heat-transfer-exercise-book

heat-transfer-exercise-book


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dt
TTd
mCTThA fsfs
)(
)(
\uf02d\uf02d\uf03d\uf02d (1) 
 
We know that nfs TTGh )( \uf02d\uf03d 
 
Where G is a constant. 
 
(Note that this relation arises from the usual Nusselt/Grashof relationship in free convection; for 
example: \uf028 \uf029 3/1Pr1.0 GrNu \uf03d in turbulent flow or \uf028 \uf029 4/1Pr54.0 GrNu \uf03d for laminar flow) 
 
Equation 1 then becomes: 
 
\uf028 \uf029
dt
TTd
A
mCTTTTG fsfs
n
fs
)(
)(
\uf02d\uf02d\uf03d\uf02d\uf02d 
 
\uf0f2 \uf0f2
\uf03d \uf02d
\uf02b\uf02d
\uf02d\uf03d\uf02d
t
t
t
t
n
fs
fs
TT
TTd
dt
mC
GA
0 0
1)(
)(
 
 
\uf028 \uf029 \uf028 \uf029 n
tfs
n
fs TTTTmC
GnAt \uf02d
\uf03d
\uf02d \uf02d\uf02d\uf02d\uf03d
0
 (2) 
 
At \uf028 \uf029 \uf028 \uf029fisfs TTTTt \uf02d\uf03d\uf02d\uf03d ,,0 
 
If we divide equation 2 by \uf028 \uf029 nfis TT \uf02d\uf02d, 
 
And use the definition 
\uf028 \uf029
\uf028 \uf029fis
fs
TT
TT
\uf02d
\uf02d\uf03d
,
\uf071 
 
We obtain 
 
\uf028 \uf029 \uf028 \uf029nfisnnfis TTmC
GnAt
TTmC
GnAt \uf02d\uf03d\uf02d\uf03d\uf02d
\uf02d
\uf02d ,
,
1\uf071 
 
Since \uf028 \uf029 ifis hTTG \uf03d\uf02d, , the heat transfer coefficient at time t = 0, then 
 
1\uf02b\uf03d\uf02d
mC
Athin\uf071 
 
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Heat Transfer: Exercises
 
31 
Conduction
Or 1\uf02b\uf03d\uf02d tnhin \uf06c\uf071 
 
 
For aluminium KkgJCmkg /870,/2750 3 \uf03d\uf03d\uf072 
 
For laminar free convection, n = ¼ 
 
kgXAm 22.0002.004.02750 \uf03d\uf0b4\uf0b4\uf03d\uf03d \uf072 
 
JKm
mC
A /101.2
87022.0
04.0 24\uf02d\uf0b4\uf03d\uf0b4\uf03d\uf03d\uf06c 
1\uf02b\uf03d\uf02d tnhin \uf06c\uf071 which gives 
 
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Heat Transfer: Exercises
 
32 
Conduction
 
\uf028 \uf029
\uf06c
\uf071
i
n
nh
t 1\uf02d\uf03d
\uf02d
 
 
When 2.0
20120
204040 \uf03d\uf02d
\uf02d\uf03d\uf0b0\uf03d \uf071CT 
 
Then 
 
\uf028 \uf029
\uf028 \uf029 st 590101.2164/1
12.0
4
4/1
\uf03d\uf0b4\uf0b4\uf0b4
\uf02d\uf03d \uf02d
\uf02d
 
 
For the equation the \uf06c\uf071 \uf02d\uf03d 
 
which assumes that the heat transfer coefficient is independent of surface-to-fluid temperature 
difference. 
 
s
h
t 479
101.216
2.0lnln
4 \uf03d\uf0b4\uf0b4\uf02d\uf03d\uf02d\uf03d \uf02d\uf06c
\uf071
 
 
Percentage error = %19100
590
479590 \uf03d\uf0b4\uf02d 
 
Example 2.11 
 
A 1 mm diameter spherical thermocouple bead (C = 400 J/kg K, \ufffd \ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd) is required to 
respond to 99.5% change of the surrounding air \ufffd \ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd \ufffd \ufffd\ufffd\ufffd \ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd , \ufffd \ufffd
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\u2044 and Pr = 0.77) temperature in 10 ms. What is the minimum air speed at which this 
will occur? 
 
 
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Heat Transfer: Exercises
 
33 
Conduction
Solution 
 
Spherical bead: \ufffd\ufffd\ufffd\ufffd \ufffd \ufffd\ufffd\ufffd\ufffd 
 
 \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd \ufffd\ufffd\ufffd\ufffd 6\u2044 
 
Assume this behaves as a lumped mass, then 
 
 
\ufffd\ufffd \ufffd \ufffd\ufffd
\ufffd\ufffd \ufffd \ufffd\ufffd \ufffd \ufffd\ufffd\ufffd\ufffd\ufffd 
 
(given) 
 
For lumped mass on cooling from temperature Ti 
 
\ufffd\ufffd \ufffd \ufffd\ufffd
\ufffd\ufffd \ufffd \ufffd\ufffd \ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd \ufffd\ufffd\ufffd\ufffd\ufffd 
 
\ufffd \ufffd \ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd 
 
\ufffd\ufffd\ufffd \ufffd \ufffd\ufffd\ufffd\ufffd\ufffd 
 
\ufffd\ufffd \ufffd \ufffd\ufffd\ufffd 
 
Which gives the required value of heat transfer coefficient 
 
\ufffd\ufffd
\ufffd\ufffd\ufffd \ufffd \ufffd\ufffd\ufffd 
 
So 
 
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Heat Transfer: Exercises
 
34 
Conduction
\ufffd \ufffd 0.\ufffd \ufffd\ufffd
\ufffd
6
\ufffd\ufffd
\ufffd\ufffd\ufffd \ufffd
0.\ufffd \ufffd \ufffd \ufffd
6 
 
\ufffd \ufffd 0.\ufffd \ufffd 10
\ufffd\ufffd \ufffd 400 \ufffd 7800
6 \ufffd 260 \ufffd \ufffd
\ufffd \ufffd\u2044 
 
\ufffd\ufffd\ufffd \ufffd \ufffd\ufffd\ufffd \ufffd
260 \ufffd 10\ufffd\ufffd
0.0262 \ufffd \ufffd.\ufffd 
 
For a sphere 
 
\ufffd\ufffd\ufffd \ufffd 2 \ufffd \ufffd0.4\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd 0.06\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd\ufffd\ufffd.\ufffd 
 
From which with Pr = 0.707 
 
\ufffd \ufffd 0.4\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd 0.06\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd \ufffd.4 \ufffd 0 
 
\ufffd\ufffd \ufffd 0.2\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd 0.04\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd 
 
Using Newton iteration 
 
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd \ufffd\ufffd \ufffd \ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd 
 
Starting with ReD = 300 
 
\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd \ufffd 300 \ufffd
\ufffd0.4\u221a300 \ufffd 0.06\ufffd300\ufffd\ufffd\ufffd\ufffd \ufffd \ufffd.4\ufffd
\ufffd 0.2\u221a300 \ufffd
0.04
300\ufffd\ufffd\ufffd\ufffd
\ufffd 300 \ufffd 0.2220.01782 
 
Which is close enough to 300 
 
From which 
 
\ufffd\ufffd \ufffd \ufffd\ufffd\ufffd\ufffd\ufffd \ufffd 4.\ufffd \ufffd\ufffd\ufffd 
 
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Heat Transfer: Exercises
 
35 
Convection
3. Convection 
 
Example 3.1 
 
Calculate the Prandtl number (Pr = \uf06dCp/k) for the following 
 
a) Water at 20\uf0b0C: \uf06d = 1.002 x 10\uf02d3 kg/m s, Cp = 4.183 kJ/kg K and k = 0.603 W/m K 
b) Water at 90\uf0b0C: \uf072 = 965 kg/m3, \uf06e = 3.22 x 10\uf02d7 m2/s, Cp = 4208 J/kg K and k = 0.676 W/m K 
c) Air at 20\uf0b0C and 1 bar: R = 287 J/kg K, \uf06e = 1.563 x 10\uf02d5 m2/s, Cp = 1005 J/kg K and 
k = 0.02624 W/m K 
d) Air at 100\uf0b0C: \uf028 \uf029T
T
\uf02b
\uf0b4\uf03d
\uf02d
110
1046.1 236\uf06d kg/m s 
 KkgkJTTC p /1098.31058.2917.0
284 \uf02d\uf02d \uf0b4\uf02d\uf0b4\uf02b\uf03d (Where T is the absolute temperature in 
K) and k = 0.03186 W/m K. 
e) Mercury at 20\uf0b0C: \uf06d = 1520 x 10\uf02d6 kg/m s, Cp = 0.139 kJ/kg K and k = 0.0081 kW/m K 
f) Liquid Sodium at 400 K: \uf06d = 420 x 10\uf02d6 kg/m s, Cp = 1369 J/kg K and k = 86 W/m K 
g) Engine Oil at 60\uf0b0C: \uf06d = 8.36 x 10\uf02d2 kg/m s, Cp = 2035 J/kg K and k = 0.141 W/m K 
 
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Heat Transfer: Exercises
 
36 
ConvectionSolution 
 
a)
 
95.6
603.0
418310002.1Pr
3
\uf03d\uf0b4\uf0b4\uf03d\uf03d
\uf02d
k
Cp\uf06d
 
 
b)
 
93.1
676.0
42081022.3965Pr
7
\uf03d\uf0b4\uf0b4\uf0b4\uf03d\uf03d\uf03d
\uf02d
k
C
k
C pp \uf072\uf06e\uf06d
 
 
c)
 
k
Cp\uf072\uf06e\uf03dPr
 
 
3/19.1
293287
100000 mkg
RT
P \uf03d\uf0b4\uf03d\uf03d\uf072 
 
712.0
02624.0
100510563.119.1Pr
5
\uf03d\uf0b4\uf0b4\uf0b4\uf03d
\uf02d
 
 
d)
 
\uf028 \uf029 smkgT
T /1018.2
373110
3731046.1
110
1046.1 52/36236 \uf02d\uf02d\uf02d \uf0b4\uf03d\uf02b
\uf0b4\uf0b4\uf03d\uf02b
\uf0b4\uf03d\uf06d
 
 
KkgJ
TTCp
/7.1007
3731098.33731058.2917.01098.31058.2917.0 284284
\uf03d
\uf0b4\uf0b4\uf02d\uf0b4\uf0b4\uf02b\uf03d\uf0b4\uf02d\uf0b4\uf02b\uf03d \uf02d\uf02d\uf02d\uf02d
 
 
689.0
03186.0
7.10071018.2Pr
5
\uf03d\uf0b4\uf0b4\uf03d
\uf02d
 
 
e)
 
0261.0
100081.0
139101520Pr 3
6
\uf03d\uf0b4
\uf0b4\uf0b4\uf03d\uf03d
\uf02d
k
Cp\uf06d
 
 
Solution 
 
a)
 
95.6
603.0
418310002.1Pr
3
\uf03d\uf0b4\uf0b4\uf03d\uf03d
\uf02d
k
Cp\uf06d
 
 
b)
 
93.1
676.0
42081022.3965Pr
7
\uf03d\uf0b4\uf0b4\uf0b4\uf03d\uf03d\uf03d
\uf02d
k
C
k
C pp \uf072\uf06e\uf06d
 
 
c)
 
k
Cp\uf072\uf06e\uf03dPr
 
 
3/19.1
293287
100000 mkg
RT
P \uf03d\uf0b4\uf03d\uf03d\uf072 
 
712.0
02624.0
100510563.119.1Pr
5
\uf03d\uf0b4\uf0b4\uf0b4\uf03d
\uf02d
 
 
d)
 
\uf028 \uf029 smkgT
T /1018.2
373110
3731046.1
110
1046.1 52/36236 \uf02d\uf02d\uf02d \uf0b4\uf03d\uf02b
\uf0b4\uf0b4\uf03d\uf02b
\uf0b4\uf03d\uf06d
 
 
KkgJ
TTCp
/7.1007
3731098.33731058.2917.01098.31058.2917.0 284284
\uf03d
\uf0b4\uf0b4\uf02d\uf0b4\uf0b4\uf02b\uf03d\uf0b4\uf02d\uf0b4\uf02b\uf03d \uf02d\uf02d\uf02d\uf02d
 
 
689.0
03186.0
7.10071018.2Pr
5
\uf03d\uf0b4\uf0b4\uf03d
\uf02d
 
 
e)
 
0261.0
100081.0
139101520Pr 3
6
\uf03d\uf0b4
\uf0b4\uf0b4\uf03d\uf03d
\uf02d
k
Cp\uf06d
 
 
Solution 
 
a)
 
95.6
603.0
418310002.1Pr
3
\uf03d\uf0b4\uf0b4\uf03d\uf03d
\uf02d
k
Cp\uf06d
 
 
b)
 
93.1
676.0
42081022.3965Pr
7
\uf03d\uf0b4\uf0b4\uf0b4\uf03d\uf03d\uf03d
\uf02d
k
C
k
C pp \uf072\uf06e\uf06d
 
 
c)
 
k
Cp\uf072\uf06e\uf03dPr
 
 
3/19.1
293287
100000 mkg
RT
P \uf03d\uf0b4\uf03d\uf03d\uf072 
 
712.0
02624.0
100510563.119.1Pr