Tài liệu Handbook of Mechanical Engineering Calculations P11 doc

11.1
SECTION 11
HEAT TRANSFER AND
HEAT EXCHANGE
Selecting Type of Heat Exchanger for a
Specific Application
11.1
Shell-and-Tube Heat Exchanger Size
11.4
Heat Exchanger Actual Temperature
Difference
11.6
Fouling Factors in Heat-Exchanger
Sizing and Selection
11.8
Heat Transfer in Barometric and Jet
Condensers
11.10
Selection of a Finned-Tube Heat
Exchanger
11.12
Spiral-Type Heating-Coil Selection
11.15
Sizing Electric Heaters for Industrial Use
11.16
Economizer Heat Transfer Coefficient
11.19
Boiler-Tube Steam-Generating Capacity
11.20
Shell-and-Tube Heat Exchanger Design
Analysis

11.21
Designing Spiral-Plate Heat Exchangers
11.37
Spiral-Tube Heat Exchanger Design
11.46
Heat-Transfer Design for Internal Steam
Tracing of Pipelines
11.54
Designing Heat-Transfer Surfaces for
External Heat Tracing of Pipelines
11.62
Air-Cooled Heat Exchangers: Preliminary
Selection
11.67
Heat Exchangers: Quick Design and
Evaluation
11.70
SELECTING TYPE OF HEAT EXCHANGER FOR A
SPECIFIC APPLICATION
Determine the type of heat exchanger to use for each of the following applications:
(1) heating oil with steam; (2) cooling internal combustion engine liquid coolant;
(3) evaporating a hot liquid. For each heater chosen, specify the typical pressure
range for which the heater is usually built and the typical range of the overall
coefficient of heat transfer U.
Calculation Procedure:
1. Determine the heat-transfer process involved
In a heat exchanger, one or more of four processes may occur: heating, cooling,
boiling, or condensing. Table 1 lists each of these four processes and shows the
usual heat-transfer fluids involved. Thus, the heat exchangers being considered here
involve (a) oil heater—heating—vapor-liquid; (b) internal-combustion engine

coolant—cooling—gas-liquid; (c) hot-liquid evaporation—boiling—liquid-liquid.
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Source: HANDBOOK OF MECHANICAL ENGINEERING CALCULATIONS
11.2
TABLE 1
Heat-Exchanger Selection Guide*
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HEAT TRANSFER AND HEAT EXCHANGE
11.3
TABLE 1
(Continued )
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HEAT TRANSFER AND HEAT EXCHANGE
11.4
PLANT AND FACILITIES ENGINEERING
2. Specify the heater action and the usual type selected
Using the same identifying letters for the heaters being selected, Table 1 shows the
action and usual type of heater chosen. Thus,
3. Specify the usual pressure range and typical U
Using the same identifying letters for the heaters being selected, Table 1 shows the
action and usual type of heater chosen. Thus,
4. Select the heater for each service
Where the heat-transfer conditions are normal for the type of service met, the type
of heater listed in step 2 can be safely used. When the heat-transfer conditions are

unusual, a special type of heater may be needed. To select such a heater, study the
data in Table 1 and make a tentative selection. Check the selection by using the
methods given in the following calculation procedures in this section.
Related Calculations. Use Table 1 as a general guide to heat-exchanger selec-
tion in any industry—petroleum, chemical, power, marine, textile, lumber, etc. Once
the general type of heater and its typical U value are known, compute the required
size, using the procedure given later in this section.
SHELL-AND-TUBE HEAT EXCHANGER SIZE
What is the required heat-transfer area for a parallel-flow shell-and-tube heat ex-
changer used to heat oil if the entering oil temperature is 60
Њ
F (15.6
Њ
C), the leaving
oil temperature is 120
Њ
F (48.9
Њ
C), and the heating medium is steam at 200 lb /in
2
(abs) (1378.8 kPa)? There is no subcooling of condensate in the heat exchanger.
The overall coefficient of heat transfer U
ϭ
25 Btu / (h
⅐ Њ
F
⅐
ft
2
) [141.9 W / (m

2
⅐ Њ
C)].
How much heating steam is required if the oil flow rate through the heater is 100
gal/min (6.3 L / s), the specific gravity of the oil is 0.9, and the specific heat of the
oil is 0.5 Btu/(lb
⅐ Њ
F) [2.84 W / (m
2
⅐ Њ
C)]?
Calculation Procedure:
1. Compute the heat-transfer rate of the heater
With a flow rate of 100 gal / min (6.3 L/s) or (100 gal/ min)(60 min/ h)
ϭ
6000
gal/h (22,710 L/h), the weight flow rate of the oil, using the weight of water of
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HEAT TRANSFER AND HEAT EXCHANGE
HEAT TRANSFER AND HEAT EXCHANGE
11.5
FIGURE 1 Temperature relations in typical parallel-flow and counterflow heat exchang-
ers.
specific gravity 1.0 as 8.33 lb/gal, is (6000 gal/h) (0.9 specific gravity)(8.33 lb /
gal)
ϭ
45,000 lb/h (20,250 kg/h), closely.
Since the temperature of the oil rises 120

Ϫ
60
ϭ
60
Њ
F (33.3
Њ
C) during passage
through the heat exchanger and the oil has a specific heat of 0.50, find the heat-
transfer rate of the heater from the general relation Q
ϭ
wc
⌬
t, where Q
ϭ
heat-
transfer rate, Btu/ h; w
ϭ
oil flow rate, lb/h; c
ϭ
specific heat of the oil, Btu / (lb
⅐ Њ
F);
⌬
t
ϭ
temperature rise of the oil during passage through the heater. Thus, Q
ϭ
(45,000)(0.5)(60)
ϭ

1,350,000 Btu/h (0.4 MW).
2. Compute the heater logarithmic mean temperature difference
The logarithmic mean temperature difference (LMTD) is found from LMTD
ϭ
(G
Ϫ
L)/ln (G/L), where G
ϭ
greater terminal temperature difference of the heater,
Њ
F; L
ϭ
lower terminal temperature difference of the heater,
Њ
F; ln
ϭ
logarithm to
the base e. This relation is valid for heat exchangers in which the number of shell
passes equals the number of tube passes.
In general, for parallel flow of the fluid streams, G
ϭ
T
1
Ϫ
t
1
and L
ϭ
T
2

Ϫ
t
2
,
where T
1
ϭ
heating fluid inlet temperature,
Њ
F; T
2
ϭ
heating fluid outlet temperature,
Њ
F; t
1
ϭ
heated fluid inlet temperature,
Њ
F; t
2
ϭ
heated fluid outlet temperature,
Њ
F.
Figure 1 shows the maximum and minimum terminal temperature differences for
various fluid flow paths.
For this parallel-flow exchanger, G
ϭ
T

1
Ϫ
t
1
ϭ
382
Ϫ
60
ϭ
322
Њ
F (179
Њ
C),
where 382
Њ
F (194
Њ
C)
ϭ
the temperature of 200-lb/in
2
(abs) (1379-kPa) saturated
steam, from a table of steam properties. Also, L
ϭ
T
2
Ϫ
t
2

ϭ
382
Ϫ
120
ϭ
262
Њ
F
(145.6
Њ
C), where the condensate temperature
ϭ
the saturated steam temperature
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HEAT TRANSFER AND HEAT EXCHANGE
11.6
PLANT AND FACILITIES ENGINEERING
because there is no subcooling of the condensate. Then LMTD
ϭ
G
Ϫ
L /ln
(G/L)
ϭ
(322
Ϫ
262)/ln(322 / 262)
ϭ

290
Њ
F (161
Њ
C).
3. Compute the required heat-transfer area
Use the relation A
ϭ
Q/U
ϫ
LMTD, where A
ϭ
required heat-transfer area, ft
2
;
U
ϭ
overall coefficient of heat transfer, Btu /(ft
2
⅐
h
⅐ Њ
F). Thus, A
ϭ
1,350,000/
[(25)(290)]
ϭ
186.4 ft
2
(17.3 m

2
), say 200 ft
2
(18.6 m
2
).
4. Compute the required quantity of heating steam
The heat added to the oil
ϭ
Q
ϭ
1,350,000 Btu/ h, from step 1. The enthalpy of
vaporization of 200-lb/in
2
(abs) (1379-kPa) saturated steam is, from the steam ta-
bles, 843.0 Btu/ lb (1960.8 kJ / kg). Use the relation W
ϭ
where W
ϭ
flow
Q/h ,
ƒg
rate of heating steam, lb/ h;
ϭ
enthalpy of vaporization of the heating steam,
h
ƒg
Btu/lb. Hence, W
ϭ
1,350,000/843.0

ϭ
1600 lb/h (720 kg/h).
Related Calculations. Use this general procedure to find the heat-transfer area,
fluid outlet temperature, and required heating-fluid flow rate when true parallel flow
or counterflow of the fluids occurs in the heat exchanger. When such a true flow
does not exist, use a sitable correction factor, as shown in the next calculation
procedure.
The procedure described here can be used for heat exchangers in power plants,
heating systems, marine propulsion, air-conditioning systems, etc. Any heating or
cooling fluid—steam, gas, chilled water, etc.—can be used.
To select a heat exchanger by using the results of this calculation procedure,
enter the engineering data tables available from manufacturers at the computed heat-
transfer area. Read the heater dimensions directly from the table. Be sure to use
the next larger heat-transfer area when the exact required area is not available.
When there is little movement of the fluid on either side of the heat-transfer
area, such as occurs during heat transmission through a building wall, the arithmetic
mean (average) temperature difference can be used instead of the LMTD. Use the
LMTD when there is rapid movement of the fluids on either side of the heat-transfer
area and a rapid change in temperature in one, or both, fluids. When one of the
two fluids is partially, but not totally, evaporated or condensed, the true mean tem-
perature difference is different from the arithmetic mean and the LMTD. Special
methods, such as those presented in Perry—Chemical Engineers’ Handbook, must
be used to compute the actual temperature difference under these conditions.
When two liquids or gases with constant specific heats are exchanging heat in
a heat exchanger, the area between their temperature curves, Fig. 2, is a measure
of the total heat being transferred. Figure 2 shows how the temperature curves vary
with the amount of heat-transfer area for counterflow and parallel-flow exchangers
when the fluid inlet temperatures are kept constant. As Fig. 2 shows, the counterflow
arrangement is superior.
If enough heating surface is provided, in a counterflow exchanger, the leaving

cold-fluid temperature can be raised above the leaving hot-fluid temperature. This
cannot be done in a parallel-flow exchanger, where the temperatures can only ap-
proach each other regardless of how much surface is used. The counterflow ar-
rangement transfers more heat for given conditions and usually proves more eco-
nomical to use.
HEAT-EXCHANGER ACTUAL TEMPERATURE
DIFFERENCE
A counterflow shell-and-tube heat exchanger has one shell pass for the heating fluid
and two shell passes for the fluid being heated. What is the actual LMTD for this
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HEAT TRANSFER AND HEAT EXCHANGE
HEAT TRANSFER AND HEAT EXCHANGE
11.7
FIGURE 2 For certain conditions, the area between the temperature curves measures the
amount of heat being transferred.
exchanger if T
1
ϭ
300
Њ
F (148.9
Њ
C), T
2
ϭ
250
Њ
F (121

Њ
C), t
1
ϭ
100
Њ
F (37.8
Њ
C), and
t
2
ϭ
230
Њ
F (110
Њ
C)?
Calculation Procedure:
1. Determine how the LMTD should be computed
When the numbers of shell and tube passes are unequal, true counterflow does not
exist in the heat exchanger. To allow for this deviation from true counterflow, a
correction factor must be applied to the logarithmic mean temperature difference
(LMTD). Figure 3 gives the correction factor to use.
2. Compute the variables for the correction factor
The two variables that determine the correction factor are shown in Fig. 3 as P
ϭ
(t
2
Ϫ
t

1
)/(T
1
Ϫ
t
1
) and R
ϭ
(T
1
Ϫ
T
2
)/(t
2
Ϫ
t
1
). Thus, P
ϭ
(230
Ϫ
100)/(300
Ϫ
100)
ϭ
0.65, and R
ϭ
(300
Ϫ

250)/(230
Ϫ
100)
ϭ
0.385. From Fig. 3, the
correction factor is F
ϭ
0.90 for these values of P and R.
3. Compute the theoretical LMTD
Use the relation LMTD
ϭ
(G
Ϫ
L)/ln(G/L), where the symbols for counterflow
heat exchange are G
ϭ
T
2
Ϫ
t
1
; L
ϭ
T
1
Ϫ
t
2
;ln
ϭ

logarithm to the base e. All
temperatures in this equation are expressed in
Њ
F. Thus, G
ϭ
250
Ϫ
100
ϭ
150
Њ
F
(83.3
Њ
C); L
ϭ
300
Ϫ
230
ϭ
70
Њ
F (38.9
Њ
C). Then LMTD
ϭ
(150
Ϫ
70)/ln (150 /
70)

ϭ
105
Њ
F (58.3
Њ
C).
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HEAT TRANSFER AND HEAT EXCHANGE
11.8
PLANT AND FACILITIES ENGINEERING
FIGURE 3 Correction factors for LMTD when the heater flow path differs
from the counterflow. (Power.)
4. Compute the actual LMTD for this exchanger
The actual LMTD for this or any other heat exchanger is
ϭ
LMTD
actual
ϭ
0.9(105)
ϭ
94.5
Њ
F (52.5
Њ
C). Use the actual LMTD to computeF(LMTD )
computed
the required exchanger heat-transfer area.
Related Calculations. Once the corrected LMTD is known, compute the re-

quired heat-exchanger size in the manner shown in the previous calculation pro-
cedure. The method given here is valid for both two- and four-pass shell-and-tube
heat exchangers. Figure 4 simplifies the computation of the uncorrected LMTD for
temperature differences ranging from 1 to 1000
Њ
F(
Ϫ
17 to 537.8
Њ
C). It gives LMTD
with sufficient accuracy for all normal industrial and commercial heat-exchanger
applications. Correction-factor charts for three shell passes, six or more tube passes,
four shell passes, and eight or more tube passes are published in the Standards of
the Tubular Exchanger Manufacturers Association.
FOULING FACTORS IN HEAT-EXCHANGER
SIZING AND SELECTION
A heat exchanger having an overall coefficient of heat transfer of U
ϭ
100 Btu /
(ft
2
⅐
h
⅐ Њ
F) [567.8 W / (m
2
⅐ Њ
C)] is used to cool lean oil. What effect will the tube
fouling have on the value of U for this exchanger?
Calculation Procedure:

1. Determine the heat exchange fouling factor
Use Table 2 to determine the fouling factor for this exchanger. Thus, the fouling
factor for lean oil
ϭ
0.0020.
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HEAT TRANSFER AND HEAT EXCHANGE
HEAT TRANSFER AND HEAT EXCHANGE
11.9
FIGURE 4 Logarithmic mean temperature for a variety of heat-
transfer applications.
2. Determine the actual U for the heat exchanger
Enter Fig. 5 at the bottom with the clean heat-transfer coefficient of U
ϭ
100 Btu/
(h
⅐
ft
2
⅐ Њ
F) [567.8 W / (m
2
⅐ Њ
C)] and project vertically upward to the 0.002 foul-
ing-factor curve. From the intersection with this curve, project horizontally to the
left to read the design or actual heat-transfer coefficient as U
a
ϭ

78 Btu/(h
⅐
ft
2
⅐
Њ
F) [442.9 W/ (m
2
⅐ Њ
C)]. Thus, the fouling of the tubes causes a reduction of the U
value of 100
Ϫ
78
ϭ
22 Btu / (h
⅐
ft
2
⅐ Њ
F) [124.9 W /(m
2
⅐ Њ
C)]. This means that the
required heat transfer area must be increased by nearly 25 percent to compensate
for the reduction in heat transfer caused by fouling.
Related Calculations. Table 2 gives fouling factors for a wide variety of ser-
vice conditions in applications of many types. Use these factors as described above;
or add the fouling factor to the film resistance for the heat exchanger to obtain the
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HEAT TRANSFER AND HEAT EXCHANGE
11.10
PLANT AND FACILITIES ENGINEERING
TABLE 2
Heat-Exchanger Fouling Factors*
total resistance to heat transfer. Then U
ϭ
the reciprocal of the total resistance. Use
the actual value U
a
of the heat-transfer coefficient when sizing a heat exchanger.
The method given here is that used by Condenser Service and Engineering Com-
pany, Inc.
HEAT TRANSFER IN BAROMETRIC AND JET
CONDENSERS
A counterflow barometric condenser must maintain an exhaust pressure of 2 lb/in
2
(abs) (13.8 kPa) for an industrial process. What condensing-water flow rate is re-
quired with a cooling-water inlet temperature of 60
Њ
F (15.6
Њ
C); of 80
Њ
F (26.7
Њ
C)?
How much air must be removed from this barometric condenser if the steam flow
rate is 25,000 lb / h (11,250 kg / h); 250,000 lb / h (112,500 kg / h)?

Calculation Procedure:
1. Compute the required unit cooling-water flow rate
Use Fig. 6 as a quick guide to the required cooling-water flow rate for counterflow
barometric condensers. Thus, entering the bottom of Fig. 6 at 2-lb /in
2
(abs) (13.8-
kPa) exhaust pressure and projecting vertically upward to the 60
Њ
F (15.6
Њ
C) and
80
Њ
F (26.7
Њ
C) cooling-water inlet temperature curves show that the required flow
rate is 52 gal / min (3.2 L / s) and 120 gal/min (7.6 L /s), respectively, per 1000 lb
/h (450 kg/h) of steam condensed.
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HEAT TRANSFER AND HEAT EXCHANGE
HEAT TRANSFER AND HEAT EXCHANGE
11.11
FIGURE 5 Effect of heat-exchanger fouling on the overall coefficient of heat transfer. (Condenser
Service and Engineering Co., Inc.)
2. Compute the total cooling-water flow rate required
Use this relation: total cooling water required, gal/min
ϭ
(unit cooling-water flow

rate, gal/ min per 1000 lb / h of steam condensed) (steam flow, lb / h) /1000. Or, total
gpm
ϭ
(52)(250,000/1000)
ϭ
13,000 gal/min (820.2 L/ s) of 60
Њ
F (15.6
Њ
C) cooling
water. For 80
Њ
F (26.7
Њ
C) cooling water, total gal /min
ϭ
(120)(250,000/1000)
ϭ
30,000 gal/ min (1892.7 L /s). Thus, a 20
Њ
F (11.1
Њ
C) rise in the cooling-water tem-
perature raises the flow rate required by 30,000
Ϫ
13,000
ϭ
17,000 gal / min (1072.5
L/s).
3. Compute the quantity of air that must be handled

With a steam flow of 25,000 lb / h (11,250 kg / h) to a barometric condenser, man-
ufacturers’ engineering data show that the quantity of air entering with the steam
is 3 ft
3
/min (0.08 m
3
/min); with a steam flow of 250,000 lb /h (112,500 kg/ h), air
enters at the rate of 10 ft
3
/min (0.28 m
3
/min). Hence, the quantity of air in the
steam that must be handled by this condenser is 10 ft
3
/min (0.28 m
3
/min).
Air entering with the cooling water varies from about 2 ft
3
/min per 1000 gal/
min of 100
Њ
F (0.06 m
3
/min per 3785 L/min of 37.8
Њ
C) water to 4 ft
3
/min per 1000
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HEAT TRANSFER AND HEAT EXCHANGE
11.12
PLANT AND FACILITIES ENGINEERING
FIGURE 6 Barometric condenser condensing-water
flow rate.
gal/min at 35
Њ
F (0.11 m
3
/min per 3785 L/ min at 1.7
Њ
C). Using a value of 3 ft
3
/
min (0.08 m
3
/min) for this condenser, we see the quantity of air that must be
handled is (ft
3
/min per 1000 gal / min)(cooling-water flow rate, gal /min)(1000, or
cfm of air
ϭ
(3)(13,000/1000)
ϭ
39 ft
3
/min at 60
Њ

F (1.1 m
3
/min at 15.6
Њ
C). At
80
Њ
F (26.7
Њ
C) ft
3
/min
ϭ
(3)(30,000/1000)
ϭ
90 ft
3
/min (2.6 m
3
/min).
Hence, the total air quantity that must be handled is 39
ϩ
10
ϭ
49 ft
3
/min (1.4
m
3
/min) with 60

Њ
F (15.6
Њ
C) cooling water, and 90
ϩ
10
ϭ
100 ft
3
/min (2.8 m
3
/
min) with 80
Њ
F (26.7
Њ
C) cooling water. The air is usually removed from the baro-
metric condenser by a two-stage air ejector.
Related Calculations. For help in specifying conditions for parallel-flow and
counterflow barometric condensers, refer to Standards of Heat Exchange
Institute—Barometric and Low-Level Jet Condensers. Whereas Fig. 6 can be used
for a first approximation of the cooling water required for parallel-flow barometric
condensers, the results obtained will not be as accurate as for counterflow con-
densers.
SELECTION OF A FINNED-TUBE HEAT
EXCHANGER
Choose a finned-tube heat exchanger for a 1000-hp (746-kW) four-cycle turbo-
charged diesel engine having oil-cooled pistons and a cooled exhaust manifold. The
heat exchanger will be used only for jacket-water cooling.
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HEAT TRANSFER AND HEAT EXCHANGE
HEAT TRANSFER AND HEAT EXCHANGE
11.13
TABLE 3
Approximate Rates of Heat Rejection to Cooling Systems*
Calculation Procedure:
1. Determine the heat-exchanger cooling load
The Diesel Engine Manufacturers Association (DEMA) tabulation, Table 3, lists
the heat rejection to the cooling system by various types of diesel engines. Table
3 shows that the heat rejection from the jacket water of a four-cycle turbocharged-
engine having oil-cooled pistons and a cooled manifold is 1800 to 2200 Btu/
(bhp
⅐
h) (0.71 to 0.86 kW/kW). Using the higher value, we see the jacket-water
heat rejection by this engine is (1000 bhp)[2200 Btu / (bhp
⅐
h)]
ϭ
2,200,000 Btu /
h (644.8 kW).
2. Determine the jacket-water temperature rise
DEMA reports that a water temperature rise of 15 to 20
Њ
F (8.3 to 11.1
Њ
C) is com-
mon during passage of the cooling water through the engine. The maximum water
discharge temperature reported by DEMA ranges from 140 to 180

Њ
F (60 to 82.2
Њ
C).
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HEAT TRANSFER AND HEAT EXCHANGE
11.14
PLANT AND FACILITIES ENGINEERING
Assume a 20
Њ
F (11.1
Њ
C) water temperature rise and a 160
Њ
F (71.1
Њ
C) water dis-
charge temperature for this engine.
3. Determine the air inlet and outlet temperatures
Refer to weather data for the locality of the engine installation. Assume that the
weather data for the locality of this engine show that the maximum dry-bulb tem-
perature met in summer is 90
Њ
F (32.2
Њ
C). Use this as the air inlet temperature.
Before the required surface area can be determined, the air outlet temperature
from the radiator must be known. This outlet temperature cannot be computed

directly. Hence, it must be assumed and a trial calculation made. If the area obtained
is too large, a higher outlet air temperature must be assumed and the calculation
redone. Assume an outlet air temperature of 150
Њ
F (65.6
Њ
C).
4. Compute the LMTD for the radiator
The largest temperature difference for this exchanger is 160
Ϫ
90
ϭ
70
Њ
F (38.9
Њ
C),
and the smallest temperature difference is 150
Ϫ
140
ϭ
10
Њ
F (5.6
Њ
C). In the smallest
temperature difference expression, 140
Њ
F (77.8
Њ

C)
ϭ
water discharge temperature
from the engine
Ϫ
cooling-water temperature rise during passage through the en-
gine, or 160
Ϫ
20
ϭ
140
Њ
F (77.8
Њ
C). Then LMTD
ϭ
(70
Ϫ
10)/[ln(70 / 10)]
ϭ
30
Њ
F (16.7
Њ
C). (Figure 4 could also be used to compute the LMTD).
5. Compute the required exchanger surface area
Use the relation A
ϭ
Q/U
ϫ

LMTD, where A
ϭ
surface area required, ft
2
; Q
ϭ
rate of heat transfer, Btu/ h; U
ϭ
overall coefficient of heat transfer, Btu/(h
⅐
ft
2
⅐
Њ
F). To solve this equation, U must be known.
Table 1 in the first calculation procedure in this section shows that U ranges
from 2 to 10 Btu / (h
⅐
ft
2
⅐ Њ
F) [56.8 W / (m
2
⅐ Њ
C)] in the usual internal-combustion-
engine finned-tube radiator. Using a value of 5 for U, we get A
ϭ
2,200,000/
[(5)(30)]
ϭ

14,650 ft
2
(1361.0 m
2
).
6. Determine the length of finned tubing required
The total area of a finned tube is the sum of the tube and fin area per unit length.
The tube area is a function of the tube diameter, whereas the finned area is a
function of the number of fins per inch of tube length and the tube diameter.
Assume that 1-in (2.5-cm) tubes having 4 fins per inch (6.35 mm per fin) are
used in this radiator. A tube manufacturer’s engineering data show that a finned
tube of these dimensions has 5.8 ft
2
of area per linear foot (1.8 m
2
/lin m) of tube.
To compute the linear feet L of finned tubing required, use the relation L
ϭ
A
/(ft
2
/ft), or L
ϭ
14,650/5.8
ϭ
2530 lin ft (771.1 m) of tubing.
7. Compute the number of individual tubes required
Assume a length for the radiator tubes. Typical lengths range between 4 and 20 ft
(1.2 and 6.1 m), depending on the size of the radiator. With a length of 16 ft (4.9
m) per tube, the total number of tubes required

ϭ
2530/16
ϭ
158 tubes. This
number is typical for finned-tube heat exchangers having large heat-transfer rates
[more than 10
6
Btu/h (100 kW)].
8. Determine the fan hp required
The fan hp required can be computed by determining the quantity of air that must
be moved through the heat exchanger, after assuming a resistance—say 1.0 in of
water (0.025 Pa)—for the exchanger. However, the more common way of deter-
mining the fan hp is by referring to the manufacturer’s engineering data.
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HEAT TRANSFER AND HEAT EXCHANGE
HEAT TRANSFER AND HEAT EXCHANGE
11.15
Thus, one manufacturer recommends three 5-hp (3.7-kW) fans for this cooling
load, and another recommends two 8-hp (5.9-kW) fans. Hence, about 16 hp (11.9
kW) is required for the radiator.
Related Calculations. The steps given here are suitable for the initial sizing
of finned-tube heat exchangers for a variety of applications. For exact sizing, it
may be necessary to apply a correction factor to the LMTD. These correction factors
are published in Kern—Process Heat Transfer, McGraw-Hill, and McAdams—Heat
Transfer, McGraw-Hill.
The method presented here can be used for finned-tube heat exchangers used
for air heating or cooling, gas heating or cooling, and similar industrial and com-
mercial applications.

SPIRAL-TYPE HEATING COIL SELECTION
How many feet of heating coil are required to heat 1000 gal / h (1.1 L/ s) of 0.85-
specific-gravity oil if the specific heat of the oil is 0.50 Btu / (lb
⅐ Њ
F) [2.1 kJ /(kg
⅐
Њ
C)], the heating medium is 65-lb/ in
2
(gage) (448.2-kPa) steam, and the oil enters
at 60
Њ
F (15.6
Њ
C) and leaves at 125
Њ
F (51.7
Њ
C)? There is no subcooling of the con-
densate.
Calculation Procedure:
1. Compute the LMTD for the heater
Steam at 65
ϩ
14.7
ϭ
79.7 lb/ in
2
(abs) (549.5 kPa) has a temperature of approx-
imately 312

Њ
F (155.6
Њ
C), as given by the steam tables. Condensate at this pressure
has the same approximate temperature. Hence, the entering and leaving tempera-
tures of the heating fluid are approximately the same.
Oil enters the heater at 60
Њ
F (15.6
Њ
C) and leaves at 125
Њ
F (51.7
Њ
C). Therefore,
the greater temperature G across the heater is G
ϭ
312
Ϫ
60
ϭ
252
Њ
F (140.0
Њ
C),
and the lesser temperature difference L is L
ϭ
312
Ϫ

125
ϭ
187
Њ
F (103.9
Њ
C).
Hence, the LMTD
ϭ
(G
Ϫ
L)/ [ln(G/L)], or (252
Ϫ
187)/[ln (252 / 187)]
ϭ
222
Њ
F
(123.3
Њ
C). In this relation, ln
ϭ
logarithm to the base e
ϭ
2.7183. (Figure 4 could
also be used to determine the LMTD.)
2. Compute the heat required to raise the oil temperature
Water weighs 8.33 lb / gal (1.0 kg / L). Since this oil has a specific gravity of 0.85,
it weighs (8.33)(0.85)
ϭ

7.08 lb / gal (0.85 kg/L). With 1000 gal/h (1.1 L /s) of oil
to be heated, the weight of oil heated is (1000 gal/h)(7.08 lb/gal)
ϭ
7080 lb/h
(0.89 kg /s). Since the oil has a specific heat of 0.5 Btu / (lb
⅐ Њ
F) [2.1 kJ/ (kg
⅐ Њ
C)]
and this oil is heated through a temperature range of 125
Ϫ
60
ϭ
65
Њ
F (36.1
Њ
C),
the quantity of heat Q required to raise the temperature of the oil is Q
ϭ
(7080
lb/h) [0.5 Btu/(lb
⅐ Њ
F) (65
Њ
F)]
ϭ
230,000 Btu/h (67.4 kW).
3. Compute the heat-transfer area required
Use the relation A

ϭ
Q /(U
ϫ
LMTD), where Q
ϭ
heat-transfer rate, Btu/ h; U
ϭ
overall coefficient of heat transfer, Btu/(h
⅐
ft
2
⅐ Њ
F). For heating oil to 125
Њ
F
(51.7
Њ
C), the U value given in Table 1 is 20 to 60 Btu / (h
⅐
ft
2
⅐ Њ
F) [0.11 to 0.34
kW/(m
2
⅐ Њ
C)]. Using a value of U
ϭ
30 Btu /(h
⅐

ft
2
⅐ Њ
F) [0.17 kW / (m
2
⅐ Њ
C)] to
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HEAT TRANSFER AND HEAT EXCHANGE
11.16
PLANT AND FACILITIES ENGINEERING
produce a conservatively sized heater, we find A
ϭ
230,000/[(30)(222)]
ϭ
33.4 ft
2
(3.1 m
2
) of heating surface.
4. Choose the coil material for the heater
Spiral-type tank heating coils are usually made of steel because this material has a
good corrosion resistance in oil. Hence, this coil will be assumed to be made of
steel.
5. Compute the heating steam flow required
To determine the steam flow rate required, use the relation S
ϭ
where S

ϭ
Q/h ,
ƒg
steam flow, lb /h;
ϭ
latent heat of vaporization of the heating steam, Btu/ lb,h
ƒg
from the steam tables; other symbols as before. Hence, S
ϭ
230,000/901.1
ϭ
256
lb/h (0.03 kg/ s), closely.
6. Compute the heating coil pipe diameter
Steam-heating coils submerged in the liquid being heated are usually chosen for a
steam velocity of 4000 to 5000 ft/min (20.3 to 25.4 m / s). Compute the heating
pipe cross-sectional area a in
2
from a
ϭ
2.4S
v
g
/V, where
v
g
ϭ
specific volume of
the steam at the coil operating pressure, ft
3

/lb, from the steam tables; V
ϭ
steam
velocity in the heating coil, ft / min; other symbols as before. With a steam velocity
of 4000 ft/min (20.3 m / s), a
ϭ
2.4(256)(5.47)/4000
ϭ
0.838 in
2
(5.4 cm
2
).
Refer to a tabulation of pipe properties. Such a tabulation shows that the internal
transverse area of a schedule 40 1-in (2.5-cm) diameter nominal steel pipe is 0.863
in
2
(5.6 cm
2
). Hence, a 1-in (2.5-cm) pipe will be suitable for this heating coil.
7. Determine the length of coil required
A pipe property tabulation shows that 2.9 lin ft (0.9 m) of 1-in (2.5-cm) schedule
40 pipe has 1.0 ft
2
(0.09 m
2
) of external area. Hence, the total length of pipe
required in this heating coil
ϭ
(33.1 ft

2
)(2.9 ft / ft
2
)
ϭ
96 ft (29.3 m).
Related Calculations. Use this general procedure to find the area and length
of spiral heating coil required to heat water, industrial solutions, oils, etc. This
procedure also can be used to find the area and length of cooling coils used to cool
brine, oils, alcohol, wine, etc. In every case, be certain to substitute the correct
specific heat for the liquid being heated or cooled. For typical values of U, consult
Perry—Chemical Engineers’ Handbook, McGraw-Hill; McAdams—Heat Trans-
mission, McGraw-Hill; or Kern—Process Heat Transfer, McGraw-Hill.
SIZING ELECTRIC HEATERS FOR
INDUSTRIAL USE
Choose the heating capacity of an electric heater to heat a pot containing 600 lb
(272.2 kg) of lead from the charging temperature of 70
Њ
F (21.1
Њ
C) to a temperature
of 750
Њ
F (398.9
Њ
C) if 600 lb (272.2 kg) of the lead is to be melted and heated per
hour. The pot is 30 in (76.2 cm) in diameter and 18 in (45.7 cm) deep.
Calculation Procedure:
1. Compute the heat needed to reach the melting point
When a solid is melted, first it must be raised from its ambient or room temperature

to the melting temperature. The quantity of heat required is H
ϭ
(weight of solid,
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HEAT TRANSFER AND HEAT EXCHANGE
HEAT TRANSFER AND HEAT EXCHANGE
11.17
lb)[specific heat of solid, Btu/ (lb
⅐ Њ
F)](t
m
Ϫ
t
t
), where H
ϭ
Btu required to raise
the temperature of the solid,
Њ
F; t
1
ϭ
room, charging, or initial temperature of the
solid,
Њ
F; t
m
ϭ

melting temperature of the solid,
Њ
F.
For this pot with lead having a melting temperature of 620
Њ
F (326.7
Њ
C)
and an average specific heat of 0.031 Btu / (lb
⅐ Њ
F) [0.13 kJ/ (kg
⅐ Њ
C)], H
ϭ
(600)(0.031)(620
Ϫ
70)
ϭ
10,240 Btu/h (3.0 kW), or (10,240 Btu / h) /(3412 Btu /
kWh)
ϭ
2.98 kWh.
2. Compute the heat required to melt the solid
The heat H
m
Btu required to melt a solid is H
m
ϭ
(weight of solid melted, lb)(heat
of fusion of the solid, Btu /lb). Since the heat of fusion of lead is 10 Btu/ lb (23.2

kJ/kg), H
m
ϭ
(600)(10)
ϭ
6000 Btu/h, or 6000/ 3412
ϭ
1.752 kWh.
3. Compute the heat required to reach the working temperature
Use the same relation as in step 1, except that the temperature range is expressed
as t
w
Ϫ
t
m
, where t
w
ϭ
working temperature of the melted solid. Thus, for this pot,
H
ϭ
(600)(0.031)(750
Ϫ
620)
ϭ
2420 Btu / h (709.3 W), or 2420/ 3412
ϭ
0.709
kWh.
4. Determine the heat loss from the pot

Use Fig. 7 to determine the heat loss from the pot. Enter at the bottom of Fig. 7
at 750
Њ
F (398.9
Њ
C), and project vertically upward to the 10-in (25.4-cm) diameter
pot curve. At the left, read the heat loss at 7.3 kWh / h.
5. Compute the total heating capacity required
The total heating capacity required is the sum of the individual capacities, or
2.98
ϩ
1.752
ϩ
0.708
ϩ
7.30
ϭ
12.74 kWh. A 15-kW electric heater would be
chosen because this is a standard size and it provides a moderate extra capacity for
overloads.
Related Calculations. Use this general procedure to compute the capacity re-
quired for an electric heater used to melt a solid of any kind—lead, tin, type metal,
solder, etc. When the substance being heated is a liquid—water, dye, paint, varnish,
oil, etc.—use the relation H
ϭ
(weight of liquid heated, lb) [specific heat of liquid,
Btu/(lb
⅐ Њ
F)] (temperature rise desired,
Њ

F), when the liquid is heated to approxi-
mately its boiling temperature, or a lower temperature.
For space heating of commercial and residential buildings, two methods used
for computing the approximate wattage required are the W /ft
3
and the ‘‘35’’
method. These are summarized in Table 4. In many cases, the results given by these
methods agree closely with more involved calculations. When the desired room
temperature is different from 70
Њ
F (21.1
Њ
C), increase or decrease the required kil-
owatt capacity proportionately, depending on whether the desired temperature is
higher than or lower than 70
Њ
F (21.1
Њ
C).
For heating pipes with electric heaters, use a heater capacity of 0.8 W/ft
2
(8.6
W/m
2
) of uninsulated exterior pipe surface per
Њ
F temperature difference between
the pipe and the surrounding air. If the pipe is insulated with 1 in (2.5 cm) of
insulation, use 30 percent of this value, or 0.24 (W / (ft
2

⅐ Њ
F) [4.7 W / (m
2
⅐ Њ
C)].
The types of electric heaters used today include immersion (for water, oil, plat-
ing, liquids, etc.), strip, cartridge, tubular, vane, fin, unit, and edgewound resistor
heaters. These heaters are used in a wide variety of applications including liquid
heating, gas and air heating, oven warming, deicing, humidifying, plastics heating,
pipe heating, etc.
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HEAT TRANSFER AND HEAT EXCHANGE
11.18
PLANT AND FACILITIES ENGINEERING
FIGURE 7 Heat losses from melting
pots. (General Electric Co.)
TABLE 4
Two Methods for Determining Wattage for Heating Buildings Electrically*
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HEAT TRANSFER AND HEAT EXCHANGE
HEAT TRANSFER AND HEAT EXCHANGE
11.19
FIGURE 8 Temperature vs surface area of economizer.
For pipe heating, a tubular heating element can be fastened to the bottom of the
pipe and run parallel with it. For large-wattage applications, the heater can be
spiraled around the pipe. For temperatures below 165

Њ
F (73.9
Њ
C), heating cable can
be used. Electric heating is often used in place of steam tracing of outdoor pipes.
The procedure presented above is the work of General Electric Company.
ECONOMIZER HEAT TRANSFER COEFFICIENT
A 4530-ft
2
(421-m
2
) heating surface counterflow economizer is used in conjunction
with a 150,000-lb / h (68,040-kg /h) boiler. The inlet and outlet water temperatures
are 210
Њ
F (99
Њ
C) and 310
Њ
F (154
Њ
C). The inlet and outlet gas temperatures are 640
Њ
F
(338
Њ
C) and 375
Њ
F (191
Њ

C). Find the overall heat transfer coefficient in Btu/(h
⅐
ft
2
⅐ Њ
F) [W/ (m
2
⅐ Њ
C)] [kJ / (h
⅐
m
2
⅐ Њ
C)].
Calculation Procedure:
1. Determine the enthalpy of water at the inlet and outlet temperatures
From Table 1, Saturation: Temperatures, of the Steam Tables mentioned under Re-
lated Calculations of this procedure, for water at inlet temperature, t
1
ϭ
210
Њ
F
(99
Њ
C), the enthalpy, h
1
ϭ
178.14 Btu/ lb (414 kJ /kg), and at the outlet temperature,
t

2
ϭ
310
Њ
F (154
Њ
C), the enthalpy, h
2
ϭ
279.81 Btu/lb
m
(651 kJ /kg).
2. Compute the logarithmic mean temperature difference between the gas and
water
As shown in Fig. 8, the temperature difference of the gas entering and the water
leaving,
⌬
t
a
ϭ
t
3
Ϫ
t
2
ϭ
640
Ϫ
310
ϭ

330
Њ
F (166
Њ
C) and for the gas leaving and
the water entering,
⌬
t
b
ϭ
t
4
Ϫ
t
1
ϭ
375
Ϫ
210
ϭ
165
Њ
F (74
Њ
C). Then, the loga-
rithmic mean temperature difference,
⌬
t
m
ϭ

(
⌬
t
a
Ϫ ⌬
t
b
)/[2.3
ϫ
log
10
(
⌬
t
a
/
⌬
t
b
)]
ϭ
(330
Ϫ
165)/[2.3
ϫ
log
10
(330/165)]
ϭ
238

Њ
F (115
Њ
C).
3. Compute the economizer heat transfer coefficient
All the heat lost by the gas is considered to be transferred to the water, hence the
heat lost by the gas, Q
ϭ
w(h
2
Ϫ
h
1
)
ϭ
UA
⌬
t
m
, where the water rate of flow, w
ϭ
150,000 lb /h (68,000 kg / h); U is the overall heat transfer coefficient; heating sur-
face area, A
ϭ
4530 ft
2
(421 m
2
); other values as before. Then, 150,000
ϫ

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HEAT TRANSFER AND HEAT EXCHANGE
11.20
PLANT AND FACILITIES ENGINEERING
FIGURE 9 Temperature vs surface area of boiler tubes.
(279.81
Ϫ
178.41)
ϭ
U(4530)(238). Solving U
ϭ
[150,000
ϫ
(279.81
Ϫ
178.14)]/(4530
ϫ
238)
ϭ
14.1 Btu/(h
⅐
ft
2
⅐ Њ
F) [80 W/(m
2
⅐ Њ
C)] [288 kJ / (h

⅐
m
2
⅐
Њ
C)].
Related Calculations. The Steam Tables appear in Thermodynamic Properties
of Water Including Vapor, Liquid, and Solid Phases, 1969, Keenan, et al., John
Wiley & Sons, Inc. Use later versions of such tables whenever available, as nec-
essary.
BOILER TUBE STEAM-GENERATING CAPACITY
A counterflow bank of boiler tubes has a total area of 900 ft
2
(83.6 m
2
) and its
overall coefficient of heat transfer is 13 Btu/(h
⅐
ft
2
⅐ Њ
F) [73.8 W / (m
2
⅐
K). The boiler
tubes generate steam at a pressure of 1000 lb/in
2
absolute (6900 kPa). The tube
bank is heated by flue gas which enters at a temperature of 2000
Њ

F (1367 K) and
at a rate of 450,000 lb/h (56.7 kg/ s). Assume an average specific heat of 0.25
Btu/(lb
⅐ Њ
F) [1.05 kJ / (kg
⅐
K)] for the gas and calculate the temperature of the gas
that leaves the bank of boiler tubes. Also, calculate the rate at which the steam is
being generated in the tube bank.
Calculation Procedure:
1. Find the temperature of steam at 1000
(
6900 kPa
)
2
lf /in
ƒ
From Table 2, Saturation: Pressures, of the Steam Tables mentioned under Related
Calculations of this procedure, the saturation temperature of steam at 1000 lb / in
2
(6900 kPa), t
s
ϭ
544.6
Њ
F (558 K), a constant value as indicated in Fig. 9.
2. Determine the logarithmic mean temperature difference in terms of the flue-
gas leaving temperature
The logarithmic mean temperature difference,
⌬

t
m
ϭ
(
⌬
t
1
Ϫ ⌬
t
2
)/{2.3
ϫ
log
10
[(t
1
Ϫ
t
s
)/(t
2
Ϫ
t
s
)]}, where
⌬
t
1
ϭ
flue gas entering temperature

ϭ
steam temperature
ϭ
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HEAT TRANSFER AND HEAT EXCHANGE
HEAT TRANSFER AND HEAT EXCHANGE
11.21
(t
1
Ϫ
t
s
)
ϭ
(2000
Ϫ
544.6);
⌬
t
2
ϭ
flue-gas leaving temperature
Ϫ
steam
temperature
ϭ
(t
2

Ϫ
t
s
)
ϭ
(t
2
Ϫ
544.6); (
⌬
t
1
Ϫ ⌬
t
2
)
ϭ
[(2000
Ϫ
544.6)
Ϫ
(t
2
Ϫ
544.6)]
ϭ
(2000
Ϫ
t
2

); [(t
1
Ϫ
t
s
)/(t
2
Ϫ
t
s
)]
ϭ
[(2000
Ϫ
544.6)/(t
2
Ϫ
544.6)]. Hence,
⌬
t
m
ϭ
[(2000
Ϫ
t
2
)/{2.3
ϫ
log
10

[(1455.4)/(t
2
Ϫ
544.6)]}.
3. Compute the flue-gas leaving temperature
Heat transferred to the boiler water, Q
ϭ
w
g
ϫ
c
p
ϫ
(t
1
Ϫ
t
2
)
ϭ
UA
⌬
t
m
, where the
flow rate of flue gas, w
g
ϭ
450,000 lb/h (56.7 kg/s); flue-gas average specific heat,
c

p
ϭ
0.25 Btu /(lb/
Њ
F) [1.05 kg/(kg
⅐
K)]; overall coefficient of heat transfer of the
boiler tubes, U
ϭ
13 Btu/(h
⅐
ft
2
⅐ Њ
F) [73.8 W/(m
2
⅐
K)]; area of the boiler tubes
exposed to heat, A
ϭ
900 ft
2
(83.6 m
2
); other values as before.
Then, Q
ϭ
450,000
ϫ
0.25

ϫ
(2000
Ϫ
t
2
)
ϭ
13
ϫ
900
ϫ
[(2000
Ϫ
t
2
)/{2.3
ϫ
log
10
[(1455.4)/(t
2
Ϫ
544.6)]}. Or, log
10
[(1455.4/(t
2
Ϫ
544.6)]
ϭ
13

ϫ
900/(2.3
ϫ
450,000
ϫ
0.25)
ϭ
0.0452. The antilog of 0.0452
ϭ
1.11, hence, [(1455.4 / (t
2
Ϫ
544.7)]
ϭ
1.11, and t
2
ϭ
(1455.4/1.11)
ϩ
544.6
ϭ
1850
Њ
F (1280 K).
4. Find the heat of vaporization of the water
From the Steam Tables, the heat of vaporization of the water at 1000 lb / in
2
(6900
kPa),
ϭ

649.5 Btu/lb (1511 kJ /kg).h
ƒg
5. Compute the steam-generating rate of the boiler tube bank
Heat absorbed by the water
ϭ
heat transferred by the flue gas, or Q
ϭ
w
s
ϫϭ
h
ƒg
w
g
ϫ
c
p
ϫ
(t
1
Ϫ
t
2
), where the mass of steam generated is w
s
in lb /h (kg /s); other
values as before. Then, w
s
ϫ
649.5

ϭ
450,000
ϫ
0.25
ϫ
(2000
Ϫ
1850)
ϭ
16.9
ϫ
10
6
Btu/h (4950 kJ/s) (4953 kW). Thus, w
s
ϭ
16.9
ϫ
10
6
/649.5
ϭ
26,000 lb / h
(200 kg/s).
Related Calculations. The Steam Tables appear in Thermodynamic Properties
of Water Including Vapor, Liquid, and Solid Phases, 1969, Keenan, et al., John
Wiley & Sons, Inc. Use later versions of such tables whenever available, as re-
quired.
SHELL-AND-TUBE HEAT EXCHANGER
DESIGN ANALYSIS

Determine the heat transferred, shellside outlet temperature, surface area, maximum
number of tubes, and tubeside pressure drop for a liquid-to-liquid shell-and-tube
heat exchanger such as that in Fig. 10, when the conditions below prevail. This
exchanger will be of the single tube-pass and single shell-pass design, with coun-
tercurrent flows of the tubeside and shellside fluids.
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HEAT TRANSFER AND HEAT EXCHANGE
11.22
FIGURE 10 Components of shell-and-tube heat exchanger. This unit has an outside-packed stuffing box. (Chemical Engineering.)
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HEAT TRANSFER AND HEAT EXCHANGE
HEAT TRANSFER AND HEAT EXCHANGE
11.23
Conditions Tubeside Shellside
Flowrate, lb / hr 307,500 32,800 kg / h 139,605 14,891
Inlet temperature,
Њ
C 105 45 F 221 45
Outlet temperature,
Њ
C unknown 90
Viscosity, cp 1.7 0.3
Specific heat, Btu / hr /
Њ
F 0.72 0.9 kJ / hr
Њ

C 3.0 3.7
Molecular weight 118 62
Specific gravity with reference to
water at 20
Њ
C (68
Њ
F) 0.85 0.95
Allowable pressure drop, psi 10 10 kPa 68.9 68.9
Maximum tube length, ft 12 12 m 5.45 5.45
Minimum tube dia., in 5 / 8 5 / 8 mm 15.9 15.9
Material of construction steel (k
ϭ
26)
Calculation Procedure:
1. Determine the heat transferred in the heat exchanger
Use the relation, heat transferred, Btu/ hr
ϭ
(flow rate, lb / hr)(outlet temperature
Ϫ
inlet temperature)(liquid specific heat)(1.8 to convert from
Њ
Cto
Њ
F). Substituting for
this heat exchanger, we have, heat transferred
ϭ
(32,800)(90
Ϫ
45)(0.9)(1.8)

ϭ
2,391,120 Btu/ hr (2522.6 kJ /hr). This is the rate of heat transfer from the hot fluid
to the cool fluid.
2. Find the shellside outlet temperature
The temperature decrease of the hot fluid
ϭ
(rate of heat transfer)/(flow rate, lb/
hr)(specific heat)(1.8 conversion factor). Or, temperature decrease
ϭ
(2,391,120)/
(307,500)(0.72)(1.8)
ϭ
6
Њ
C (10.8
Њ
F). Then, the shellside outlet temperature
ϭ
105
Ϫ
6
ϭ
99
Њ
C (210.2
Њ
F). Then, the LMTD
ϭ
(54
Ϫ

15)/ln (54 / 15)
ϭ
30.4
Њ
C
(86.7
Њ
F)
ϭ ⌬
T
m
.
3. Make a first trial calculation of the surface area of this exchanger
For a first-trial calculation, the approximate surface can be calculated using an
assumed overall heat-transfer-coefficient, U, of 250 Btu/(hr) (ft
2
)(
Њ
F) (44.1 W/
m
2
Њ
C). The assumed value of U can be obtained from tabulations in texts and
handbooks and is used only to estimate the approximate size for a first trial:
22
A
ϭ
2,391,000/(250
ϫ
30.4

ϫ
1.8)
ϭ
175 ft (16.3 m )
Since the given conditions specify a maximum tube length of 12 ft and a min-
imum tube diameter of
5
⁄
8
in, the number of tubes required is:
n
ϭ
175 (12
ϫ
0.1636)
ϭ
89 tubes
and the approximate shell diameter will be:
0.47
D
ϭ
1.75
ϫ
0.625
ϫ
89
ϭ
9 in (228.6 cm)
o
With the exception of baffle spacing, all preliminary calculations have been made

for the quantities to be substituted into the dimensional equations. For the first trial,
we may start with a baffle spacing equal to about half the shell diameter. After
calculating the shellside pressure drop, we may adjust the baffle spacing. Also, it
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HEAT TRANSFER AND HEAT EXCHANGE
11.24
PLANT AND FACILITIES ENGINEERING
is advisable to check the Reynolds number on the tubeside to confirm that the
proper equations are being used.
4. Find the maximum number of tubes for this heat exchanger
To find the maximum number of tubes in parallel that still permits flow in(n )
max
the turbulent region (N
Re
ϭ
12,600), a convenient relationship is n
max
ϭ
W
i
/(2d
i
Z
i
). In this example, n
max
ϭ
307.5/(2

ϫ
0.495
ϫ
1.7)
ϭ
183. For any number of
tubes less than 183 tubes in parallel, we are in the turbulent range and can use Eq.
(1).
From Table 6, the appropriate expressions for rating are: Eq. (1) for tubeside,
Eq. (11) for shellside, Eq. (18) for tube wall, and Eq. (19) for fouling. Eq. (21)
and (25) respectively are used for tubeside and shellside pressure drops.
5. Compute the tubeside and shellside heat transfer
Using the equations from Table 6, Tubeside, Eq. (1):
0.467 0.222 0.2 0.8
⌬
T 1.7
ϫ
118 307.5
ϫ
6 0.495
i
ϭ
10.43
ϫ
ͫͬͫͬͫͬ
0.89 0.2
⌬
T 0.85 30.4 89
ϫ
12

M
ϭ
10.43
ϫ
4.27
ϫ
0.621
ϫ
0.0193
ϭ
0.535
Shellside, Eq. (11):
0.267 0.222 0.4 0.282 0.6
⌬
T 0.3
ϫ
62 32.8
ϫ
45 1
ϫ
5
o
ϭ
4.28
ϫ
ͫͬͫͬͫͬ
0.89 0.718
⌬
T 0.95 30.4 89
ϫ

12
M
ϭ
4.28
ϫ
1.89
ϫ
5.98
ϫ
0.00872
ϭ
0.424
Tube Wall, Eq. (18):
⌬
T 0.72 307.5
ϫ
6 0.625
Ϫ
0.495
w
ϭ
159
ͫͬͫ ͬͫ ͬ
⌬
T 26 30.4 89
ϫ
0.625
ϫ
12
M

ϭ
159
ϫ
0.0277
ϫ
60.7
ϫ
0.000195
ϭ
0.052
Fouling, Eq. (19):
⌬
T 0.72 307.5
ϫ
61
s
ϭ
3,820
ͫͬͫ ͬͫ ͬ
⌬
T 1,000 30.4 89
ϫ
0.625
ϫ
12
M
ϭ
3,820
ϫ
0.00072

ϫ
60.7
ϫ
0.00150
ϭ
0.250
(SOP)*
ϭ
0.535
ϩ
0.424
ϩ
0.052
ϩ
0.250
ϭ
1.261
Because SOP is greater than 1, the assumed exchanger is inadequate. The surface
area must be increased by adding tubes or increasing the tube length, or the per-
formance must be improved by decreasing the baffle spacing. Since the maximum
tube length is fixed by the conditions given, the alternatives are increasing the
number of tubes and / or adjusting the baffle spacing. To estimate assumptions for
the next trial, pressure drops are calculated.
*Sum of the Product—see Related Calculations for data.
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HEAT TRANSFER AND HEAT EXCHANGE
11.25
TABLE 5

Design Features of Shell-and-Tube Heat Exchangers
Design Features
Fixed
Tubesheet
Return Bend
(U-Tube)
Outside-Packed
Stuffing Box
Outside-Packed
Lantern Ring
Pull-Through
Bundle
Inside
Split-Backing Ring
Is tube bundle removable? No Yes Yes Yes Yes Yes
Can spare bundles be used? No Yes Yes Yes Yes Yes
How is differential thermal
expansion relieved?
Expansion joint
in shell
Individual tubes
free to expand
Floating head Floading head Floating head Floating head
Can individual tubes be replaced? Yes Only those in
outside rows
without special
designs
Yes Ye s Yes Yes
Can tubes be chemically cleaned,
both inside and outside?

Yes Yes Yes Yes Yes Yes
Can tubes be physically cleaned on
inside?
Yes With special tools Yes Yes Yes Yes
Can tubes be physically cleaned on
outside?
No With square or
wide triangular
pitch
With square or
wide triangular
pitch
With square or
wide triangular
pitch
With square or
wide triangular
pitch
With square or
wide triangular
pitch
Are internal gaskets and bolting
required?
No No No No Yes Yes
Are double tubesheets practical? Yes Yes Yes No No No
What number of tubeside passes are
available?
Number limited
by number of
tubes

Number limited by
number of U-
tubes
Number limited by
number of tubes
One or two Number limited by
number of
tubes.
Odd number of
passes requires
packed joint or
expansion joint
Number limited by
number of tubes.
Odd number of
passes requires
packed joint or
expansion joint
Relative cost in ascending order,
least expensive
ϭ
1
21 4 3 5 6
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HEAT TRANSFER AND HEAT EXCHANGE