heat-transfer-exercise-book-winter 2018

Chris Long & Naser Sayma
Heat Transfer: Exercises
2
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Heat Transfer: Exercises
© 2010 Chris Long, Naser Sayma & Ventus Publishing ApS
ISBN 978-87-7681-433-5
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Heat Transfer: Exercises
Contents
Contents
Preface
5
1.
Introduction
6
2.
Conduction
11
3.
Convection
35
4.
Radiation
60
5.
Heat Exchangers
79
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Preface
Heat Transfer: Exercises
Preface
Worked examples are a necessary element to any textbook in the sciences, because they reinforce the
theory (i.e. the principles, concepts and methods). Once the theory has been understood, well chosen
examples can be used, with modification, as a template to solve more complex, or similar problems.
This work book contains examples and full solutions to go with the text of our e-book (Heat Transfer,
by Long and Sayma). The subject matter corresponds to the five chapters of our book: Introduction to
Heat Transfer, Conduction, Convection, Heat Exchangers and Radiation. They have been carefully
chosen with the above statement in mind. Whilst compiling these examples we were very much aware
of the need to make them relevant to mechanical engineering students. Consequently many of the
problems are taken from questions that have or may arise in a typical design process. The level of
difficulty ranges from the very simple to challenging. Where appropriate, comments have been added
which will hopefully allow the reader to occasionally learn something extra. We hope you benefit
from following the solutions and would welcome your comments.
Christopher Long
Naser Sayma
Brighton, UK, February 2010
5
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Introduction
Heat Transfer: Exercises
1. Introduction
Example 1.1
The wall of a house, 7 m wide and 6 m high is made from 0.3 m thick brick with k  0.6 W / m K .
The surface temperature on the inside of the wall is 16oC and that on the outside is 6oC. Find the heat
flux through the wall and the total heat loss through it.
Solution:
For one-dimensional steady state conduction:
q  k
q
dT
k
  Ti  To 
dx
L
0 .6
16  6  20 W / m 2
0 .3
Q  qA  20  6  7   840 W
The minus sign indicates heat flux from inside to outside.
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Introduction
Heat Transfer: Exercises
Example 1.2
A 20 mm diameter copper pipe is used to carry heated water, the external surface of the pipe is
subjected to a convective heat transfer coefficient of h  6 W / m 2 K , find the heat loss by convection
per metre length of the pipe when the external surface temperature is 80oC and the surroundings are at
20oC. Assuming black body radiation what is the heat loss by radiation?
Solution
qconv  h Ts  T f   680  20  360 W / m 2
For 1 metre length of the pipe:
Qconv  q conv A  qconv  2 r  360  2    0.01  22.6 W / m
For radiation, assuming black body behaviour:

q rad   Ts4  T f4


q rad  5.67  10 8 353 4  293 4

q rad  462 W / m 2
For 1 metre length of the pipe
Qrad  q rad A  462  2    0.01  29.1 W / m 2
A value of h = 6 W/m2 K is representative of free convection from a tube of this diameter. The heat
loss by (black-body) radiation is seen to be comparable to that by convection.
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Introduction
Heat Transfer: Exercises
Example 1.3
A plate 0.3 m long and 0.1 m wide, with a thickness of 12 mm is made from stainless steel (
k  16 W / m K ), the top surface is exposed to an airstream of temperature 20oC. In an experiment,
the plate is heated by an electrical heater (also 0.3 m by 0.1 m) positioned on the underside of the plate and
the temperature of the plate adjacent to the heater is maintained at 100oC. A voltmeter and ammeter are
connected to the heater and these read 200 V and 0.25 A, respectively. Assuming that the plate is
perfectly insulated on all sides except the top surface, what is the convective heat transfer coefficient?
Solution
Heat flux equals power supplied to electric heater divided by the exposed surface area:
q
V I V I
200  0.25


 1666.7 W / m 2
A
W L
0 .1  0 .3
This will equal the conducted heat through the plate:
k
T2  T1 
t
1666.7  0.012   98.75C
qt
T1  T2   100 
16
k
q
(371.75 K)
The conducted heat will be transferred by convection and radiation at the surface:

q  hT1  T f    T14  T f4
h

q   T14  T f4
T
1  Tf


  1666.7  5.67  10 371.75
8
371.75  293
4
 293 4
  12.7 W / m
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2
K
Introduction
Heat Transfer: Exercises
Example 1.4
An electronic component dissipates 0.38 Watts through a heat sink by convection and radiation (black
body) into surrounds at 20oC. What is the surface temperature of the heat sink if the convective heat
transfer coefficient is 6 W/m2 K, and the heat sink has an effective area of 0.001 m2 ?
Solution
q
Q
 hTs  T    Ts4  T4
A


0.38
 6Ts  293  5.67  10 3 Ts4  293 4
0.001


5.67  10 8 Ts4  6Ts  2555.9  0
This equation needs to be solved numerically. Newton-Raphson’s method will be used here:
f  5.67  10 8 Ts4  6Ts  2555.9
df
 22.68  10 8 Ts3  6
dTs
T
n 1
s
5.67  10 8 Ts4  6Ts  2555.9
T 
T 
22.68Ts3  6
 df 


 dTs 
n
s
f
n
s
Start iterations with Ts0  300 K
Ts1  300 
5.67  10 8  300 4  6  300  2555.9
 324.46 K
22.68  300 3  6
5.67  10 8  324.46 4  6  324.46  2555.9
T  324.46 
 323 K
22.68  324.46 3  6
2
s
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Introduction
Heat Transfer: Exercises
The difference between the last two iterations is small, so:
Ts0  323 K  50C
The value of 300 K as a temperature to begin the iteration has no particular significance other than
being above the ambient temperature.
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Conduction
Heat Transfer: Exercises
2. Conduction
Example 2.1
Using an appropriate control volume show that the time dependent conduction equation in cylindrical
coordinates for a material with constant thermal conductivity, density and specific heat is given by:
 2T 1 T  2T 1 T



r 2 r r z 2  t
Were  
k
is the thermal diffusivity.
c
Solution
Consider a heat balance on an annular control volume as shown the figure above. The heat balance in
the control volume is given by:
Heat in + Heat out = rate of change of internal energy
Q r  Q z  Q r  r  Q z  z 
Q r  r  Q r 
Q
r
r
Q z  z  Q z 
Q
z
z
u
t
(2.1)
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Conduction
Heat Transfer: Exercises
u  mcT
Substituting in equation 2.1:

Q
Q
 (mcT )
r
z
r
z
t
(2.2)
Fourier’s law in the normal direction of the outward normal n:
Q
T
 k
A
n
Qr   kA
T
T
  k  2 r z
r
r
( A  2 r z )
Q z   kA
T
T
  k  2 r r
z
z
( A  2 r r )
Equation 2.1 becomes

 
T 
 
T 
T
 k  2 r z
 r   k  2 r r
 z  mc
r 
r 
z 
z 
t
(2.3)
Noting that the mass of the control volume is given by:
m   2 r r z
Equation 2.3 becomes
  T 
  T 
T
k r
 r  k r
 z  cr
r  r 
z 
z 
t
Dividing by r, noting that r can be taken outside the brackets in the second term because it is not a
function of z. Also dividing by k since the thermal conductivity is constant:
1   T   2T c T

r

r r  r  z 2
k t
Using the definition of the thermal diffusivity:  
1    2T T r   2T 1 T


r

r r  r 2 r r  z 2  t
k
and expanding the first term using the product rule:
c
which gives the required outcome:
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Conduction
Heat Transfer: Exercises
 2T 1 T  2T 1 T



r 2 r r z 2  t
Example 2.2
An industrial freezer is designed to operate with an internal air temperature of -20oC when the external
air temperature is 25oC and the internal and external heat transfer coefficients are 12 W/m2 K and 8
W/m2 K, respectively. The walls of the freezer are composite construction, comprising of an inner
layer of plastic (k = 1 W/m K, and thickness of 3 mm), and an outer layer of stainless steel (k = 16
W/m K, and thickness of 1 mm). Sandwiched between these two layers is a layer of insulation material
with k = 0.07 W/m K. Find the width of the insulation that is required to reduce the convective heat
loss to 15 W/m2.
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Conduction
Heat Transfer: Exercises
Solution
q  UT
U
where U is the overall heat transfer coefficient given by:
q
15

 0.333W / m 2 K
T 25  (20)
 1 L p Li Ls 1 
U  
 
 
 hi k p k i k s ho 
1
 0.333
 1 L p Li Ls 1 
1
 
 
 
 hi k p k i k s ho  0.333
 1
 1 L p Ls 1  
 1
 1 0.003 0.001 1  
Li  k i 
 

    0.07 
 

 
1
16
8 
 0.333  hi k p k s ho  
 0.333 12
Li  0.195m
(195 mm)
Example 2.3
Water flows through a cast steel pipe (k = 50 W/m K) with an outer diameter of 104 mm and 2 mm
wall thickness.
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Conduction
Heat Transfer: Exercises
i.
ii.
Calculate the heat loss by convection and conduction per metre length of uninsulated pipe
when the water temperature is 15oC, the outside air temperature is -10oC, the water side heat
transfer coefficient is 30 kW/m2 K and the outside heat transfer coefficient is 20 W/m2 K.
Calculate the corresponding heat loss when the pipe is lagged with insulation having an
outer diameter of 300 mm, and thermal conductivity of k = 0.05 W/m K.
Solution
Plain pipe
Q  2 r1 Lhi Ti  T1  
Q
2Lk T1  T2 
lnr2 / r1 
Q  2 r2 Lho T2  To 
Ti  T1 


Q
2 r1 Lhi
T2  T1 
Q
2 Lk / ln(r2 / r1 )
T2  To 
Q
2 r2 Lho
Adding the three equations on the right column which eliminates the wall temperatures gives:
Q
Q

L
2LTi  To 
ln r2 / r1 
1
1


hi r1
k
ho r2
2 15  (10) 
 163.3W / m
1
ln0.052 / 0.05
1


30000  0.05
50
20  0.052
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Conduction
Heat Transfer: Exercises
Insulated pipe
2 Ti  To 
Q

lnr2 / r1  ln(r3 / r2 )
1
1
L



hi r1
k
k ins
ho r3
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Conduction
Heat Transfer: Exercises
Q

L
2 15  (10) 
 7.3W / m
1
ln0.052 / 0.05 ln(0.15 / 0.052)
1


30000  0.05
50
0.05
20  0.15
For the plain pipe, the heat loss is governed by the convective heat transfer coefficient on the outside,
which provides the highest thermal resistance. For the insulated pipe, the insulation provides the
higher thermal resistance and this layer governs the overall heat loss.
Example 2.4
Water at 80oC is pumped through 100 m of stainless steel pipe, k = 16 W/m K of inner and outer radii
47 mm and 50 mm respectively. The heat transfer coefficient due to water is 2000 W/m2 K. The outer
surface of the pipe loses heat by convection to air at 20oC and the heat transfer coefficient is 200 W/m2
K. Calculate the heat flow through the pipe. Also calculate the heat flow through the pipe when a layer
of insulation, k = 0.1 W/m K and 50 mm radial thickness is wrapped around the pipe.
Solution
The equation for heat flow through a pipe per unit length was developed in Example 2.3:
Q
2LTi  To 
ln r2 / r1 
1
1


hi r1
k
ho r2
Hence substituting into this equation:
Q
2  10080  20
 0.329  10 6 W
1
ln50 / 47 
1


0.047  2000
16
0.05  200
For the case with insulation, we also use the equation from Example 2.3
Q
Q
2LTi  To 
lnr2 / r1  ln(r3 / r2 )
1
1



hi r1
k
k ins
ho r3
2  10080  20 
 5.39  10 3 W
1
ln50 / 47  ln(100 / 50)
1



0.047  2000
16
0.1
0.1  200
Notice that with insulation, the thermal resistance of the insulator dominates the heat flow, so in the equation
above, if we retain the thermal resistance for the insulation and ignore all the other terms, we obtain:
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Conduction
Heat Transfer: Exercises
Q
2LTi  To  2  10080  20

 5.44  10 3 W
ln(100 / 50)
ln(r3 / r2 )
0.1
k ins
This has less than 1% error compared with the full thermal resistance.
Example 2.5
A diagram of a heat sink to be used in an electronic application is shown below. There are a total of 9
aluminium fins (k = 175 W/m K, C = 900 J/kg K,   2700kg / m 3 ) of rectangular cross-section,
each 60 mm long, 40 mm wide and 1 mm thick. The spacing between adjacent fins, s, is 3 mm. The
temperature of the base of the heat sink has a maximum design value of Tb  60C , when the external
air temperature T f is 20oC. Under these conditions, the external heat transfer coefficient h is 12 W/m2
K. The fin may be assumed to be sufficiently thin so that the heat transfer from the tip can be
neglected. The surface temperature T, at a distance, x, from the base of the fin is given by:
T  Tf 
T
b
 T f cosh m( L  x)
sinh mL
where m 2 
hP
and Ac is the cross sectional area.
kAc
Determine the total convective heat transfer from the heat sink, the fin effectiveness and the
fin efficiency.
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Conduction
Heat Transfer: Exercises
Solution
Total heat fluxed is that from the un-finned surface plus the heat flux from the fins.
Q  Qu  Q f
Qu  Au h (Tb  T f )  w  s  N  1) h Tb  T f

Qu  0.04  0.0039  1)   12 60  20  0.461 W
For a single fin:
 dT 
Q f  kAc 

 dx  x 0
Where Ac is the cross sectional area of each fin
Since
T  Tf 
T
b
 T f cosh m( L  x)
sinh mL
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Conduction
Heat Transfer: Exercises
Then
dT  sinh m( L  x)

m Tb  T f
cosh mL
dx

Thus
 dT 
  sinh mL 
Q f  kAc 
  kAc 
 m Tb  T f
 dx  x 0
 cosh mL 
Q f  kAc m Tb  T f  tanh(mL)  hpkAc 
1/ 2
T
b

 T f  tanh mL 
Since
 hP
m  
 kAc
1
2


P  2( w  t )  2(0.04  0.001)  0.082 m
Ac  w  t  0.04  0.0001  40  10 6 m 2
1
 12  0.082  2
 11.856 m 1
m
6 
 175  40  10 
mL  11.856  0.06  0.7113
tanhmL   tanh0.7113  0.6115

Q f  12  0.082  175  40  10 6

1/ 2
 60  20  0.6115  2.03 W / fin
So total heat flow:
Q  Qu  Q f  0.461  9  2.03  18.7 W
Finn effectiveness
 fin 
Qf
Fin heat transfer rate

Heat transfer rate that would occur in the absence of the fin hAc Tb  T f
 fin 
2.03
 106
12  40  10 6 60  20 
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
Conduction
Heat Transfer: Exercises
Fin efficiency:
Actual heat transfer through the fin
Heat that would be transferred if all the fin area were at the base temperature
 fin 
Qf
 fin 
hAs Tb  T f

As  wL  wL  Lt  Lt  2 L( w  t )
As  2  0.06(0.04  0.001)  4.92  10 3 m 2
 fin 
2.03
 0.86
12  4.92  10 3 60  20 
Example 2.6
For the fin of example 4.5, a fan was used to improve the thermal performance, and as a result, the
heat transfer coefficient is increased to 40 W/m2 K. Justify the use of the lumped mass approximation
to predict the rate of change of temperature with time. Using the lumped mass approximation given
below, calculate the time taken,  , for the heat sink to cool from 60oC to 30oC.
T  T   T
f
i
 hA  
 T f exp  s 
 mC 
Solution
Consider a single fin (the length scale L for the Biot number is half the thickness t/2)
Bi 
hL h  t / 2 40  0.0005


 10  4
k
k
175
Since Bi  1 , we can use he “lumped mass” model approximation.
T  T 
 hA  
 exp  

T  T 
 mC 
f
i
 
s
f
mC  T  T f
ln
hAs  Ti  T f




21
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Conduction
Heat Transfer: Exercises
m  As t / 2
 
 T  Tf
ln
2h  Ti  T f
Ct

   2700  900  0.001 ln 30  20   42 seconds

2  40
 60  20 

Example 2.7
The figure below shows part of a set of radial aluminium fins (k = 180 W/m K) that are to be fitted to
a small air compressor. The device dissipates 1 kW by convecting to the surrounding air which is at
20oC. Each fin is 100 mm long, 30 mm high and 5 mm thick. The tip of each fin may be assumed to be
adiabatic and a heat transfer coefficient of h = 15 W/m2 K acts over the remaining surfaces.
Estimate the number of fins required to ensure the base temperature does not exceed 120oC
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Conduction
Heat Transfer: Exercises
Solution
Consider a single fin:
P  2( w  t )  2(0..005  0.03)  0.07 m
Ac  w  t  0.005  0.03  150  10 6 m 2
 hP
m  
 kAc
1
1
 2  15  0.07
2
  
 6.2361 m 1
6 
 180  150  10 

mL  6.2361 0.1  0.62361
tanhmL   0.5536
Q f  hPkAc 
1/ 2

T
b
 T f  tanh(mL)
Q f  15  0.07  180  150  10 6

1/ 2
(From example 2.5)
 120  20  0.5536  9.32 W
So for 1 kW, the total number of fins required:
N
1000
 108
9.32
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Conduction
Heat Transfer: Exercises
Example 2.8
An air temperature probe may be analysed as a fin. Calculate the temperature recorded by a probe of length
L = 20 mm, k = 19 W/m K, D = 3 mm, when there is an external heat transfer coefficient of h = 50 W/m2K,
an actual air temperature of 50oC and the surface temperature at the base of the probe is 60oC.
Solution
The error should be zero when Ttip  T . The temperature distribution along the length of the probe
(from the full fin equation) is given by:
cosh m( L  x) 
Tx  T

Tb  T
 hP 
m

 kA 
cosh mL 
1/ 2
htip
mk
htip
mk
sinh m( L  x)
sinh mL
A   D 2 / 4,
P  D
At the tip, x  L , the temperature is given by ( cosh( 0)  1 , sinh( 0)  0 ):
Ttip  T
Tb  T

cosh mL 
1
htip
mk

sinh mL
Where  is the dimensionless error:,
  0,
Ttip  T
(no error)
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Conduction
Heat Transfer: Exercises
  1,
TL  Tb
For L  20mm,
T  50C ,
A   D 2 / 4,
 hP 
m

 kA 
1/ 2
(large error)
k  19W / m K ,
D  3mm,
h  htip  50 W / m 2 K
Tb  60C
P  D
 h D  4 

 
2 
 k D 
1/ 2
 4h 


 kD 
1/ 2
 4  50 


 19  0.003 
1/ 2
 59.235 m 1
mL  59.235  0.02  1.185
50
h

 0.0444
mk 59.235  19
Tx  T
1

 0.539
Tb  T cosh 1.185  0.0444  sinh 1.185
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Conduction
Heat Transfer: Exercises
Ttip  0.539Tb  T   T
Ttip  0.53960  50  50  55.39C
Hence error  5.39C
Example 2.9
A design of an apartment block at a ski resort requires a balcony projecting from each of the 350
separate apartments. The walls of the building are 0.3 m wide and made from a material with k = 1
W/m K. Use the fin approximation to examine the implications on the heat transfer for two separate
suggestions for this design. In each case, the balcony projects 2 m from the building and has a length
(parallel to the wall) of 4 m. Assume an inside temperature of 20oC and an outside temperature of 5oC; the overall (convective + radiative) heat transfer coefficient on the inside of the building is 8
W/m2 K and on that on the outside of the building is 20 W/m2 K
a) A balcony constructed from solid concrete and built into the wall, 0.2 m thick, k = 2 W/m K.
b) A balcony suspended from 3 steel beams, k = 40 W/m K, built into the wall and projecting out by 2
m each of effective cross sectional area Ae  0.01 m 2 , perimetre P  0.6 m (The actual floor of
the balcony in this case may be considered to be insulated from the wall
c) No balcony.
Solution
a)
For the concrete balcony
Treat the solid balcony as a fin Bi 
ho t / 2
kb
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Conduction
Heat Transfer: Exercises
Bi 
20  0.1
1
2
Not that Bi is not << 1, thus 2D analysis would be more accurate. However, treating it as a fin will
give an acceptable result for the purpose of a quick calculation.
P  2 ( H  t )  2 ( 4  0 .2)  8 .4 m
Ac  H  t  4  0.2  0.8 m 2
To decide if the fin is infinite, we need to evaluate mL (which is in fact in the notation used here is mW)
 hP 
mW  

 kA 
1/ 2
 20  8.4 
W 

 2  0 .8 
1/ 2
 2  20.5
This is large enough to justify the use of the fin infinite equation.
Qb  ho Pk b Ac 
1/ 2
T2  To 

1
ho Pk b Ac 1 / 2 T2  To    ho Pk b
qb 
Ac
 Ac



1/ 2
T2  To 
(1)
Also assuming 1-D conduction through the wall:
qb  hi (Ti  T1 )
qb 
(2)
kb
(T1  T2 )
L
(3)
Adding equations 1, 2 and 3 and rearranging:
qb 
(To  Ti )
1 L  Ac
 
hi k b  ho Pk b



(4)
1/ 2
This assumes 1D heat flow through the wall, the concrete balcony having a larger k than the wall may
introduce some 2-D effects.
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Conduction
Heat Transfer: Exercises
From (4)
qb 
20  (5)
1 0 .3 
0 .8




8 2  20  8.4  2 
1/ 2
 77.2 W / m 2
Compared with no balcony:
qb 
(To  Ti )
20  (5)

 52.6 W / m 2
1
L
1
1 0.3 1




hi k w ho 8 1 20
The difference for one balcony is Ac (77.2  52.6)  0.8  24.6  19.7 W
For 350 apartments, the difference is 6891 W.
For the steel supported balcony where Ac  0.01 m 2 and P  0.6 m
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Conduction
Heat Transfer: Exercises
As before, however, in this case Bi << 1 because k s  k b
 hP 
mW  

 kA 
1/ 2
 20  6 
w

 40  0.1 
1/ 2
 2  11
mW  2 , so we can use the infinite fin approximation as before
qb 
(To  Ti )
1 L  Ac
 
hi k s  ho Pk s



1/ 2

20  ( 5)
1 0.3  0.01 



8 40  20  6  40 
1/ 2
 182 W / m 2
Qb  Ac qb  0.01  182  1.82 W / beam
For 350 apartments, Qb  1915 W
Example 2.10
In free convection, the heat transfer coefficient varies with the surface to fluid temperature difference
T
s
 T f  . Using the low Biot number approximation and assuming this variation to be of the form
h  G Ts  T f
 Where G and n are constants, show that the variation of the dimensionless
n
temperature ratio with time will be given by
  n  1  nhinit   t
Where

T
s
 Tf
Tinit  T f

, 
Area
Mass  Specific Heat Capacity
and hinit = the heat transfer coefficient at t = 0. Use this expression to determine the time taken for an
aluminium motorcycle fin (   2750 kg / m 3 , C  870 J / kgK ) of effective area 0.04 m2 and
thickness 2mm to cool from 120oC to 40oC in surrounding air at 20oC when the initial external heat
transfer coefficient due to laminar free convection is 16 W/m2 K. Compare this with the time estimated
from the equation (   e  ht ) which assumes a constant value of heat transfer coefficient.
Solution
Low Biot number approximation for free convection for Bi  1
Heat transfer by convection = rate of change of internal energy
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Conduction
Heat Transfer: Exercises
hA(Ts  T f )  mC
d (Ts  T f )
(1)
dt
n
We know that h  G (Ts  T f )
Where G is a constant.
(Note that this relation arises from the usual Nusselt/Grashof relationship in free convection; for
example: Nu  0.1Gr Pr 
1/ 3
in turbulent flow or Nu  0.54Gr Pr 
1/ 4
for laminar flow)
Equation 1 then becomes:
G Ts  T f
 (T
n
 Tf )  
s
mC d (Ts  T f )
A
dt
t
d (Ts  T f )
 GA
dt

t 0 mC
t 0 (Ts  T f ) n1
t
GnAt
 Ts  T f
mC
At t  0,
T
s

n
 Ts  T f

n
(2)
t 0
 T f   Ts ,i  T f


If we divide equation 2 by Ts ,i  T f
And use the definition

T
T
s
s ,i

n

T 
 Tf
f
We obtain
GnAt
mC Ts ,i  T f

n

  n  1 
GnAt
Ts ,i  T f
mC

n

Since G Ts ,i  T f  h i , the heat transfer coefficient at time t = 0, then
 n 
hi At
1
mC
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Conduction
Heat Transfer: Exercises
Or
  n  nhi  t  1
For aluminium   2750 kg / m 3 ,
C  870 J / kg K
For laminar free convection, n = ¼
m   A X  2750  0.04  0.002  0.22 kg

 n
0.04
A

 2.1 10  4 m 2 K / J
mC 0.22  870
 nhi  t  1 which gives
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Conduction
Heat Transfer: Exercises
t

n
 1
nhi 
When
T  40C

40  20
 0 .2
120  20
Then
t
0.2
 1
 590 s
1 / 4  16  2.1  10 4
1 / 4
For the equation
  e  h t
which assumes that the heat transfer coefficient is independent of surface-to-fluid temperature
difference.
t
ln 
ln 0.2

 479 s
 h  16  2.1  10  4
Percentage error =
590  479
 100  19%
590
Example 2.11
A 1 mm diameter spherical thermocouple bead (C = 400 J/kg K, � � ���������� ) is required to
respond to 99.5% change of the surrounding air � � ��������� � � � ��� � ���� ����������� , � �
������� � ⁄��� and Pr = 0.77) temperature in 10 ms. What is the minimum air speed at which this
will occur?
32
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Conduction
Heat Transfer: Exercises
Solution
Spherical bead:
���� � ��� �
������� � � �� � ⁄6
Assume this behaves as a lumped mass, then
�� � ��
� �����
�� � ��
(given)
For lumped mass on cooling from temperature Ti
�� � ��
� ��������� � �����
�� � ��
��
��
������������ � �������
��
��� � �����
�� � ���
Which gives the required value of heat transfer coefficient
��
� ���
���
So
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Conduction
Heat Transfer: Exercises
� � 0.�
��
�� � ��
0.� � � �
�
�
6
6 ��
0.� � 10�� � 400 � 7800
� 260 � ⁄�� �
6
��� �
�� 260 � 10��
�
� �.�
�
0.0262
For a sphere
���
��� � 2 � �0.4���
���
� 0.06��� � �� �.�
From which with Pr = 0.707
���
���
� � 0.4��� � 0.06���
� �.4 � 0
����
����
�� � 0.2���
� 0.04���
Using Newton iteration
� ����� � � � �
����
�
����
Starting with ReD = 300
���
��� � 300 �
�0.4√300 � 0.06�300���� � �.4�
0.222
� 300 �
0.2
0.04
0.01782
�
�
�
√300 300���
Which is close enough to 300
From which
�� �
���
� 4.� ���
��
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Convection
Heat Transfer: Exercises
3. Convection
Example 3.1
Calculate the Prandtl number (Pr = Cp/k) for the following
a)
b)
c)
Water at 20C:  = 1.002 x 103 kg/m s, Cp = 4.183 kJ/kg K and k = 0.603 W/m K
Water at 90C:  = 965 kg/m3,  = 3.22 x 107 m2/s, Cp = 4208 J/kg K and k = 0.676 W/m K
Air at 20C and 1 bar: R = 287 J/kg K,  = 1.563 x 105 m2/s, Cp = 1005 J/kg K and
k = 0.02624 W/m K
d)
Air at 100C:

1.46  10 6 T 3 2
kg/m s
110  T 
C p  0.917  2.58  10 4 T  3.98  10 8 T 2 kJ / kg K (Where T is the absolute temperature in
e)
f)
g)
K) and k = 0.03186 W/m K.
Mercury at 20C:  = 1520 x 106 kg/m s, Cp = 0.139 kJ/kg K and k = 0.0081 kW/m K
Liquid Sodium at 400 K:  = 420 x 106 kg/m s, Cp = 1369 J/kg K and k = 86 W/m K
Engine Oil at 60C:  = 8.36 x 102 kg/m s, Cp = 2035 J/kg K and k = 0.141 W/m K
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Convection
Heat
Transfer: Exercises
Solution
a)
Solution
Solution
 Cp
1.002  10 3  4183
 6.95
0.603
k
CCp 1.002  10 33 4183
1.002  10  4183 6.95
p
Pr

b) Pr  k 
 6.95
00.603
.603
k
a)
a)
Pr 
 Cp

 C p
965  3.22  10 7  4208
 1.93
0.676
k
k
C p 965  3.22  10 77 4208
CCp 
 4208  1.93
p   C p  965  3.22  10
Pr

c) Pr  k
 k

 1.93
0
.
676
0.676
k
k
b)
b)
Pr 
c)
c)
Pr 


 C p
k

CCpp
Pr

Pr  Pk 100000
 1.19 kg / m 3

k
RT 287  293
100000
P
kg
  P  100000 511.19
//mm33
kg
.
19



1
.
19
1
.
563
10
1005
RT 287  293
Pr  RT 287  293
 0.712
0.02624
5
11.19
.1911.563
.56310
10 51005
1005  0.712
Pr

d) Pr 
 0.712
00.02624
.02624
d)
d)
1.46  10 6 T 3 2 1.46  10 6  3733 / 2

 2.18  10 5 kg / m s
110  T 
110  373
6 3 2
6
3/ 2
11.46
10
11.46
10
6T 3 2
6 373 3 / 2
.
46
10
.
46
10
373
T

 4
 
 2.18  10 55 kg
mmss4
8 2  2.18  10
kg//10
373
C p  0.110
110
3.98
 10 T  0.917  2.58
 373  3.98  10 8  3732
917
110T2T.58  10 T 110
373

 1007.7 J / kg K
4
8 2
4
8
2
CCp 00.917
.91722.58
.5810
10 4TT 33.98
.9810
10 8TT 2 00.917
.91722.58
.5810
10 4373
37333.98
.9810
10 8373
3732
p
 1007
J /5kg 1007
K .7
2.18.710
Pr  1007.7 J / kg K
 0.689
0.03186
5
22.18
.1810
10 51007
1007.7.7  0.689
Pr

e) Pr 
 0.689
00.03186
.03186
e)
e)
 Cp
1520  10 6  139
 0.0261
k
0.0081  10 3
CCp 1520  10 66 139
1520  10  139
p
Pr

Pr  k  0.0081  10 33 00.0261
.0261
k
0.0081  10
Pr 

36
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Convection
Heat Transfer: Exercises
f)
f)
Pr 
g)
Pr 
g)
Pr 
 Cp
k
 Cp
k
 Cp
k
 Cp
Pr 
Comments:
k
420  10 6  1369

 0.0067
86

420  10 6  1369
 0.0067
86
8.36  10 2  2035

 1207
0.141

8.36  10 2  2035
 1207
0.141
 Large temperature dependence for water as in a) and b);
Comments:
 small temperature dependence for air as in c) and d);
 use of Sutherland’s law for viscosity as in part d);
 Large temperature dependence for water as in a) and b);
 difference between liquid metal and oil as in e), f) and g);
 small temperature dependence for air as in c) and d);
 units of kW/m K for thermal conductivity;
 use of Sutherland’s law for viscosity as in part d);
 use of temperature dependence of cp as in part a).
 difference between liquid metal and oil as in e), f) and g);
 units of kW/m K for thermal conductivity;
Example 3.2
 use of temperature dependence of cp as in part a).
Calculate the appropriate Reynolds numbers and state if the flow is laminar or turbulent for
Example 3.2
the following:
Calculate the appropriate Reynolds numbers and state if the flow is laminar or turbulent for
a) A 10 m (water line length) long yacht sailing at 13 km/h in seawater  = 1000 kg/m3 and
the following: 3
 = 1.3 x 10 kg/m s,
b) A compressor disc of radius 0.3 m rotating at 15000 rev/min in air at 5 bar and 400C and
a) A 10 m (water line length) long yacht sailing at 13 km/h in seawater  = 1000 kg/m3 and
6 3 2
3
1.46
 10
T s,
=1.3
x 10
kg/m
kg/m s
110 disc
T  of radius 0.3 m rotating at 15000 rev/min in air at 5 bar and 400C and
b) A compressor
6 3 2dioxide gas at 400 K flowing in a 20 mm diameter pipe. For the viscosity
c) 0.05 1kg/s
of10
carbon
.46 
T

6 kg/m
32 s
1
.
56

110  T10
 T kg/m s
take  
233 dioxide
T
c) 0.05 kg/s of carbon
gas at 400 K flowing in a 20 mm diameter pipe. For the viscosity
3
5
d) The roof of
a coach
1.56
 10 66Tm3 2long, travelling at 100 km/hr in air ( = 1.2 kg/m and  = 1.8 x 10
take
kg/m s
kg/ms)
233  T 
over
e) The flow of exhaust gas (p = 1.1 bar, T = 500ºC, R = 287 J/kg K and  = 33.56 x 105 kg/m s)
d) The roof of a coach 6 m long, travelling at 100 km/hr in air ( = 1.2 kg/m and  = 1.8 x 105
a valve guide of diameter 10 mm in a 1.6 litre, four cylinder four stroke engine running at
kg/m s)
an exhaust
3000 rev/min (assume 100% volumetric efficiency an inlet density of 1.2 kg/m3 and
e) The flow of exhaust gas (p = 1.1 bar, T = 500ºC, R = 287 J/kg K and  = 3.56 x 105 kg/m s) over
port diameter of 25 mm)
a valve guide of diameter 10 mm in a 1.6 litre, four cylinder four stroke engine running at
3000 rev/min (assume 100% volumetric efficiency an inlet density of 1.2 kg/m3 and an exhaust
port diameter of 25 mm)
37
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Convection
Heat Transfer: Exercises
Solution
uL


13  10 3
 10
3600
 2.78  10 7
1.3  10 3
10 3 
a)
Re 
b)
T  400  273  673 K
1.46  10 6  6733 2

 3.26  10 5
110  673

(turbulent)
kg / m s
15000
 2  1571 rad / s
60
u   r  1571  0.3  471 .3 m / s

100000
P

 2.59 kg / m 3
RT 287  673
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Convection
Heat Transfer: Exercises
Characteristic length is r not D
Re 
c)
uD 2.59  471.3  3

 1.12  10 7
5

3.26  10
m   uA  u 
u
D 2
4
4m
D 2
Re 
uD   4m D 4m



D 2  D
1.56  10 6  400 3 2

 1.97  10 5
233  400
Re 
d)
u
kg / m s
4  0.05
 1.6  10 5
5
  0.02  1.97  10
(turbulent)
100  10 3
 27.8 m / s
3600
Re 
e)
(turbulent)
uL 1.2  27.8  6

 11.1  10 7

1.8  10 5
(turbulent)
 be the mass flow through the exhaust port
Let m
m = inlet density X volume of air used in each cylinder per
second
m  1.2 
u
1.6  10 3 3600 1

  0.012 kg / s
4
60 2
4m
 D2
Re d 
ud

39
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Convection
Heat Transfer: Exercises
Re 
4  0.01  0.012
 6869
  3.56  10 5  0.025
(laminar)
Comments:


Note the use of D to obtain the mass flow rate from continuity, but the use of d for the
characteristic length
Note the different criteria for transition from laminar flow (e.g. for a pipe Re  2300 plate
Re  3  10 5 )
Example 3.3
Calculate the appropriate Grashof numbers and state if the flow is laminar or turbulent for the following:
a) A central heating radiator, 0.6 m high with a surface temperature of 75C in a room at 18C ( =
1.2 kg/m3 , Pr = 0.72 and  = 1.8 x 105 kg/m s)]
b) A horizontal oil sump, with a surface temperature of 40C, 0.4 m long and 0.2 m wide containing
oil at 75C ( = 854 kg/m3 , Pr = 546,  = 0.7 x 103 K1 and  = 3.56 x 102 kg/m s)
c) The external surface of a heating coil, 30 mm diameter, having a surface temperature of 80C in
water at 20C ( = 1000 kg/m3, Pr = 6.95,  = 0.227 x 103K1 and  = 1.00 x 10-3kg/m s)
d) Air at 20ºC ( = 1.2 kg/m3 , Pr = 0.72 and  = 1.8 x 105 kg/m s) adjacent to a 60 mm diameter
vertical, light bulb with a surface temperature of 90C
Solution
a)
Gr 
 2 g  T L3
2
T  75  18  57 K

1
1
1


K 1
T 18  273 291
Gr 
1.2 2  9.81  57  0.6 3
291  1.8  10

3 2
 1.84  10 9
Gr Pr  1.84  10 9  0.72  1.3  10 9
b)
L
(mostly laminar)
Area
0.4  0.2

 0.0667 m
Perimeter 2  0.4  0.2 
40
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Convection
Heat Transfer: Exercises
T  75  40  35 K
 2 g  T L3 854 2  9.81  0.7  10 3  35  0.0667 3

 4.1  10 4
Gr 
2
2

2

3.56  10 
Gr Pr  4.1  10 4  546  2.24  10 7
Heated surface facing downward results in stable laminar flow for all Gr Pr
c)
41
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Convection
Heat Transfer: Exercises
T  80  20  60 K
 2 g  T L3 1000 2  9.81  0.227  10 3  60  0.033

 3.6  10 6
Gr 
2
2

3

1 10 
Gr Pr  3.6  10 6  6.95  25  10 6
L
d)
(laminar)
Area
D 2 D


Perimeter 4D 4
T  90  20  70 K

1
1
1


K 1
T 20  273 293
 2 g  T L3 1.2 2  9.8  70  0.0153

 3.5  10 4
Gr 
2
2

5

293  1.8  10 
Gr Pr  3.5  10 4  0.72  2.5  10 4
(laminar)
Comments:

Note evaluation of  for a gas is given by   1 / T

For a horizontal surface L  A / p
Example 3.4
Calculate the Nusselt numbers for the following:
a) A flow of gas (Pr = 0.71,  = 4.63 x 105 kg/m s and Cp = 1175 J/kg K) over a turbine blade of
chord length 20 mm, where the average heat transfer coefficient is 1000 W/m2 K.
b) A horizontal electronics component with a surface temperature of 35C, 5 mm wide and 10 mm
long, dissipating 0.1 W by free convection from one side into air where the temperature is 20C
and k = 0.026 W/m K.
c) A 1 kW central heating radiator 1.5 m long and 0.6 m high with a surface temperature of 80ºC
dissipating heat by radiation and convection into a room at 20C (k = 0.026 W/m K assume black
body radiation and  = 56.7 x 109 W/m K4)
d) Air at 4C (k = 0.024 W/m K) adjacent to a wall 3 m high and 0.15 m thick made of brick with k =
0.3 W/m K, the inside temperature of the wall is 18C, the outside wall temperature 12C
42
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Convection
Heat Transfer: Exercises
Solution
a)
 Cp
Pr 
Solutionk
 C p  C4p.63  10 5  1175
ka)  Pr  k
 0.0766 W / m K
Pr
0.71
 C p 4.63  10 5  1175
  0.02
 0.0766 W / m K
1000
 261
Nu 

Pr
0.71
0.0766
k
k h L
b)
Nu
h L
Nu 
k
h L 1000  0.02
q L0.0766  261
k
T k
hL
q L
b) Nu 


k
0 .1 T k
Q
q
A

0.01  0.005
 2000 W / m 2
0 .1
Q
q 
 2000 W / m 2
A 0.01  0.005
T  35  20  15 C
T  35  20  15 C
L
50 5
Area

 mm  0.001667 m
Perimeter
Area 30 50 3 5
L
Perimeter

30

3
mm  0.001667 m
h L 2000  0.001667
 8 .5

2000
0.001667
h L 15
.026
0
k

 8 .5
Nu 

Nu 
k
c)
15  0.026
qc L
q L
Nu
T k c
Nu 
c)
T k
In thisIncase,
q must be the convective heat flux – radiative heat flux
this case, q must be the convective heat flux – radiative heat flux
Ts  T80
 273  353 K
s  80  273  353 K
 273
 293
T T20
273
 293
KK
 20
4
4
9
4
9
 .56
 10
35344 
Q R QRATs4ATsT4 T 56
7 .710
 1.15.500.6.6353
 293
293 4416
416WW
T  80  20  60 K
T  80  20  60 K
Qc  Q  QR  1000  416  584 W
Qc  Q  QR  1000  416  584 W
43
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Convection
Heat Transfer: Exercises
Qc
584

 649 W / m 2
A 1.5  0.6
q L 649 0.6
Nu  c


 249
T k
60 0.026
qc 
T  12  4  8 K
d)
q
k b T1  T2 
 60 C
W
(assuming 1-D conduction)
0.318  12 
 12 W / m 2
0.18
q L 12
3
Nu  c
 
 188
T k
8 0.024
q
Comments:



Nu is based on convective heat flux; sometimes the contribution of radiation can be significant
and must be allowed for.
The value of k is the definition of Nu is the fluid (not solid surface property).
Use of appropriate boundary layer growth that characterises length scale.
44
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Convection
Heat Transfer: Exercises
Example 3.5
In forced convection for flow over a flat plate, the local Nusselt number can be represented by the
general expression Nu x  C1 Re nx . In free convection from a vertical surface the local Nusselt number
is represented by Nu x  C 2 Grxm , where C1, C2, n and m are constants
a) Show that the local heat transfer coefficient is independent of the surface to air temperature
difference in forced convection, whereas in free convection, h, depends upon (Ts  T)m
b) In turbulent free convection, it is generally recognised that m = 1/3. Show that the local heat
transfer coefficient does not vary with coordinate x.
Solution
a)
Nu x 
hx
k
Re x 
ux

For forced convection: Nu x  C1 Re nx
Hence
k ux

h  C1 
x   
n
This shows that the heat transfer coefficient for forced does not depend on temperature difference.
For free convection
Grx 
Hence
Nu x  C 2 Grxm
 2 g  T x 3
2
k   2 g  T x 3 

h  C 2 
x 
2

m
(1)
So for free convection, heat transfer coefficient depends on T m
b)
From (1), with m = 1/3 for turbulent free convection:
45
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Convection
Heat Transfer: Exercises
k   2 g  T x 3 

h  C 2 
x 
2

k   2 g  T
h  C 2 
x 
2
  2 g  T
h  kC 2 
2







1/ 3
1/ 3
x
1/ 3
Hence the convective heat transfer coefficient does not depend on x
Example 3.6
An electrically heated thin foil of length L = 25 mm and width W = 8 mm is to be used as a wind
speed metre. Wind with a temperature T and velocity U  blows parallel to the longest side. The foil
is internally heated by an electric heater dissipating Q (Watts) from both sides and is to be operated in
5
2
air with T  20C , C p  1.005 kJ / kg K ,   1.522  10 m / s   1.19 kg / m 3 and Pr  0.72
. The surface temperature, T of the foil is to be measured at the trailing edge – but can be assumed to
be constant. Estimate the wind speed when T  32C and Q  0.5 W .
Solution
Firstly, we need to
estimate if the flow
laminar
or
turbulent.
is
Assuming a critical (transition) Reynolds number of Re  3  10 5 the velocity required would be:
u turb 
3  10 5
3  10 3 3  10 5  1.522  10 5


 304 m / s
L
L
25  10 3
Wind speed is very unlikely to reach this critical velocity, so the flow can be assumed to be laminar.
Nu x  0.331 Re 1x/ 2 Pr 1 / 3
Nu av  0.662 Re1L/ 2 Pr 1 / 3 
q av L
Ts  T k
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Convection
Heat Transfer: Exercises
Re1L/ 2 
q av 
q av L
Ts  T k  0.662 Pr 1 / 3
0 .5 / 2
 1250 W / m 2
0.025  0.008
Re1L/ 2 
1250  0.025
32  20  0.0253  0.662  0.721 / 3
 173.5
Re L  3  10 4
u 
Re L  3  10 4  1.522  10 5

 18.3 m / s
L
25  10 3
.
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Convection
Heat Transfer: Exercises
Example 3.7
The side of a building of height H = 7 m and length W = 30 m is made entirely of glass. Estimate the
heat loss through this glass (ignore the thermal resistance of the glass) when the temperature of the air
inside the building is 20C, the outside air temperature is -15C and a wind of 15 m/s blows parallel to
the side of the building. Select the appropriate correlations from those listed below of local Nusselt
numbers to estimate the average heat transfer coefficients. For air take: ρ= 1.2 kg / m3, μ = 1.8 x 10-5
kg / m s, Cp = 1 kJ / kg K and Pr = 0.7.




Free convection in air, laminar (Grx < 109): Nux = 0.3 Grx1/4
Free convection in air, turbulent (Grx > 109): Nux = 0.09 Grx1/3
Forced convection, laminar (Rex < 105): Nux = 0.33 Rex0.5 Pr1/3
Forced convection, turbulent (Rex > 105): Nux = 0.029 Rex0.8 Pr1/3
Solution
Pr 
 Cp
k
gives: k 
 Cp
Pr

1.8  10 5  1000
 0.026 W / m K
0.7
First we need to determine if these flows are laminar or turbulent.
For the inside (Free convection):

1
1
1


T 20  273 293
K 1
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Convection
Heat Transfer: Exercises
Gr 
 2 g  T L3 1.2 2  9.81  T  7 3

2
1.8  10 5 2  293
Gr  5.1  1010 T
(Flow will be turbulent over most of the surface for all reasonable values of T )
For the outside (Forced convection)
Re L 
 u  L 1.2  15  30

 3  10 7
5

1.8  10
(Flow will be turbulent for most of the surface apart from the first 0.3 m)
Hence we use the following correlations:
On the inside surface:
Nu x  0.09Gr 1 / 3
On the outside surface:
Nu x  0.029 Re 0x.8 Pr 1 / 3
For the inside:
  2 g  Ti  Ts  x 3 
hx

 0.09 
Nu x 
2
k



x 
h  constant 
3 1/ 3
x
1/ 3
 constant
Hence heat transfer coefficient is not a function of x
hav  hx  L
(1)
For the outside:
ux
hx

Nu x 
 0.029
k
  
h  constant 
x 0.8
x
0.8
Pr 1 / 3
 C x 0.2
49
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Convection
Heat Transfer: Exercises
xL
C
1
hav   h dx 
L
L x 0
xL
x
 0.2
dx 
x 0
hx  L
0.8
(2)
Write a heat balance:
Assuming one-dimensional heat flow and neglecting the thermal resistance of the glass
q  hi Ti  Ts 
q  ho Ts  To 
hi Ti  Ts   ho Ts  To 
(3)
From equation 1
  2 g Ti  Ts  H 3 
hi H

 0.09 

k
 2  Ti


1/ 3
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Convection
Heat Transfer: Exercises
 1.2 2  9.81  T1  Ts  
  0.026
hi  0.09 
 1.8  10 5 2  293 


hi  1.24 Ti  Ts 
1/ 3
(4)
From equation 2:
ho W 0.029   u W


k
0.8  
ho 



0.8
Pr 1 / 3
0.026 0.029  1.2  15  30 



30
0.8  1.8  10 5 
0. 8
 0 .7 1 / 3
ho  26.7 W / m 2 K
(5)
From (3) with (4) and (5)
1.24 Ti  Ts 
4/3
1.24 20  Ts 
4/3
 26.7 Ts  To 
 26.7 Ts  15
Ts  0.0464 20  Ts 
4/3
 15
(6)
To solve this equation for Ts an iterative approach can be used
First guess: Ts  10C
Substitute this on the right hand side of equation 6:
Ts  0.0464 20   10
4/3
 15  10.7C
For the second iteration we use the result of the first iteration:
Ts  0.0464 20   10.7 
4/3
 15  10.6C
The difference between the last two iterations is 0.1C , so we can consider this converged.
Ts  10.6C
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Convection
Heat Transfer: Exercises
From which:
q  ho Ts  To   26.7  10.6  15  117 W / m 2
Q  qA  117  30  7  24600 W  24.6 kW
Example 3.8
The figure below shows part of a heat exchanger tube. Hot water flows through the 20 mm diameter
tube and is cooled by fins which are positioned with their longest side vertical. The fins exchange heat
by convection to the surrounds that are at 27C.
Estimate the convective heat loss per fin for the following conditions. You may ignore the contribution
and effect of the cut-out for the tube on the flow and heat transfer.
a) natural convection, with an average fin surface temperature of 47C;
b) forced convection with an air flow of 15 m / s blowing parallel to the shortest side of the fin and
with an average fin surface temperature of 37C.
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Convection
Heat Transfer: Exercises
The following correlations may be used without proof, although you must give reasons in support of
your choice in the answer.
Nux = 0.3 Rex1/2 Pr1/3
Nux = 0.02 Rex0.8 Pr1/3
Nux = 0.5 Grx1/4 Pr1/4
Nux = 0.1 Grx1/3 Pr1/3
Rex < 3 x 105
Rex  3 x 105
Grx < 109
Grx  109
For air at these conditions, take: Pr = 0.7, k = 0.02 W / m K, μ = 1.8 x 10-5 kg /m s and ρ = 1.0 kg / m3
Solution
On the outside of the water tube, natural convections means that we need to evaluate Gr number to see
if flow is laminar ot turbulent
Gr 
 2 g  T L3
2
T  47  27  20 K

1
1

27  273 300
Gr 
K 1
12  9.81  20  0.13
1.8  10 
5 2
 300
 2  10 6
(Laminar)
(L here is height because it is in the direction of the free convection boundary layer)
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Convection
Heat Transfer: Exercises
So we use:
Nu x  0.5 Grx Pr 
1/ 4
hav 
hav 
L
L
1
h dx  constant  x 1 / 4 dx
L 0
0
hx  L 
3/ 4
Nu av 
2
GrL Pr 1 / 4
3
Nu av 
1/ 4
2

2  10 6  0.7   23
3
hav 
Nu av k 23  0.02

 4.6 W / m 2 K
0.1
L
q av  hav T
Q  q av A  hav TA  4.6  20  0.1  0.05  2
(Last factor of 2 is for both sides)
Q  0.92 W
For forced convection, we need to evaluate Re to see if flow is laminar or turbulent
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Convection
Heat Transfer: Exercises
Re 
 u L 1  15  0.05

 4.17  10 4
5

1.8  10
(Laminar)
(L here is the width because flow is along that direction)
Nu x  0.3 Re1x/ 2 Pr 1 / 3
L
h
1
hav   hdx  x  L
L0
1/ 2

Nu av  0.6 Re1L/ 2 Pr 1 / 3  0.6  4.17  10  4
hav 

1 / 2
 0.71 / 3  109
Nu av k 109  0.02

 43.5 W / m 2 K
0.05
L
T  10C 
Q  q av A  hav TA  43.5  10  0.1  0.05  2
Q  4.35 W
Example 3.9
Consider the case of a laminar boundary layer in external forced convection undergoing transition to a
turbulent boundary layer. For a constant fluid to wall temperature difference, the local Nusselt
numbers are given by:
Nux = 0.3 Rex1/2 Pr1/3 (Rex < 105)
Nux = 0.04 Rex0.8 Pr1/3 (Rex ≥ 105)
Show that for a plate of length, L, the average Nusselt number is:
Nuav = (0.05 ReL0.8 - 310) Pr1/3
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Convection
Heat Transfer: Exercises
Solution
Nu av 
hav k
L
Where for a constant surface-to-fluid temperature:
hav 
x
L

1  L

h
dx
hturbulent dx 
  laminar

L  0

xL
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Convection
Heat Transfer: Exercises
Since for laminar flow ( Re x  10 5 ):
Nu x  0.3 Re1x/ 2 Pr 1 / 3
hlam
k   u
 0.3  
x 



hlam
  u
 0.3  k 
 



1/ 2
x1 / 2 Pr 1 / 3
1/ 2
Pr 1 / 3 x 1 / 2  C lam x 1 / 2
Where C lam does not depend on x
Similarly:
hturb  C turb x 0.2
Where
Cturb
  u
 0.04  k 
 



0.8
Pr 1 / 3
Hence
x
L

1  L
1 / 2
hav    Clam x dx   C turb x 0.2 dx 
L  0

xL
xL
L
 x1 / 2 
 x 0.8  
1 
hav  Clam 
  Cturb 
 
L
1/ 2  0

 0.8  xL 

Nu av 
hav k
L
Nu av 
Clam 1 / 2 C turb 0.8
L  x L0.8
2xL 
k
0.8k
Nu av
  u
 0.6 
 




1/ 2
x
1/ 2
L
Pr
1/ 3

  u L  0.8   u x

 L
  
 0.05
 
  
57
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0.8
  1/ 3
  Pr
 
Convection
Heat Transfer: Exercises
But
 u xL
 10 5

(The transition Reynolds number)
So

 
Nu av  Pr 1 / 3 0.6  10 5

1/ 2
 
 0.05 Re 0L.8  0.05  10 5
0. 8


Nu av  0.05 Re 0L.8  310 Pr 1 / 3
Example 3.10
A printed circuit board dissipates 100 W from one side over an area 0.3m by 0.2m. A fan is used to
cool this board with a flow speed of 12 m / s parallel to the longest dimension of the board. Using the
average Nusselt number relationship given in Example 3.9 to this question, calculate the surface
temperature of the board for an air temperature of 30 ºC.
Take an ambient pressure of 1 bar, R = 287 J / kg K,
Cp = 1 kJ / kg K, k = 0.03 W / m K and μ = 2 x 10-5 kg/m s
Solution
q av 
Q
100

 1666.7 W / m 2
A 0.2  0.3
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Convection
Heat Transfer: Exercises
Pr 
C P
k
2  10 5  10 3
 0.667
0.03
 u L

Re L 


P
10 5

 1.15 kg / m 3
RT 287  303
Re L 
1.15 12  0.3
 2.07  10 5
5
2  10
Using the formula for Nusselt Number obtained in Example 3.9:


Nu av  0.05 Re 0L.8  310 Pr 1 / 3


Nu av  0.05  2.07  10 5
Nu av 
T 

0 .8

 310  0.667 
1/ 3
 511
hav k q av L

L
Tk
q av L 1666.7  0.3

 32.6C
Nu av k
511  0.03
Ts  T  T
Ts  30  32.6  62.6C
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Radiation
Heat Transfer: Exercises
4. Radiation
Example 4.1
In a boiler, heat is radiated from the burning fuel bed to the side walls and the boiler tubes at the top.
The temperatures of the fuel and the tubes are T1 and T2 respectively and their areas are A1 and A2.
a) Assuming that the side walls (denoted by the subscript 3) are perfectly insulated show that the
temperature of the side walls is given by:
 A1 F13T14  A2 F23T24
T3  
 A2 F23  A1 F13



14
where F13 and F23 are the appropriate view factors.
b) Show that the total radiative heat transfer to the tubes, Q2, is given by:

AF A F 
Q 2   A1 F12  1 13 2 23  T14  T24
A2 F23  A1 F13 



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Radiation
Heat Transfer: Exercises
c)
Calculate the radiative heat transfer to the tubes if T1 = 1700C, T2 = 300C, A1 = A2 = 12m2 and
the view factors are each 0.5?
Solution
a)
Q 2  Q 1 2  Q 3 2
(1)
Since the walls are adiabatic
Q 3 2  Q 13
(2)
From (2)
 A3 F32 T34  T24    A1 F13 T14  T34 
A1 F13 T14  A3 F32 T24
T 
A3 F32  A1 F13
4
3
 A F T 4  A2 F23 T24 

T3   1 13 1

A2 F23  A1 F13


b)
1/ 4
since Ai Fij  A j F ji
From (1)



Q 2   A1 F12 T14  T24   A3 F32 T34  T24

Q 2   A1 F12 T14  T24    A2 F23 T34  T24 
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Radiation
Heat Transfer: Exercises
 A1 F13 T14  A3 F32 T24

4
4

Q2   A1 F12 T1  T2   A2 F23 
 T24 
A3 F32  A1 F13




 A1 F13 T14  A3 F32 T24  A3 F32T24  A1 F13T24 
4
4


Q2   A1 F12 T1  T2   A2 F23 
A3 F32  A1 F13




 A1 F13 T14  A1 F13T24
4
4

Q2   A1 F12 T1  T2   A2 F23 
 A3 F32  A1 F13







A1 F13

Q 2   A1 F12 T14  T24   A2 F23 T14  T24 
 A3 F32  A1 F13 




 A F A F 
Q 2   A1 F12 T14  T24   T14  T24  2 23 1 13 
 A3 F32  A1 F13 





A F A F 
Q 2   T14  T24  A1 F12  2 23 1 13 
A2 F23  A1 F13 


c)

T34 
A1 F13 T14  A3 F32 T24 A1 F13 T14  A2 F23 T24

A3 F32  A1 F13
A2 F23  A1 F13
T34 
12  0.5  1973 4  12  0.5  573 4
 1662 K
12  0.5  12  0.5
66 

6
Q 2  56.7  10 9 1973 4  573 4  6 
  7.68  10 W
66



Example 4.2
Two adjacent compressor discs (Surfaces 1 and 2) each of 0.4 m diameter are bounded at the periphery
by a 0.1 wide shroud (Surface 3).
a) Given that F12 = 0.6, calculate all the other view factors for this configuration.
b) The emissivity and temperature of Surfaces 1 and 2 are 1 = 0.4, T1 = 800 K, 2 = 0.3, T2 = 700K
and Surface 3 can be treated as radiatively black with a temperature of T3 = 900 K. Apply a grey
body radiation analysis to Surface 1 and to Surface 2 and show that:
2.5 J1 – 0.9 J2 = 45545
W/m2
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Radiation
Heat Transfer: Exercises
and
W/m2.
3.333 J2 – 1.4 J1 = 48334
The following equation may be used without proof:
E B ,i  J i N
  Fi , j ( J i  J j )
1  i
j 1
i
c) Determine the radiative heat flux to Surface 2
Solution
a)
r1  r2  r  0.2 m
a  0 .1 m
r2 0.2

2
a 0.1
a 0 .1

 0 .5
r1 0.2
F12  0.6 (Although this is given in the question, it can be obtained from appropriate tables
with the above parameters)
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Radiation
Heat Transfer: Exercises
F11  0 (As surface 1 is flat, it cannot see itself)
F13  1  0.6  0.4 (From the relation
F
ij
 1 in an enclosure)
F21  0.6 (Symmetry)
F22  0
F23  0.4
F31 
A1
  0.2 2
F13 
 0.4  0.4
A3
2    0.2  0.1
F32  0.4 (Symmetry)
F33  1  0.4  0.4  0.2
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Radiation
Heat Transfer: Exercises
b)
n
E b ,i  J i
  J i  J j Fij
1 i
j 1
i
Apply to surface 1, (i = 1)
Let
E
b ,1
1  1
 1
1
 J 1   1 F12  J 1  J 2   F13  J 1  J 3 
E b ,1  J 1  1  1 F12  1 F13   1 F12 J 2  1 F13 J 3
Eb,1   T14
J 3   T34 (Radiatively black surface)
1 
1  1
1

1  0.4
 1.5
0.4
 T14  2.5 J 1  0.9 J 2  0.6  T34
56.7  10 9  800 4  2.5  J 1  0.9  J 2  0.6  56.7  10 9  900 4
2.5 J 1  0.9 J 2  45545 W / m 2
Applying to surface 2
(1)
(i = 2)
E b , 2  J 2  1   2 F21   2 F23    2 F21 J 1   2 F23 J 3
Eb, 2   T24
2 
1 2
2

1  0.3
 2.333
0.3
 T24  3.333 J 2  1.4 J 1  0.9333  T34
3.333 J 2  1.4 J 1  48334 W / m 2
(2)
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Radiation
Heat Transfer: Exercises
c)
From (2):
J1 
3.333 J 2  48334
1.4
Substituting in (1)
2.5 
3.333 J 2  48334
 0.9 J 2  45545 W / m 2
1.4
J 2  26099 W / m 2
The net radiative flux to surface 2 is given by
q2 
E b , 2  J 2 56.7  10 9  700 4  26099

 5.351  10 3 W / m 2
1 2
1  0 .3
0 .3
2
The minus sign indicates a net influx of radiative transfer as would be expected from
consideration of surface temperatures.
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Radiation
Heat Transfer: Exercises
Example 4.3
The figure below shows a simplified representation of gas flame inside a burner unit. The gas flame is
modelled as a cylinder of radius r1 = 10 mm (Surface 1). The burner comprises Surface 2 (a cylinder of
radius r2 = 40 mm and height h = 40 mm), concentric with Surface 1 and a concentric base (Surface 3),
of radius r3 = 40 mm. The end of the cylinder, Surface 4, opposite to the base is open to the
surrounding environment.
a)
Given that F21 = 0.143 and F22 = 0.445 use the dimensions indicated on the diagram to calculate
all the other relevant view factors.
b)
The flame, base and surroundings can be represented as black bodies at constant temperatures T1,
T3 and T4, respectively. The emissivity of the inside of Surface 2 is ε2 = 0.5. Apply a grey body
radiation analysis to Surface 2 and show that the radiosity is given by:
J2 
 (T24  F21T14  F23T34  F24T44 )
1  F21  F23  F24
The following equation may be used without proof:
E b ,i  J i
1   i   i
c)
N
  Fij J i  J j 
j 1
The temperatures T1 and T3 are found to be: T1 = 1800K and T3 = 1200K, and the surrounds are at
500 K. Estimate the temperature T2, using a radiative heat balance on the outer surface of Surface
2, where the emissivity is ε0 = 0.8
Solution
a)
A1  2  r1 h
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Heat Transfer: Exercises
A2  2  r2 h
A3  A4   r22  r12 
F11  0
F13  F14
F11  F12  F13  F14  1
but
A1 F12  A2 F21
F12 
A2
r
40
F21  2 F21 
 0.14338  0.57352
A1
r1
10
Thus
F13  F14 
1  0.57352
 0.21324
2
F21  F22  F23  F24  1
F23  F24 
1  F21  F22 1  0.14338  0.44515

 0.20574
2
2
F31  F32  F33  F34  1
F33  0
A1 F13  A3 F31
F31 
A1
2  r1 h
2  0.01  0.04
F13 
F13 
 0.21324  0.11373
2
2
A3
 r2  r1
0.04 2  0.012


A2 F23  A3 F32
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Heat Transfer: Exercises
F32 
A2
2  r2 h
2  0.04  0.04
F23 
F23 
 0.20574  0.43891
2
2
A3
 r2  r1
0.04 2  0.012


F34  1  0.11373  0.43891  0.44736
Similarly (using symmetry)
F41  F31  0.11373
F42  F32  0.43891
F43  F34  0.44736
F44  0
b)
n
E b ,i  J i
  J i  J j Fij
1 i
j 1
i
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Radiation
Heat Transfer: Exercises
For surface 2, i = 2, j = 1, 3, 4
Eb, 2  J 2
 F12  J 2  J 1   F23  J 2  J 3   F24  J 2  J 4 
1 2
2
 2  0.5 ,
1  0.5
1
0.5
J 1  E b ,1 , J 3  E b ,3 , J 4  E b , 4
(1, 3, 4 are black)
E b , 2  J 2  F12 J 2  E b ,1   F23 J 2  E b ,3   F24 J 2  E b , 4 
J 2 F21  F23  F24  1   T24   T14 F21   T34 F23   T44 F24
J2 
c)
J2 
 T24  T14 F21  T34 F23  T44 F24 
F21  F23  F24  1
56.7  10 9 T24  1800 4  0.57352  1200 4  0.20574  500 4  0.20574
0.57352  20574  0.20574  1
J 2  36.47  10 9 T24  70913
On the outside of surface 2:

 q 2    2,0 T24  T44

Also
q2 
Eb, 2  J 2
  T24  36.47  10 9 T24  70913
1 2
2
20.23  10 9 T24  70913  56.7  10 9  0.8 T24  500 4 
T2  1029 K
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Radiation
Heat Transfer: Exercises
Example 4.4
The figure below shows a schematic diagram, at a particular instant of the engine cycle, of a cylinder
head (Surface 1), piston crown (Surface 2) and cylinder liner (Surface 3).
a)
Using the dimensions indicated on the diagram, and given that F12 = 0.6, calculate all the other
relevant view factors.
b)
The cylinder head can be represented as a black body at a temperature T1 = 1700 K and the
emissivity of the piston crown is  2  0.75 . Apply a grey body radiation analysis to the piston
crown (Surface 2) and show that the radiosity is given by:
J2 = 42.5 x 10-9 T24 + 71035 + 0.1 J3
The following equation may be used without proof:
Eb,i  J i
1   i   i
c)
N
  Fij J i  J j 
j 1
Similar analysis applied to the cylinder liner gives:
J3 = 107210 + 0.222 J2
If the surface temperature of the piston crown is, T2 = 600 K, calculate the radiative heat flux into
the piston crown.
d)
Briefly explain how this analysis could be extended to make it more realistic
Solution
a)
A1  A2   r 2    50 2  2500 
mm 2
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Heat Transfer: Exercises
A3   DL    100  25  2500 
mm 2
F11  0 (Flat surface)
F12  0.6 (Given)
F13  1.0  F12  1.0  0.6  0.4
By Symmetry:
F21  F12  0.6
F23  F32  0.4
F22  0
F31 
A1
F13  0.4 Since A1  A3
A3
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Radiation
Heat Transfer: Exercises
F32  0.4 (By symmetry)
F33  1.0  F31  F32  1.0  0.4  0.4  0.2
b)
For surface 2, i = 2
Eb,2  J 2
 F21 J 2  J 1   F23  J 2  J 3 
1 2
2
J 1   T14 (Black body)
1  0.75 1

0.75
3
 2  0.75 ,
Eb, 2   T24
 T24  J 2
1/ 3
J2 
J2 
 F21 J 2   T14   F23  J 2  J 3 
1
F21 T14  F23 J 3
3
1
1  F21  F23 
3
 T24 

56.7  10 9  T24 

1
0.6  56.7  10 9 1700 4  0.4 J 3
3
1
1  0.6  0.4
3

J 2  42.5  10 9 T24  71035  0.1 J 3
We are also given that
J 3  107210  0.222 J 2
0.1 J 3  10721  0.0222 J 2
Hence
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
Radiation
Heat Transfer: Exercises
J 2  42.5  10 9  600 4  71035  10721  0.0222 J 2
0.97778 J 2  5508  81756
J 2  89247 W / m 2
Also
q2 
E b , 2  J 2 56.7  10 9  600 4  89247

 246  10 3 W / m 2
1 2
1/ 3
2
Negative sign indicates J 2  E b , 2  E 2 , so net flux is into the piston crown.
c)
To make the analysis more realistic, it needs to be extended by including convection from the
piston crown, and cylinder liner. Radiation from the piston underside also needs to be included.
We then carry out analysis over a complete engine cycle.
Example 4.5
The figure below shows the variation of view factor Fi,j with geometric parametres h / L and W / L for
the case of two rectangular surfaces at right angles to each other. This plot is to be used to model the
radiative heat transfer between a turbocharger housing and the casing of an engine management
system. The horizontal rectangle, W = 0.12 m and L = 0.2 m, is the engine management system and is
denoted Surface 1. The vertical rectangle, h = 0.2 m and L = 0.2 m, is the turbocharger casing and
denoted by Surface 2. The surrounds, which may be approximated as a black body, have a temperature
of 60C.
a)
Using the graph and also view factor algebra, evaluate the view factors: F 1,2, F2,1, F1,3 and F2,3
b)
By applying a grey-body radiation analysis to Surface 1 with ε1 = 0.5, show that the radiosity
J1 is:
J1 = 28.35 x 10-9 T14 + 0.135 J2 + 254 (W/m2)
The following equation may be used without proof:
Eb,i  J i
1   i   i
N
  Fij J i  J j 
j 1
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Heat Transfer: Exercises
c)
A similar analysis is applied to Surface 2 with ε2 = 0.4 obtained the result:
J2 = 22.7 x 10-9 T24 + 0.097 J1 + 350 (W/m2).
Use this to estimate the surface temperature of the engine management system when the turbocharger
housing has a surface temperature of T2 = 700K.
Solution
h 0. 2

 1,
L 0. 2
W 0.12

 0 .6
L
0. 2
From the figure: F12  0.27
A1 F12  A2 F21
F21 
A1
w
0.12
F12  F12 
 0.27  0.162
A2
h
0.2
F11  F12  F13  1
F11  0
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Radiation
Heat Transfer: Exercises
F13  1  F12  1  0.27  0.73
F21  F22  F23  1
F22  0
F23  1  F21  1  0.162  0.838
For a grey body radiative heat transfer in an enclosure (n surfaces)
n
E b ,i  J i
  J i  J j Fij
1 i
j 1
i
Applying for surface 1, i = 1 (the casing)
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Radiation
Heat Transfer: Exercises
E b ,1  J 1
 F12  J 1  J 2   F13  J 1  J 3 
1  1
1
Eb,1   T14
J 3   T34
1  1
1

1  0.5
 1.0
0.5
So
J1 
J1 
 T14  F12 J 2  F13 T34
1  F12  F13
56.7  10 9 T14  0.27 J 2  0.73  56.7  10 9  3334
1  0.27  0.73
J 1  28.35  10 9 T14  0.135 J 2  254
W / m2
(1)
c)
Given: J 2  22.68  10 9 T24  0.0972 J 1  350
J 2  22.68  10 9  700 4  0.0972 J 1  350
W / m2
W / m2
J 2  5796  0.0972 J 1
(2)
Substituting from equation 2 into equation 1:
J 1  28.35  10 9 T14  0.135 5796  0.0972 J 1   254
W / m2
Which gives:
J 1  28.7  10 9 T14  1050
W / m2
Applying a heat balance to surface 1
qin  qout
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Heat Transfer: Exercises


E  J 
b ,1
1
  57.9  10 9 T14  28.7  10 9 T14  1050
qin   


1

1 
 

1


q in  28.  10 9 T14  1050
q out   1 T14  T4   0.5  56 .7  10 9 T14  333 4 
Combining and solving for T1, gives:
T1  396 K
Note that qin = - q since q is out of the surface when q > 0 .
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Heat Exchangers
Heat Transfer: Exercises
5. Heat Exchangers
5. Heat Exchangers
Example 5.1
Example 5.1
A heat exchanger consists of numerous rectangular channels, each 18 mm wide and 2.25 mm high. In
anheat
adjacent
pair ofconsists
channels,
are two
streams: channels,
water k = each
0.62518W/m
and and
air k2.25
= 0.0371
W/mIn
A
exchanger
of there
numerous
rectangular
mmKwide
mm high.
K,
separated
by
a
18
mm
wide
and
0.5
mm
thick
stainless
steel
plate
of
k
=
16
W/m
K.
The
fouling
an adjacent pair of channels, there are two streams: water k = 0.625 W/m K and air k = 0.0371 W/m
4
2
m2thick
K/W stainless
and 5 x 10
K/W,
and
thefouling
Nusselt
resistances
andmm
water
areand
2 x0.5
104mm
K, separatedfor
byair
a 18
wide
steelmplate
of krespectively,
= 16 W/m K.
The
2 hydraulic diameter.
= 5.95are
where
the4 subscript
'Dh' 5refers
to m
the
number
given
m2 K/W and
x 104
K/W, respectively, and the Nusselt
resistances
forby
airNu
andDhwater
2 x 10
number given by NuDh = 5.95 where the subscript 'Dh' refers to the hydraulic diameter.
a) Calculate the overall heat transfer coefficient ignoring both the thermal resistance of the
separatingthe
wall
and the
two
foulingcoefficient
resistances.
a) Calculate
overall
heat
transfer
ignoring both the thermal resistance of the
b) Calculate
the
overall
heat
transfer
coefficient
with
separating wall and the two fouling resistances. these resistances.
c)
is the
transfer
coefficient?
b) Which
Calculate
the controlling
overall heatheat
transfer
coefficient
with these resistances.
c) Which is the controlling heat transfer coefficient?
Solution:
Solution:
Hydraulic Diameter = 4 x Area / Wetted perimetre
Hydraulic Diameter = 4 x Area / Wetted perimetre
2.25  10 3  18  10 3
Dh  4 
 4  10 3
3 3
3
(2.2510 18)18
 10
2.25
 10
Dh  4 
 4  10 3
(2.25  18)  10 3
Nu D k
h
DDh k
Nu
h
Dh
5.95  0.625
(a)
hwater 
 930W / m 2 K
3
4 10
5.95
0.625
(a)
hwater 
 930W / m 2 K
3
4  10
5.95  0.0371
hair 
 55.186W / m 2 K
3
4 10
5.95
0.0371
hair 
 55.186W / m 2 K
3
4  10
1
1 
 1
U 

W / m2 K
1  52.1

930
55
.
186
1
1


U 

 52.1 W / m 2 K

 930 55.186 
1

 0.5  10 3
1
1
U 

 2  10  4 
 5  10  4  1  50.2
3
930
55.1186

 0.5 16
1
 10
b) U  

 2  10  4 
 5  10  4   50.2
930
55.186

 16
W / m2 K
W / m2 K
b)
c) The controlling heat transfer coefficient is the air heat transfer coefficient.
c) The controlling heat transfer coefficient is the air heat transfer coefficient.
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Heat Transfer: Exercises
Example 5.2
A heat exchanger tube of D = 20 mm diameter conveys 0.0983 kg/s of water (Pr = 4.3, k = 0.632 W/m
K,  = 1000 kg/m3,  = 0.651 x 103 kg/ms) on the inside which is used to cool a stream of air on the
outside where the external heat transfer coefficient has a value of ho = 100 W/m2 K. Ignoring the
thermal resistance of the tube walls, evaluate the overall heat transfer coefficient, U, assuming that the
internal heat transfer coefficient is given by the Dittus-Boelter relation for fully developed turbulent
pipe flow:
.
Nu D  0.023 Re 0D.8 Pr 0.4
Solution:
m  VA
V 
m
A
Re D 
4  0.0983
VD 4m


 9613

D   0.02  0.651  10 3
Nu D  0.023  9613 0.8  4.3 0.4  63
Nu D 
h
hD
k
Nu D k 63.3  0.632

 2000W / m 2 K
D
0.02
1 
 1

U 
 2000 100 
1
 95.2W / m 2 K
Example 5.3
a)
Show that the overall heat transfer coefficient for a concentric tube heat exchanger is given by the
relation:
r  r  r
1
U o   o ln o   o  
 k  ri  hi ri ho 
-1
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Heat Exchangers
Heat Transfer: Exercises
With the terminology given by the figure below
b)
A heat exchanger made of two concentric tubes is used to cool engine oil for a diesel engine. The
inner tube is made of 3mm wall thickness of stainless steel with conductivity k = 16 W/m K . The
inner tube radius is 25mm and has a water flow rate of 0.25 kg/s. The outer tube has a diameter of
90mm and has an oil flow rate of 0.12 kg/s. Given the following properties for oil and water:
oil:
C p  2131 J/kg K,
  3.25  10 2 kg/m s,
k  0.138 W/m K
Water:
C p  4178 J/kg K,
  725  10 6 kg/m s,
k  0.625 W/m K
Using the relations:
Nu D  5.6
Nu D  0.023 Re 0D.8 Pr 0.4
Re D  2300
Re D  2300
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ssolve p
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Heat Exchangers
Heat Transfer: Exercises
Calculate the overall heat transfer coefficient.
Which is the controlling heat transfer coefficient?
If the heat exchanger is used to cool oil from 90oC to 55oC, using water at 10oC calculate the length of
the tube for a parallel flow heat exchanger
Solution:
a)
For the convection inside
Q  Ai hi (Ti  T1 )
Q  2 ri Lhi (Ti  T1 )
(1)
For the convection outside
Q  Ao ho (To  T1 )
Q  2 ro Lho (To  T1 )
(2)
For conduction through the pipe material
Q  2 r k
dT
dr
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Heat Exchangers
Heat Transfer: Exercises
 Q  dr

dT  
 2 r L  r
(3)
Integrating between 1 and 2:
 Q   ro 
 ln 
T2  T1  
 2 r L   ri 
(4)
From 1 and 2
 Q 

Ti  T1  
 2 ri Lhi 
(5)


Q

T2  To  
 2 ro Lho 
(6)
Adding 4, 5 and 6
Ti  To 
Q  lnro / ri  1
1 




2L 
k
hi ri ho ro 
Rearranging
Q
Ti  To

 U o Ti  To 
2Lro  ro  ro  ro

1
 ln  

 k  r  hr  h 
i
i
i
o




Therefore, overall heat transfer coefficient is
r r  r
1
U o   o ln o   o  
 k  ri  hi ri ho 
1
b)
i) To calculate the overall heat transfer coefficient, we need to evaluate the convection heat transfer
coefficient both inside and outside.
83
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Heat Exchangers
Heat Transfer: Exercises
Re 
Vm Dh

For water:
D 2
Vm 
m
,
A
Re 
4m
4  0.25

 8781
D   0.05  725 106
Pr 
 Cp
k
Re > 2300
A

4
725  10 6  4178
 4.85
0.625
(turbulent flow)
Therefore: Nu D  0.023 Re 0D.8 Pr 0.4  0.023  87810.8  4.850.4  62
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Heat Exchangers
Heat Transfer: Exercises
From which:
hi 
Nu D k 62  0.625

 775 W / m 2 K
D
0.05
For oil:
Dh 
4 (rb2  ra2 )
4 Area

 2(rb  ra )  2(0.045  0.025)  0.034 m
Perimeter 2 (rb  ra )
Re 
2m
2  0.12
Vm Dh 2m ( rb  ra )



 33
2
2

 ( rb  ra )   (rb  ra )    0.045  0.028 3.25  10 2
Re < 2300 (Laminar flow)
Therefore: Nu D  5.6
ho 
Nu D k 5.6  0.138

 22.7
Dh
0.034
W/m2 K
1
 0.028  28 
0.028
1 

U o  
ln  
  21.84
 25  725  0.025 22.7 
 16
W/m2 K
ii) The controlling heat transfer coefficient is that for oil, ho because it is the lower one. Changes in
ho will cause similar changes in the overall heat transfer coefficient while changes in hi will cause
little changes. You can check that by doubling one of them at a time and keep the other fixed and
check the effect on the overall heat transfer coefficient.
iii) Thi  90C , Tci  10C , Tho  55C
Tco is unknown. This can be computed from an energy balance
For the oil side:
Q  m hC ph (Thi  Tho )  0.12  2131(90  35)  8950 W
Q  m cC pc (Tco  Tci )  0.25  4178(Tco  10)  8950 W
Therefore Tco  18.56C
Evaluate LMTD
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Heat Exchangers
Heat Transfer: Exercises
T1  90  10  80C
T2  55  18.56  36.44C
Tlm 
T2  T1
36.44  80

 56.1C
ln(T2 / T1 ) ln(36.44 / 80)
Q  UATlm  U o  2 ro LTlm
L
Q
U o  2 ro Tlm

8950
 41.5m
21.84  2  0.028  56.1
Example 5.4
Figure (a) below shows a cross-sectional view through part of a heat exchanger where cold air is
heated by hot exhaust gases. Figure (b) shows a schematic view of the complete heat exchanger which
has a total of 50 channels for the hot exhaust gas and 50 channels for the cold air. The width of the
heat exchanger is 0.3m
Using the information tabulated below, together with the appropriate heat transfer correlations,
determine:
i.
ii.
iii.
iv.
v.
the hydraulic diameter for each passage;
the appropriate Reynolds number;
the overall heat transfer coefficient;
the outlet temperature of the cold air;
and the length L
Use the following relations:
Using the relations:
Nu D  4.6
Nu D  0.023 Re 0D.8 Pr 1 / 3
Re D  2300
Re D  2300
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Heat Exchangers
Heat Transfer: Exercises
Data for example 4.4
Hot exhaust inlet temperature
Hot exhaust outlet temperature
Cold air inlet temperature
Hot exhaust total mass flow
Cold air total mass flow
Density for exhaust and cold air
Dynamic viscosity, exhaust and cold air
Thermal conductivity, exhaust and cold air
Specific heat capacity, exhaust and cold air
Heat exchanger wall thickness
Heat Exchanger wall thermal conductivity
Hot exhaust side fouling resistance
Cold air side fouling resistance
100oC
70oC
30oC
0.1 kg/s
0.1 kg/s
1 kg/m3
1.8x10-5 kg/m s
0.02 W/m K
1 kJ/kg K
0.5 mm
180 W/m K
0.01 K m2/W
0.002 K m2/W
Solution:
Re 
VL

L  Dh
Dh 
(Hydraulic diameter)
4  cross sectional area 4  w  H 4  0.003  0.3

 5.94 mm

perimenter
2w  H  20.003  0.3
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Heat Exchangers
Heat Transfer: Exercises
For a single passage:
V 
m / 50  0.1 / 50  2.22
H  w 0.003  0.3  1
Re 
m/s
1  2.22  5.94  10 3
 733
1.8  10 5
Re  2300 (laminar flow)
Nu D  4.6
h
Nu D k
4.6  0.02

 15.5 W / m 2 K
3
Dh
5.98  10
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Heat Exchangers
Heat Transfer: Exercises
Since the thermal properties are the same and the mass flow rate is the same then the hot stream and
cold stream heat transfer coefficients are also the same.
1

t 1
U    R f , h    R f ,c 
k hc
 hh

1
 1

0.5  10 3
1

 0.01 

 0.002
180
15.5
15.5

1
 7.1 W / m 2 K
Note that if the third term in the brackets that includes the resistance through the metal is neglected, it
will not affect the overall heat transfer coefficient because of the relatively very small thermal
resistance.
Q  m C p (Th ,i  Th ,o )  m C p (Tc ,i  Tc ,o )
Tc ,o  Tc ,i  (Th,i  Th ,o )  30  (100  70)  60 o C
Also
Q  UATlm
Tlm is constant in a balanced flow heat exchanger
Tlm  100  60  70  30  40C
Q  m C p Th ,i  Th ,o  
0 .1
 1000100  70   60 w / passage
50
Area of passage:
A
Q
60

 0.211 m 2
UTlm 7.1  40
And since: A  w  L
L
0.211
 0.704 m
0 .3
89
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