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

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?

Heat Transfer: Exercises

33
Conduction
Solution

\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

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

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```