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4.15 DARCY FRICTION FACTOR, F 149

Figure 4-12-(continued)

where derive f D directly and thus requires an iterative solution (e.g., trial
u’ = kinematic viscosity =
viscosity -
-
E and error). Colebrook [9] also proposed a direct solution equation
density p that is reported [ 111 to have
D = Pipe internal diameter
u = Liquid velocity.
(7)
-2
Note that the term E / D is the relative roughness from Figure 4-13. f = 1.8 loglo (4-34)
Equation (4-33) is implicit in fo, as it cannot be rearranged to
150 FLUID FLOW

WI 4
I

Figure 4-13a Relative roughness of pipe materials and friction factors for complete turbulence. (ReprintdAdapted with permission from "Flow of
Fluids Through Valves, Fittings, and Pipe", Technical Paper No. 410, Crane Co. All rights reserved.)
 4.15 DARCY FRICTION FACTOR, f 151

Pipe Diameter, in inches

1 2 3 4 5 6 8 1 0 20 30 40 5060 80 100 200 300

. .07

. .06

. .05

.04

,035

.03

,025
0
a,
n
h
c
0)
.02 =
2

a,
+
-
a,
P

,014 0
Eo
8
LL
I
*-

,012

.01

,009

,008

Pipe Diameter, in millimetres - d

(Absolute Roughness E is in millimetres)

Figure 4-13b Relative roughness of pipe materials and friction factors for complete turbulence. (ReprintedAdapted with permission from “Flow of
Fluids Through Valves, Fittings, and Pipe”, Technical Paper No. 4 LOM, Crane Co. All rights reserved.)
152 FLUID FLOW

TABLE 4-2 Reynolds Number

Numerator Denominator
Reynolds Second Third Fourth Fifth
Number, Re Coefficient First Symbol Symbol Symbol Symbol Symbol

- ft ft/s 1b/ft3 Ib massifts -

DVPlP
124dvplz 124 in. ftis Ib/ft3 CP -
50.7Gpj d z 50.7 gpm Ib/ft3 - in. CP
6.32Wldz 6.32 Ibih - - in. CP
35.5Bpldz 35.5 bbl/h Ib/ft3 - in. CP
7,742dvlk 7,742 in. ftis - - CP
3,162Gldk 3,162 gpm - - in. CP
2,214Bfdk 2,214 bblih - - in. CP
22,735qpldz 22,735 ft3lS Ib/ft3 - in. CP
378.9Qpldz 378.9 ft3/min Ib/ft3 - in. CP

TABLE 4-3 Equivalent Roughness of Various Surfaces

Material Condition Roughness Range Recommended

Drawn brass, New 0.01-0.0015 m m 0.002 m m
copper, stainless (0.0004-0.00006 in.) (0.00008 in.)
commercial steel New 0.1-0.02 m m 0.045 mm
(0.004-0.0008 in.) (0.0018in.)
Light rust 1.O-0.15 m m 0.3 mm
(0.04-0.006 in.) (0.015 in.)
General rust 3.0-1 .O m m 2.0 m m
Iron Wrought, new 0.046 m m 0.046 m m
(0.002 in.) (0.002 in.)
Cast, new 1.0-0.25 m m 0.30 m m
(0.04-0.01 in.) (0.025 in.)
Galvanized 0.15-0.025 m m 0.15 m m
(0.006-0.001 in.) (0.006 in.)
Asphalt-coated 1.0-0.1 m m 0.15mm
(0.04-0.004 in.) (0.006 in.)
Sheet metal Ducts 0.1-0.02 m m 0.03 m m
Smooth joints (0.004-0.0008 in.) (0.0012 in.)
Concrete Very smooth 0.18-0.2 m m 0.04 m m
(0.007-0.001 in.) (0.0016 in.)
Wood floated, 0.8-0.2 m m 0.3 m m
brushed (0.03-0.007 in.) (0.012 in.)
Rough, visible 2.5-0.8 m m 2.0 m m
form marks (0.1-0.03 in.) (0.08 in.)
Wood Stave, used 1.0-0.25 m m 0.5 m m
(0.035-0.0 1 in.) (0.02 in.)
Glass or plastic Drawn tubing 0.01-0.0015 m m 0.002 m m
(0.0004-0.00006 in.) (0.00008 in.)
Rubber Smooth tubing 0.07-0.006 m m 0.01 m m
(0.003-0.00025 in.) (0.0004 in.)
Wire-reinforced 4.0-0.3 m m 1.0 m m
(0.15-0.01 in.) (0.04in.)
~

(Source: Darby [51.)

An explicit equation for calculating the friction factor ( f c ) as where

proposed by Chen [ 121 is
A =-&ID +
3.7
(E) 0.9

1 E = epsilon, absolute pipe roughness. in. (mm)

(4-35) D = pipe inside diameter, in. (mm).
 4.15 DARCY FRICTION FACTOR, F 153

-
BELL MOUTH
INLET OR REDUCER

NOTE K DECREASES WITH

INCREASING WALL THICKNESS OF
PIPE AND ROUNDING OF EDGES

h = K e FEET OF FLUID
2g

Figure 4-14a Resistance coefficients for fittings. (Reprinted by permission from Hydraulic Institute, Engineering Data Book, 1st ed., 1979,
Cleveland, OH.)

The relationship between Chen friction factor and Darcy fric- where
tion factor is fc = l / j f o .
1
Churchill [13] developed a single expression for the friction
factor in both laminar and turbulent flows as: (7/Re)' + 0.27 E / D
f=2[(A) 12
154 FLUID FLOW

GATE
JALVE

AND COUPLING

USED AS REDUCER Kz0.05- 2 0

SEE ALSO IllB-7
USED AS INCREASER LOSS IS UP
TO 40% MORE THAN THAT CAUSED
BY A SUDDEN ENLARGEMENT

SUDDEN ENLARGEMENT

SEE ALSO EQUATION (5)

IF A2=m SO THAT V2=0
VI
h=- FEET OF FLUID
2g
U I
V2
h= K- FEET OF FLUID
29

Figure 4-14b Resistance coefficients for valves and fittings. (Reprinted by permission from Hydraulic Institute, Engineering Data Book, 1st ed., 1979,
Cleveland, OH.)

Equation (4-36) is an explicit equation, and adequately satisfactory friction factor in comparison to others as shown in
represents the Fanning friction factor over the entire range of Table 4-4.
Reynolds numbers within the accuracy of the data used to
construct the Moody diagram, including a reasonable estimate
for the intermediate or transition region between laminar and 4*16 FRICTloNHEAD Loss (RESISTANCE) IN ‘IPEr
turbulent flow. FITTINGS, AND CONNECTIONS
Gregory and Fogarasi [I41 have provided a detailed review Friction head loss develops as fluids flow through the various
of other explicit equations for determining the friction factor, pipes, elbows, tees, vessel connections, valves, and so on. These
and concluded that Chen’s friction factor equation is the most losses are expressed as loss of fluid static head in feet (meter) of
 TABLE 4-4 Explicit Equations for Calculating t h e Friction Factor for Rough Pipe

Moody (1947) Wood (1966) Altshul (1975) Swamee and Jain (1976)

(i-+- 3°’25
1 6.97
f=0.11

i where
Jain (1976)

where
A, = 0.094 ( ; ) 0 ’ 2 2 5 +0.53 );(
E = absolute pipe roughness A, = 0.88 (
D = inside pipe diameter
Re = Reynolds number. A, = I5 2 (i)””‘
It is valid for Re > 10,000 and
< E/D i0.04

Churchill (1977) Chen (1979) Round (1980) Zigrang and Sylvester (1982)

12 1 5.0452 Re 1

[(k) + (A4+A,J3/’-
1 11’12
-=-21og
47
~ -
(3%5
_
Re
_ log
_A,
0.135Re(~/D)+6.5
log

where where where

16
A (E/ D) 1’1098 c/D 13
cn A, = A8=-+-
cn 2.8257 3.7 Re

Chen (1991) Zigrang and Sylvester (1982)

37,530
A 5 -( R e)
l6
1
- = -410g
8
(g - Re logA7
5’02
1
_-
8
- -410g (g - 5.02 log A,

where where
E/D 6.7 O.’
A7 = 3.7 + (z)
Harland (1983) Serghides (1984)

1 = -3.6log
-
8
[ + (g)] 1.1,

(
f = Ai,-

where
where
E= absolute pipe roughness
D = inside pipe diameter
A, = -2 log (g+ g)
Re = Reynolds number. A,, = - 2 1 0 9 ( % + 2.51A1,
7)

A, = -210g (g+ 7 ) 2.51A11

156 FLUID FLOW
fluid flowing. (Note: This is actually energy dissipated per unit of Substituting Eq. (4-42) into Eq. (4-40) gives
fluid mass.)
FLU PLQ
A P = 0.000668- = 0.000273 __
d2 d4
PRESSURE DROP IN STRAIGHT PIPE:
FLW (4-45)
INCOMPRESSIBLE FLUID =0.000034-, lb,/in2.
Pd4
The frictional resistance or pressure drop due to the flow in the
fluid, h,, is expressed by the Darcy equation: In SI units,

hf = f D (">
D
5 ft of fluid, resistance
2g'
(4-37) f --Ep
D-
64
Re
64,000~
dvp
(4-46)

fLV2 fLQ2
hf=0.1863- =0,0311- Substituting Eq. (4-46) into Eq. (4-41) gives
d ds
fLW2
WLQ FLW
=0.000483- p2 d5 ' ft
(4-38) A P = 0.32-F L U = 6.79 -= 113.2 -, bar (4-47)
d2 d4 P d4
where
Note: These values for hf and APf are the losses between (differ-
L = length of pipe, ft entials from) point (1) upstream to point (2) downstream, separated
D = pipe internal diameter, ft by a length, L. (Note that h, is a PATH function, whereas A P is a
f = Moody friction factor POINT function.) These are not absolute pressures, and cannot be
h, = loss of static pressure head due to fluid flow, ft. meaningful if converted to such units (but they can be expressed
as a pressure DIFFERENCE).
(Note: The units of d , u, Q, p are given above.)
Feet of fluid, h,, can be converted to pounds per square
In SI units,
inch by:

hf=51- f L v 2 =22,950- f L Q 2
d d5 h, = ____ ft for any fluid (4-48)
P
fLW2
= 6,376,000-, m (4-39)
p2d5 Referenced to water, convert psi to feet of water:
where
[(l lb/in.2)] (144in.*/ft2)
hf = = 2.31 ft (4-49)
L = length of pipe, m 62.3 lb/ft3
D = pipe internal diameter, m
f = Moody friction factor. For conversion, 1psi = 2.31 ft of water head.
(Note: The units of d , u, Q, p are given above.) This represents a column of water at 60"F, 2.31 ft high. The
The frictional pressure loss (drop) is defined by: bottom pressure is one pound per square inch (psi) on a gauge. The
pressure at the bottom as psi will vary with the density of the fluid.

Apf = Z
1
f D (2) p-,
V2

2gc
resistance loss, Ib,/in2. (4-40)
In SI units,

AP
hf = 10,200--, m (4-50)
In SI units, P

AP, = f o (k) p g , resistance loss, N/m2 (4-41)

where
A P = pressure difference (drop) in bar,
64 = -64p p = fluid density, kg/m3.

(s)
For Re < 2000, f,,= - (4-42)
Re 124dup For fluids other than water, the relationship is
h, = 0.0962 -
= 0.0393 @Q
Pd4 1psi = 2.31/(sp gr related to water), ft fluid (4-51)
=O.O049O-,ft.
FLW
p2d4
(4-43) In SI units,

In SI units, AP
h, = 10,200 m (4-52)
(sp gr related to water) '
k , =3263 ($) - = 69,220-P L Q
Pd4 With extreme velocities of liquid in a pipe, the downstream pressure
may fall to the vapor pressure of the liquid and cavitation with
=1,154,000-, F L w m (4.44) erosion will occur. Then the calculated flow rates or pressure drops
p2d4 are not accurate or reliable.
 4.20 L/D VALUES IN LAMINAR REGION 157
4.17 PRESSURE DROP IN FITTINGS, VALVES, Therefore, it can be seen that K , the loss coefficient for a given
AND CONNECTIONS fitting type divided by the Moody friction factor f is equal to the
INCOMPRESSIBLE FLUID ratio of the “length” of the fitting divided by its inside diameter, or

The resistance to flow through the various “piping” components

that make up the system (except vessels, tanks, pumps - items
which do not necessarily provide frictional resistance to flow) such
(k) (T)=
(4-57b)

as valves, fittings, and connections into or out of equipment (not the The K for a given fitting type is a direct function of the friction
loss through the equipment) are established by test and presented in factor, so that L I D for that type of fitting is the same dimensionless
the published literature, but do vary depending on the investigator. constant for all pipe diameters. The right ordinate of Figure 4-13
Resistance to fluid flow through pipe and piping components gives the Moody friction factor for complete turbulent flow as a
is brought about by dissipation of energy caused by (1) pipe compo- function of the inside pipe diameter for clean steel pipe. These
nent internal surface roughness along with the density and viscosity factors are used with the K values found in the Hydraulic Institute
of the flowing fluid, (2) directional changes in the system through or Crane data to determine LID. The L I D value multiplied by the
the piping components, (3) obstructions in the path to flow, and inside diameter* of the pipe’ gives the so-called equivalent length
(4) changes in system component cross section and shape, whether of pipe corresponding to the resistance induced by the fitting.
gradual or sudden. This equation defines the loss coefficient K by The Cameron tables show the L I D constant in addition to the
K values for a wide range of fittings for common pipe sizes. The
hf = K (u2/2g), ft (m) of the fluid flowing (4-53) fittings are shown as pictorial representations similar to Figure 4-6
to assist the user in selecting the fitting most nearly corresponding
to the actual one to be incorporated in the piping. Table 4-5 summa-
4.18 VELOCITY AND VELOCITY HEAD rizes the L I D constants for the various fittings discussed in the
Cameron Hydraulics Data book. The constants for valves are based
The average or mean velocity is determined by the flow rate divided on full-port openings in clean steel pipe.
by the cross section area for flow in feet per second (meters per
second), v. The loss of static pressure head due to friction of fluid
flow is defined by 4.20 LID VALUES IN LAMINAR REGION

h, = hf = v2/2g, termed velocity head, ft (m) (4-54) The L I D values in Table 4-5 are valid for flows in the transient
and turbulent regions. However, it has been found good practice
Note that the static reduction (friction loss or energy dissipated) to modify L I D for use in the laminar region where Re < 1000. In
due to fluid flowing through a system component (valve, fitting, this area, it is suggested that equivalent length can be calculated
etc.) is expressed in terms of the number of corresponding velocity according to the expression
head, using the resistance coefficient, K , in the equation above.
This K represents the loss coefficient or the energy dissipated in
terms of the number of velocity heads (lost) due to flow through the
respective system component. It is always associated with diameter
(k)‘,, = i&
(4-58)

for flow, hence velocity through the component. (Note: K can be However, in no case should the equivalent length so calculated be
defined on the basis of a specified velocity in Eq. (4-53), e.g., the less than the actual length of fitting. For various components’ K
velocity either into or out of a reducer, etc.). values, see Figures 4-14a-4-18, and Tables 4-6 and 4-7.
The pipe loss coefficient is related to (from) the Darcy friction Pressure drop through line systems containing more than
factor for straight pipe length by equation [4]: one pipe size can be determined by (a) calculating the pressure
drop separately for each section at assumed flows, or (b) deter-
mining the K totals for each pipe size separately using the 2
K = (h;) (4-55) or 3-K method. Flow then can be determined for a fixed head
system by
Head loss through a pipe, h,, is

(4-56)
In SI units,
Head loss through a valve, h,, is (for instance)

h L = K (v2/2g) (4-57)

Of course, by selecting the proper equation, flows for vapors and

4.19 EQUIVALENT LENGTHS OF FITTINGS gases can be determined in the same way, as the K value is for the
Instead of obtaining the friction loss directly and separately for fitting or valve and not for the fluid.
each fitting at each pipe size, the equivalent length of a fitting may
be found by equating Eq. (4-53) with Eq. (4-56) the Darcy friction
loss formula for fluids in turbulent flow, such that * The diameter for the conversion is usually taken as feet, but if K is divided
by 12, then L I D has the dimension of equivalent feet per inch of diameter.
V2 L v2 ‘Use of nominal pipe size instead of actual inside diameter is usually
h -K- (4-57a) sufficient for most engineering purposes.
f - 2g = f ( 5 ) 2g
158 FLUID FLOW

TABLE 4-5 Equivalent Length-to-Diameter Ratios for Fittings

[ L/dj [ L/dj
Fitting (Fully Open) ( d i n feet) d i n inches Remarks

Gate valve 8 0.67 Correct for constriction

Globe valve 340 28 Correct for constriction
Angle valve 55 4.6 Plug or Y type; correct for constriction
Angle valve 150 12.5 Globe type; correct for constriction
Ball valve 3 0.25 Correct for constriction
Butterfly valve 45 3.8 Correct for constriction
Plug valve 18 1.5 Correct for constriction
Three-way plug valve 30 2.5 Through flow; correct for constriction
Three-way plug valve 90 7.5 Branch flow; correct for constriction
Standard tee 20 1.7 Through flow
Standard tee 60 5.0 Branch flow
Standard 45" elbow 16 1.3
Standard 90" elbow 30 2.5 r i d = 0.5
Long-radius 90" elbow 16 1.3 r i d = 1.O
90" bend 20 1.7 r i d = 1.O
90" bend 12 1.o r i d = 2.0
90" bend 14 1.2 r i d = 4.0
90" bend 30 2.5 r / d = 10
90" bend 50 4.2 r / d = 20
30" miter bend 8 0.67
45" miter bend 15 1.25
60" miter bend 25 2.1
90" miter bend 60 5.0
close return bend 50 1.2
Stop check valve (vertical disk 400 33 Minimum velocity= 55/,,9; correct for
rise, straight flow) constriction
Stop check valve (vertical disk 200 16.7 Minimum velocity=75/fi; correct for
rise, right-angle flow) constriction
Stop check valve (disk at 45", 350 25 Minimum velocity=60/,,9; correct for
right-angle flow) constriction
Swing check valve 50 12.2 Minimum velocity= 50/fi; correct for
constriction

(Source: Adapted from Hydraulic Institute's Engineering Data Book.)

The head loss has been correlated as a function of the velocity 4.22 LAMINAR FLOW
head equation using K as the resistance (loss) coefficient in the
equation. When the Reynolds number is below a value of 2000, the flow
region is considered laminar. The pipe friction factor is defined as
given in Eq. (4-42).
Between Re of 2000 and 4000, the flow is considered unsteady
or unstable or transitional where laminar motion and turbulent
(4-61) mixing flows may alternate randomly [4]. K values can still be
calculated from the Reynolds number for straight length of pipe.
In SI units, and either the 2 or 3-K method for pipe fittings.

l,- KQ2 K q'

11 i -- K - = K-
2g 19.62
= 22.96-
d4
= 8265 x lo7-
d4 K=f ($) (4-55)
KW2
= 6377 p?dl. m (4-62)

(4-62a)
4.21 VALIDITY OF K VALUES
Equation (4-55) is valid for calculating the head loss due to valves and
and fittings for all conditions of flows: laminar, transition, and

g) (g)
turbulent [4]. The K values are a related function of the pipe system
component internal diameter and the velocity of flow for v2/2g. h, = (f , ft (m) fluid of pipe
The values in the standard tables are developed using standard
ANSI pipe, valves, and fittings dimensions for each schedule or
class [4]. The K value is for the sizehype of pipe, fitting, or valve
and not for the fluid, regardless of whether it is liquid or gas/vapor.
h, = K
(3
- , ft (m) fluid for valves and fittings (4-57)
 4.22LAMINAR FLOW 159
0.6
I I I

0.5

0.4

0.3

I
0.2

-
&
D
0.1
"

0.0
0 1 2 3 4

--
R
D
5

D
6 7 8

Note: 1. Use 0.00085 ft for &ID uncoated cast iron and cast steel elbows.
9 10

2. Not reliable when RID < 1 .O.

3. R= radius of elbow, ft

Figure 4-15a Resistance coefficients for 90" bends of uniform diameter for water. (Reprinted by permission from Hydraulic Institute, Engineering Datu
Book, 1st ed., 1979, Cleveland. OH.)

0.30

t
0.25

0.20

K 0.15

0.10

0.05

a" Lc
c

Figure 4-15b Resistance coefficients for bends of uniform diameter and smooth surface at Reynolds number = 2.25 x 105. (Reprinted by permission
from Hydraulic Institute, Engineering Dura Book, 1st ed., 1979. Cleveland, OH.)
 +
160 FLUID FLOW

K, = 1.129
K, = 0.471
K,= 0.684 K,= 1.265

K, = 0.400 K, = 0.400
K, = 0.534 K,= 0.601

%
1.23 0.157 0.30C
30" 1.67 0.156 0.37E
2.37 0.143 0.264

30"

I
Note:
I I
K,= RESISTANCE COEFFICIENT FOR SMOOTH SURFACE
K,= RESISTANCE COEFFICIENT FOR ROUGH SURFACE, E I 0.0022
D
'OPTIMUM VALUE OF a INTERPOLATED

Figure 4-16 Resistance coefficients for miter bends at Reynolds number = 2.25 x lo5 for water. (Reprinted by permission from Hydraulic Institute,
Engineering Datu Book. 1st ed., 1979, Cleveland, OH.)

In SI units,

AP/lOOeq.m' =32 (5) (%) = 679 , bar/lOOeq.m

i:::0.1
= 11320 ($) (4-64)

0.0 Turbulent Flow ( R e > 2000)

1.o 1.5 2.0 2.5 3.0 3.5 4.0
English Engineering units

fPV2
A P / lOOeq.ft* =0.1294- = 0.0216@, psi/lOOeq.ft
d d5
Figure 4-17 Resistance coefficients for reducers for water. (Reprinted by
permission from Hydraulic Institute, Engineering Datu Book, 1 st ed., 1979, =0.000336-f w 2
Cleveland, OH.) (4-65)
Pd5
For Re < 2000 In SI units,
English Engineering units

AP/lOOeq.ft* = 0.0668 (e)= 0.0273 AP/lOOeq.m* = O S E

d
= 225 - bar/100eq.m
d5
d2

= 0.0034 (g) (4-63)

= 62,530 f- w2
P d5
(4-66)

KpQ2
A P =O.O001078Kp~~= 0.00001799-
d4 *Equivalent feet (m) of straight pipe; that is, straight pipe plus equivalents
KW2 for valves, fittings, other system components (except vessels, etc.). There-
= 0.00000028- (4-63a) fore, APl100eq. ft (m) = pressure drop (friction) per 100 equivalent feet
Pd4 (m) of straight pipe.
 4.23 LOSS COEFFICIENT 161

L=12"
L=lBM-.-.-.-
-I-- I. -
tan 812 = (D2- 0,)/2
L

0 10 20 30 40 50 60
Q IN DEGREES

Figure 4-18 Resistance coefficients for increasers and diffusers for water. (Reprinted by permission from Hydraulic Institute, Engineering Dura Book,
1st ed., 1979, Cleveland, OH.)

and the head loss equation can be expressed as:

A P = 5 x 10-6Kpv2 = 2.25 x -
K p Q 2 ,bar
d4 hE=-=
g
( f D- i+ x K )$ (4-70)
KW2
= 0.6253 -
Pd4 (4-67) where
D = internal diameter of pipe, ft (m)
(4-68)
\ I
fn = Darcy friction factor
.-
K = loss coefficient
The total pressure drop equation including the elevation L = pipe length, ft (m)
change can be expressed as: u = fluid velocity, ft/s ( d s )
English Engineering units AP = pressure drop, psi.(N/m2)
_

p Av2 pAz g h, = head loss, ft (m)

-AP = (PI - P2) = --+ -- p = fluid density, lb/ft3 (kg/m3)
144 2g, 144 g, ft - lbf
e, = energy dissipated by friction, -(Nm/kg).
lbm

(4-69)
4.23 LOSS COEFFICIENT
In SI units,
K is a dimensionless factor defined as the loss coefficient in a pipe
pAv2 fitting, and expressed in velocity heads. The velocity head is the
-AP = (PI - P2) = 2+pgAz amount of kinetic energy contained in a stream or the amount of
potential energy required to accelerate a fluid to its flowing velocity.
N/m2 (4-69a) Most published K values apply to fully turbulent flow, because at
high Reynolds number, K is found to be independent of Reynolds
162 FLUID FLOW

TABLE 4-6 "K" Factor Table: Representative Resistance Coefficients ( K )for Valves and Fittings. Pipe Friction Data
for Clean Commercial Steel Pipe with Flow in Zone of Complete Turbulence

Nominal size (in.) 'I2 3/3 1 1 1 2 2 '12, 3 4 5 6 8-10 12-16 18-24

Friction factor, fT 0.027 0.025 0.023 0.022 0.21 0.019 0.018 0.017 0.016 0.015 0.014 0.013 0.012

Formulas for Calculating "K" Factors for Valves and Fittings with Reduced Port.

Formula 1 Formula 6

e
0.8 sin-(I - $) -
K2 K1 + Formula 2 + Formula 4
Kt - 2

P -I

h u l a2 A¶ - P

knnuk 7

Formula 3 Kz - 5P + 0 (Formula + Formula 2 4) when 6.- 180'

e
2.6sin- ( I - 9)'
K, - 2
P Kz =
K1+@[0.S(I- n + ( I - b ) * ]
d
h u l a4

Fwmula 5

K2= $' + Formula I + Formula 3

Subscript 1 defines dimensions
e
KL+ sin-[0.8 - jP) + 2.6 ( I - $)'I
and coefficients with reference to

K¶ - 2
(I

d
the smaller diameter.
Subscript 2 refers to the larger
diameter.

Sudden and gradual contraction

I Sudden and gradual enlargement

- -
If: 8 7 45'. . . .
8 > 45'
. . . . .Kz Formula 1
7 I 80". -
. .KI Formula 2 I If: e 7 4 9 . . . . . . , . .Kz Formula 3
e , 45' 7 180'. . .K2= Formula 4
(Continued)
 4.23 LOSS COEFFICIENT 163

TABLE 4-6-4continued)
Gate values Swing check values
wedge disc, double disc, or plug type

K - IOofT K = 5of~
I f : @ - I , 0 - 0 . . . . . . . . . . . . . .K i - 8 fT
-
@ < I and 0 7 4 5 ’ . . . . . . . . .Kz Formula 5
Minimum pipe velocity
(fps) for full disc l i f t
Minimum pipe velocity
(fps) for full disc l ift
-
@ < I and e > 45O 7 180’. . .K2 Formula 6 - 3 5 fl - 48 fl

Lift check values

Globe and angle values

If: @ = 1 , . .I% 6oofT

-
E

@ < I . . .K2 Formula 7

Minimum pipe velocity (fps) for full disc lift
-40 Ist fl

If: 6
6<
- I . .
I . .
.K1-55 fT
-
.K, Formula 7
Minimum pipe velocity (fps) for full disc lift
= 140 PdF

lilting disc check values

d - 5 O d - IS0
sizes 2 t O 8”.. . K = 40fT IzofT
sizes IO to 14”.. .K 30 fT
E 90 f T
All globe and angle valves, Sizes 16 to 48”. . . K = 20 f T 60 f T
whether reduced seat or throttled, Minimum pipe velocity
(fps) for full disc lift = 80 fl 3 0 fl
-
If: B < I , . .K2 Formula 7
164 FLUID FLOW

TABLE 4-64cont in u e d )
Stop-check values Foot values with strainer
(Globe and angle types)
Poppet disk Hinged disk

-+-
Minimum pipe velocity Minimum pipe velocity
Minimum pipe velocity Minimum pipe velocity (fps) for full disc lift (fps) for full disc lift
for full disc lift
- 55 j 3 z d T -
for full disc lift
- 7 5 /?' 4 7 -
- I5 dr- -35 fl

Ball values

If: P = I , e - O . . . . . . . . . , . . . .K , ; . J f T
Minimum pipe velocity (fps) for full disc lift B c I and 8 7 45'. . . . . . . , . K2= Formula 5
-6OPdT- B < I and 0 > 4 5 O 3 1 8 0 ~. .. K o = Formula 6

Butterfly values

Sizes 2 to 8". . .K - 45 ,fT

Minimum pipe velocity (fps) for full disc lift
- Sizes I O to 14". . . K -
-
35fT
I40 flzG Sizes 16 to 24". . . K 25 f T

(continued
 4.23 LOSS COEFFICIENT 165

TABLE 4-&(continued)

Plug values and cocks Standard elbows

Straight-way 3-way 90" 45"

If:
Ki
/3 =
18fT
1,
- -
If: B I ,
Ki 3 0 f ~
If: /3
Ki = W f T
- I,
Standard tees
If: 8 c 1...K*-Formula6

Mitre bends

V) \ld
r
4s-

75'
15 fr
25 h
40 fr
Flow thm run. . . . . . . K
Flow thm branch. . . . K -- 60 f~
20
fT

Pipe entrance
90" Pipe bends and
flanged or butt-welding 90" elbows
Inward Flush
I r/d 1 K llr/d I K 1 projecting

The resistance coefficient, Ko, for pipe bends other

than cpomay be determined as follows:
0.25
r
nfr;i+o.5 K
-f
K - 0.78
0.04
0.06
0.1 0
0.15 & up
'Sharp-deed
0.24

0.04 E For K ,
see table

n = number of 90" bends

K = resistance coefficient for one 90" bend (per table)
Pipe exit
Projecting Sharp-edged rounded
Close pattern return bends

K - 5of~ K - 1.0 K - 1.0 K - I.0

166 FLUID FLOW

TABLE 4-7 Resistance Coefficients for Valves and Fittings

Approximate Range of Variations for K

Fitting Range of Variation

90" elbow Regular screwed 120% above 2-in. size

Regular screwed 540% below 2-in. size
Long radius, screwed 125%
Regular flanged 135%
Long radius, flanged 530%
45" elbow Regular screwed 510%
Long radius, flanged 110%
180" bend Regular screwed 125%
Regular flanged 135%
Long radius, flanged &30%
Tee Screwed, line or branch flow +25%
Flanged, line or branch flow 135%
Globe valve Screwed &25%
Flanged 125%
Gate valve Screwed 525%
Flanged 150%
Check valve Screwed 130%
Flanged
:1_;+
Sleeve check valve Multiply flanged values by 0.2-0.5
Tilting check valve Multiply flanged values by 0.13-0.19
Drainage gate check Multiply flanged values by 0.03-0.07
Angle valve Screwed 120%
Flanged &50%
Basket strainer 550%
Foot valve 150%
Couplings 150%
Unions &50%
Reducers 150%

(Source: Reprinted by permission from Hydraulic Institute, Engineering Data Handbook,

1st ed., 1979, Cleveland, OH.)
Notes on the use of Figures 4-12a and b, and Table 4-7
1. The values of D given in the charts is nominal IPS (Iron Pipe Size).
2. For velocities below 15ft/s, check valves and foot valves will be only partially open
and will exhibit higher values of K than that shown in the charts.

number. However, the 2-K technique includes a correction factor where

for low Reynolds number. Hooper [ 15, 161 gives a detailed analysis K, = K for the fitting at Re = 1
of his method compared to others and has shown that the 2-K K, = K for a large fitting at Re = 00
method is the most suitable for any pipe size. In general, the 2-K
method is independent of the roughness of the fittings, but it is
__
ID = internal diameter of attached pipe, in. or mm.
The Hooper,s method is valid over a much wider range of Reynolds
a function Of number and Of the geometry Of the numbers than the other methods. However, the effect of the pipe
fitting. The method can be expressed as: size (e.g., 1nD) in Eq. (4-71) does not accurately reflect data
over a wide range of sizes for valves and fittings as reported in
sources [l, 4 and 71. Hooper's method and that given in Crane tend
K to under-predict the friction loss for pipes of larger diameters. The
K,= L + K , (4-71)
Re disadvantage of the 2-K method is that it is limited to the number
of values of K, and K, available as shown in Table 4-8. For other
fittings, approximations must be made from data in Table 4-8.
In SI units, Darby [5, 171 recently developed a 3-K equation which repre-
sents various valves, tees, and elbows and is expressed by

K ~ =K
L+K=(I +->25.4
Re ID mm (4-73)
 4.24 SUDDEN ENLARGEMENT OR CONTRACTION 167
TABLE 4-8 2-K Constants for Loss Coefficients for Valves and Fittings

Fitting Type

Elbows 90" Standard (RID = 1) Screwed 800 0.40

Flanged/welded 800 0.25
Long-radius (RID = 1.5) All types 800 0.20
Mitered ( R / D = 1.5) 1 weld (90" angle) 1000 1.15
2 weld (45" angle) 800 0.35
3 weld (30"angle) 800 0.30
4 weld (22.5" angle) 800 0.27
5 weld (18" angle) 800 0.25
45" Standard (RID = 1) All types 500 0.20
Long-radius (RID = 1.5) All types 500 0.15
Mitered (R/D= 1.5) 1 weld (45" angle) 500 0.25
2 weld (22.5" angle) 500 0.15
Standard (RID = 1) Screwed 1000 0.70
180" Standard (RID = 1) Flanged/welded 1000 0.35
Long-radius (RID = 1.5) All types 1000 0.30
Used as Standard Screwed 500 0.70
Elbow Long-radius Screwed 800 0.40
Standard Flanged/welded 800 0.80
Stub-in type branch 1000 1.oo
Tees Run Screwed 200 0.10
Through Flanged or welded 250 0.05
Tee Stub-in type branch 100 0.00
Valves Gate Full line size p = 1.0 300 0.10
Ball Reduced trim p = 0.9 500 0.15
Plug Reduced trim p = 0.8 1000 0.25
Globe Standard 1500 4.00
Angle or Y-type 1000 2.00
Diaphragm Dam type 1000 2.00
Butte rf Iy 800 0.25
Check Lift 2000 10.00
Swing 1500 1.50
Tilting-disc 1000 0.50

(Source: Hooper, Chern. Eng., Aug 24, 1981, pp. 96-100.)

Note: For pipe bends of angle 45"-180" with RID = 5, use K values for R I D = 1.5. For flow
through crosses, use the appropriate tee values. Inlet, flush, K = 160/Re+0.5; Inlet, intruding,
K=160/Re=1.0;Exit,K=1.0.K=1.0,K=K,/Re+K, ( l + l / I D ) , w i t h ID in inches.

where 4.24 SUDDEN ENLARGEMENT OR CONTRACTION [2]

D, = nominal pipe diameter in (mm) For sudden enlargements in a pipe system when there is an abrupt
In SI units, change from a smaller pipe flowing into a larger pipe, the resistance
coefficient (or loss coefficient) K is given by:
For sudden enlargement:
(4-74)

where K , are mostly those of the Hooper 2-K method, and K , were K , = (1 -O?/D:)' = (1 - /3')z (4-75)
mostly determined from the Crane data. The values of Kd are all
very close to 4.0, and this can be used to scale known values of K ,
for a given pipe size to apply to other sizes. Darby's 3-K method where subscripts 1 and 2 refer to the smaller (upstream) and larger
is the most accurate method for all Reynolds numbers and fitting pipes respectively [4] or using Eq. (4-57), h, is given by
sizes, since it scales with pipe size more accurately than the 2-K
method.
The conversion between equivalent pipe length and the resis-
tance coefficient, K,, can be expressed as
h,=K, [1- (::)']'
- (2) , ft (m) of fluid (4-76)

(4-55)
K , = (1 - d?/d:)' (4-77)
Table 4-9 lists values of the 3-K method, and Table 4-10
shows a comparison between the methods used to calculate the loss
coefficient, Kf, for pipe fittings and valves. (Note: This applies for fully turbulent flow.)
168 FLUID FLOW

TABLE 4-9 3-Kconstants for Loss Coefficients for Valves and Fittings

Fitting K= + K ~(1 + +)*

Dn inch
(LID),, K, Kd
-
Elbows -90" Threaded, standard r/D= 1 30 800 0.14 4.0
Threaded, long radius r / D = 1.5 16 800 0.071 4.2
Flanged, welded, bends r/D = 1 20 800 0.091 4.0
r/D =2 12 800 0.056 3.9
r/D =4 14 800 0.066 3.9
r/D = 6 17 800 0.075 4.2
Mitered 1 weld, 90" 60 1000 0.27 4.0
2 welds, 45" 15 800 0.068 4.1
3 welds, 30" 8 800 0.035 4.2
Elbows -45" Threaded, standard r/D=l 16 500 0.071 4.2
Long radius r / D = 1.5 500 0.052 4.0
Mitered, 1 weld 45" 15 500 0.086 4.0
Mitered, 2 welds 22.5" 6 500 0.052 4.0
Elbows -180" Threaded,
Close return bend r/D= 1 50 1000 0.23 4.0
Flanged r/D= 1 1000 0.12 4.0
AI I r / D = 1.5 1000 0.10 4.0
Tees Through-branch
(as elbow)
Threaded rJD = 1 60 500 0.274 4.0
r / D = 1.5 800 0.14 4.0
Flanged r/D= 1 20 800 0.28 4.0
Stub-in branch 1000 0.34 4.0
Run through threaded r/D= 1 20 200 0.091 4.0
Flanged r/D= 1 150 0.017 4.0
Stub-in branch 100 0 0
Valves Angle valve - 45" Full line size, p = 1 55 950 0.25 4.0
Angle valve - 90" Full line size, p = 1 150 1000 0.69 4.0
Globe valve Standard, p = 1 340 1500 1.70 3.6
Plug valve Branch flow 90 500 0.41 4.0
Plug valve Straight through 18 300 0.084 3.9
Plug valve Three-way (flow through) 30 300 0.14 4.0
Gate valve Standard, p = 1 8 300 0.037 3.9
Ball valve Standard, p = 1 3 300 0.017 4.0
Diaphragm Dam-type 1000 0.69 4.9
Swing checkt v,,, = 35p-'12 100 1500 0.46 4.0
Lift checkt V,,, = 4Op-''' 600 2000 2.85 3.8

(Source: Darby, Chem. Eng., July, 1999, pp. 101-104.)

*See Equation
'Units of p are Ib,/ft3
D, = nominal pipe size, in.

For sudden pipe system contraction as represented in Figures vessels are additive as long as they are on the same size
4-14a4-18, the values of the resistance or loss coefficient, K , can (velocity) basis (see Table 4-6 and Figures 4-14a4-18). Thus
be read from the charts. For more details for various angles of the resistance equation is applicable to calculate the head or pres-
enlargements and contractions, see [2, 41. sure loss through the specific system when the combined K value
For sudden contractions: is used.
K , = 0.5 (1 - d : / d z ) = 0.5 (1 - p') (4-78)
h&($) (4-57)
This applies for fully turbulent flow, where subscripts 1 and 2
indicate small and large pipes respectively.
Then or

hf = f (g) (d) (4-37)

Table 4-1 1 shows how K varies with changes in pipe size.
where K =summation of all K values in a specific system,
4.25 PIPING SYSTEMS when all are on the same size (internal flow) basis. See discus-
sion in "Common Denominator" section. (Note: The frictional
The K coefficient values for each of the items of pipe, bends, energy loss, or head loss, is additive even if the velocities
valves, fittings, contractions, enlargements, entrance/exits into/from change).
 TABLE 4-10 Comparison Between the Loss Coefficients (velocity heads), Kf, for Pipe Fittings and Valves
The Equivalent length (LID) K-factor Method N e w Crane Method 2-K (Hooper's) method 3-K(Darby's) method
Method

The equivalent length adds some The excess loss in a fitting The latest version of the equivalent K is a dimensionless factor defined Darby's 3-K method represents
hypothetical length of pipe to the i s normally expressed in a length method given in Crane as the excess head loss in pipe improved features over the widest
actual length of the fitting. dimensionless " K " factor. Technical Paper 410 (Crane Co. fitting, expressed in velocity heads. range of Reynolds number and
However, the drawback is that the 1991) requires the use of t w o K does not depend o n the fitting size, and is expressed by
The excess head loss (AH) is less
equivalent length for a given fitting friction factors. The first is the roughness of the fitting (or the
than the total by the amount of
is not constant, but depends on actual friction factor for flow in the attached pipe) or the size of the
frictional loss that would be
Reynolds number, pipe roughness, straight pipe (f), and the second is system, but it is a function of the
experienced by straight pipe of the Reynolds number and the exact
pipe size, and geometry. a standard friction factor for the
same physical length. Or in SI units
Kf = f (k) The loss coefficient K f , depends on
the Reynolds number of the flow.
particular fitting (fT) given i n
Table 4-..
geometry of the fitting. The 2-K
method accounts for these
dependencies b y the following
D The values of Kf at low Re can be equation.
Le, = Kf .
significantly greater than those at
~

f where
Every equivalent length has a
specific friction factor. The method
assumes that
high Re.
Additionally, valves and fittings do
not scale exactly. e.g., the loss
Kf = 4 f (k)
coefficient for a 1/4in. valve is not
the same as that for a 4in. valve.
( 1 ) Sizes of fittings of a given
type can be scaled by the
The value o f f is determined from
the Colebrook equation
K
Kt = 2 +K, (1
Re
+ &) Where
D, = nominal pipe diameter.
a corresponding pipe diameter. ~ 0.0625 Or i n SI units The values of K are shown in Table
4-10 for various valves and fittings.
$ (2) The influence of Reynolds
number on the friction loss on
[log ( 3 . 7 0 / ~ ) ] '
These values were determined
the fitting is the same as the where from combinations of literature
where
pipe loss. E is the pipe roughness (0.0018in., values from references, and follow
K, = K for the fitting at Re = 1 the scaling law given in the 3-K
However, neither of the above 0.045mm) for commercial steel
K, = K for a large fitting at Re = m. method equation.
assumption is accurate. pipe.
The Crane paper gives fT for a wide The values of K, are mostly those
variety of fittings, valves, etc. of the Hooper 2-K method, and the
This method gives satisfactory ID = Internal pipe diameter, in. values of Ki were determined from
Furthermore, the nature of the the Crane data. Kd values are very
results for high turbulence levels (mm).
laminar or turbulent flow field close to 4.0.
(e.g., high Reynolds number) but is
within a valve or a fitting is The ID correction i n the two K
less accurate at low Reynolds The 3-K method is highly
generally quite different from that expression accounts for the size
number. recommended because it accounts
in a straight pipe. differences. K is higher for small
This method provides a better sizes, but nearly constant for larger directly for the effect of both
Therefore, there is an uncertainty Reynolds number and fitting size
estimate for the effect of geometry, sizes.
when determining the effect of on the loss coefficient. Also, it more
but does not reflect any Reynolds
Reynolds number on the loss However, the effect of pipe size accurately reflects the scale effect
number dependence.
coefficients. This method does not (e.g., I/ID) does not accurately of fitting size than the 2-K method.
properly account for the lack of reflect data over a wide range of
exact scaling for valves and fittings. sizes for valves and fittings. Further
Hooper's scaling factor is not
consistent with the Crane values at
high Reynolds numbers and is
especially inconsistent for larger
fitting sizes.
170 FLUID FLOW

TABLE 4-11 Excess Head Loss K Correlation for Changes in Pipe Size

Type Fitting Inlet N R ~ K based on inlet velocity head

N R 5~2500
[
K = 1.2 + e][(2)4 - I]

1 K = [0.6 + 0.48fDl
Square reduction N R z~2500

Multiply K from Type 1 b y

/sinfor45' < e < 1800

"1

All
2
Tapered reduction

m
N RI
~2500

3 NRe
Thin, sharp orifice
N R >~ 2500

N R 5~ 4000
K = 2 [I - ($)4]

4 -
Square expansion N R >~ 4000

"2 If 0 > 45", use K from Type 4, otherwise

multiply K from Type 4 by

5
NRe

YI
All [2.6 sin (i)]
I
Tapered expansion
.. Rounded

All K= .I':;
[OI+- [(2)4-1]
6
Pipe reducer

IfL/DZ > 5, use Case A and Case F; Otherwise multiply

K from Case D b y
7 %
Thick orifice
All

{0.584+[ (L/D)'.5
0'0936
+0.225 I)
(continued)
 4.27 FLOW COEFFICIENTS FOR VALVES, C, 171
TABLE 4-1 I-(continued)

Type Fitting Inlet N R ~ K based on inlet velocity head

4

8 All Use the K for Case F

Pipe reducer

(Source: Hooper, W.B. Chern. Eng., Nov 7, 1988, pp. 89-92.)

4.26 RESISTANCE OF VALVES At 0 5 180”, for sudden and gradual enlargements:

Figure 4-14b and Table 4-6 present several typical valves and
connections, screwed and flanged, for a variety of sizes or internal (4-83)
diameters. These do not apply for mixtures of suspended solids in
liquids; rather specific data for this situation are required (see [2]). At 0 5 180”, for gradual contraction:
Reference [4] presents data for specific valves.
Valves such as globes and angles generally are designed with
changes in flow direction internally and, thereby, exhibit rela- K , = [ { O S (sin 0/2)1/2] (1 - p 2 ) ] / p 4 (4-84)
tively high flow resistances. These same types of valves exhibit
even greater resistances when they are throttled down from the The use of these equations requires some assumptions or judg-
“wide open” position for control of flow to a smaller internal flow ment regarding the degree of opening for fluid flow. Even so,
path. For design purposes, it is usually best to assume a I/, or l/4 this is better than assuming a wide open or full flow condition
open position, rather than wide open. Estimated K values can be which would result in too low a resistance to flow for the design
determined [4] by reference to Figures 4- 14a-4- 18 and Tables 4-6 situation.
and 4-7.

where 4.27 FLOW COEFFICIENTS FOR VALVES, C,

K , = refers to coefficient for smaller diameter Flow coefficients (not resistance) for valves are generally available
K , = refers to coefficient for larger diameter from the manufacturer. The C, coefficient of a valve is defined as
/3 = ratio of diameters of smaller to larger pipe size the flow of water at 60”F, in gallons per minute, at a pressure drop
0 = angles of convergence or divergence in enlargements or of one pound per square inch across the valve [4], regardless of
contractions in pipe systems, degrees. whether the valve ultimately will be flowing liquid or gaseshapors
in the plant process. (Manufacturers give values of Cg or C1,the
coefficient for gas flow, for valves flowing gas or vapor.) It is
From [4], K values for straight-through valves, such as gate and
expressed:
ball (wide open), can also be calculated. These types of valves are
not normally used to throttle flow, but are either open or closed.
For sudden and gradual (Note: Subscript 1 =smaller pipe; C, = 29.9 d2/(K)’/‘ (4-85)
Subscript 2 =larger pipe)
(4-86)
K2 = Kl /P4 (4-80) (4-87)

For 0 5 45”, as enlargements: = 7.9OCv [ A P , / p ] ” ’ (4-87a)

K , = 2.6 [(sin 0/2) (1 - p2)’]/p4 (4-81)

For 6 5 45”, as contractions: where

d = internal pipe diameter, in.
C, = flow coefficient for valves; expresses flow rate in gallons
K , = [0.8 (sin 6/2) (1 -p2)]/p4 (4-82)
per minute of 60” F water with 1.O psi pressure drop across
valve
For higher resistance valves, such as globes and angles, the losses K = resistance (loss) coefficient
are less than sudden enlargements or contractions situations. For Q = flow rate, gpm
these reduced seat valves the resistance or loss coefficient K can A P = pressure drop across the control valve, psi
be calculated as [4]: p = fluid density, lb/ft3.
172 FLUID FLOW
4.28 NOZZLES AND ORIFICES [4] A = cross-sectional area of orifice, nozzle, or pipe, ft2(m’)
h = static head loss, ft (m) of fluid flowing
The piping items shown in Figures 4-19 and 4-20 are important AP = differential static loss, l b / h 2 ( N / m 2 )of fluid flowing,
pressure drop or head loss items in a system and must be accounted under conditions of h, above
for to obtain the total system pressure loss. For liquids (Note: The
1 P in these equations is NOT a “loss” pressure). /3 = ratio of small to large diameter orifices and nozzles and
contractions or enlargements in pipes.
q = C’ A J2g (144) ( A P ) / p = C’ A (2ghL)1’2 (4-89) For discharging incompressible fluids to atmosphere, take C, values
from Figure 4-19 or 4-20 if h, or A P is taken as upstream head or
In SI units. gauge pressure.
For flow of compressible fluids use the net expansion factor

where
q=C‘A
i”:’
__ - C’ A (2gh,)1/2 (4-90)
Y (see later discussion) [4]:

q = YC’A [2g (144) ( A P ) / p ] ” ’ (4-92)

q = ft3/s (m3/s) of fluid at flowing conditions In SI units.

C‘ = flow coefficient for nozzles and orifices.

C‘ = C , / J v . corrected for velocity of approach

q = YC’A -
d2iP (4-93)

(4-9 1) where Y = net expansion factor for compressible flow through

orifices, nozzles, and pipe.
Note: The expansion factor Y is a function of

C’ = C, for Figures 4-19 and 4-20, corrected for velocity of 1. The specific heat ratio, y
approach. 2. The ratio ( p ) of orifice or throat diameter to inlet diameter
C, = discharge coefficient for nozzles and orifices 3. Ratio of downstream to upstream absolute pressures.
h, = differential static head or pressure loss across flange taps
when C or C‘ values come from Figures 4-19 and 4-20, ft C’ = flow coefficient from Figure 4-19 or 4-20, when discharging
of fluid. Taps are located one diameter upstream and 0.5 to atmosphere
diameter down from the device. P = inlet gauge pressure (also see critical flow discussion).

Flow -
cd
c=d-
Example: The flow coeffi-
cient C for a diameter ratio
3 of 0.60 at a Reynolds
number of 20,000 (2 x I 04)
equals 1.03.

2 4 68104 2 4 68105 2 4 68106 2

Re - Reynolds Number based on d2

Figure 4-19 Flow coefficients “C” for nozzles. C based on the internal diameter of the upstream pipe. (By permission from Crane Co. [4].Crane
reference [18] is to Fluid Meters, American Society of Mechanical Engineers, Part 1-6th ed., 1971. Data used to construct charts. Chart not copied from
ASME reference.)
 4.28 NOZZLES AND ORIFICES 173

3 4 6810 20 406080102 2 4 68103 2 4 68104

Re - Reynolds Number based on dl

Figure 4-20 Flow coefficients "C" for square-edged orifices (By permission from Crane Co. [4], Technical Paper No. 410, Engineering Div. (1976) and
Fluid Meters, Their Theory and Application Part 1, 6th ed., 1971, ASME and, Tuve, G.L. and R.E. Sprinkle, "Orifice Discharge Coefficients for Viscous
Liquids," Znstruments Nov, 1993, p. 201.)

EXAMPLE 4-1 - (20 gpm) (8.33 lb/gal) (0.8 1 Sp gr)

Pipe Sizing Using Resistance Coefficients, K -
(62.3 x 0.81)(3.355 in.*)(6Os/min)/144
A plant decides to add a nitrogen blanket (at 5psig) to a storage
tank holding up to 25,000gal of a hydrocarbon mixture having = 1.91 ft/s
kerosene-like properties (Sg = 0.81 and viscosity = 1.125cP) and u2 (1.91)'
pumps this material into a process reactor operating at 30psig Velocity head, - = -= 0.05664 ft of fluid
2g 2(32.2)
(Figure 4-21). Liquid tank elevation is l o f t .
The flow rate needs t o be 20gpm. Connections of pipe and 50.6Qp
Reynolds number, Re - (4-27)
valve are flanged, with the 6 x 90" elbows added in the line. dP
Solution -
-
50.6(20)(0.81 x 62.3)
Pump suction velocity = 2 ft/s (selected in accordance with good (2.067) ( 1.125 cP)
pump suction practice, from Table 4-12 or 4-13). Re = 21961 (turbulent)
Estimated flow velocity for assumed 2in. Sch. 40 pipe (see
Appendix D- 16) E / D= 0.0018/2.067 = 0.00087

(continued)
174 FLUID FLOW

/30 psig

Normal operating level

1.-
Check valve

6-90' ell

f T
in system,
welded

10,ft

L
Storage tank
b
Sharp exit

15ft
.i
Centrifugal pump
valve

/
1
A Reactor

Figure 4-21 Pipe sizing using resistance coefficients, K illustration for Example 4-1

EXAMPLE 4-l-(conrinued) Loss through pump suction fittings:

Using Chen's explicit Eq. (4-35) to calculate the friction factor, fc:
09
A = - +E/D
(:) a. Square-edged inlet (tank to pipe), K = 0.5, Figure 4-14a
3.7 b. Gate valve flanged, open, in suction line, from Table 4-6, with
/3=1,K=8fT
0.00087
where fT = 0.019 (Table 4-6)
K = 8 (0.019) = 0.152
= 9.2059 x
Frictional head loss (fittings and pipe entrance) is
-1- --410g E 5.02
& 3.70 Re
0.00087 5.02 h, = K v 2 / 2 g = (0.5 +0.152) (1.91)2/(2)(32.2)
= -4l0g,,
{ --
3.7 21,961
log,, (9.2059 x
~

= 0.0370 ft fluid
= 12.1277

fc = 6.798934 x Total suction pipe side friction loss:

f o = 4fc
x h F =0.134+0.0369=0.1709ft kerosene
= 4(6.798934 x
= 0.0272
Pressure drop per lOOft from Eq. (4-65) is
Head loss due to friction (in pipe only-friction also in fitting and
values) is calculated using Eq. (4-37).
15 (1.91)' hPpsi/lOOft = 0.0216fp Q2/di (4-65)
h, = 0.0272 ~ ___
(2.067/12) ( 2 ) (32.2) = 0.0216 (0.0272) (0.81 x 62.3) (202)/2.0675
= 0.134 ft of kerosene fluid, pipe friction for 15 ft = 0.314psi/lOOft

(continued)
 Next Page

4.28 NOZZLES AND ORIFICES 175

TABLE 4-12 Suggested Fluid Velocities in Pipe and Tubing: Liquids, Gases, and Vapors at LowIModerate Pressure
to 50 psig and 50" to 100"F

Fluid Suggested Trial Pipe Material Fluid Suggested Trial Pipe Material
Velocity Velocity
Acetylene (observe Sodium Hydroxide
pressure limitations) 4000 fpm Steel 0-30% 6 Steel
Air, 0-30 psig 4000 fpm Steel 30-50% 5 and
Ammonia 50-73% 4 Nickel
Liquid 6 fps Steel Sodium Chloride Solution
Gas 6000 fps Steel No Solids 5 Steel
Benzene 6 fps Steel With Solids (6 min .-
Bromine 15 max.) Monel or nickel
Liquid 4 fps Glass 7.5
Gas 2000 fpm Glass Perch lo ret hy le ne 6 Steel
Calcium chloride 4 fps Steel Steam
Carbon tetrachloride 6 fps Steel 0-30 psi Saturated* 4,000-6,OOQ Steel
Chloride (dry) 30-1 50 psi Saturated or
Liquid 5 fps Steel, Sch. 80 Superheated* 6,000-1 0,000
Gas 2000-5000 fpm Steel, Sch. 80 15Opsi up
Chloroform Superheated 6,500-1,500
Liquid 6 fps Copper and steel *Short lines 15,000
Gas 2000 fpm Steel (max.)
Ethylene Gas 6000 fpm Steel Sulfuric Acid
Ethylene dibromide 4 fps Glass 88-93% 4 S.S.-N316,
Lead
Ethylene dichloride 6 fps Steel 93-1 00% 4 Cast Iron and
Steel
Ethylene glycol 6 fps Steel Sch. 80
Hydrogen Steel Sulfur Dioxide 4,000 Steel
Liquid 5 fps Rubber Liquid Styrene 6 Steel
Gas 4000 fpm R.L., Saran Trichlorethylene 6 Steel
Haveg Vinyl Chloride 6 Steel
Methyl chloride Vinylidene Chloride 6 Steel
Liquid 6 fps Steel Water
Gas 4000 fpm Steel Average service 3-8 (avg 6) Steel
Natural gas 6000 fpm Steel Boiler feed 4-1 2 Steel
Oils, lubricating 6 fps Steel Pump suction lines 1-5 Steel
Oxygen 1800fpm Max. Steel (300 psig Max.) Maximum economical
(ambient 4000 fpm Type 304 SS (usual) 7-1 0 Steel
temperature)
(Low temperature) Sea and brackish R.L., concrete
Propylene glycol 5 fps Steel water, lined pipe 5-8fps 3 Asphalt-line,
saran-lined,
Concrete 5-12fps (min.) transite

Note: R. L. Rubber-lines steel.

The velocities are suggestive only and are to be used to approximate line size as a starting point for pressure drop calculations.
The final line size should be such as to give an economical balance between pressure drop and reasonable velocity.

EXAMPLE
4-l-(continued)
Using Darby's 3-K method, Eq. (4-73) L L
K . =4fF-=f -
R e = 21961 (turbulent) pipe D DD
Pipe nominal diameter = 2 in. 15
= 0.0272 = 2.369
(2.067/12)
+ 2.369 = 3.044
~~

Fittings /I Kq nK1 4 nKi Kd K, Sum K ' s = 0.675

Gate valve 300 300 0.037 0.037 3.9 Frictional head loss (fittings and pipe entrance) is:
Square-edged inlet 0.507
+
Kf = (160/Re Km)
hf = K u 2 / 2 g = (0.507 + 0.168) (1.91)2/(2) (32.174)
K, = 0.5
Total 0.675 = 0.038ft of fluid

(continued)