ME3122
Heat Transfer
Tutorial
Tutorial Solution – Solution – Heat Exchangers
1. Hot exhaust gases used in a finned‐tube cross‐flow heat exchanger heat 2.5 kg/s of water from 35 to 85C. The gases [cp = 1.09 kJ/kgC] enter at 200C and leave at 93C. The overall 2
heat‐transfer coefficient is 180 W/m C. Calculate the area of the heat exchanger using (a) 2
the LMTD approach and (b) the effectiveness‐NTU method. (37.8 m ) (c) If the water flow rate is reduced by half, while the gas flow rate is maintained constant along with the fluid inlet temperatures. Calculate the percentage reduction in heat transfer as a result of this reduced flow rate. Assume that the overall heat transfer coefficient remains the same. (15%) Solution:
(a) T LMTD
P R
Hot gases T 1 = 200C
T1 T 2
ln T1
200 85 93 35 ln 115 58
Water
83.3C
t 1
to
t i
Ti
t i
Ti
T o
to
t i
T 2
85 35 200 35
200 93 85 35
0.30
35C
w m
U
t 2
= 180 W/m2K
2.5 kg/s
2.14
T 2
= 93C
T
From crossflow hx chart (Figure 3 of Handbook), 200
F 0.92
wcwt qm
UA
F TL MTD
2.5 4180 50 180 A 0.92 83.3
85
Hot gases
A 37.8 m 2
93 Water 35 A 1
2
= 85C
(b) Energy balance:
g cg Tg m
wcw Tw m
g cg 200 93 2.5 4.180 85 35 522.5 kW m g cg m w cw m C min C max
4.883 kW/ C C min gas
2.5 4.180 10.450 kW/ C C max water 4.883 10.450
0.47
200 93
0.65 200 35 Using -NTU chart (Figure 9 of Handbook),
NTU A
UA C min
1.4
1.4 4883 180
38.0 m 2
(c)
g cg m
w cw m
NTU C min C max
4.883 kW/C C min 2.5
4.180
2 UA
C min
1.4
0.467 0.5
10.450 2
5.225 kW/C C max
(no change)
0.93
Using -NTU chart (Figure 9 of Handbook),
0.55 T g
200 35 Tg 90.75 200 T g ,exit T g ,exit 109.25C
q Cg T g
q a q c q c
4.883 90.75 443.1 kW
522.5 443.1 522.500
15%
2. A shell and tube heat exchange operates with two shell passes and four tube passes. The shell fluid is ethylene glycol, which enters at 140C and leaves at 80C with a flow rate of 4500 kg/h. Water flows in the tubes, entering at 35C and leaving at 85C. The 2
overall heat‐transfer coefficient for this arrangement is 850 W/m C. Calculate the flow 2
rate of water required and the area of the heat exchanger. (0.984 kg/s, 5.24 m ) The flow rate of glycol to the exchanger is reduced in half with the entrance temperatures of both fluids remaining the same. What is the water exit temperature under these new conditions, and by how much is the heat‐transfer rate reduced? (70.9C, 28.2%) Solution:
(a) Energy balance:
wcw Tw m
g c g Tg m
w 4.180 85 35 m
4500 3600
2.742
140 80 205.7 kW
w m
0.984 kg/s
wcw m
0.984 4.180 4.113 kW/ C C max
g cg m C min C max
4500
140 2.742 3.428 kW/ C C min
3600 3.428 4.113
T
85
Ethylene glycol
0.83
140 80
0.57 140 35 Using -NTU chart (Figure 8 of Handbook),
80 Water 35
NTU 1.3 UA C min
A 1.3
A
1
1.3 3.428 103 850
5.24 m 2
2
(b)
g cg m
4500 2
2.742
3.428
3600 2 wcw 4.113 kW/C C max m UA
NTU
C min C max
C min
0.833 2
1.3 0.5
1.714
kW/C
=2.6
0.42
Using -NTU chart (Figure 8 of Handbook),
0.82 T g
140 35 Tg 86.1 140 T g 2 T g 2
53.9C
q C g T g
q a q c q c
147.6
3.428 2
86.1 147.6
205.7 147.6 205.7
35.9 T w1 35 4.113 T w1 35 35.9 70.9C Tw
kW
28.2%
C min
3. A small steam condenser is designed to condense 0.76 kg/min of steam at 85 kPa with cooling water at 10C. The exit water temperature is not to exceed 57C. The overall 2
heat‐transfer coefficient is 3400 W/m C. Calculate the area required for a double‐pipe 2
heat exchanger. Use both LMTD and effectiveness – NTU methods. (0.145 m ) Solution: T
(a)
Steam Tsat = 95C
Steam is condensed at 85 kPa h fg
2270 kJ/kg
q
s h fg m
T LMTD
0.76 60
2270
57
28.75 kW
T1 T 2
ln T1
T 2
Water
95 10 95 57 ln 85 38
10
58.3C
A 1
q UAT LMTD A
28.75 10 3 3400 58.3
0.145 m 2
(b) C max
steam
C min
water
q C min 57 10 28.75 103 C min
611.7 W/ C
57 10
0.553 95 10 For heat exchanger with steam condensing,
1 e
NTU A
NTU
UA C min
0.553
0.805
0.805 611.7 3400
0.145 m 2
2
4. A shell‐and‐tube exchanger operates with one shell pass and two tube passes with a 2
total surface area of 47.5 m . Water (the tube‐side fluid) enters the heat exchanger at 15C and 6.5 kg/s and is heated by exhaust gas entering at 200C and 5 kg/s. The gas may be assumed to have the properties of atmospheric air, and the overall heat transfer 2
coefficient is approximately 200 W/m . What are the gas and water outlet temperatures? (55.7C, 41.6C) Solution:
wcw m
6.5 4.200 27.3 kW/C C max
g cg m
5 1.005 5.025 kW/ C C min
C min C max
NTU
5.025 27.3 UA C min
T 200
0.184 Tw1
200 47.5 5.025 10
3
Gas
1.89
Using -NTU chart (Figure 7 of Handbook),
Tg2
0.78
Water 15
200 T g 2
0.78 200 15 T g 2 55.7C
Energy balance: Cw Tw
C g Tg
27.3 T w1 15 5.025 200 55.7 T w1
41.6C
A 1
2
5. A single‐pass cross‐flow heat exchanger uses hot exhaust gases (mixed) to heat water (unmixed) from 30 to 80C at a rate of 3 kg/s. The exhaust gases, having thermophysical properties similar to air, enter and exit the exchanger at 225 and 100C, respectively. If 2
the overall heat transfer coefficient is 200 W/m K, estimate the required surface area. 2
(33.2 m ) Solution:
LMTD Method
P R
225 80 100 30 103C T 2 ln 145 70
T1 T 2
T LMTD
ln T1
to
t i
Ti
t i
Ti
T o
to
t i
80 30
225 30
225 100
80 30
0.26 T
2.5 225
From crossflow hx chart (Figure 4 of Handbook), F 0.92
80
wcwTw qm
UA
F TLMTD
Gas
3 4200 50 200 A 0.92 103 A
33.2 m 2
100 Water
-NTU Method
30
Energy balance:
g cg Tg m
A
wcwTw m
1
g cg 125 3 4.200 50 m g cg m wcw m
C mixed C unmixed
5.04 kW/C C min
3 4.200 12.6 kW/C C max
5.04 12.6
0.4
225 100
0.64 225 30 Using -NTU chart (Figure 10 of Handbook),
NTU 1.3 UA C min A
1.3
1.3 5.04 103 200
32.8 m 2
2
6. The oil in an engine is cooled by air in a cross‐flow heat exchanger where both fluids are unmixed. Atmospheric air enters at 30C and 0.53 kg/s. Oil at 0.026 kg/s enters at 75C and flows through a tube of 10 mm diameter. If the overall convection coefficient is 53 2
2
W/m K and the total heat transfer area is 1m , determine the effectiveness. What are the exit temperatures of the oil and water? (46.2C, 32.9C) Solution:
oco m
0.026 2047 53.2 W/ C C min
a ca m
0.53 1005 533 W/ C C max
C min C max
53.2 533 UA
NTU
C min
T
0.10 75
53 1 53.2
1.0
Ta1
Using -NTU chart (Figure 9 of Handbook),
Oil
0.64
75 T o 2
To2
0.64
Air
75 30 T o 2 46.2C
30
Energy balance: Ca Ta1 T a 2 Co To1 T o2
533 T a1 30 53.2 75 46.2 T a1
32.9C
A 1
2
