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Chris Long & Naser Sayma
Heat Transfer: Exercises
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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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Heat Transfer: Exercises Preface
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
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Heat Transfer: Exercises Introduction
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:
dT k
q k Ti To
dx L
0 .6
q 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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Heat Transfer: Exercises Introduction
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 680 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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Heat Transfer: Exercises Introduction
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:
V I V I 200 0.25
q 1666.7 W / m 2
A W L 0 .1 0 .3
This will equal the conducted heat through the plate:
k
q T2 T1
t
qt
T1 T2 100
1666.7 0.012 98.75C (371.75 K)
k 16
The conducted heat will be transferred by convection and radiation at the surface:
q hT1 T f T14 T f4
h
q T14 T f4 1666.7 5.67 10 371.75
8 4
293 4 12.7 W / m 2
K
T 1 Tf 371.75 293
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Heat Transfer: Exercises Introduction
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
A
hTs T Ts4 T4
0.38
0.001
6Ts 293 5.67 10 3 Ts4 293 4
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
n 1 n f 5.67 10 8 Ts4 6Ts 2555.9
n
T s T s T s
df 22.68Ts3 6
dT
s
Start iterations with Ts0 300 K
5.67 10 8 300 4 6 300 2555.9
Ts1 300 324.46 K
22.68 300 3 6
2 5.67 10 8 324.46 4 6 324.46 2555.9
T 324.46
s 323 K
22.68 324.46 3 6
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Heat Transfer: Exercises Introduction
The difference between the last two iterations is small, so:
Ts0 323 K 50C
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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Heat Transfer: Exercises Conduction
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
k
Were 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
u
Q r Q z Q r r Q z z (2.1)
t
Q
Q r r Q r r
r
Q
Q z z Q z z
z
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Heat Transfer: Exercises Conduction
u mcT
Substituting in equation 2.1:
Q Q (mcT )
r z (2.2)
r z t
Fourier’s law in the normal direction of the outward normal n:
Q T
k
A n
T T
Qr kA k 2 r z ( A 2 r z )
r r
T T
Q z kA k 2 r r ( A 2 r r )
z z
Equation 2.1 becomes
T T T
k 2 r z r k 2 r r z mc (2.3)
r r z z t
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
k
Using the definition of the thermal diffusivity: and expanding the first term using the product rule:
c
1 2T T r 2T 1 T
r which gives the required outcome:
r r r 2 r r z 2 t
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Heat Transfer: Exercises Conduction
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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Heat Transfer: Exercises Conduction
Solution
q UT where U is the overall heat transfer coefficient given by:
q 15
U 0.333W / m 2 K
T 25 (20)
1
1 L p Li Ls 1
U 0.333
hi k p k i k s ho
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
0.333 hi k p k s ho
0.333 12 1 16 8
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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Heat Transfer: Exercises Conduction
i. 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.
ii. 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
Q 2 r1 Lhi Ti T1 Ti T1
2 r1 Lhi
2Lk T1 T2 Q
Q T2 T1
lnr2 / r1 2 Lk / ln(r2 / r1 )
Q
Q 2 r2 Lho T2 To T2 To
2 r2 Lho
Adding the three equations on the right column which eliminates the wall temperatures gives:
2LTi To
Q
1 ln r2 / r1 1
hi r1 k ho r2
Q 2 15 (10)
163.3W / m
L 1 ln0.052 / 0.05 1
30000 0.05 50 20 0.052
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Heat Transfer: Exercises Conduction
Insulated pipe
Q 2 Ti To
L 1 lnr2 / r1 ln(r3 / r2 ) 1
hi r1 k k ins ho r3
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Heat Transfer: Exercises Conduction
Q 2 15 (10)
7.3W / m
L 1 ln0.052 / 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:
2LTi To
Q
1 ln r2 / r1 1
hi r1 k ho r2
Hence substituting into this equation:
2 10080 20
Q 0.329 10 6 W
1 ln50 / 47 1
0.047 2000 16 0.05 200
For the case with insulation, we also use the equation from Example 2.3
2LTi To
Q
1 lnr2 / r1 ln(r3 / r2 ) 1
hi r1 k k ins ho r3
2 10080 20
Q 5.39 10 3 W
1 ln50 / 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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Heat Transfer: Exercises Conduction
2LTi To 2 10080 20
Q 5.44 10 3 W
ln(r3 / r2 ) ln(100 / 50)
k ins 0.1
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 60C , 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:
Tb T f cosh m( L x) hP
T Tf where m 2 and Ac is the cross sectional area.
sinh mL kAc
Determine the total convective heat transfer from the heat sink, the fin effectiveness and the
fin efficiency.
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Heat Transfer: Exercises Conduction
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 0.0039 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
Tb T f cosh m( L x)
T Tf
sinh mL
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Heat Transfer: Exercises Conduction
Then
dT sinh m( L x)
m Tb T f
dx cosh mL
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 T T f tanh mL
1/ 2
b
Since
1
hP 2
m
kA
c
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
m 6
11.856 m 1
175 40 10
mL 11.856 0.06 0.7113
tanhmL tanh0.7113 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 heat transfer rate Qf
fin
Heat transfer rate that would occur in the absence of the fin hAc Tb T f
2.03
fin 106
12 40 10 6 60 20
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Heat Transfer: Exercises Conduction
Fin efficiency:
Actual heat transfer through the fin
fin
Heat that would be transferred if all the fin area were at the base temperature
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
2.03
fin 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.
hA
T T T
f i T f exp s
mC
Solution
Consider a single fin (the length scale L for the Biot number is half the thickness t/2)
hL h t / 2 40 0.0005
Bi 10 4
k k 175
Since Bi 1 , we can use he “lumped mass” model approximation.
T T f hA s
exp
T T
i f mC
mC T T f
ln
hAs Ti T f
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Heat Transfer: Exercises Conduction
m As t / 2
Ct T Tf
ln 2700 900 0.001 ln 30 20 42 seconds
2h Ti T f
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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Heat Transfer: Exercises Conduction
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
1 1
hP 2 15 0.07 2
m
kA
6
6.2361 m 1
c 180 150 10
mL 6.2361 0.1 0.62361
tanhmL 0.5536
Q f hPkAc
1/ 2
T
b T f tanh(mL) (From example 2.5)
Q f 15 0.07 180 150 10 6
1/ 2
120 20 0.5536 9.32 W
So for 1 kW, the total number of fins required:
1000
N 108
9.32
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Heat Transfer: Exercises Conduction
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:
htip
cosh m( L x) sinh m( L x)
Tx T mk
Tb T htip
cosh mL sinh mL
mk
1/ 2
hP
m A D 2 / 4, P D
kA
At the tip, x L , the temperature is given by ( cosh( 0) 1 , sinh( 0) 0 ):
Ttip T 1
Tb T htip
cosh mL sinh mL
mk
Where is the dimensionless error:,
0, Ttip T (no error)
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Heat Transfer: Exercises Conduction
1, TL Tb (large error)
For L 20mm, k 19W / m K , D 3mm, h htip 50 W / m 2 K
T 50C , Tb 60C
A D 2 / 4, P D
1/ 2 1/ 2 1/ 2 1/ 2
hP h D 4 4h 4 50
m k D 2
59.235 m 1
kA kD 19 0.003
mL 59.235 0.02 1.185
h 50
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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Heat Transfer: Exercises Conduction
Ttip 0.539Tb T T
Ttip 0.53960 50 50 55.39C
Hence error 5.39C
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
ho t / 2
Treat the solid balcony as a fin Bi
kb
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Heat Transfer: Exercises Conduction
20 0.1
Bi 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)
1/ 2 1/ 2
hP 20 8.4
mW W 2 20.5
kA 2 0 .8
This is large enough to justify the use of the fin infinite equation.
Qb ho Pk b Ac T2 To
1/ 2
1/ 2
qb
1
ho Pk b Ac 1 / 2 T2 To ho Pk b
A
T2 To (1)
Ac c
Also assuming 1-D conduction through the wall:
qb hi (Ti T1 ) (2)
kb
qb (T1 T2 ) (3)
L
Adding equations 1, 2 and 3 and rearranging:
(To Ti )
qb 1/ 2
(4)
1 L Ac
hi k b ho Pk b
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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Heat Transfer: Exercises Conduction
From (4)
20 (5)
qb 1/ 2
77.2 W / m 2
1 0 .3 0 .8
8 2 20 8.4 2
Compared with no balcony:
(To Ti ) 20 (5)
qb 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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Heat Transfer: Exercises Conduction
As before, however, in this case Bi << 1 because k s k b
1/ 2 1/ 2
hP 20 6
mW w 2 11
kA 40 0.1
mW 2 , so we can use the infinite fin approximation as before
(To Ti ) 20 ( 5)
qb 1/ 2
1/ 2
182 W / m 2
1 L Ac 1 0.3 0.01
hi k s ho Pk s
8 40 20 6 40
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 Area
,
Tinit T f 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 ht ) 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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Heat Transfer: Exercises Conduction
d (Ts T f )
hA(Ts T f ) mC (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 0.1Gr Pr in turbulent flow or Nu 0.54Gr Pr
1/ 3 1/ 4
for laminar flow)
Equation 1 then becomes:
mC d (Ts T f )
G Ts T f (T
n
s Tf )
A dt
t
GA
t
d (Ts T f )
0 mC
t
dt
t 0 (Ts T f )
n 1
GnAt
Ts T f Ts T f
n n
(2)
mC t 0
At t 0, T s T f Ts ,i T f
If we divide equation 2 by Ts ,i T f
n
T s
Tf
And use the definition
T s ,i T f
We obtain
GnAt GnAt
n 1 Ts ,i T f
n
mC Ts ,i T f mC
n
Since G Ts ,i T f h i , the heat transfer coefficient at time t = 0, then
hi At
n 1
mC
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30
Heat Transfer: Exercises Conduction
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
A 0.04
2.1 10 4 m 2 K / J
mC 0.22 870
n nhi t 1 which gives
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31
Heat Transfer: Exercises Conduction
t
n
1
nhi
40 20
When T 40C 0 .2
120 20
Then
t
0.2 1
1 / 4
590 s
1 / 4 16 2.1 10 4
For the equation e h t
which assumes that the heat transfer coefficient is independent of surface-to-fluid temperature
difference.
ln ln 0.2
t 479 s
h 16 2.1 10 4
590 479
Percentage error = 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?
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32
Heat Transfer: Exercises Conduction
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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33
Heat Transfer: Exercises Conduction
�� � �� 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
�0.4√300 � 0.06�300���� � �.4� 0.222
��� � 300 � � 300 �
���
0.2 0.04 0.01782
� � �
√300 300���
Which is close enough to 300
From which
���
�� � � 4.� ���
��
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34
Heat Transfer: Exercises Convection
3. Convection
Example 3.1
Calculate the Prandtl number (Pr = Cp/k) for the following
a) Water at 20C: = 1.002 x 103 kg/m s, Cp = 4.183 kJ/kg K and k = 0.603 W/m K
b) Water at 90C: = 965 kg/m3, = 3.22 x 107 m2/s, Cp = 4208 J/kg K and k = 0.676 W/m K
c) Air at 20C and 1 bar: R = 287 J/kg K, = 1.563 x 105 m2/s, Cp = 1005 J/kg K and
k = 0.02624 W/m K
1.46 10 6 T 3 2
d) Air at 100C: 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
K) and k = 0.03186 W/m K.
e) Mercury at 20C: = 1520 x 106 kg/m s, Cp = 0.139 kJ/kg K and k = 0.0081 kW/m K
f) Liquid Sodium at 400 K: = 420 x 106 kg/m s, Cp = 1369 J/kg K and k = 86 W/m K
g) Engine Oil at 60C: = 8.36 x 102 kg/m s, Cp = 2035 J/kg K and k = 0.141 W/m K
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35
Heat Transfer: Exercises
Solution Convection
a)
Solution
Solution
a) Cp
1.002 10 3 4183
a) Pr 6.95
k 0.603
C p 1.002 10 3 4183
3
Pr C p 1.002 10 4183 6.95
b) Pr k 6.95
k 0.603
0.603
b) Cp C p
965 3.22 10 7 4208
b) Pr 1.93
k k 0.676
C p C p 965 3.22 10 7 4208
7
Pr C p C p 965 3.22 10 4208 1.93
c) Pr k k 1.93
k k 0.676
0.676
c) C p
c) Pr
k
C p
Pr C p
Pr Pk 100000
k 1.19 kg / m 3
RT 287 293
P 100000
P 100000 51.19 kg //m 33
RT19 1.563 10 1.19 kg m
1. 287 293 1005
Pr RT 287 293 0.712
0.02624
5
1.19 1.563 10 5 1005
Pr 1.19 1.563 10 1005 0.712
d) Pr 0.712
0.02624
0.02624
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 6
1.46 10 6T 3322 1.46 10 6 37333/ /22
1.46 10 T 1.46 10 373 2.18 10 5 kg / m s
5
0110 T 58 10 4 T 110.98 10 8 T 2 2..18 102.58 /10 s 373 3.98 10 8 3732
Cp .110 2.
917 T 3 373
110 373 0 917 kg m 4
1007.7 J / kg K
4 8 4 8
C p 0.917 2.58 10 4T 3.98 10 8T 22 0.917 2.58 10 4 373 3.98 10 8 37322
C p 0.917 2.58 10 T 3.98 10 T 0.917 2.58 10 373 3.98 10 373
2.18 7 J K
1007. 10/ 5kg 1007.7
Pr 1007.7 J / kg K 0.689
0.03186
5
2.18 10 5 1007.7
Pr 2.18 10 1007.7 0.689
e) Pr 0.689
0.03186
0.03186
e) Cp 1520 10 6 139
e) Pr 0.0261
k 0.0081 10 3
C p 1520 10 6 139
6
Pr C p 1520 10 3139 0.0261
Pr k 0.0081 10 3 0.0261
k 0.0081 10
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36
Heat Transfer: Exercises Convection
f)
Cp 420 10 6 1369
f) Pr 0.0067
k 86
Cp 420 10 6 1369
g) Pr 0.0067
k 86
Cp 8.36 10 2 2035
g) Pr 1207
k 0.141
Cp 8.36 10 2 2035
Pr
Comments: 1207
k 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 400C and
a) A 10 m (water line length) long yacht sailing at 13 km/h in seawater = 1000 kg/m3 and
6 3 2
1.3 x 1010 T
= 1.46 3 kg/m s, kg/m s
110 disc
b) A compressor T of radius 0.3 m rotating at 15000 rev/min in air at 5 bar and 400C and
kg/s of carbon
c) 0.05 1.46 10 6 T 3 2dioxide gas at 400 K flowing in a 20 mm diameter pipe. For the viscosity
1.56 T 6 kg/m s
32
take 110 10 T kg/m s
233 dioxide gas at 400 K flowing in a 20 mm diameter pipe. For the viscosity
c) 0.05 kg/s of carbon T
3 5
1. a 10 6 3 long, travelling at 100 km/hr in air ( = 1.2 kg/m and = 1.8 x 10
d) The roof of56coach6 Tm 2
take
kg/m s) kg/m s
233 T
e) The flow of exhaust gas (p = 1.1 bar, T = 500ºC, R = 287 J/kg K and = 3 3.56 x 105 kg/m s) over
d) The roof of a coach 6 m long, travelling at 100 km/hr in air ( = 1.2 kg/m and = 1.8 x 105
a valve guide of diameter 10 mm in a 1.6 litre, four cylinder four stroke engine running at
kg/m s)
3000 rev/min (assume 100% volumetric efficiency an inlet density of 1.2 kg/m3 and an exhaust
e) The flow of exhaust gas (p = 1.1 bar, T = 500ºC, R = 287 J/kg K and = 3.56 x 105 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)
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37
Heat Transfer: Exercises Convection
Solution
13 10 3
10 3 10
uL 3600
a) Re 2.78 10 7 (turbulent)
1.3 10 3
b) T 400 273 673 K
1.46 10 6 6733 2
3.26 10 5 kg / m s
110 673
15000
2 1571 rad / s
60
u r 1571 0.3 471 .3 m / s
P 100000
2.59 kg / m 3
RT 287 673
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38
Heat Transfer: Exercises Convection
Characteristic length is r not D
uD 2.59 471.3 3
Re 1.12 10 7 (turbulent)
3.26 10 5
D 2
c) m uA u
4
4m
u
D 2
uD 4mD 4m
Re
D D
2
1.56 10 6 400 3 2
1.97 10 5 kg / m s
233 400
4 0.05
Re 1.6 10 5 (turbulent)
0.02 1.97 10 5
100 10 3
d) u 27.8 m / s
3600
uL 1.2 27.8 6
Re 11.1 10 7 (turbulent)
1.8 10 5
e) Let m be the mass flow through the exhaust port
m = inlet density X volume of air used in each cylinder per
second
1.6 10 3 3600 1
m 1.2
0.012 kg / s
4 60 2
4m
u
D2
ud
Re d
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39
Heat Transfer: Exercises Convection
4 0.01 0.012
Re 6869 (laminar)
3.56 10 5 0.025
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 75C in a room at 18C ( =
1.2 kg/m3 , Pr = 0.72 and = 1.8 x 105 kg/m s)]
b) A horizontal oil sump, with a surface temperature of 40C, 0.4 m long and 0.2 m wide containing
oil at 75C ( = 854 kg/m3 , Pr = 546, = 0.7 x 103 K1 and = 3.56 x 102 kg/m s)
c) The external surface of a heating coil, 30 mm diameter, having a surface temperature of 80C in
water at 20C ( = 1000 kg/m3, Pr = 6.95, = 0.227 x 103K1 and = 1.00 x 10-3kg/m s)
d) Air at 20ºC ( = 1.2 kg/m3 , Pr = 0.72 and = 1.8 x 105 kg/m s) adjacent to a 60 mm diameter
vertical, light bulb with a surface temperature of 90C
Solution
2 g T L3
a) Gr
2
T 75 18 57 K
1 1 1
K 1
T 18 273 291
1.2 2 9.81 57 0.6 3
Gr 1.84 10 9
291 1.8 10
3 2
Gr Pr 1.84 10 9 0.72 1.3 10 9 (mostly laminar)
Area 0.4 0.2
b) L 0.0667 m
Perimeter 2 0.4 0.2
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Heat Transfer: Exercises Convection
T 75 40 35 K
2 g T L3 854 2 9.81 0.7 10 3 35 0.0667 3
Gr 4.1 10 4
2
3.56 10 2 2
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)
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Heat Transfer: Exercises Convection
T 80 20 60 K
2 g T L3 1000 2 9.81 0.227 10 3 60 0.033
Gr 3.6 10 6
2
1 10
3 2
Gr Pr 3.6 10 6 6.95 25 10 6 (laminar)
Area D 2 D
d) L
Perimeter 4D 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
Gr 3.5 10 4
293 1.8 10
2 5 2
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 105 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 35C, 5 mm wide and 10 mm
long, dissipating 0.1 W by free convection from one side into air where the temperature is 20C
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 20C (k = 0.026 W/m K assume black
body radiation and = 56.7 x 109 W/m K4)
d) Air at 4C (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 18C, the outside wall temperature 12C
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42
Heat Transfer: Exercises Convection
Solution
Cp
a) Pr
k
Solution
Cp
a) C p 4.63 10 5 1175
k Pr k 0.0766 W / m K
Pr 0.71
C p 4.63 10 5 1175
khL
1000 0.02
0.0766 W / m K
Nu
Pr 0.71 261
k 0.0766
h L 1000 0.02
Nu L
h q L0.0766 261
b) Nu k
k T k
hL q L
b) Nu
Q k 0 .1 T k
q 2000 W / m 2
A 0.01 0.005
Q 0 .1
q 2000 W / m 2
A 0.01 0.005
T 35 20 15 C
T 35 20 15 C
Area 50 5
L mm 0.001667 m
Perimeter 30 50 3 5
Area
L mm 0.001667 m
Perimeter 30 3
h L 2000 0.001667
Nu 8 .5
Nuk h L 15 0.026
2000 0.001667
8 .5
k 15 0.026
qc L
c) Nu q L
c) Nu T k c
T k
q must be be convective heat flux radiative heat flux
In this case,case, q mustthe the convective heat flux––radiative heat flux
In this
80 273 353 K
Ts Ts 80 273 353 K
273 293 K
T T20 20 273 293 K
Q R ATs4 TsT4 .56 1010 .1.5 06 3534 293 4416 WW
Q R A 4 T4 56 7 .7 9 9 1 5 0. .6 353 4 2934 416
T 80 20 60 K
T 80 20 60 K
Qc Q QR 1000 416 584 W
Qc Q QR 1000 416 584 W
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Heat Transfer: Exercises Convection
Qc 584
qc 649 W / m 2
A 1.5 0.6
q L 649 0.6
Nu c 249
T k 60 0.026
d) T 12 4 8 K
k b T1 T2
q 60 C
W
(assuming 1-D conduction)
0.318 12
q 12 W / m 2
0.18
q L 12 3
Nu c 188
T k 8 0.024
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.
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44
Heat Transfer: Exercises Convection
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 n . In free convection from a vertical surface the local Nusselt number
x
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
hx
a) Nu x
k
ux
Re x
For forced convection: Nu x C1 Re n
x
n
k ux
Hence h C1
x
This shows that the heat transfer coefficient for forced does not depend on temperature difference.
For free convection Nu x C 2 Grxm
2 g T x 3
Grx
2
m
k 2 g T x 3
Hence h C2
(1)
x 2
So for free convection, heat transfer coefficient depends on T m
b) From (1), with m = 1/3 for turbulent free convection:
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45
Heat Transfer: Exercises Convection
1/ 3
k 2 g T x 3
h C2
x 2
1/ 3
k 2 g T
h C2
x
x 2
1/ 3
2 g T
h kC 2
2
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
air with T 20C , C p 1.005 kJ / kg K , 1.522 10 m / s 1.19 kg / m 3 and Pr 0.72
5 2
. 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 32C and Q 0.5 W .
Solution
Firstly, we need to
estimate if the flow is
laminar or
turbulent.
Assuming a critical (transition) Reynolds number of Re 3 10 5 the velocity required would be:
3 10 5 3 10 3 3 10 5 1.522 10 5
u turb 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
q av L
Nu av 0.662 Re1 / 2 Pr 1 / 3
L
Ts T k
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46
Heat Transfer: Exercises Convection
q av L
Re1 / 2
L
Ts T k 0.662 Pr 1 / 3
0 .5 / 2
q av 1250 W / m 2
0.025 0.008
1250 0.025
Re1 / 2 173.5
L
32 20 0.0253 0.662 0.721 / 3
Re L 3 10 4
Re L 3 10 4 1.522 10 5
u 18.3 m / s
L 25 10 3
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47
Heat Transfer: Exercises Convection
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 20C, the outside air temperature is -15C 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
Cp Cp 1.8 10 5 1000
Pr gives: k 0.026 W / m K
k Pr 0.7
First we need to determine if these flows are laminar or turbulent.
For the inside (Free convection):
1 1 1
K 1
T 20 273 293
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Heat Transfer: Exercises Convection
2 g T L3 1.2 2 9.81 T 7 3
Gr
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)
u L 1.2 15 30
Re L 3 10 7
1.8 10 5
(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 0.8 Pr 1 / 3
x
For the inside:
2 g Ti Ts x 3
1/ 3
hx
Nu x 0.09
k 2
h constant
x 3 1/ 3
constant
x
Hence heat transfer coefficient is not a function of x
hav hx L (1)
For the outside:
0.8
hx ux
Nu x 0.029
Pr 1 / 3
k
h constant
x 0.8 C x 0.2
x
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Heat Transfer: Exercises Convection
xL xL
1 C hx L
hav h dx x
0.2
dx (2)
L x 0 L x 0
0.8
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
1/ 3
hi H
0.09
k 2 Ti
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Heat Transfer: Exercises Convection
1.2 2 9.81 T1 Ts
hi 0.09 0.026
1.8 10 5 2 293
hi 1.24 Ti Ts
1/ 3
(4)
From equation 2:
0.8
ho W 0.029 u W
Pr 1 / 3
k 0.8
0. 8
0.026 0.029 1.2 15 30
ho 0 .7 1 / 3
30 0.8 1.8 10 5
ho 26.7 W / m 2 K (5)
From (3) with (4) and (5)
1.24 Ti Ts 26.7 Ts To
4/3
1.24 20 Ts 26.7 Ts 15
4/3
Ts 0.0464 20 Ts
4/3
15 (6)
To solve this equation for Ts an iterative approach can be used
First guess: Ts 10C
Substitute this on the right hand side of equation 6:
Ts 0.0464 20 10
4/3
15 10.7C
For the second iteration we use the result of the first iteration:
Ts 0.0464 20 10.7
4/3
15 10.6C
The difference between the last two iterations is 0.1C , so we can consider this converged.
Ts 10.6C
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Heat Transfer: Exercises Convection
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 27C.
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 47C;
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 37C.
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Heat Transfer: Exercises Convection
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 Rex < 3 x 105
Nux = 0.02 Rex0.8 Pr1/3 Rex 3 x 105
Nux = 0.5 Grx1/4 Pr1/4 Grx < 109
Nux = 0.1 Grx1/3 Pr1/3 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
2 g T L3
Gr
2
T 47 27 20 K
1 1
K 1
27 273 300
12 9.81 20 0.13
Gr 2 10 6 (Laminar)
1.8 10
5 2
300
(L here is height because it is in the direction of the free convection boundary layer)
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Heat Transfer: Exercises Convection
So we use:
Nu x 0.5 Grx Pr
1/ 4
L L
1
L
hav h dx constant x 1 / 4 dx
0 0
hx L
hav
3/ 4
2
Nu av GrL Pr 1 / 4
3
2
Nu av
3
2 10 6 0.7 1 / 4 23
Nu av k 23 0.02
hav 4.6 W / m 2 K
L 0.1
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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Heat Transfer: Exercises Convection
u L 1 15 0.05
Re 4.17 10 4 (Laminar)
1.8 10 5
(L here is the width because flow is along that direction)
Nu x 0.3 Re1x/ 2 Pr 1 / 3
L
1 h
hav hdx x L
L0 1/ 2
Nu av 0.6 Re1 / 2 Pr 1 / 3 0.6 4.17 10 4
L
1 / 2
0.71 / 3 109
Nu av k 109 0.02
hav 43.5 W / m 2 K
L 0.05
Q q av A hav TA 43.5 10 0.1 0.05 2 T 10C
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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Heat Transfer: Exercises Convection
Solution
hav k
Nu av
L
Where for a constant surface-to-fluid temperature:
1L
x L
hav hlaminar dx hturbulent dx
L 0
xL
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Heat Transfer: Exercises Convection
Since for laminar flow ( Re x 10 5 ):
Nu x 0.3 Re1x/ 2 Pr 1 / 3
1/ 2
k u
hlam 0.3 x1 / 2 Pr 1 / 3
x
1/ 2
u
hlam 0.3 k
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
0.8
u
Cturb 0.04 k
Pr 1 / 3
Hence
1L
x L
hav Clam x dx C turb x 0.2 dx
1 / 2
L 0
xL
1 x 0.8
xL L
x1 / 2
hav Clam Cturb
L 1/ 2 0 0.8 xL
hav k
Nu av
L
Clam 1 / 2 C turb 0.8
Nu av
k
2xL
0.8k
0
L x L.8
u
1/ 2
1/ 2 1/ 3
u L 0.8 u x 1/ 3
0.8
Nu av 0.6
x L Pr 0.05
L
Pr
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Heat Transfer: Exercises Convection
u xL
But 10 5 (The transition Reynolds number)
So
Nu av Pr 1 / 3 0.6 10 5
1/ 2
0.05 Re 0.8 0.05 10 5
L
0. 8
Nu av 0.05 Re 0.8 310 Pr 1 / 3
L
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 100
q av 1666.7 W / m 2
A 0.2 0.3
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Heat Transfer: Exercises Convection
C P 2 10 5 10 3
Pr 0.667
k 0.03
u L
Re L
P 10 5
1.15 kg / m 3
RT 287 303
1.15 12 0.3
Re L 5
2.07 10 5
2 10
Using the formula for Nusselt Number obtained in Example 3.9:
Nu av 0.05 Re 0.8 310 Pr 1 / 3
L
Nu av 0.05 2.07 10 5
0 .8
310 0.667
1/ 3
511
hav k q av L
Nu av
L Tk
q av L 1666.7 0.3
T 32.6C
Nu av k 511 0.03
Ts T T
Ts 30 32.6 62.6C
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59
Heat Transfer: Exercises Radiation
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:
14
A1 F13T14 A2 F23T24
T3
A F AF
2 23 1 13
where F13 and F23 are the appropriate view factors.
b) Show that the total radiative heat transfer to the tubes, Q2, is given by:
360°
.
AF A F
Q2 A1 F12 1 13 2 23 T14 T24
A2 F23 A1 F13
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Heat Transfer: Exercises Radiation
c) Calculate the radiative heat transfer to the tubes if T1 = 1700C, T2 = 300C, A1 = A2 = 12m2 and
the view factors are each 0.5?
Solution
a)
Q2 Q1 2 Q3 2 (1)
Since the walls are adiabatic
Q3 2 Q13 (2)
From (2)
A3 F32 T34 T24 A1 F13 T14 T34
4 A1 F13 T14 A3 F32 T24
T
3
A3 F32 A1 F13
1/ 4
A F T 4 A2 F23 T24
T3 1 13 1
since Ai Fij A j F ji
A2 F23 A1 F13
b) From (1)
Q2 A1 F12 T14 T24 A3 F32 T34 T24
Q2 A1 F12 T14 T24 A2 F23 T34 T24
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Heat Transfer: Exercises Radiation
A F T 4 T 4 A F A1 F13 T1 A3 F32 T2 T 4
4 4
Q2 1 12 1 2
2 23
A3 F32 A1 F13
2
A F T 4 T 4 A F A1 F13 T1 A3 F32 T2 A3 F32T2 A1 F13T2
4 4 4 4
Q2 1 12 1 2
2 23
A3 F32 A1 F13
A F T 4 T 4 A F A1 F13 T1 A1 F13T2
4 4
Q2 1 12 1 2
2 23
A3 F32 A1 F13
A1 F13
Q2 A1 F12 T14 T24 A2 F23 T14 T24
A F A F
3 32 1 13
A F A F
Q2 A1 F12 T14 T24 T14 T24 2 23 1 13
A F A F
3 32 1 13
A F A F
Q2 T14 T24 A1 F12 2 23 1 13
A2 F23 A1 F13
A1 F13 T14 A3 F32 T24 A1 F13 T14 A2 F23 T24
c) T34
A3 F32 A1 F13 A2 F23 A1 F13
12 0.5 1973 4 12 0.5 573 4
T34 1662 K
12 0.5 12 0.5
66
Q2 56.7 10 9 1973 4 573 4 6
66
6
7.68 10 W
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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62
Heat Transfer: Exercises Radiation
and
3.333 J2 – 1.4 J1 = 48334 W/m2.
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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63
Heat Transfer: Exercises Radiation
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
A1 0.2 2
F31 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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Heat Transfer: Exercises Radiation
E b ,i J i n
b) J i J j Fij
1 i j 1
i
Apply to surface 1, (i = 1)
1 1
Let 1
1
E b ,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 0.4
1 1.5
1 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 (1)
Applying to surface 2 (i = 2)
E b , 2 J 2 1 2 F21 2 F23 2 F21 J 1 2 F23 J 3
Eb, 2 T24
1 2 1 0.3
2 2.333
2 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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65
Heat Transfer: Exercises Radiation
c) From (2):
3.333 J 2 48334
J1
1.4
Substituting in (1)
3.333 J 2 48334
2.5 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
E b , 2 J 2 56.7 10 9 700 4 26099
q2 5.351 10 3 W / m 2
1 2 1 0 .3
2 0 .3
The minus sign indicates a net influx of radiative transfer as would be expected from
consideration of surface temperatures.
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Heat Transfer: Exercises Radiation
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:
(T24 F21T14 F23T34 F24T44 )
J2
1 F21 F23 F24
The following equation may be used without proof:
E b ,i J i N
Fij J i J j
1 i i j 1
c) 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 Radiation
A2 2 r2 h
A3 A4 r22 r12
F11 0
F13 F14
F11 F12 F13 F14 1
but
A1 F12 A2 F21
A2 r 40
F12 F21 2 F21 0.14338 0.57352
A1 r1 10
Thus
1 0.57352
F13 F14 0.21324
2
F21 F22 F23 F24 1
1 F21 F22 1 0.14338 0.44515
F23 F24 0.20574
2 2
F31 F32 F33 F34 1
F33 0
A1 F13 A3 F31
A1 2 r1 h 2 0.01 0.04
F31 F13 F13 0.21324 0.11373
A3
r2 r1
2
2
0.04 2 0.012
A2 F23 A3 F32
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Heat Transfer: Exercises Radiation
A2 2 r2 h 2 0.04 0.04
F32 F23 F23 0.20574 0.43891
A3
r2 r1
2 2
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
E b ,i J i n
b) J i J j Fij
1 i j 1
i
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Heat Transfer: Exercises Radiation
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
1 0.5
2 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
T24 T14 F21 T34 F23 T44 F24
J2
F21 F23 F24 1
56.7 10 9 T24 1800 4 0.57352 1200 4 0.20574 500 4 0.20574
c) J2
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
Eb, 2 J 2
q2 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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Heat Transfer: Exercises Radiation
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 N
Fij J i J j
1 i i j 1
c) 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 Radiation
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
A1
F31 F13 0.4 Since A1 A3
A3
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Heat Transfer: Exercises Radiation
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
2 0.75 ,
0.75 3
Eb, 2 T24
T24 J 2
F21 J 2 T14 F23 J 2 J 3
1/ 3
1
T24
3
F21 T14 F23 J 3
J2
1
1 F21 F23
3
1
56.7 10 9 T24
3
0.6 56.7 10 9 1700 4 0.4 J 3
J2
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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73
Heat Transfer: Exercises Radiation
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
E b , 2 J 2 56.7 10 9 600 4 89247
q2 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 60C.
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 N
Fij J i J j
1 i i j 1
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Heat Transfer: Exercises Radiation
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 W 0.12
1, 0 .6
L 0. 2 L 0. 2
From the figure: F12 0.27
A1 F12 A2 F21
A1 w 0.12
F21 F12 F12 0.27 0.162
A2 h 0.2
F11 F12 F13 1
F11 0
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Heat Transfer: Exercises Radiation
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)
E b ,i J i n
J i J j Fij
1 i j 1
i
Applying for surface 1, i = 1 (the casing)
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Heat Transfer: Exercises Radiation
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 0.5
1.0
1 0.5
So
T14 F12 J 2 F13 T34
J1
1 F12 F13
56.7 10 9 T14 0.27 J 2 0.73 56.7 10 9 3334
J1
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 W / m2
J 2 22.68 10 9 700 4 0.0972 J 1 350 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 Radiation
E J
qin 57.9 10 9 T14 28.7 10 9 T14 1050
b ,1 1
1 1
1
q in 28. 10 9 T14 1050
q out 1 T14 T4 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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78
Heat Transfer: Exercises Heat Exchangers
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
A heat exchanger consists of there are two streams: channels, each 18 mmK and and k = 0.0371 W/m
an adjacent pair of channels, numerous rectangular water k = 0.625 W/m wide air 2.25 mm high. In
K, adjacent pair a 18 mm wide and are two streams: water k = 0.625 W/m = 16 W/mk = The fouling
an separated by of channels, there 0.5 mm thick stainless steel plate of k K and air K. 0.0371 W/m
resistances for air 18 mm wide andx0.5 4 m2thick stainless 104 m2 K/W, k = 16 W/m K. The fouling
K, separated by a and water are 2 10 mm K/W and 5 x steel plate of respectively, and the Nusselt
resistances for air and = 5.95 where the m2 K/W 'Dh refers to the hydraulic diameter.
number given by NuDhwater are 2 x 104 subscript and' 5 x 104 m2 K/W, respectively, and the Nusselt
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
separating wall and the two fouling resistances.
a) Calculate the overall heat transfer coefficient ignoring both the thermal resistance of the
b) Calculate the overall heat transfer coefficient with these resistances.
separating wall and the two fouling resistances.
c) Which is the controlling heat transfer coefficient? these resistances.
b) Calculate the overall heat transfer coefficient with
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 3 3
4 10 3
(2 25
2.25. 10 18)181010
3
Dh 4 4 10 3
(2.25 18) 10 3
Nu D k
h
Dh
Nu D k
h
Dh
5.95 0.625
(a) hwater 3
930W / m 2 K
4 10
5.95 0.625
(a) hwater 3
930W / m 2 K
4 10
5.95 0.0371
hair 3
55.186W / m 2 K
4 0.
5.95 100371
hair 3
55.186W / m 2 K
4 10
1
1 1
U 1 52.1 W / m2 K
930 55.1
1 186 2
U 52.1 W / m K
930 55.186
1
0.5 10 3 1 1
U 2 10 4 5 10 4 1 50.2 W / m2 K
0.5 16
10 3
930
1 186
55.1
b) U
2 10 4 5 10 4 50.2
W / m2 K
b) 16 930 55.186
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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79
Heat Transfer: Exercises Heat Exchangers
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 103 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 0.8 Pr 0.4
D
Solution:
m VA
m
V
A
VD 4m 4 0.0983
Re D 9613
D 0.02 0.651 10 3
Nu D 0.023 9613 0.8 4.3 0.4 63
hD
Nu D
k
Nu D k 63.3 0.632
h 2000W / m 2 K
D 0.02
1
1 1
U 95.2W / m 2 K
2000 100
Example 5.3
a) Show that the overall heat transfer coefficient for a concentric tube heat exchanger is given by the
relation:
-1
r r r 1
U o o ln o o
r hr h
k i i i o
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80
Heat Transfer: Exercises Heat Exchangers
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 Re D 2300
Nu D 0.023 Re 0.8 Pr 0.4
D Re D 2300
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81
Heat Transfer: Exercises Heat Exchangers
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
dT
Q 2 r k
dr
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82
Heat Transfer: Exercises Heat Exchangers
Q dr
2 r L r
dT (3)
Integrating between 1 and 2:
Q ro
2 r L ln r
T2 T1 (4)
i
From 1 and 2
Q
2 r Lh
Ti T1 (5)
i i
Q
2 r Lh
T2 To (6)
o o
Adding 4, 5 and 6
Q lnro / ri 1 1
Ti To
2L k hi ri ho ro
Rearranging
Q Ti To
U o Ti To
2Lro ro ro ro 1
ln
k r hr h
i ii o
Therefore, overall heat transfer coefficient is
1
r r r 1
U o o ln o o
k r hr h
i ii o
b)
i) To calculate the overall heat transfer coefficient, we need to evaluate the convection heat transfer
coefficient both inside and outside.
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83
Heat Transfer: Exercises Heat Exchangers
Vm Dh
Re
For water:
m D 2
Vm , A
A 4
4m
4 0.25
Re 8781
D 0.05 725 106
Cp 725 10 6 4178
Pr 4.85
k 0.625
Re > 2300 (turbulent flow)
Therefore: Nu D 0.023 Re 0.8 Pr 0.4 0.023 87810.8 4.850.4 62
D
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84
Heat Transfer: Exercises Heat Exchangers
Nu D k 62 0.625
From which: hi 775 W / m 2 K
D 0.05
For oil:
4 Area 4 (rb2 ra2 )
Dh 2(rb ra ) 2(0.045 0.025) 0.034 m
Perimeter 2 (rb ra )
Vm Dh 2m( rb ra )
2m 2 0.12
Re 33
( rb ra ) (rb ra ) 0.045 0.028 3.25 10 2
2 2
Re < 2300 (Laminar flow)
Therefore: Nu D 5.6
Nu D k 5.6 0.138
ho 22.7 W/m2 K
Dh 0.034
1
0.028 28 0.028 1
Uo
16 ln 25 725 0.025 22.7 21.84
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 90C , Tci 10C , Tho 55C
Tco is unknown. This can be computed from an energy balance
For the oil side:
Q mhC ph (Thi Tho ) 0.12 2131(90 35) 8950 W
Q mcC pc (Tco Tci ) 0.25 4178(Tco 10) 8950 W
Therefore Tco 18.56C
Evaluate LMTD
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85
Heat Transfer: Exercises Heat Exchangers
T1 90 10 80C
T2 55 18.56 36.44C
T2 T1 36.44 80
Tlm 56.1C
ln(T2 / T1 ) ln(36.44 / 80)
Q UATlm U o 2 ro LTlm
Q 8950
L 41.5m
U o 2 ro Tlm 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. the hydraulic diameter for each passage;
ii. the appropriate Reynolds number;
iii. the overall heat transfer coefficient;
iv. the outlet temperature of the cold air;
v. and the length L
Use the following relations:
Using the relations:
Nu D 4.6 Re D 2300
Nu D 0.023 Re 0.8 Pr 1 / 3
D Re D 2300
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86
Heat Transfer: Exercises Heat Exchangers
Data for example 4.4
Hot exhaust inlet temperature 100oC
Hot exhaust outlet temperature 70oC
Cold air inlet temperature 30oC
Hot exhaust total mass flow 0.1 kg/s
Cold air total mass flow 0.1 kg/s
Density for exhaust and cold air 1 kg/m3
Dynamic viscosity, exhaust and cold air 1.8x10-5 kg/m s
Thermal conductivity, exhaust and cold air 0.02 W/m K
Specific heat capacity, exhaust and cold air 1 kJ/kg K
Heat exchanger wall thickness 0.5 mm
Heat Exchanger wall thermal conductivity 180 W/m K
Hot exhaust side fouling resistance 0.01 K m2/W
Cold air side fouling resistance 0.002 K m2/W
Solution:
VL
Re
L Dh (Hydraulic diameter)
4 cross sectional area 4 w H 4 0.003 0.3
Dh 5.94 mm
perimenter 2w H 20.003 0.3
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87
Heat Transfer: Exercises Heat Exchangers
For a single passage:
V
m / 50 0.1 / 50 2.22
m/s
H w 0.003 0.3 1
1 2.22 5.94 10 3
Re 733
1.8 10 5
Re 2300 (laminar flow)
Nu D 4.6
Nu D k 4.6 0.02
h 15.5 W / m 2 K
Dh 5.98 10 3
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88
Heat Transfer: Exercises Heat Exchangers
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 1
1 t 1 1 0.5 10 3 1
U R f , h R f ,c 0.01 0.002
hh k hc 15.5 180 15.5
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 mC p (Th ,i Th ,o ) mC p (Tc ,i Tc ,o )
Tc ,o Tc ,i (Th,i Th ,o ) 30 (100 70) 60 o C
Also
Q UATlm
Tlm is constant in a balanced flow heat exchanger
Tlm 100 60 70 30 40C
0 .1
Q mC p Th ,i Th ,o
1000100 70 60 w / passage
50
Area of passage:
Q 60
A 0.211 m 2
UTlm 7.1 40
And since: A w L
0.211
L 0.704 m
0 .3
89
