1. in a counter flow double pipe heat exchanger, oil is cooled … year/heat and mass... ·...
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1. In a counter flow double pipe heat exchanger, oil is cooled from 85˚C to 55˚C by water
entering at 25˚C. The mass flow rate of oil is 9,800 kg/h and specific heat f oil is 2000 J/kg K.
The mass flow rate of water is 8,000 kg/h and specific heat of water is 4180 J/kg K. Determine
the heat exchanger area and heat transfer rate for an overall heat transfer co-efficient of 280
W/m2 K.
Given:
Hot fluid – oil, Cold fluid - water --------------------------
(T1, T2) (t1, t2) Water
--------------------------
Entry temperature of oil, T1 = 85˚C Oil
--------------------------
Exit temperature of oil T2 = 55˚C Water
--------------------------
Mass flow rate of oil (Hot fluid), mh = 9,800 kg/h
skg /3600
800,9
Specific heat of oil, Cph = 2000 J/kg K
Mass flow rate of water (cold fluid), mc = 8,000 kg/h
skg /3600
000,8
Specific heat of water, Cpc = 4180 J/kg K
Overall heat transfer co-efficient, U = 280 W/ m2K
mh = 2.72 kg/s
mc = 2.22 kg/s
To find:
1. Heat exchanger area, (A)
2. Heat transfer rate, (Q)
Solution:
We know that,
Heat lost by oil (Hot fluid) = Heat gained by water (cold fluid)
Qh = Qc
Ct
t
tt
ttCmTTCm pccphh
5.42
109.2316.9279102.163
418022.25585200072.2
2
3
2
3
12
1221
Heat transfer,
WQ
Q
TTCmorttCmQ phhpcc
3
2112
10162
255.42418022.2
We know that,
Heat transfer, mTUAQ …….. (1)
Where
mT - Logarithmic Mean Temperature Difference. (LMTD)
For Counter flow,
Exit temperature of water, Ct 5.422
2555
5.4285
25555.4285
12
21
1221
In
T
tT
tTIn
tTtTT
m
m
Substitute mT U and Q values in Equn (1)
8.3528010162
)1(
3
A
TUAQ m
==>
Result:
1. Heat exchanger area, A = 16.16 m2
2. Heat transfer, Q = 162 103 W
2. Water flows at the rate of 65 kg/min through a double pipe, counter flow heat exchanger.
Water is heated from 50˚C to 75˚C by oil flowing through the tube. The specific heat of the oil
is 1.780 kJ/kg K. The oil enters at 115˚C and leaves at 70˚C. The overall heat transfer co-
efficient is 340 W/m2K. Calculate the following
1. Heat exchanger area
2. Rate of heat transfer
Given:
Hot fluid – oil, Cold fluid - water
(T1, T2) (t1, t2)
Mass flow rate of oil (cold fluid), mc = 65 kg/min
skg /60
65
mT = 35. 8˚C
A = 16.16 m2
mc = 1.08 kg/s
Entry temperature of water, t1 = 50˚C
Exit temperature of water, t2 = 75˚C
Specific heat of oil (Hot fluid) Cph = 1.780 kJ/kg K
= 1.780103 J/m
2K
Entry temperature of oil, T1 = 115˚C
Exit temperature of oil, T2 = 70˚C
Overall heat transfer co-efficient, U = 340 W/m2K
To find:
1. Heat exchanger area, (A)
2. Heat transfer rate, (Q)
Solution:
We know that,
Heat transfer, 2112 TTCmorttCmQ phhpcc
5075418608.1
12
Q
TTCmQ pch
We know that,
Heat transfer, mTAUQ …….(1)
Where
mT - Logarithmic Mean Temperature Difference. (LMTD)
For Counter flow,
Specific heat of water, Cpc = 4186 J/kgK
5070
75115
507075115
12
21
1221
In
T
tT
tTIn
tTtTT
m
m
Substitute mT U and Q and U values in Equn (1)
8.2834010113
)1(
3
A
TUAQ m
==>
Result:
1. Heat exchanger area, A = 11.54 m2
2. Heat transfer, Q = 113 103 W
3. In a counter flow single pass heat exchanger is used to cool the engine oil from 150˚C to
55˚C with water, available at 23˚C as the cooling medium. The specific heat of oil is 2125 J/kg
K. The flow rate of cooling water through the inner tube of 0.4 m diameter is 2.2 kg/s. The
flow rate of oil through the outer tube of 0.75 m diameter is 2.4 kg/s. If the value of the overall
heat transfer co-efficient is 240 W/m2 K, how long must the heat exchanger be to meet its
cooling requirement?
Given:
Hot fluid – oil, Cold fluid - water
(T1, T2) (t1, t2)
Entry temperature of oil, T1 = 150˚C
Exit temperature of oil, T2 = 55˚C
Entry temperature of water, T1 = 23˚C
Specific heat of oil (Hot fluid) Cph = 2125 kJ/kg K
mT = 28. 8˚C
A = 11.54 m2
Inner diameter, D1 = 0.4 m
Flow rate of water (cooling fluid), mc = 2.2 kg/s
Outer diameter, D2= 0.75 m
Flow rate of oil (Hot fluid), mh = 2.4 kg/s
Overall heat transfer co-efficient, U = 240 W/m2K
To find:
Length of the heat exchanger, L
Solution:
We know that,
Heat lost by oil (Hot fluid) = Heat gained by water (cold fluid)
2341862.25515021254.2 2
1221
t
ttCmTTCm
pccphh
ch
3
2
3 102112.9209105.484 t
Heat transfer, 2112 TTCmorttCmQ phhpcc
236.7541862.2 Q
We know that,
Heat transfer mTUAQ …… (1)
Where
mT - Logarithmic Mean Temperature Difference. (LMTD)
[ Specific heat of water, Cpc = 4186 J/kgK]
WQ 3104.484
For Counter flow,
2355
6.75150
23556.75150
12
21
1221
In
T
tT
tTIn
tTtTT
m
m
Substitute mT , U and Q values in Equn (1)
2.5024010104.484
)1(
33
A
TUAQ m
==>
We know that,
Area, A = LD 1
40.20 = 0.4 L
==>
Result:
Length of the heat exchanger, L = 31.9 m.
4. In an oil cooler for a lubrication system, oil is cooled from 70˚C to 40˚C by using a cooling
water flow at 25˚C. The mass flow rate of oil is 900 kg/h and the mass flow rate of water is 700
kg/h. Give your choice for a parallel flow or counter flow heat exchanger, with reasons. If the
overall heat transfers co-efficient is 20 W/m2K, find the area of the heat exchanger. Take
specific heat of oil is 2 kJ/ kg˚C.
Given:
Hot fluid – oil, (T1, T2) Cold fluid - water (t1, t2)
Entry temperature of oil, T1 = 70˚C
Exit temperature of oil, T2 = 40˚C
mT = 50.2˚C
A = 40.20 m2
L = 31.9 m
Entry temperature of water, T1 = 25˚C
The mass flow rate of oil, hkgmh /900
skg
skg
/25.0
/3600
900
The mass flow rate of water, hkgmc /700
skg
skg
/197.0
/3600
700
Overall heat transfer co-efficient, U = 20 W/m2K
Specific heat of oil, Cph = 2 kJ/kg˚C
= 2 103 J/kg˚C
To find:
1. Choice of heat exchanger (Whether parallel flow or counter flow)
2. Area of heat exchanger.
Solution:
We know that,
Heat lost by oil (Hot fluid) = Heat gained by water (Cold fluid)
254186194.0407010225.0 2
3
1221
t
ttCmTTCm
pccphh
ch
10.302,2008.812000,15 2 t
==> t2 = 43.47˚C
[ Specific heat of water, Cpc = 4186 J/kg K]
> T2
Since t2 > T2, counter flow arrangement should be used.
We know that,
Heat transfer mTUAQ …… (1)
For Counter flow,
569.0
53.11
2540
47.4370
254047.4370
12
21
1221
In
T
tT
tTIn
tTtTT
m
m
…….. (2)
We know that,
Heat transfer,
407010225.0 3
21
1221
TTCmQ
ttCormTTCmQ
phh
pccphh
…….. (3)
Substitute, Q, U, and mT values in equation (1)
26.2020000,15
A
TUAQ m
==>
mT = 20.26˚C
Q = 15,000 J/s
A = 37.02 m2
Exit temperature of water, t2 = 43.47˚C
Result:
1. Choice of heat exchanger – counter flow arrangement
2. Surface area, A = 37.02 m2.
5. In a refrigerating plant water is cooled from 20˚C to 7˚C by brine solution entering at - 2˚C
and leaving at 3˚C. The design heat load is 5500 W and the overall heat transfer co-efficient is
800 W/m2 K. What area required when using a shell and tube heat exchanger with the water
making one shell pass and the brine making two tube passes.
Given:
Hot fluid – water, Cold fluid – water
(T1, T2) (t1, t2)
Entry temperature of water, T1 = 20˚C
Exit temperature of water, T2 = 7˚C
Entry temperature of water, T1 = - 2˚C
Exit temperature of brine solution, t2 = 3˚C
Heat load, Q = 5500 W
Overall heat transfer co-efficient, U = 800 W/m2K
To find:
Area required (A)
Solution:
Shell and tube heat exchanger – One shell pass and two tube passes
For shell and tube heat exchanger (or) cross flow heat exchanger.
mTFUAQ [Counter flow] …….. (1)
Where
F – Correction factor
mT - Logarithmic mean temperature difference for counter flow.
For Counter flow,
27
320
27320
12
21
1221
In
tT
tTIn
tTtTT m
From graph,
22
5
220
23,
11
12
tT
ttPvalueX axis
Curve value, 5
13
23
720
12
21
tt
TTP
Xaxis value is 0.22, curve value is 2.6, and corresponding Y axis value is 0.94.
Substitute mT Q , U and F value in Equn (1)
57.1280094.05500
)1(
A
TFUAQ m
==>
Result:
Area of heat exchanger, A = 0.58 m2
mT = 12.57˚C
P = 0.22
A = 0.58 m2
R = 2.6
6. A parallel flow heat exchanger is used to cool 4.2 kg/min of hot liquid of specific heat 3.5
kJ/kg K at 130˚C. A cooling water of specific heat 4.18 kJ/kg K is used for cooling purpose at
a temperature of 15˚C. The mass flow rate of cooling water is 17 kg/min. calculate the flowing.
1. Outlet temperature of liquid
2. Outlet temperature of water
3. Effectiveness of heat exchanger
Take
Overall heat transfer co-efficient is 1100 W/m2K.
Heat exchanger area is 0.30 m2
Given:
Mass flow rate of hot liquid, mh = 4.2 kg/min
Specific heat of hot liquid, Cph = 3.5 kJ/kg K
Inlet temperature of hot liquid, T1 = 130˚C
Specific heat of water, Cpc = 4.18 kJ/kg K
Inlet temperature of cooling water, t1 = 15˚C
Mass flow rate of cooling water, mc = 17 kg/min
Overall heat transfer co-efficient, U = 1100 W/m2K
Area, A = 0.30 m2
To find:
1. Outlet temperature of liquid, (T2)
2. Outlet temperature of water, (t2)
mh = 0.07 kg/s
kgKJC ph /105.3 3
kgKJC pc /1018.4 3
mc = 0.28 kg/s
3. Effectiveness of heat exchanger,
Solution:
Capacity rate of hot liquid, phh CmC
3105.307.0
……….. (1)
Capacity rate of water, phc CmC
31018.428.0
………… (2)
From (1) and (2),
Cmin = 245 W/K
Cmax = 1170.4 W/K
209.04.1170
245
max
min C
C
……………. (3)
Number of transfer units, NTU = minC
UA
245
03.01100 NTU
…….. (4)
From graph,
Xaxis NYU = 1.34
C = 245 W/K
C = 1170.4 W/K
max
min
C
C= 0.209
NTU = 1.34
Curve 209.0max
min C
C
Corresponding Yaxis value is 64%
i.e.,
Maximum possible heat transfer
Qmax = Cmin (T1 – t1)
= 245 (130 – 15)
Actual heat transfer rate
175,2864.0
max
We know that,
Heat transfer, 12 ttCmQ pcc
64.0
Qmax = 28,175 W
Q = 18,032 W
Ct
t
tt
40.30
175564.11700323,18
1018.428.0032,18
2
2
52
3
We know that,
Heat transfer, 21 TTCmQ pch
Ct
T
T
4.56
245318500323,18
130105.307.0032,18
2
2
2
3
Result:
1. T2 = 56.4˚C
2. t2 = 30.40˚C
3. = 0.64
7. In a counter flow heat exchanger, water at 20˚C flowing at the rate of 1200 kg/h. it is heated
by oil of specific heat 2100 J/kg K flowing at the rate of 520 kg/h at inlet temperature of 95˚C.
Determine the following
1. Total heat transfer
2. Outlet temperature of water
3. Outlet temperature of oil
Take
Overall heat transfer co-efficient is 100 W/m2 K.
Heat exchangers are is 1 m2.
Given:
Capacity rate of hot oil, C = phh Cm
Outlet temperature of cold water, Ct 40.302
Outlet temperature of hot liquid, CT 4.562
2100144.0
………… (1)
Capacity rate of water, C = phh Cm
418633.0
………… (2)
Inlet temperature of hot liquid, T1 = 130˚C
Specific heat of water, Cpc = 4.18 kJ/kg K
From Equn (1) and (2),
Cmin = 302.4 W/K
Cmax = 1381.3 W/K
218.03.1381
4.302
max
min C
C
……………. (3)
Number of transfer units, NTU = minC
UA
4.302
11100 NTU
…….. (4)
KWC /3.1381
kgKJC pc /1018.4 3
max
min
C
C= 0.218
NTU = 3.3
C = 302.4 W/K
From graph,
Xaxis NYU = 3.3
Curve 218.0max
min C
C
Corresponding Yaxis value is 0.95
i.e.,
218.0max
min C
C
Maximum possible heat transfer
Qmax = Cmin (T1 – t1)
= 302.4 (95 – 20)
Actual heat transfer rate
680,2295.0
max
95.0
Qmax = 22,680 W
Q = 21,546 W
We know that,
Heat transfer, 12 ttCmQ pcc
20418633.0546,21 2 t
kgKJC pc /4186
6.627,2738.1381546,21 2 t
==>
We know that,
Heat transfer, 21 TTCmQ pch
2
2
4.302728,28546,21
952100144.0546,21
T
T
==>
Result:
1. T2 = 21,546W
2. T2 = 23.75˚C
3. t2 = 35.5˚C
8. In a counter flow heat exchanger, water is heated from 20˚C to 80˚C by an oil with a
specific heat of 2.5 kJ/kg – K and mass flow rate of 0.5 kg/s. The oil is cooled from 110˚C to
40˚C. If the overall heats transfer co-efficient is 1400 W/m2K, find the following by using
NTU method
1. Mass flow rate of water
2. Effectiveness of heat exchanger
3. Surface area
Outlet temperature of water, Ct 5.352
Outlet temperature of oil, CT 75.232
t2 = 35.5˚C
T2 = 23.75˚C
Given:
Hot fluid – oil, Cold fluid - water
(T1, T2) (t1, t2)
Inlet temperature of water, t1 = 20˚C
Outlet temperature of water, t2 = 80˚C
Specific heat of oil, Cph = 2.5 kJ/kg - K
= 2.5 103 J/kg - K
The mass flow rate of oil, skgmh /5.0
Inlet temperature of oil, T1 = 110˚C
Outlet temperature of oil, T2 = 40˚C
Overall heat transfer co-efficient, U = 1400 W/m2 K
To find:
1. Mass flow rate of water, mc
2. Effectiveness of heat exchanger,
3. Surface area, A
Solution:
We know that,
Heat lost by oil = Heat gained by water
2080418640110105.25.0 3
1221
c
pccphh
ch
m
ttCmTTCm
==> mc = 0.348 kg/s
Mass flow rate of oil (Hot fluid), C = mhCph
[ Specific heat of water, Cpc = 4186 J/kg K]
3105.25.0
……… (1)
Capacity rate of water (Cold fluid), C = mcCpc
4186348.0
………… (2)
From Equn (1) and (2),
Cmin = 1250 W/K
Cmax = 1456.73 W/K
858.073.1456
1250
max
min C
C
………… (3)
We know that,
Effectiveness, 11
21
TT
TT
20110
40110
From graph,
Xaxis 77.0
Curve 858.0max
min C
C
C = 1250 W/K
C = 1456.73 W/K
max
min
C
C= 0.858
77.0