AUTONOMOUS UNIVERSITY OF COAHUILA Brayan Fernando Aguilar Medrano

FACULTY OF CHEMICAL SCIENCES 16180198

Heat transfer
Exercises of exchangers of
heat

1.- To cool oil, a double tube heat exchanger is desired to be used

countercurrent. The inner tube of copper has an internal diameter of 25mm and a
thickness of 1mm. The inner diameter of the outer tube is 45mm. The oil flows through
the exterior with a flow rate of 0.1 kg/s and the water on the inside at a rate of 0.2 kg/s. If the
The inlet water temperature is 30ºC and the oil temperature is 100ºC. Calculate the length of
exchanger to cool the water to a temperature of 60ºC. Note: Coefficient
transfer global referred to the outside area Ue=42W/(m²K).
Properties
D125mm = 0.025m
2.131 kJ
D2=27mm=0.027m Oil at 80°C=
kgK
D345mm = 0.045m
4.183kJ
Water Chamber Cp=
Mass flow rate of water = 0.2 kg/s kgK
oil mass flow = 0.1 kg/s
TFE30°C
TCE=100°C TCS60°C
0.1 kg 2.131 kJ
Oil Q=mCp∆T=
Taking out the TFSWater
( s )( kgK )
( 100−60 )=8.524kW
Q 8.524 kW
Q=mCpΔT=T FS= +T =( )+303.15K=
mCp FE 0.2kg 4.183kJ

∆ T E −∆ T S
( s )(
kgK )
∆ T lm= =43.2044° C
∆TE
ln
( )
∆TS
Q=UeAe∆T
Q 8524W
L= = =55.39
Ueπ D2D e l t a T lm W
( 42 2 )(π )(0.027m)(43.2K)
mK
AUTONOMOUS UNIVERSITY OF COAHUILA Brayan Fernando Aguilar Medrano
FACULTY OF CHEMICAL SCIENCES 16180198

2.- A condenser operates with condensing vapor on the shell side at 27°C. The water
Cooling enters at 5°C and exits at 10°C. If the total heat transfer coefficient
it is 5000 W/m2 -°C based on the outer surface of the tube. Determine the
heat transfer per square meter of the outer tube surface.
q
q=UA ∆Tlm =U∆Tlm
A
∆ T E −∆ T S
∆ T lm=
∆TE
ln
( )
∆TS
∆ T E = ( 27° C−5° C )=22° C

Δ T S = ( 27°C−10° C )=17° C
22−17
∆ T lm= =19
22
ln
( )
17
q
= ( 500W
A

3.- In a tubular heat exchanger steam-water, the dry steam and water
saturated with a pressure of 3.5 x 10^5 Pa, condense on the outer surface of the
tubes. The water, which circulates through the tubes, is heated from 20°C to 90°C. Determine:

a) The logarithmic mean temperature in this heat exchanger.

b) The steam flow rate in the steam-water heat exchanger if the water flow rate is 8 t/h.
Assume that there is no subcooling of the condensate.
P=3.5x105Pa
At this pressure, we have a temperature of T=138.86°C from Cengel 6th edition.

∆ T E −D e l t a T S
∆ T lm=
∆TE
ln
( )
∆TS
∆ T E = ( 138.86°C−20° C ) =118.86°C

∆ T S = ( 138.86°C−90° C ) =48.86°C
a) 118.86−48
∆ T lm=
118.86
b) Liquid
ln
(48.86
water at 55° Cp=4.18kJ/kgK

m l Cp ( Ts−You) =m vh fg

m l Cp ( Ts−You) (800
mv = =
hfg
AUTONOMOUS UNIVERSITY OF COAHUILA Brayan Fernando Aguilar Medrano
FACULTY OF CHEMICAL SCIENCES 16180198

mv =1089.91 kg/h
4.- A shell and tube heat exchanger heats 3.783 kg/s of water from 37.78
C up to 54.44°C. Water is used as the hot fluid, with a mass flow rate of 1.892 kg/s,
with a step in the casing and enters the exchanger at 93.33 °C. The coefficient
The global heat transfer is 1419 W/m2-°C, and the average speed of the water in
the pipes with a diameter of 1.905 cm are at 0.366 m/s. Due to space limitations, the
the length of the tube must not exceed 2.438 m. Calculate the number of pipe steps, the
number of tubes per step, and the length of the tubes compatible with this restriction.
Assuming a one-way pipe
The hot water outlet temperature:

q=mfCpfΔ T f=mcCpc∆ T c =
q 263632.65W
T h.s= +T he.=93.33− =60.01°C
c Cpc 1.892kg4183j

(93.33−54
( s )( kgK )
D e l t a T lm=
(93
ln
(
(6
q 263632.65W
q=UA∆Tlm=A= = =6.23m2
U∆Tlm1419W
( 2 )(29.78° C)
m °C
The density of hot water at 46.09°C:
990.88 kg
ρ=
m3
Then the flow area is given by:
3.783kg
mc s
Af= = =0.0104m3
( ρ)(v) 990.88kg 0.366m
( )(
This area is the prodm3reduction of the
)
s number of tubes and the flow area through the tubes:
2
pdi
A f =n T =T=36.48
4

To find the length of the tube we use the total surface area:

6.23m2=nTπdL=L=2.853m
This length is greater than the allowed 2.438m, so we need more than 1 step.
We obtained a correction value with R=.2 and P.3 in a one-pipe system.
F is approximately 0.88 and we increase one step in the formula:
q 263632.65W
q=UAF∆Tlm=A= = =7.089m2
U∆FTlm 1419W
( 2 )(.88)(29.78° C)
m ° C
7.089m2=2 n TπdL=L=1.623m
AUTONOMOUS UNIVERSITY OF COAHUILA Brayan Fernando Aguilar Medrano
FACULTY OF CHEMICAL SCIENCES 16180198

This length meets the permitted limit then

nt=36.49
number of steps=2
1.623m

A heat exchanger is used to heat water, as described in the

example. Using the same input temperatures of the fluids as in exercise 4,
calculate the outlet temperature of the water when only 40 kg/min of water is heated,
but the same amount of oil is used. Calculate the heat transfer as well
total with these new conditions.
Calculating the Cmin using the data from exercise 4, we have:
1.9 kJ/kg°C

m cC c = ( 1.892 ) ( 4183 )=7914.24W

m fC f = ( 40/60 ) ( 1900
) =12666.6W
The hot fluid is minimized because the maximum heat difference it will receive, the fluid
cold is also capable of receiving it.
Cmin 7914.2
= 0.6248
max12666.6
UA ( 1419) ( 6.23 )
NTUmax=
C min
=
7914.2 =1.12
Where the area and the overall coefficient were taken from exercise 4.
Where the effectiveness is:

N=NTUmax =1.12
ϵ=
1−exp(−N ( 1−C ) )
=
1−exp(−1.12 ( 1−.62
)
) =.58
C
C= min=.62
1−Cexp−N
( ( 1−C ) ) 1−.62exp(−1.12 1−.62
( ) )
Cmax
∆ T water ∆ T water
∈= = =0.58
Δ T max93.33−60
D e l t a T water=19.33
T agua. salida=37.78° C+ 19.33° C=57.11°C
So the heat transfer under the new conditions will be:

q=m c C c ∆ T c = (1.892 ) ( 4183 ) ( 19.33)=152982.18W=152.98 kW

AUTONOMOUS UNIVERSITY OF COAHUILA Brayan Fernando Aguilar Medrano
FACULTY OF CHEMICAL SCIENCES 16180198

6.- To heat 2.36 m3/s of air at 1 atm from 15.5 °C to 29.44 °C, a
fin tube heat exchanger with fins as shown in Figure 10.5. The
hot water enters the tubes at 82.22 °C, and air circulates transversely to the tubes,
with an average global heat transfer coefficient of 227 W/m2-°C. The total area
The surface area of the heat exchanger is 9.29 m2. Calculate the water temperature at the outlet and
the heat flow.
The density of the incoming air is:
P 1.0132x1051.223 kg
ρ= = =
RT(287)( 288.7) m3
The airflow is given by:
2.887 kg
m f= ( 2.36 ) ( 1.223
) s=
q=mf Cpf∆ T f = ( 2.887 ) ( 1006 ) ( 29.44−15.55 )=40341.03W=40.34 kW
Assuming that air is the minimum fluid

2904.32W
m fCp= ( 2.887 ) ( 1006
) = °C
UA ( 227 ) ( 9.29 )
NTUmax=
C min (
=
2904.32 )
=.726

∆ T air 29.44−15.55
∈= = =0.208
∆ T max82.22−15.55

In tables these values do not match so the hot fluid is the minimum

2904.32W UA ∆Th ∆Th

C max=m fCf= NTUmax= ϵ= =
°C C min ∆ T max82.22−15.55
40340W
∆ T h=
C
m in
Iterating in the efficiency table in cross flow from .5 we found an efficiency of .222
which gives us a Cmin=645W/°C

Interpolating the Cp of hot water at 82.22°C from appendix 2.11

4.212 kJ/kgK

645W 645 0.153kg

m cC c = =m c = =
°C 4212 s

The water outlet temperature is:

40340
T h .s =82.22− =19.67° C
645
The heat flow is:
q=40340W
AUTONOMOUS UNIVERSITY OF COAHUILA Brayan Fernando Aguilar Medrano
FACULTY OF CHEMICAL SCIENCES 16180198

7.- A shell and tube heat exchanger is used as an ammonia condenser. The vapor
Ammonia enters the shell as saturated vapor at 50°C. Water enters a
single step tube device at 20°C and the total heat transfer required is
200 kW. The overall heat transfer coefficient is estimated at 1,000 W/m2-°C.
Determine the area needed to achieve the performance of the heat exchanger.
of 60 percent, with an outlet water temperature of 40°C. What percentage of
heat transfer reduction would be achieved if the mass flow of water were reduced to the
half, keeping the heat exchanger area and the value of U equal?
q 200 2.391 kg
q=m fCpf∆ T f=m f = = =
CpfD e l t a ( 4.182
T f ) (40−20 ) s
C min=mfCpf = ( 2.391 ) ( 4.182=10kW
)
ϵ =1−e NTU
e NTU =1−ϵ
NTU=−l n ( 1−ϵ )=−ln ( 1−.6 )=.9162
C minNTU ( 10.9162
)( )
A= = =9.162m2
U 1

If the water flow were reduced to half

mf(2.391kg/s)/2=1.1955kg/s

( 1.000 ) ( 9.162 ) 1832.5

NTU= = =1.832
( 1.1955 ) ( 4.182 ) 1000

ϵ =1−e−1.832=0.839
∆ T =ΔTmax = 0.839
( ) ( 50−20 )=25.17° C
q=C min∆ T watera = ( 5 ) ( 25.17)=125.85 kW

8.- A long steel pipe with an inner diameter (ID) of 5 cm and a thickness of 3.2 mm.
a wall crosses a large room that is kept at 30 °C and atmospheric pressure;
At one end of the pipe, 0.6 kg/s of hot water at 82°C enters. If the pipe is 15 m long.
length, calculate the outlet temperature of the water, considering losses from the
exterior of the pipe, both by natural convection and by radiation.

82−30 41 7.27W
h= (1.32() )=
0.0564 m2 ° C
q withv =( 7.27 ) π ( 0.0564 ) ( 15 ) ( 82−30)=1005W
ε≈0.8
−8 4 4
)
q rad=( 5.669x10π(0.0564)(15)(0.8)( 355−303)=898W
q total=1005W+898W=(0.6)(4187) ∆ T water
∆ T water=0.76°C
T water. exit=82° C−0.76°C=81.24°C
AUTONOMOUS UNIVERSITY OF COAHUILA Brayan Fernando Aguilar Medrano
FACULTY OF CHEMICAL SCIENCES 16180198

9.- In a tube with an inner diameter (ID) of 2.5 cm, air enters at 207 kPa, 200 °C, and 6 m/s.
The tube is made of copper and has a thickness of 0.8 mm and a length of 3 m. In a direction normal to
Outside the tube, ambient air circulates at 1 atm and 20°C, with a free flow velocity.
at 12 m/s. Calculate the air temperature at the exit of the tube. What would be the effect of
reduce the flow of hot air by half?
Inside the tube.
473K-5
P 20
ρ= =
RT (287

0.0385

m= ( 1.525 ) π ( 0.01252) )=4.491x10

(6
−3
kg/
(1.525)(6)(0.025)
R= =8866
2.58x10−5
0.0385 45.4W
hi =
( 0.025 )
( 8866)8 (0.681)0.3 =
m2° C
the resistance by conduction is negligible
Out of the tube

Tf=200+20/2=100°C=373K vf=25.15x10-6kf0.0324

Prf0.69 do0.025 + 0.0016 = 0.0266m

(12)(0.0266)
R= =12691
25.15x10−6

Por lo tanto C=0.193 n=0.618

1
(0.0324)(0.193) 71.36W
h o=
( 0.0266 )
(12,691)0.618(0.69) 3 =
m2° C
Taking U based on Ai
1 28.41W
U I= =
1 0.025 m2° C
( 45.4 )+( ( 0.0266) (71.36 ) ) Cmin
=0
4.626W
C min= ( 4.491x10−3 ) ( 1030
) = °C
Cmax

28.41 ) ( 0.236 )
=1. 42
(
NTU=
4.626
ϵ =0.7
∆ T i= ( 0.78 ) ( 200−20)=140.4° C
AUTONOMOUS UNIVERSITY OF COAHUILA Brayan Fernando Aguilar Medrano
FACULTY OF CHEMICAL SCIENCES 16180198
T exit=200−140.4=59.6° C
If we reduce the flow by half

NTU=( 2)(1.45)=2.9 epsilon=0.95

∆ T i=(0.95)(200−20)=171°C
T exit=(200−171)=29°C

10.- In a tube exchanger with cross-flow fins, hot gases are used.
escape to heat 2.5 kg/s of water from 35 °C to 85 °C. The gases [c = 1.09 kJ/kg-°C]
They enter at 200 °C and exit at 93 °C. The overall heat transfer coefficient is 180 W/m2.
°C. Calculate the area of the heat exchanger using the NTU method.
q=mgCpg∆ T g =magCpand∆ T ag

mg Cpg (200−93)= (2.5 ) ( 4179 ) ( 85−35 )

C min=mgCpg=4882
C max=m agCpag= 2.5
( ) ( 4179 )=10447.5
For performance
ΔT
ϵ=
D e l t a T max
(200−93)
ϵ= =0.65
(200−35)
Cmin4882
= =0.467
Cmax10447.5

Cmin
Locating the value of NTU at a =0.467 yϵ =¿0.65
Cmax
NTU=1.4
C minNTU ( 4882 ) ( 1.4 )
A=
U
=
180 =37.97m2
11.- A mass flow of water of 230 kg/h at 35 °C is available for use as
Refrigerant in a heat exchanger with a total surface area of 1.4 m2. It is going to
use water to cool oil [c, = 2.l kJ/kg-°C] from an initial temperature of
120°C. Due to other circumstances, it is not possible to have a water outlet temperature.
greater than 99 °C. The outlet temperature of the oil must not be lower than 60 °C.
The overall heat transfer coefficient is 280 W/m2-°C. Estimate the mass flow rate.
maximum oil that can be cooled, assuming that the mass flow of water has been fixed at
230 kg/h.
AUTONOMOUS UNIVERSITY OF COAHUILA Brayan Fernando Aguilar Medrano
FACULTY OF CHEMICAL SCIENCES 16180198
239 267W
C min=
q=m
max (
CpΔT=
3600
ag
(
and
4180
ag)
maximum temperature3600
) 230
=( )
°C ( 4180 ) ( 99−35 )=17090W
differences

D e l t a =T( oil
120−60=60°C
)
∆ T water= ( 99−35 )=64° C
280 ) ( 1.4 )
=1.47
(
NTU=
267
64
ϵ= =0.753
120−35
Cmin
For counter current and double tube this can be fulfilled with ≈0.2
Cmax
267 2100J
C oil =
( )
0.2
=1335=m oilCpoil=moil (
kg° C
)

0.636kg
m oil =
17090
( 1335 )
T accepte . salida=120−
Which is more than the 60 degrees allowed.
=107.2° C

q=
(2500 ) ( 2202
3600
)( ) 1000 =1.53MW
90−80
Δ t lm= =84.9° C
90
ln( )
80 1,530,000
A= =383m2
(84.9)(47)

12.- A small casing and tube exchanger with a tube flow [A = 4.64 m2 and V = 280
W/m2-°C], will be used to heat water at high pressure, initially at 20°C, with air
initially heated to 260 °C. If the outlet temperature of the water does not exceed 93 °C
and the mass flow of air is 0.45 kg/s, calculate the mass flow of water.
q=(0.45)( 1009)(260−Exit) =Q1

q=(280)(4.64)¿ =Q2

Temperature °C Q1 Q2
50 95350.5 103675.891
60 90810 115455.001
40 99891 89990.0396
45 97620.75 97143.2008
AUTONOMOUS UNIVERSITY OF COAHUILA Brayan Fernando Aguilar Medrano
FACULTY OF CHEMICAL SCIENCES 16180198

45.5 97393.725 97821.1555

45.6 97348.32 97956.0291
45.4 97439.13 97686.0447
45.3 97484.535 97550.6954
45.2 97529.94 97415.106
45.25 97507.2375 97482.9308
45.26
45.27 97498.1565 97510.0439
45.265 97500.4268 97503.2665
45.266 97499.9727 97504.622
45.264 97500.8808 97501.911
45.263 97501.3349 97500.5554
45.2635, 97501.1078, 97501.2332

Iterating gives us an output temperature of = 45.26350C

( 0.45 ) ( 1009 ) ( 260−45.2635 ) 0.32kg

mwater= =
(4175)(93−20)

A shell and tube heat exchanger operates with two shell passes and four tube passes.
tube steps, The fluid of the casing is ethylene glycol, which enters at 140°C and exits at 80°C, with a
mass flow rate of 4500 kg/h. Water circulates through the pipes, entering at 35 °C and exiting at 85 °C.
The overall heat transfer coefficient for this device is 850 W/m2-°C. Calculate it.
the mass flow of water needed and the area of the heat exchanger.
ethylene glycol Cp=2742J/kgK
4500
q=( )(2742)(140−80)=205650W
3600

q=m water4175
( ) ( 85−35 )
0.985kg
m water=
s

C max=m waterCp=(0.985)(4175)=4112.4
4500) ( 2742 )
=3428
(
C min=
3600

Cmin
=0.8634
Cmax
140−80
ϵ= =0.571
1 4 0 −35
NTU=1.3
(3428)(1.3)
A= =5.24m2
850
AUTONOMOUS UNIVERSITY OF COAHUILA Brayan Fernando Aguilar Medrano
FACULTY OF CHEMICAL SCIENCES 16180198

14.- A supply water heater uses a shell and tube heat exchanger with
water vapor that condenses at 120 °C in a shell passage. Water enters the tubes at 30
C and take four steps, so that the global value of U is 2,000 W/m-°C. Calculate the
changing area for a mass flow of water of 2.5 kg/s, with an outlet temperature
from water at 100 °C.
90−2 0
D e l t a t lm= =4 6 . 5 4 ° C
90
ln( )
20
5
q=m waterCp∆Twater= 2.5( 4180
)( 100−30=7.315x10W
)( )
q 7.315x105
A= = =7.86m2
U∆tlm(2000)(46.45)

15.- In a large air conditioning installation, 1500 m3/min needs to be heated.

from air at 1 atm and 10°C in a finned tube heat exchanger, using water
hot water enters the exchanger at 80 °C. The overall heat transfer coefficient is 50
W/m2-°C. Calculate the area of the heat exchanger needed for a temperature of
air outlet temperature of 35 °C and a water outlet temperature of 50 °C.

P 1.0132x105 1.2467 kg
ρ= = =
RT(287)( 283.15) m3

For the flow of the cold fluid:

31.167
m f= ( 1500/60 ) ( 1.2467
) = s
q=mf Cpf∆ T f = ( 31.1675 ) ( 1006 ) ( 35−10=783862.62W=783.862kW
)
(80−35)−(
∆ T lm=
80−
ln
(
50−
Th1−Th2 80−50
R= = =1.2
Tc2−Tc135−10
Tc2−Tc1 35−10
P= = =0.35
Th1−Tc180−10
From the tables we obtain that the correction factor is approximately F=0.91

∆ T lm=(42.45°C)(0.91)=38.6295° C
From the equation:

UA ∆ T lm solving for A
q 783.86
A= = =405.8349m2
U∆Tlm .05
( )( 38.6295 )