
⎛ ΔL ΔL ⎞ ⎛ E Nb3 Sn EBr ⎞
σ Nb Sn =⎜ − ⎟⎜ ⎟ ABr
3
⎝ L Nb3 Sn L Br ⎠ ⎝ EA Nb 3Sn
+ EABr ⎠
≈ 1%
1
Normalized Ic [-]
0.8
Irreversible strain limit
0.6
0.4
Normalized Ic @ RT
0.2
Normalized Ic @ 77 K
0
0.0 60.0 120.0 180.0 240.0 300.0 360.0
Stress [MPa]
Critical stresses/strains ~ 200 MPa/0.4% (RT), 300 MPa/0.6% (77 K)