Coker A.K. Fortran Programs for Chemical Process Design, Analysis, and Simulation
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Heat Transfer 667
and laid on 1-in. square pitch. The bundles are arranged for two tube
passes with segmental baffles spaced 5 in. apart.
Physical property data:
Physical property Acetone Acetic Acid
Viscosity, cP 0.20 0.85
Density, lb/ft 3 49.296 66.768
Specific heat, Btu/lb.~ 0.63 0.51
Thermal conductivity,
Btu/lb.ftZ(~ 0.095 0.098
Exchanger:
Solution
Hot Fluid (Acetone)
Shell side
ID=21 1/4 in.
Baffle space = 5in.
Passes = 1
Cold Fluid (Acetic Acid)
Tube side
Number and length = 270, 16'0"
OD, BWG, pitch = 3/4in., 14 BWG, 1 in. square.
Passes = 2
Heat Balance:
Heat lost by acetone to storage = heat gained by acetic acid from storage
that is,
W.Cp.(T 2 - T,) - W.Cp.(t 2 -t~)
(60,000)(0.57)(250 - 100) = w(0.51)(150 - 90)
w -- 168,000 lb/hr
The computer program PROG82 rates the shell and tube heat exchanger
using 168,000 lb/hr of acetic acid in the tube side. Table 8-22 gives the
input data and computer output of the heat exchanger design. A mini-
mum of three shell passes is required to permit the transfer of heat with
the temperatures given by the process. The overall heat transfer coeffi-
cient is 79 Btu/hrftZ~ and the heat transfer surface is 2545 ft 2. Three
exchangers are sufficient for the heat transfer, although the pressure
668 Fortran Programs for Chemical Process Design
Table 8-22
Input Data and Computer Output for the Rating
of a Shell and Tube Heat Exchanger
DATA82.DAT
ACETIC ACID
270.0 0.584 0.75 1.0
S
16.0 2.0 0.001 90.0
150.0 0.51 0.85 66.768
0.098 168000.0
ACETONE
]..0 21.25 38.0 5.0
0.003 250.0 i00.0 0.63
0.20 49.296 0.095 60000.0
*SIIELL AND TUBE HEAT EXCHANGER RATING*
FLUID TYPE:
FLUID FLOW RATE , Ib/hr."
SPECIFIC HEAT CAPACITY, Btu/ib.oF:
FLUID DENSITY, ib/ft^3:
FLUID VISCOSITY, cP:
FLUID THERMAL COND. Btu/ft.h.oF:
FLUID INLET TEMPERATURE,
oF:
FLUID OUTLET TEMPERATURE,
oF:
INSIDE DIAMETER, inch:
OUTSIDE DIAMETER, inch:
TUBE PITCH, inch:
TUBE LENGTH, ft.:
NUMBER OF TUBES:
NUMBER OF PASSES:
NUMBER OF BAFFLES:
BAFFLE SPACING, inch:
FLUID VELOCITY, ft/sec.:
REYNOLDS NUMBER:
PRANDTL NUMBER:
FRICTION FACTOR (ft^2/in^2):
HEAT COEFF., Btu/hr.ft^2.oF:
PRESSURE DROP,
psi:
FOULING FACTOR"
APPROACH FACTOR, F:
LOG MEAN TEMP. DIFF.
oF:
CORRECTED MEAN TEMP. DIFF. oF:
OVERALL HEAT COEFF., Btu/hr.ft^2.oF:
OUTSIDE AREA OF UNIT, ft^2:
HEAT LOAD ON UNIT, Btu/hr:
HEAT TRANSFERRED, Btu/hr:
.............
TUBE SIDE SHELL SIDE
ACETIC ACID ACETONE
168000. 60000.
.510 .630
66.768 49.296
.850 .200
.098 .095
9O. 250.
150. i00.
.584 21.250
.750
1.000 SQUARE PITCH
16.000
270.
6. 3.
38.
5.0
2.783 1.833
15828. 53072.
10.7048 3.2097
.00024 .00153
211.8 253.4
1.5682 3.4464
.0010 .0030
.8749
39.09
34.20
79.0
2544.7
5140800.
6870014.
drop on the shell side is higher than the permissible pressure drop. Fewer
exchangers will be inadequate for the service.
Problem 8-3
Petroleum distillate oil (28 ~ API) is to be cooled in a double pipe finned
tube exchanger with cooling-tower water under the following conditions:
Heat Transfer 669
Tube side: Water flow = 23,310 lb/hr
Tin =
80~
Tou t =
120~
Fol = 0.003 la = 0.72 cP
K = 0.366 Btu/fthr~
Cp = 1 Btu/lb ~
Density = 62.3 lb/ft 3
Number of passes = 5
Shell side: Oil flow = 18,000 lb/hr
Tin--
250~ Tou , = 150~
Fol = 0.001 la = 2.45cP
K = 0.074 Btu/ft hr~
Cp - 0.518 Btu/lb~
Density = 45 lb/ft 3
Number of passes = 1
Exchanger size:
Shell side: 3 in. (3.068 in. ID, 3.5 in. OD)
Tube side: 1 1/2 in. (1.61 in. ID, 1.9 in. OD)
Tube length: 12 ft
Fins: 24, 0.5 in. high x 0.035 in. wide
Thermal conductivity of tube material
K = 25 Btu/flhr~
Pressure drops of 10 psi are allowable on both streams, and the fouling
factors of 0.001 for the distillate oil and 0.003 for the water are
required. Calculate the overall heat transfer coefficient, and the exchanger
fin efficiency.
Exchanger:
Solution
Hot Fluid (Oil)
Shell side
ID = 3.068 in.
Passes = 1
Cold Fluid (Water)
Tube side
ID = 1.61 in. OD = 1.9 in.
Tube length = 12"0"
Passes = 5
670
Fortran Programs for Chemical Process Design
Physical property data:
Physical property
Viscosity, cP
Density, lb/ft 3
Specific heat, Btu/lb.~
Thermal conductivity, Btu/lbftZ(~
Oil
2.45
45.0
0.518
0.074
Water
0.72
62.3
1.00
0.366
Heat Balance:
Heat lost by oil = heat gained by water
that is,
Q = W.Cp.(Y,- Y2) = w.cp.(t 2 - t,)
Q = (18000)(0.518)(250- 150) = w(1)(120- 80)
932400 = 40w
w = 233 l0 lb/hr
The flow of water required to cool the oil is 23310 lb/hr. The computer
program PROG83 rates the double pipe heat exchanger using either fin
tubes or bare tubes. Table 8-23 lists the input data and output of the
computer program. The fin efficiency is 43.3 percent and the computed
overall heat transfer coefficient is 27.3 Btu/hrftZ~ The pressure drops
have not been exceeded, and the double pipe exchanger will be satisfac-
tory for the service.
Problem 8-4
Calculate a preliminary air-cooler design using the following data.
A light hydrocarbon liquid is to be cooled from 170~ to 135~ in an
air cooler. The heat transfer rate is 5.0 MM Btu/hr, and the estimated
overall heat transfer is 60 Btu/hr
ftZ~
The design dry bulb air tempera-
ture is 85~ and the tube length is 40ft.
Solution
The computer program PROG84 calculates the air cooled heat
exchanger area, the face area and the fan horsepower. Table 8-24 shows
both the input data and computer output of the design calculations. The
exchanger area is 1602 ft 2 and the LMTD is 52~ Face area of the bundle
Heat Transfer 671
Table 8-23
Input Data and Computer Output for the Double Pipe
or Longitudinal Finned Tube Heat Exchanger
DATA83.DAT
WATER
LIQUID
1.61 1.9 23310.0
1.0 0.72 0.366
80.0 120.0 0.003
62.3 12.0 5
OIL
FINS
LIQUID
24
3.068 3.5 18000.0
0.518 2.45 0.074
250.0 150.0 0.001
0.5 0.035
25.0 45.0 1
DOUBLE PIPE HEAT EXCHANGER RATING
USING BARE-TUBES/LONGITUDINAL FINNED TUBES
FORCED CONVECTION WITH NO CHANGE OF PHASE
TUBE SIDE SHELL SIDE
FLUID NAME:
TYPE OF PHASE FLOW:
FLUID FLOW RATE, ib/hr.:
SPECIFIC HEAT CAPACITY, Btu/Ib.oF:
FLUID DENSITY, ib/ft^3.:
FLUID VISCOSITY, cP:
FLUID THERMAL COND., Btu/Ib.oF:
FLUID INLET TEMPERATURE,
oF:
FLUID OUTLET TEMPERATURE,
oF:
INSIDE DIAMETER, inch:
OUTSIDE DIAMETER, inch:
FOULING FACTOR:
TUBE LENGTH, ft.:
NUMBER OF TUBE PASSES:
NUMBER OF FINS:
FIN HEIGHT, inch:
FIN THICKNESS, inch:
THERMAL COND. TUBE MATERIAL, Btu/hr.ft.oF:
FLUID VELOCITY, ft/sec.:
REYNOLDS NUMBER:
PRANDTL NUMBER:
FRICTION FACTOR (ft^2/in^2):
PRESSURE DROP,
psi.:
FIN MODULUS:
FIN EFFICIENCY, %:
EFFECTIVE SURFACE EFFICIENCY, %:
LOG MEAN TEMP. DIFF., oF:
HEAT TRANSFER FILM COEFF. CORRECTED
FOR FOULING:, Btu/hr.ft^2.oF:
OVERALL HEAT TRANSFER COEFF., Btu/hr.ft^2.
HEAT LOAD ON UNIT, Btu/hr.:
HEAT TRANSFER SURFACE ft^2.:
WATER OIL
LIQUID LIQUID
23310. 18000.
1.000 .518
62.300 45.000
.720 2.450
.366 .074
80.000 250.000
120.000 150.000
1.610 3.068
1.900 3.500
.003 .001
12.000
5
24
.500
.035
25.000
7.351 3.867
126958. 3759.
4.7607 41.503
.000031 .000042
2.9424 .5943
2.258
43.335
44.486
96.92
264.33
27.26
932400.0
352.932
107.04
is 283ft 2, and the fan horsepower is 25.2 bhp. The required air flow is
164266 std ft3/min.
Problem 8-5
Determine the number of tracers required to maintain 100 ft of 8-in.
(Schedule 40) process line (ID - 7.981 in.) at 500~ Hot tracing
672 Fortran Programs for Chemical Process Design
Table 8-24
Input Data and Computer Output for
An Air-Cooled Heat Exchanger Design
DATA84.DAT
170.0 135.0 85.0
60.0 5000000.0 40.0
AN AIR-COOLED HEAT EXCHANGER DESIGN
INLET PROCESS FLUID TEMPERATURE, oF:
OUTLET PROCESS FLUID TEMPERATURE, oF:
INLET AIR TEMPERATURE, oF:
OVERALL HEAT-TRANSFER COEFFICEINT, Btu/hr.ft^2.oF:
HEAT LOAD, Btu/hr.:
TUBE LENGTH, ft.:
NUMBER OF TUBE ROWS:
RATIO OF BARE-TUBE SURFACE TO FACE AREA OF BUNDLE:
FACE VELOCITY OF AIR, ft./min.:
ESTIMATED AIR OUTLET TEMPERATURE, oF:
EFFECTIVE LOG MEAN TEMPERATURE DIFFERENCE, oF:
BARE-TUBE SURFACE AREA BASED ON TUBE OD, ft^2:
FACE AREA OF BUNDLE, ft^2:
CALCULATED OUTLET AIR TEMPERATURE, oF:
THE DIFFERENCE BETWEEN THE CALCULATED AIR
OUTLET TEMPERATURE AND THE ESTIMATED AIR OUTLET
TEMPERATURE IS GREATER THAN 0.5
EFFECTIVE LOG MEAN TEMPERATURE DIFFERENCE, oF:
BARE-TUBE SURFACE AREA BASED ON TUBE OD, ft^2:
FACE AREA OF BUNDLE, ft^2:
CALCULATED OUTLET AIR TEMPERATURE, oF:
CALCULATED BARE-TUBE SURFACE AREA
BASED ON TUBE OUTSIDE DIAMETER, ft^2:
FACE AREA OF BUNDLE, ft^2:
AIR FLOW OVER TUBES, std.ft^3/min.:
FAN HORSEPOWER, bhp:
AIR-COOLER WEIGHT, ib:
TUBE BUNDLE WIDTH, ft.:
170.0
135.0
85.0
60.0
5000000.
40.0
4.5
5.7
580.740
105.250
57.058
1460.5
257.9
115.916
TAO-TAOI >0.5
52.015
1602.1
282.9
113.184
1602.1
282.9
164266.
25.2
22169.8
7.1
medium is available at 630~ and has a heat capacity of 0.53 Btu/lb~
The process line is covered with 2.5 in. of insulation, and this insulation
has a thermal conductivity of 0.033 Btu/(hr)(ftz)(~ Design for 0~
air temperature and 20 mph winds. Use 1/2 in. tracers, the hot medium
outlet temperature is 550~ Allow approximately 1 1/4 in. between pipe
and insulation to accommodate a 1/2 in. in tracer line and heat transfer
cement. Assuming the trial film heat transfer coefficient is 4.0 Btu/hrft 2
~ and the thermal conductances tracer to 1/2 in. pipe are a = 0.393 and
b = 4.58 Btu/hr ~ of pipe).
Solution
The computer program PROG85 calculates the heat tracer require-
ments and heat loss for an insulated pipe line either with tracing or with-
out tracing. Table 8-25 lists the input data and computer output of the
Heat Transfer 673
Table 8-25
Input Data and Computer Output for
Tracer Requirements for Pipelines
DATA85.DAT
I00.0 500.0 630.0
550.0 0.0 7.981
1.25 2.5 0.033
0.53 4.0 0.393
4.58
HEAT TRACER REQUIREMENTS FOR PIPELINES AND
HEAT LOSS FROM AN INSULATED PIPELINE WITH TRACING
*****************************************************************************
TOTAL PIPELINE LENGTH, ft.:
TEMPERATURE IN PIPE,
oF:
HOT-MEDIUM INLET TEMPERATURE,
oF:
HOT MEDIUM OUTLET TEMPERATURE, oF:
AIR TEMPERATURE,
oF:
OUTSIDE DIAMETER OF PIPE, in.:
ALLOWANCE FOR TRACER DIAMETER, in.:
INSULATION THICKNESS, in.:
THERMAL CONDUCTIVITY OF INSULATION, Btu/hr.ft^2(oF/ft
SPECIFIC HEAT OF HOT MEDIUM, Btu/Ib.oF:
FILM HEAT TRANSFER COEFFICIENT TO AIR
CORRECTED FOR WIND, Btu/ft^2.hr.oF:
THERMAL CONDUCTANCE TRACER TO PIPE
WITHOUT CEMENT, Btu/hr.oF.ft of
pipe:
THERMAL CONDUCTANCE TRACER TO PIPE
WITH CEMENT, Btu/hr.oF.ft of
pipe:
OUTSIDE SURFACE TEMPERATURE OF INSULATION, oF:
THE COMBINED CONVECTIVE AND RADIATIVE
HEAT TRANSFER COEFFICIENT, Btu/h.oF.ft^2:
WIND FACTOR:
HEAT TRANSFER COEFFICIENT TO AIR
CORRECTED FOR THE WIND, Btu/hr.ft^2.oF:
HEAT LOSS PER FOOT OF PIPE, Btu/hr.ft:
TOTAL HEAT LOSS FROM PIPELINE, Btu/hr.:
FLOW RATE OF HOT MEDIUM,
ib/hr.:
THE NUMBER OF TRACERS WITHOUT HEAT TRANSFER CEMENT:
THE NUMBER OF TRACERS WITH HEAT TRANSFER CEMENT:
i00.0
500.0
630.0
550.0
.0
7.981
1.25
2.50
.033
.530
4.000
.393
4.580
16.972
1.330
2.807
3.734
252.934
25293.430
596.543
7.15
.61
HEAT LOSS FROM AN INSULATED PIPE WITHOUT TRACING
*****************************************************************************
OUTSIDE SURFACE TEMPERATURE OF INSULATION, oF: 15.202
THE COMBINED CONVECTIVE AND RADIATIVE
HEAT TRANSFER COEFFICIENT, Btu/h.oF.ft^2: 1.344
WIND FACTOR: 2.808
HEAT TRANSFER COEFFICIENT TO AIR
CORRECTED FOR THE WIND, Btu/hr.ft^2.oF: 3.774
HEAT LOSS PER FOOT OF PIPE, Btu/hr.ft: 206.652
TOTAL HEAT LOSS FROM PIPELINE, Btu/hr.: 20665.240
problem. The approximate number of tracers without the heat transfer
cement of an 8-inch (Schedule 40) process line is 7. The heat transfer
cement reduces the number required to 1. The circulation rate of the
tracing medium is 596.5 lb/hr, and the heat lost from 100 ft. of pipe line
is 25293.4 Btu/hr.
674 Fortran Programs for Chemical Process Design
Problem 8-6
(Internal Coil, Isothermal Heating Medium)
A tank containing 55,000 lb (24947.6 kg) of material with a specific
heat of 0.5 Btu/lb ~ (2.093 kJ/kg.K) is heated from 72~ (295.4K) to
260~ (399.8K). The tank contains a heating coil with a heat transfer
surface of 110
ft 2 (10.22m2),
and the overall heat transfer coefficient
from the coil to the tank contents is 150 Btu/(hr)(ft2)(~ W/mZ.K).
Calculate the time required to heat the tank contents with steam con-
densing at 350~ (449.8K).
Solution
When heating a batch reactor with an internal coil under isothermal
condition, Equation 8-135 applies"
T I - t I ] _ UA
In TI t~ Mc
(8- 35)
The time, 0, required to heat the tank contents with steam condensing at
350~ is,
T I -
t t ]
In TI-t
0__ 2
hr
UA '
ln( 350- 72 /
350- 260
150 x 110 ]
55,000 • 615
= 1.87 hr
Problem 8-7
Batch Cooling: External Heat Exchanger (Counterflow),
Non-isothermal Cooling Medium
For the tank described in PROBLEM 8-6, calculate the time required
to cool the batch from 260~ (399.8K) to 150~ (338.7K), if an external
Heat Transfer
675
heat exchanger with a heat transfer surface of
250ft 2
(23.23m 2) is avail-
able. The batch material is circulated through the exchanger at the rate
of 30,000 lb/hr (13608 kg/h). Cooling water is available at a tempera-
ture of 85~ (302.6K) and a flow rate of 10,000 lb/hr (4536 kg/h). The
overall heat transfer coefficient in the heat exchanger is 250 Btu/hr
ft 2
~ (1419.5
W/m 2
K).
Solution
Equation 8-169 applies when cooling a batch with an external heat
exchanger and a non-isothermal cooling medium.
/T tl)/Ks
In TI- t2 - M
WbWcCc /0
KsW~C ~
-
~rbC
(8-169)
where
K5 exp{uA/' ')t
Wbc W~Cc
Using the values in Table 8-26, K 5 is calculated as follows"
K 5 - exp{250 • 250 / 1
30,000 • 0.5
10,000 • 1.0
= 0.1245
The time, 0, required to cool the batch is determined from Equation 8-169,
ln{260- 85} _/0"125 - 1 /
260 - 85 55,000
3 o, ooo( ~ o, ooo)( ~. o) /o
o. 1245(10, 0-00)iil 0)- i5()-i000)(0.5)
0 - 2.854 hr
Table 8-26
A =250
W b -
30,000 M
=55,000
C =0.5
T 2 -
150~
U =250 Wc- 10,000 Cc- 1.0
T1 -
260~ tl - 85~
676
Fortran Programs for Chemical Process Design
Problem
8-8
8000 gal (30.28m 3) of liquid hydrocarbon under pressure at 350~
(449.8K) is required for a batch extraction process. The storage tem-
perature of the liquid is 100~ (310.9K). Available as a heating medium
is a 12,000 lb/hr (5443.2 kg/h) 28 ~ API oil stream at a temperature of
400~ (477.6K). A pump connected to the batch is capable of circulat-
ing 50,000 lb/hr (22680.0 kg/h) of the hydrocarbon. Available for the
service is 400 ft 2 (37.16m 2) of clean double pipe exchanger surface that
gives a U c of 50 Btu/hrft2~ (283.9 W/m2.K) in counterflow streams.
Calculate for the flow rates:
1. The time required to heat the agitated batch using the double pipe
countercurrent flow exchanger.
2. The time required using a 1-2 exchanger with the same surface
and coefficient.
3. The time required using a 2-4 exchanger with the same surface
and coefficient.
Physical property data:
Liquid hydrocarbon
Specific gravity: 0.88
Specific heat capacity: 0.48 Btu/lb ~ (2.01kJ/kg.K)
28 ~ API oil:
Specific heat capacity: 0.6 Btu/lb ~ (2.51 kJ/kg.K)
Solution
Volume of liquid hydrocarbon in the batch - 8000 gal
Specific gravity - 0.88
Mass (M) of liquid hydrocarbon in the batch - 8000 x 0.88 x 8.34
= 58713.6
59,000 lb.
Table 8-27
V - 8000 W h - 12,000 C h - 0.6 A - 400 t~ - 100~
Wb - 50,000 Uc - 50 c - 0.48 T1 - 400~ t2 - 350~