8.1
SECTION 8
PIPING AND FLUID FLOW
PRESSURE SURGE IN FLUID PIPING
SYSTEMS
8.2
Pressure Surge in a Piping System
From Rapid Valve Closure
8.2
Piping Pressure Surge with Different
Material and Fluid
8.5
Pressure Surge in Piping System with
Compound Pipeline
8.6
PIPE PROPERTIES, FLOW RATE, AND
PRESSURE DROP
8.8
Quick Calculation of Flow Rate and
Pressure Drop in Piping Systems
8.8
Fluid Head-Loss Approximations for
All Types of Piping
8.10
Pipe-Wall Thickness and Schedule
Number
8.11
Pipe-Wall Thickness Determination by
Piping Code Formula
8.12
Determining the Pressure Loss in

Sizing a Water Meter
8.42
Equivalent Length of a Complex
Series Pipeline
8.43
Equivalent Length of a Parallel Piping
System
8.44
Maximum Allowable Height for a
Liquid Siphon
8.45
Water-Hammer Effects in Liquid
Pipelines
8.47
Specific Gravity and Viscosity of
Liquids
8.47
Pressure Loss in Piping Having
Laminar Flow
8.48
Determining the Pressure Loss in Oil
Pipes
8.49
Flow Rate and Pressure Loss in
Compressed-Air and Gas Piping
8.56
Flow Rate and Pressure Loss in Gas
Pipelines
8.57
Selecting Hangers for Pipes at

Steam Accumulator Selection and
Sizing
8.100
Selecting Plastic Piping for Industrial
Use
8.102
Analyzing Plastic Piping and Lining
for Tanks, Pumps and Other
Components for Specific
Applications
8.104
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Source: HANDBOOK OF MECHANICAL ENGINEERING CALCULATIONS
8.2
PLANT AND FACILITIES ENGINEERING
Friction Loss in Pipes Handling Solids
in Suspension
8.111
Desuperheater Water Spray Quantity
8.112
Sizing Condensate Return Lines for
Optimum Flow Conditions
8.114
Estimating Cost of Steam Leaks from
Piping and Pressure Vessels
8.116
Quick Sizing of Restrictive Orifices in
Piping

lb/in
2
(1723 MPa), flows at the rate of 40 gal /min (2.5 L/s) through stainless steel
pipe. The pipe is 40 ft (12.2 m) long, 1.5 in (38.1 mm) O.D., 1.402 in (35.6 mm)
I.D., 0.049 in (1.24 mm) wall thickness, and has a modulus of elasticity, E,of
29
ϫ
10
6
lb/in
2
(199.8 kPa
ϫ
10
6
). Normal static pressure immediately upstream
of the valve in the pipe is 500 lb/ in
2
(abs) (3445 kPa). When the flow of the oil
is reduced to zero in 0.015 s by closing a valve at the end of the pipe, what is: (a)
the velocity of the pressure wave; (b) the period of the pressure wave; (c) the
amplitude of the pressure wave; and (d) the maximum static pressure at the valve?
Calculation Procedure:
1. Find the velocity of the pressure wave when the valve is closed
(a) Use the equation
68.094
a
ϭ
͙
␥

8.3
5,000
4,000
3,000
2,000
1524
1219
914
610
Bulk modulus K = 250,000 psi
Specific weight = 52 lb per cu ft
␥
S
t
a
i
n
l
e
s
s
s
t
e
e
l
p
i
p
e

6
p
s
i
A
l
u
m
i
n
u
m
p
i
p
e
,
E
=
1
0
.
7
x
1
0
6
p
s
i

ϭ
2(40)/4228
ϭ
0.0189 s, and the period of the
pressure wave is: 2(2L/ a)
ϭ
2(0.0189)
ϭ
0.0378 s.
3. Calculate the pressure surge for rapid valve closure
c) Since the time of 01015 s for valve closure is less than the internal time
2L/a equal to 0.0189 s, the pressure surge can be computed from:
⌬
p
ϭ
␥
aV/144g
for rapid valve closure.
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PIPING AND FLUID FLOW
8.4
PLANT AND FACILITIES ENGINEERING
The velocity of flow, V
ϭ
[(40)(231)(4)]/[(60)(
␲
)(1.402
2

893.5 lb /in
2
(abs) (6156.2 kPa).
Related Calculations. In an industrial hydraulic system, such as that used in
machine tools, hydraulic lifts, steering mechanisms, etc., when the velocity of a
flowing fluid is changed by opening or closing a valve, pressure surges result. The
amplitude of the pressure surge is a function of the rate of change in the velocity
of the mass of fluid. This procedure shows how to compute the amplitude of the
pressure surge with rapid valve closure.
The procedure is the work of Nils M. Sverdrup, Hydraulic Engineer, Aerojet-
General Corporation, as reported in Product Engineering magazine. SI values were
added by the handbook editor.
Notation
a
ϭ
velocity of pressure wave, ft/s (m /s)
a
E
ϭ
effective velocity of pressure wave, ft/s (m/s)
A
ϭ
cross-sectional area of pipe, in
2
(mm
2
)
A
o
ϭ

2
(kPa)
L
ϭ
length of pipe, ft (m)
m
ϭ
mass, slugs
N
ϭ
T/(2L/ a )
ϭ
number of pressure wave intervals during time of valve clo-
sure
p
ϭ
normal static fluid pressure immediately upstream of valve when the fluid
velocity is V, lb/in
2
(absolute) (kPa)
⌬
p
ϭ
amplitude of pressure wave, lb/in
2
(kPa)
p
max
ϭ
maximum static pressure immediately upstream of valve, lb/ in

A
ϭ
air volume, in
3
(mm
3
)
V
ϭ
normal velocity of fluid flow in pipe with valve wide open, ft/ s (m/s)
V
E
ϭ
equivalent fluid velocity, ft / s (m/ s)
V
n
ϭ
velocity of fluid flow during interval n, ft/s (m/s)
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PIPING AND FLUID FLOW
PIPING AND FLUID FLOW
8.5
W
ϭ
work, ft
⅐
lb (W)
␥

puted as in step 1 in the previous procedure.
2. Compute the time for one interval of the pressure wave
As before, in the previous procedure, 2 L / a
ϭ
2 (40 /3655)
ϭ
0.02188 s.
3. Calculate the pressure rise in the pipe
Since the time of 0.015 s for the valve closure is less than the interval time of 2
L/a equal to 0.02188, the pressure rise can be computed from:
⌬
p
ϭ
␥
aV/144g
or,
52
ϫ
3655
ϫ
8.3
2
⌬
p
ϭϭ
340.2 lb/in (2343.98 kPa)
144
ϫ
32.2
4. Find the maximum static pressure in the line

4,000
3,000
2,000
1524
1219
914
610
a, Velocity of pressure wave, ft per sec
Velocity, m/sec
0 14012010080604020
D / t; I.D. of pipe / wall thickness
See Fig. 1 for SI values
Bulk modulus K = 250,000 psi
Specific weight = 52 lb per cu ft
␥
S
t
a
i
n
l
e
s
s
s
t
e
e
l
p

x
1
0
6
p
s
i
A
l
u
m
i
n
u
m
p
i
p
e
,
E
=
1
0
.
7
x
1
0
6

ϭ
500
ϩ
463.4
ϭ
963.4 lb/ in
2
(abs) (6637.8
kPa).
Related Calculations. This procedure is the work of Nils M. Sverdrup, as
detailed in the previous procedure.
PRESSURE SURGE IN PIPING SYSTEM WITH
COMPOUND PIPELINE
A compound pipeline consisting of several stainless-steel pipes of different diam-
eters, Fig. 3, conveys 40 gal/min (2.5 L / s) of water. The length of each section of
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PIPING AND FLUID FLOW
PIPING AND FLUID FLOW
8.7
FIGURE 3 Compound pipeline
consists of pipe sections having dif-
ferent diameters. (Product Engineer-
ing.)
pipe is: L
1
ϭ
25 ft (7.6 m); L
2

1
ϭ
0.4085(G
n
/(D
n
)
2
, where the symbols are as shown
below. Substituting, V
1
ϭ
0.4085(40)/(1.402)
2
ϭ
8.31 ft /s (2.53 m/ s).
Using these two computed values, enter Fig. 2 to find the velocity of the pressure
wave in pipe 1 as 4147 ft / s (1264 m/ s).
2. Find the fluid velocity and pressure-wave velocity in the second pipe
The D
2
/t
2
ratio for the second pipe
ϭ
1.152/0.049
ϭ
23.51. Using the same ve-
locity equation as in step 1, above V
2

ϭ
E
L
ϩ
L
ϩ ⅐⅐⅐ ϩ
L
12 n
to find the equivalent fluid velocity in the compound pipe. Substituting,
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PIPING AND FLUID FLOW
8.8
PLANT AND FACILITIES ENGINEERING
25
ϫ
8.3
ϩ
15
ϫ
12.3
ϩ
10
ϫ
20.1
V
ϭ
E
25

g
(25/4147)
ϩ
(15/4234)
ϩ
(10/4326)
ϭ
4209 ft/s (1282.9 m/s)
Thus, equivalent fluid velocity and effective velocity of the pressure wave in the
compound pipe are both less than either velocity in the individual sections of the
pipe.
Related Calculations. Compound pipes find frequent application in industrial
hydraulic systems. The procedure given here is useful in determining the velocities
produced by sudden closure of a valve in the line.
L
1
, L
2
,...,L
n
ϭ
length of each section of pipe of constant diameter, ft (m)
a
1
, a
2
,...,a
n
ϭ
velocity of pressure wave in the respective pipe sections, ft /s

nnn
This procedure is the work of Nils M. Sverdrup, as detailed earlier.
Pipe Properties, Flow Rate, and
Pressure Drop
QUICK CALCULATION OF FLOW RATE AND
PRESSURE DROP IN PIPING SYSTEMS
A 3-in (76-mm) Schedule 40S pipe has a 300-gal / min (18.9-L/ s) water flow rate
with a pressure loss of 8 lb / in
2
(55.1 kPa)/100 ft (30.5 m). What would be the
flow rate in a 4-in (102-mm) Schedule 40S pipe with the same pressure loss? What
would be the pressure loss in a 4-in (102-mm) Schedule 40S pipe with the same
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PIPING AND FLUID FLOW
PIPING AND FLUID FLOW
8.9
flow rate, 300 gal /min (18.9 L /s)? Determine the flow rate and pressure loss for a
6-in (152-mm) Schedule 40S pipe with the same pressure and flow conditions.
Calculation Procedure:
1. Determine the flow rate in the new pipe sizes
Flow rate in a pipe with a fixed pressure drop is proportional to the ratio of (new
pipe inside diameter / known pipe inside diameter)
2.4
. This ratio is defined as the
flow factor, F. To use this ratio, the exact inside pipe diameters, known and new,
must be used. Take the exact inside diameter from a table of pipe properties.
Thus, with a 3-in (76-mm) and a 4-in (102-mm) Schedule 40S pipe conveying
water at a pressure drop of 8 lb/ in

4.8
.
For the first situation given above, P
ϭ
(3.068/4.026)
4.8
ϭ
0.27134. Then, the
pressure drop, PD
N
, in the new 4-in (102-mm) Schedule 40S pipe with a 300-gal/
min (18.9-L/s) flow will be PD
N
ϩ
P(PD
K
), where PD
K
ϭ
pressure drop in the
known pipe size. Substituting, PD
N
ϭ
0.27134(8)
ϭ
2.17 lb / in
2
/100 ft (14.9 kPa/
30.5 m).
For the 6-in (152-mm) pipe, using the same approach, PD

12/ 10 1.542 10/12 0.421
When computing such a listing, the actual inside diameter of the pipe, taken from
a table of pipe properties, must be used when calculating F or P.
The F and P values are useful when designing a variety of piping systems for
chemical, petroleum, power, cogeneration, marine, buildings (office, commercial,
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PIPING AND FLUID FLOW
8.10
PLANT AND FACILITIES ENGINEERING
residential, industrial), and other plants. Both the F and P values can be used for
pipes conveying oil, water, chemicals, and other liquids. The F and P values are
not applicable to steam or gases.
Note that the ratio of pipe diameters is valid for any units of measurement—
inches, cm, mm—provided the same units are used consistently throughout the
calculation. The results obtained using the F and P values usually agree closely
with those obtained using exact flow or pressure-drop equations. Such accuracy is
generally acceptable in everyday engineering calculations.
While the pressure drop in piping conveying a liquid is inversely proportional
to the fifth power of the pipe diameter ratio, turbulent flow alters this to the value
of 4.8, according to W. L. Nelson, Technical Editor, The Oil and Gas Journal.
FLUID HEAD-LOSS APPROXIMATIONS FOR ALL
TYPES OF PIPING
Using the four rules for approximating head loss in pipes conveying fluid under
turbulent flow conditions with a Reynolds number greater than 2100, find: (a)A
4-in (101.6-mm) pipe discharges 100 gal/min (6.3 L /s); how much fluid would a
2-in (50.8-mm) pipe discharge under the same conditions? (b) A 4-in (101.6-mm)
pipe has 240 gal/min (15.1 L/s) flowing through it. What would be the friction
loss in a 3-in (76.2-mm) pipe conveying the same flow? (c) A flow of 10 gal /min

5
(b) We have a 4-in (101.6-mm) pipe conveying 240 gal /min (15.1 L / s). If we
reduce the pipe size to 3 in (76.2 mm) the friction will be greater because the flow
area is smaller. The head loss
ϭ
(flow rate, gal /min or L/s)(larger pipe diameter
to the fifth power)/ (smaller pipe diameter to the fifth power). Or, head
ϭ
(240)(4
5
)/(3
5
)
ϭ
1011 ft /1000 ft of pipe (308.3 m / 304.8 m of pipe).
Again, using this rule you can quickly and easily find the friction in a different
size pipe when the capacity or flow rate remains constant. With the easy availability
of handheld calculators in the field and computers in the design office, the fifth
power of the diameter is easily found.
3. Use the rule: At constant diameter, head is proportional to gal / min
(
L/s
)
2
(c) We know that a flow of 100 gal/ min (6.3 L/s) produces 50-ft (15.2-m) friction,
h, in a pipe. The friction, with a new flow will be, h
ϭ
(friction, ft or m, at known
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0.5
)/(actual friction, ft or m
0.5
). Or,
c
ϭ
5000(200
0.5
)/(84
0.5
)
ϭ
7714 gal /min (486.6 L /s).
As before, a simple calculation, the ratio of the square roots of the friction heads
times the capacity will quickly give the new flow rates.
Related Calculations. Similar laws for fans and pumps give quick estimates
of changed conditions. These laws are covered elsewhere in this handbook in the
sections on fans and pumps. Referring to them now will give a quick comparison
of the similarity of these sets of laws.
PIPE-WALL THICKNESS AND
SCHEDULE NUMBER
Determine the minimum wall thickness t
m
in (mm) and schedule number SN for a
branch steam pipe operating at 900
Њ
F (482.2
Њ
C) if the internal steam pressure is
1000 lb/ in

ϭ
steam velocity, ft/ min (m/min). The only unknown in this equation, other
than the pipe area, is the steam velocity V. Use Table 1 to find a suitable steam
velocity for this branch line.
Table 1 shows that the recommended steam velocities for branch steam pipes
range from 6000 to 15,000 ft/min (1828 to 4572 m/min). Assume that a velocity
of 12,000 ft/min (3657.6 m /min) is used in this branch steam line. Then, by using
the steam table to find the specific volume of steam at 900
Њ
F (482.2
Њ
C) and 1000
lb/in
2
(abs) (6894 kPa), a
ϭ
2.4(72,000)(0.7604)/12,000
ϭ
10.98 in
2
(70.8 cm
2
).
The inside diameter of the pipe is then d
ϭ
2(a/
␲
)
0.5
ϭ

2
, from Piping Code. Table 2 shows
typical allowable stress values for pipe in power piping systems. For this pipe,
assuming that seamless ferritic alloy steel (1% Cr, 0.55% Mo) pipe is used with
the steam at 900
Њ
F (482
Њ
C), SN
ϭ
(1000)(1014.7)/13,100
ϭ
77.5. Since pipe is
not ordinarily made in this schedule number, use the next highest readily available
schedule number, or SN
ϭ
80. [Where large quantities of pipe are required, it is
sometimes economically wise to order pipe of the exact SN required. This is not
usually done for orders of less than 1000 ft (304.8 m) of pipe.]
3. Determine the pipe-wall thickness
Enter a tabulation of pipe properties, such as in Crocker and King—Piping Hand-
book, and find the wall thickness for 4-in (101.6-mm) SN 80 pipe as 0.337 in (8.56
mm).
Related Calculations. Use the method given here for any type of pipe—steam,
water, oil, gas, or air—in any service—power, refinery, process, commercial, etc.
Refer to the proper section of B31.1 Code for Pressure Piping when computing
the schedule number, because the allowable stress S varies for different types of
service.
The Piping Code contains an equation for determining the minimum required
pipe-wall thickness based on the pipe internal pressure, outside diameter, allowable

1. Determine the constants for the thickness equation
Pipe-wall thickness to meet ANSA Code requirements for power service is com-
puted from t
m
ϭ
{DP/[2(S
ϩ
YP)]}
ϩ
C, where t
m
ϭ
minimum wall thickness, in;
D
ϭ
outside diameter of pipe, in; P
ϭ
internal pressure in pipe, lb/in
2
(gage);
S
ϭ
allowable stress in pipe material, lb /in
2
; Y
ϭ
temperature coefficient; C
ϭ
end-
condition factor, in.

Since pipe mills do not fabricate to precise wall thicknesses, a tolerance above
or below the computed wall thickness is required. An allowance must be made in
specifying the wall thickness found with this equation by increasing the thickness
by 12
1
⁄
2
percent. Thus, for this pipe, wall thickness
ϭ
0.367
ϩ
0.125(0.367)
ϭ
0.413 in (10.5 mm).
Refer to the Code to find the schedule number of the pipe. Schedule 60 8-in
(203-mm) pipe has a wall thickness of 0.406 in (10.31 mm), and schedule 80 has
a wall thickness of 0.500 in (12.7 mm). Since the required thickness of 0.413 in
(10.5 mm) is greater than schedule 60 but less than schedule 80, the higher schedule
number, 80, should be used.
3. Check the selected schedule number
From the previous calculation procedure, SN
ϭ
1000 P
i
/S. From this pipe,
SN
ϭ
1000(900)/12,500
ϭ
72. Since piping is normally fabricated for schedule

9500 lb/in
2
(65.5 MPa)] and 700
Њ
F (371
Њ
C) [
ϭ
9000 lb/in
2
(62.0 MPa)]. Interpolate thus: allowable stress at 680
Њ
F (360
Њ
C)
ϭ
[(700
Њ
F
Ϫ
680
Њ
F)/(700
Њ
F
Ϫ
650
Њ
F)](9500
Ϫ

(700
Ϫ
650)](9500
Ϫ
9000)
ϭ
9200 lb /in
2
(63.4 MPa).
DETERMINING THE PRESSURE LOSS IN
STEAM PIPING
Use a suitable pressure-loss chart to determine the pressure loss in 510 ft (155.5
m) of 4-in (101.6-mm) flanged steel pipe containing two 90
Њ
elbows and four 45
Њ
bends. The schedule 40 piping conveys 13,000 lb /h (5850 kg /h) of 20-lb/in
2
(gage)
(275.8-kPa) 350
Њ
F (177
Њ
C) superheated steam. List other methods of determining
the pressure loss in steam piping.
Calculation Procedure:
1. Determine the equivalent length of the piping
The equivalent length of a pipe L
e
ft

C), and project
vertically downward until the 40-lb /in
2
(gage) (275.8-kPa) superheated steam pres-
sure curve is intersected. From here, project horizontally to the right until the outer
border of the chart is intersected. Next, project through the steam flow rate, 13,000
lb/h (5900 kg/h) on scale B, Fig. 5, to the pivot scale C. From this point, project
through 4-in (101.6-mm) schedule 40 pipe on scale D, Fig. 5. Extend this line to
intersect the pressure-drop scale, and read the pressure loss as 7.25 lb/ in
2
(50
kPa)/100 ft (30.4 m) of pipe.
Since the equivalent length of this pipe is 550 ft (167.6 m), the total pressure
loss in the pipe is (550 / 100)(7.25)
ϭ
39.875 lb/ in
2
(274.9 kPa), say 40 lb/ in
2
(275.8 kPa).
3. List the other methods of computing pressure loss
Numerous pressure-loss equations have been developed to compute the pressure
drop in steam piping. Among the better known are those of Unwin, Fritzche, Spitz-
glass, Babcock, Guttermuth, and others. These equations are discussed in some
detail in Crocker and King—Piping Handbook and in the engineering data pub-
lished by valve and piping manufacturers.
Most piping designers use a chart to determine the pressure loss in steam piping
because a chart saves time and reduces the effort involved. Further, the accuracy
obtained is sufficient for all usual design practice.
Figure 3 is a popular flowchart for determining steam flow rate, pipe size, steam

min), and project along this velocity line until the 120-lb/in
2
(gage) (827.3-kPa)
pressure line is intersected. From this intersection, project horizontally until the
8000-lb/h (3600-kg/h) vertical line is intersected. Read the nearest pipe size as 4
in (101.6 mm) on the nearest pipe-diameter curve.
Example. What is the steam velocity in a 6-in (152.4-mm) pipe delivering
20,000 lb /h (9000 kg/h) of steam at 85 lb /in
2
(gage) (586 kPa)?
Solution. Enter the bottom of the chart, Fig. 6, at the flow rate of 20,000
lb/h (9000 kg/ h), and project vertically upward until the 6-in (152.4-mm) pipe
curve is intersected. From this point, project horizontally to the 85-lb/in
2
(gage)
(586-kPa) curve. At the intersection, read the velocity as 7350 ft / min (2240.3 m/
min).
Table 3 shows typical steam velocities for various industrial and commercial
applications. Use the given values as guides when sizing steam piping.
PIPING WARM-UP CONDENSATE LOAD
How much condensate is formed in 5 min during warm-up of 500 ft (152.4 m) of
6-in (152.4-mm) schedule 40 steel pipe conveying 215-lb/in
2
(abs) (1482.2-kPa)
saturated steam if the pipe is insulated with 2 in (50.8 mm) of 85 percent magnesia
and the minimum external temperature is 35
Њ
F (1.7
Њ
C)?

warm-up time, min.h
ƒg
A table of pipe properties shows that this pipe weighs 18.974 lb/ft (28.1 kg/
m). The steam table shows that the temperature of 215-lb/in
2
(abs) (1482.2-kPa)
saturated steam is 387.89
Њ
F (197.7
Њ
C), say 388
Њ
F (197.8
Њ
C); the enthalpy
ϭ
837.4 Btu /lb (1947.8 kJ/kg). The specific heat of steel pipe s
ϭ
0.144 Btuh
ƒg
/(lb
⅐ Њ
F) [0.6 kJ / (kg
⅐ Њ
C)]. Then C
h
ϭ
60(500
ϫ
18.974)(388

For any pipe, C
r
ϭ
(L)(A)(
⌬
t)(H)/ , where L
ϭ
length of pipe, ft; A
ϭ
external
h
ƒg
area of pipe, ft
2
/ft of length; H
ϭ
heat loss through bare pipe or pipe insulation,
Btu/(ft
2
⅐
h
⅐ Њ
F), from the piping or insulation tables. This 6-in (152.4-mm) schedule
40 pipe has an external area A
ϭ
1.73 ft
2
/ft (0.53 m
2
/m) of length. The heat loss

of pressure-regulating or shutoff valves usually have a safety factor of 3. With a
safety factor of 3 for this pipe, the steam trap should have a capacity of at least
3(5522.1)
ϭ
16,566.3 lb /h (7454.8 kg /h), say 17,000 lb / h (7650.0 kg/ h).
Related Calculations. Use this method to find the warm-up condensate load
for any type of steam pipe—main or auxiliary—in power, process, heating, or
vacuum service. The same method is applicable to other vapors that form
condensate—Dowtherm, refinery vapors, process vapors, and others.
STEAM TRAP SELECTION FOR INDUSTRIAL
APPLICATIONS
Select steam traps for the following four types of equipment: (1) the steam directly
heats solid materials as in autoclaves, retorts, and sterilizers; (2) the steam indirectly
heats a liquid through a metallic surface, as in heat exchangers and kettles, where
the quantity of liquid heated is known and unknown; (3) the steam indirectly heats
a solid through a metallic surface, as in dryers using cylinders or chambers and
platen presses; and (4) the steam indirectly heats air through metallic surfaces, as
in unit heaters, pipe coils, and radiators.
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PIPING AND FLUID FLOW
PIPING AND FLUID FLOW
8.21
TABLE 5
Use These Specific Heats to Calculate Condensate Load
TABLE 4
Factors P
ϭ
(T

ϭ
factor
from Table 4. Thus, for this application, C
ϭ
(2000)(1.0)(0.193)
ϭ
386 lb (173.7
kg) of condensate. Note that P is based on a temperature rise of 240
Ϫ
60
ϭ
180
Њ
F
(100
Њ
C) and a steam pressure of 25 lb/in
2
(gage) (172.4 kPa). For the retort, using
the specific heat of steel from Table 5, C
ϭ
(4000)(0.12)(0.193)
ϭ
92.6 lb of
condensate, say 93 lb (41.9 kg). The total weight of condensate formed in 15 min
is 386
ϩ
93
ϭ
479 lb (215.6 kg). In 1 h, 479(60 /15)

gal of liquid heated;
w
ϭ
weight of liquid, lb/ gal. Substitute the appropriate values as follows:
C
ϭ
(500)(8.33)(1.0)
ϫ
(0.154)
ϭ
641 lb (288.5 kg), or (641)(60/ 3)
ϭ
1282 lb/h
(621.9 kg/ h). With a safety factor of 3, the trap capacity
ϭ
(3)(1282)
ϭ
3846
lb/h (1731 kg/h), say 3900 lb / h (1755 kg/ h).
b(2). Submerged heating surface and an unknown quantity of liquid. How much
condensate is formed in a coil submerged in oil when the oil is heated as quickly
as possible from 50 to 250
Њ
F (10 to 121
Њ
C) by 25-lb / in
2
(gage) (172.4-kPa) steam
if the coil has an area of 50 ft
2

ϭ
(3)(214)
ϭ
642 lb / h (289
kg/h), say 650 lb / h (292.5 kg /h).
b(3). Submerged surfaces having more area than needed to heat a specified
quantity of liquid in a given time with condensate withdrawn as rapidly as formed.
Use Table 7 instead of step b (1) or b(2). Find the condensate rate by multiplying
the submerged area by the appropriate factor from Table 7. Use this method for
heating water, chemical solutions, oils, and other liquids. Thus, with steam at 100
lb/in
2
(gage) (689.4 kPa) and a temperature of 338
Њ
F (170
Њ
C) and heating oil from
50 to 226
Њ
F (10 to 108
Њ
C) with a submerged surface having an area of 500 ft
2
(46.5
m
2
), the mean temperature difference (Mtd)
ϭ
steam temperature minus the average
liquid temperature

formed in a chamber dryer when 1000 lb (454 kg) of cereal is dried to 750 lb (338
kg) by 10-lb/in
2
(gage) (68.9-kPa) steam? The initial temperature of the cereal is
60
Њ
F (15.6
Њ
C), and the final temperature equals that of the steam.
For this condition, C
ϭ
970(W
Ϫ
D)/
ϩ
WP, where D
ϭ
dry weight of theh
ƒg
material, lb;
ϭ
enthalpy of vaporization of the steam at the trap pressure,h
ƒg
Btu/lb. From the steam tables and Table 4, C
ϭ
970(1000
Ϫ
750)/952
ϩ
(1000)(0.189)

Њ
C) and pro-
jecting to a steam pressure of 10 lb / in
2
(gage) (68.9 kPa). Read the condensate
formed as 122 lb/ h (54.9 kg/ h) per 1000 ft
3
/min (28.3 m
3
/min). Since 10,000
ft
3
/min (283.1 m
3
/min) of air is being heated, the condensate rate
ϭ
(10,000/
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8.23
TABLE 6
Ordinary Ranges of Overall Coefficients of Heat Transfer
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PIPING AND FLUID FLOW
8.24
PLANT AND FACILITIES ENGINEERING

Њ
C), 1.05 lb/h
(0.47 kg /h) of condensate will be formed per ft
2
(0.09 m
2
) of heating surface in
still air. With forced-air circulation, the condensate rate is (5)(1.05)
ϭ
5.25 lb / (h
⅐
ft
2
) [25.4 kg/(h
⅐
m
2
)] of heating surface.
Unit heaters have a standard rating based on 2-lb/ in
2
(gage) (13.8-kPa) steam
with entering air at 60
Њ
F (15.6
Њ
C). If the steam pressure or air temperature is dif-
ferent from these standard conditions, multiply the heater Btu /h capacity rating by
the appropriate correction factor form, Table 10. Thus, a heater rated at 10,000
Btu/h (2931 W) with 2-lb/ in
2

h)
[kg/(m
2
⅐
h)]
TABLE 10
Unit-Heater Correction Factors
factured by Armstrong Machine Works. Table 11 shows typical capacities of im-
pulse traps manufactured by the Yarway Company.
To select a trap from Fig. 7, when the condensate rate is uniform and the pressure
across the trap is constant, enter at the left at the condensation rate, say 8000
lb/h (3600 kg/h) (as obtained from step 1). Project horizontally to the right to the
vertical ordinate representing the pressure across the trap [
ϭ ⌬
p
ϭ
steam-line pres-
sure, lb/in
2
(gage)
Ϫ
return-line pressure with with trap valve closed, lb/in
2
(gage)].
Assume
⌬
p
ϭ
20 lb/in
2

intersection of the 30-lb/in
2
(gage) (207-kPa) ordinate and the dashed curve ex-
tended from the
3
⁄
8
-in (9.53-mm) solid curve.
To select an impulse trap from Table 11, enter the table at the trap inlet pressure,
say 125 lb /in
2
(gage) (862 kPa), and project to the desired capacity, say 8000 lb /
h (3600 kg/ h), determined from step 1. Table 11 shows that a 2-in (50.8-mm) trap
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