me3122e - tutorial solution 4
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ME3122E - Tutorial Solution 4TRANSCRIPT
1
Problem Set 4-Solutions 1. Hot exhaust gases used in a finned-tube cross-flow heat exchanger heat 2.5kg/s of water from 35 to 85oC. The gases [cp = 1.09 kJ/kg.K] enter at 200oC and leave at 93oC. The overall heat-transfer coefficient is 180W/m2K. Calculate the area of the heat exchanger using (a) the LMTD approach and (b) the effectiveness – NTU method. (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. [Ans: (a) 37.8 m2 (b) 37.8 m2 (c) 15%.] Solution: a)
1 2
1 2
2 1
1 1
1 2
2 1
ln( )
(200 85) (93 35)
ln(115 / 58)
83.3
85 350.303
(200 35)
200 932.14
(85 35)
From crossflow hx chart : 0.92
. .
2.5 4180 50 0.92 180 83.3
LMTD
LMTD
T TT
T T
C
t tP
T t
T TR
t t
F
Q mC t UA F T
A
A
237.8m
U=180W/m2K
Water t1=350C m =2.5
Hot gases T1=2000C
t2 =850C
T2 =930C
A
T
Th1 200
Th2 93
Tc1 85
Tc2 35
1 2
2
b)
min
max
min
max
min
2
:
(200 93) 2.5 4180 (85 35)
4883 ( )
2.5 4180 10450 ( )
48830.467
2.5 4180
200 930.648
200 35
using chart 1.4,
1.4 488337.98
180
g g
g g
w w
g
Energy balance
m c
m c C gas
m c C water
C
C
UANTU NTU
C
A m
c)
min
max
min
min
max
2
2
4883
2.5 4180 / 2 5225
1.4 (no change)
4883/ 5225 0.935
0.55
200 3590.75 200
109.25
4883 90.75
( ) ( ) 4.18 2.5 50 488
( )
g g
w w
g
g g
g
g g
m c C
m c C
UANTU
C
C
C
Cross flow hx chart
T
T T
T C
Q C T
Q a Q c
Q a
3 90.7515%
4.18 2.5 50
3
2. A shell-and-tube heat exchanger operates with two shell passes and four tube passes. The shell fluid is ethylene glycol, which enters at 140oC and leaves at 80oC with a flow rate of 4500 kg/h. Water flows in the tubes, entering at 35oC and leaving at 850C. The overall heat-transfer coefficient for this arrangement is 850 W/m2K. Calculate the flow rate of water required and the area of the heat exchanger. [Ans: 0.984 kg/s, 5.24 m2].
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? [Ans: 70.9oC, 28.2%]
Solution:
3max
min
min
max
) 4500 / 3600 / , 2.742 /
45002.742 (140 80) 205.7
36004.180 (85 35)
0.984 /
0.984 4.180 10 4113
45002742 3428
3600
34280.833
4113
140 800.
140 35
g g
w
w
w
g
g
i m kg s c kJ kgK
Q kW
m
m kg s
C C
C C
C
C
min
2o
2
571
/ 1.3
850W/m C.
1.30 3428 / 850 5.24
Hence NTU chart gives
NTU UA C
given U
A m
A
T
140
80
85
35
1 2
Ethylene glycol
water
4
min
min
min
max
1 2
2
2
) 4500 /(2) 3600
3428 / 2
1.302.6
0.5
0.833/ 2 0.417
0.82
0.82;140 35
140 86.1
53.9
(3428 / 2) 86.1 147.6
( ) ( ) 205.7 147.
( )
g
g
g gg
g
og
g g
ii m
C C
UANTU
C
C
C
chart
T T
T
T C
Q C T kW
Q i Q ii
Q i
1
1
628.2%
205.7
147.635.9 35
0.984 4.18
35 35.9 70.9
water w
w
T T
T C
5
3. A small steam condenser is designed to condense 0.76 kg/min of steam at 85 kN/m2 with cooling water at 10oC. The exit water temperature is not to exceed 57oC. The overall heat-transfer coefficient is 3400 W/m2K. Calculate the area required for a double-pipe heat exchanger. Use both LMTD and effectiveness-NTU methods.
[Ans: 0.145m2].
Solution:
g
2
fg
f
1 2LMTD
1 2
2
i) Steam is condensed at 85KN/m
h =2.27 MJ/kg
0.76 Q = m h 2270 = 28.75kW
60
Tln( / )
(95-10)-(85-57) = 58.3
ln(85 / 38)
Q=UA LMTD
28.75 A= 0.145
3.4 58.3
T T
T T
C
m
2 1
1 1
max min
c c
c
min
min
min
ii) C ,
T T 57 10 = 0.553
T T 95 10
For heat exchanger with steam condensing, 1
=0.553=1 , giving NTU =0.805
UANTU= 0.805,
Q= (57 10) 28753
611.77
Su
h
NTU
NTU
steam C water
e
e
C
C
C
2bstituting qives A=0.145m
10
57
Steam Tsat =950C
A
1 2
water
AA
T
6
4. A shell-and-tube heat exchanger consists of 135 thin-walled tubes in a double-pass arrangement, each of 12.5 mm diameter with a total surface area 47.5 m2. Water (the tube-side fluid) enters the heat exchanger at 15oC and 6.5 kg/s and is heated by exhaust gas entering at 200oC and 5 kg/s. The gas may be assumed to have the properties of atmospheric air, and the overall heat transfer coefficient is approximately 200 W/m2. What are the gas and water outlet temperatures? Assuming fully developed flow, what is the tube-side convection coefficient? [Ans: To(water) = 41.6oC, h = 2320 W/m2K.]
Solution: i)
2 2
min
max
min
max
5 / ; 1005 /
6.5 / ; 4200 /
200 / ; 47.5
5 1005
6.5 4200
5 10050.184
6.5 4200
200 47.51.89
5 1005Shell and tube Heat Exchanger
-NTU chart
g pg
w pw
g
w
m kg s c J kgK
m kg s c J kgK
U W m K A m
C C
C C
C
C
NTU
2 2
2
1
1
g
min
0.78
C (200 ) 2000.78
(200 15) 200 15
55.7
5 1005 (200 55.7) 6.5 4200 ( 15)
41.6
g g
g
w
w
T T
C
T C
Q T
T C
A
T 2000C
Tg2
Tw1
15
1 2
Gas
water
7
ii)
bw
2
3 6
0.8 0
w
Water in tube: properties are evaluated at
15 41.6T 28.3
2
uRe ;
4
4 4 (6.5 /135)Re
(12.5 10 ) 855 10
=5736>2000 turbulent tube flow
hdNu= 0.023Re Pr
k
d
d
C
d dm u
m
d
.4
w
2
Substituting for Pr=5.83, k 0.613 / , 0.0125
gives h=2320W/m
W mK d m
K
8
5. A single-pass cross-flow heat exchanger uses hot exhaust gases (mixed) to heat water (unmixed) from 30oC to 80oC at a rate of 3 kg/s. The exhaust gases, having thermophysical properties similar to air, enter and exit the exchanger at 225oC and 100oC, respectively. If the overall heat transfer coefficient is 200W/m2K, estimate the required surface area. [Ans: 35.2m2].
Solution:
2 1
LMTD
3 / ; 4.2 / ,
1.005 /
Use LMTD Method, counter flow equivalent
all the temperatures are known
q=m ( ) 3 4.20 (80 30) 630kJ/s
(225-80)-(100-30)T = 103
ln(145/70)
R(or Z)=
w w
g
w w w w
m kg s c KJ kgK
c KJ kgK
c T T
C
LMTD
2
min
225-1002.5
80-3080 30
0.26225 30
for single pass, crossflow hx, 1mix, 1unmixed,
correction factor F=0.92
Hence
q=630=UAF T =0.200 A 0.92 103
giving A=33.2m
method just as straight forward,
C
P
NTU
1 2
1 2
max
min
max
,
C
CCalculate
C
225 1100.64
225 30
, ,
gas g g
water w w
g g
g w
C m c
C m c
T T
T T
from chart get NTU hence A
A
T 2250C
100
80
30
1 2
Gas
water
9
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 30oC and 0.53 kg/s. Oil at 0.026 kg/s enters at 75oC and flows through a tube of 10 mm diameter. Assuming fully developed flow and constant wall heat flux, estimate the oil-side heat transfer coefficient. If the overall convection coefficient is 53 W/m2K and the total heat transfer area is 1 m2, determine the effectiveness. What is the exit temperature of the oil? [Ans: 46.2oC]
Solution:
2 2
o2
cross-flow hx, fluids unmixed
oil inside tube, 75 ( ), 0.026kg/s
air, 30 (inlet), 0.53kg/s
U=53W/m , 1
Constant heat flux for tube,
find h(oil side)
Another case of T (outlet) unknown
assume
C inlet
C
K A m
o
oil
o2
b
b
3
4 2
T 45
T (75 45) / 2 60
(Other procedure may lead to non-integer value of T )
Tables: engine oil @ 60 :
864 / , 2.047 / ,
0.839 10 / , 0.14 /
Pr 1050
p
C
C
C
kg m C kJ kgK
m s k W mK
10
4
2
h h min
c c m
4Re
4 0.026
0.01 864 (0.839 10 )
45.7 2000
flow inside tube is laminar, constant heat flux
hd 0.01Nu= 4.36 = ;
k 0.14
61.04 /
oil: m = 0.026×2047=53.2=C
air: m = 0.53×1005=533=C
oil
ud m
d
h
h W m K
c
c
1
1 1
ax
min
max
min
min
min
C0.10 ;
C
53 11.0
C 53.2
From ε-NTU chart for cross flow hx,both fluid unmixed
=0.64
C ( ) 75=0.64=
C ( ) 75 30
75 0.64 45 46.2
(not far from assumed value of 45)
o o
o
h h h
h c
h
UANTU
T T T
T T
T C
11
7. Water flows over a 3-mm-diameter sphere at 6 m/s. The free-stream temperature is
38oC, and the sphere is maintained at 93oC. Calculate the heat-transfer rate.
[Ans: 84.9W].
Solution:
o
3 3
at T =38 C, properties for water
=993kg/m , 0.682 10 /
0.63 / , Pr 4.55
kg ms
k W mK
Should not be evaluated at Tf =38+93/2=65.50C
because Nu eqn is evaluated at free steam condition
for liquid (water)
Water u=6m/s T=380C
d=3mm
Ts=930C
3
0.25
-0.3 0.54
4w
2
2
993 0.003 6Re 26, 210 2000
0.682 10
appropriate Nu expression is
NuPr 1.2 0.53Re 130.09
μ evaluated at 93 , 3.06 10 /
250.4
0.63 250.42 52589 /
0.003
(4 ) (93
w
wC kg ms
hdNu
k
h W m K
Q r h
38) 84.9W
12
8. Air at 90oC and 1 atm flows past a heated 1/16-in-diameter wire at a velocity at 6 m/s. The wire is heated to a temperature of 150oC. Calculate the heat transfer per unit length of wire. [Ans: 59.86 W/m].
Solution:
f
53
Air at 1 atm
T 90 , 6 /
First determine Re of the flow
Air properties are evaluated at the film temperature
90 150T 120 393
2 2
1.01325 100.898 /
287 393
1.013 / ; 2.25f
w
ff
p f
C u m s
T TC K
Pkg m
RT
c kJ kgK
5
3
d 5
1/3 0.466 1/3
6 10 /
0.03314 / ; Pr 0.690
0.898 6 (1.5875 10 )Re 379.1
2.256 10
: 0.683, 0.466
Pr 0.683(Re ) Pr
convection heat transfer coefficient
.0.
f
n
d d
kg ms
k W mK
u d
Fromtable C n
u dhdNu C
k
kh
d
0.466 1/3
0.466 0.333
2
"w
3
683(Re ) Pr
0.03314 0.683 (379.1) 0.690
1.5875 10
200.5 /
Heat transfer from wire to air
( ) .( )( )
.( )(T )
200.5 (1.5875 10 )(150 90)
60.0 /
d
w w
W m K
q hA T T h dl T T
qq h d T
l
W m