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Chris Long & Naser Sayma
Heat Transfer: Exercises
2
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Heat Transfer: Exercises
1st edition
© 2014 Chris Long, Naser Sayma & bookboon.com
ISBN 978-87-7681-433-5
3
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Heat Transfer: Exercises
Contents
Contents
Preface
5
1
Introduction
6
2
Conduction
11
3
Convection
36
4
Radiation
64
5
Heat Exchangers
84
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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 2014
5
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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:
G7
G[
T
N
T
4
T$
N
7L 7R /
: P u u : The minus sign indicates heat flux from inside to outside.
6
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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
2
to a convective heat transfer coefficient of h = 6 W / m 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
TFRQY
K 7V 7 I
: P For 1 metre length of the pipe:
4FRQY
T FRQY $
T FRQY u S U
u u S u : P For radiation, assuming black body behaviour:
(
q rad = σ Ts4 − T f4
T UDG
)
u q rad = 462 W / m 2
For 1 metre length of the pipe
4UDG
T UDG $
u u S u : P 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.
7
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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
(N
: P . ), 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:
T
9 u,
$
9 u,
: u/
u u : P This will equal the conducted heat through the plate:
q=
k
(T2 − T1 )
t
7
7 TW
N
u q& (371.75 K)
The conducted heat will be transferred by convection and radiation at the surface:
(
q = h(T1 − T f ) + σ T14 − T f4
K
T V 7 7 I
7 7 I
)
u 8
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: P . 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
= h(Ts − T∞ ) + σ Ts4 − T∞4
A
(
)
7V u 7V u 7V 7V This equation needs to be solved numerically. Newton-Raphson’s method will be used here:
I
u 7V 7V GI
G7V
u 7V 7VQ 7VQ I
§ GI
¨¨
© G7V
·
¸¸
¹
7VQ u 7V 7V 7V Start iterations with Ts0 = 300 K
u u u u 7V
7V
.
u u u u 9
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.
Heat Transfer: Exercises
Introduction
The difference between the last two iterations is small, so:
7V
.
q&
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 α ∂t
Were α =
k
is the thermal diffusivity.
ρc
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 +δ r − Q z +δ z =
Q r +δ r = Q r +
∂Q
δr
∂r
Q z +δ z = Q z +
∂Q
δz
∂z
∂u
(2.1)
∂t
u = mcT
11
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Heat Transfer: Exercises
Conduction
Substituting in equation 2.1:
−
∂Q
∂Q
∂ (mcT )
δr−
δz=
∂r
∂z
∂t (2.2)
Fourier’s law in the normal direction of the outward normal n:
Q
∂T
= −k
A
∂n
4U
N$
w7
wU
N u S UG ]
w7
( A = 2π rδ z )
wU
4]
N$
w7
w]
N u S UG U
w7
( A = 2π rδ r )
w]
Equation 2.1 becomes
w ­
w7 ½
w ­
w7 ½
® N u S UG ]
¾G U ® N u S UG U
¾G ]
wU ¿
wU ¯
w] ¯
w] ¿
PF
w7
(2.3)
wW
Noting that the mass of the control volume is given by:
m = ρ 2π rδ rδ z Equation 2.3 becomes
w ­ w7 ½
w ­ w7 ½
®N U
¾G U ®N U
¾G ]
wU ¯ wU ¿
w] ¯
w] ¿
w7
wW
UFU
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 ρc ∂T
=
r
+
r ∂r  ∂r  ∂z 2
k ∂t
Using the definition of the thermal diffusivity: α =
rule:
k
and expanding the first term using the product
ρc
1 ∂  ∂ 2T ∂T ∂r  ∂ 2T 1 ∂T
+
=
which gives the required outcome:
r
+
r ∂r  ∂r 2 ∂r ∂r  ∂z 2 α ∂t
∂ 2T 1 ∂T ∂ 2T 1 ∂T
+
+
=
∂r 2 r ∂r ∂z 2 α ∂t
12
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Heat Transfer: Exercises
Conduction
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 = U∆T where U is the overall heat transfer coefficient given by:
8
T
'7
: P .
 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
/L
­° ª / S /V º ½°
NL ®
« »¾
°̄ ¬« KL N S N V KR »¼ °¿
­ ª º ½
®
« »¾ ¼¿
¯ ¬
Li = 0.195m (195 mm)
14
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Heat Transfer: Exercises
Conduction
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.
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
4
S U /KL
4
S U /KL 7L 7 → 7L 7
4
S/N 7 7
→ 7 7
OQ U U
4
S U /KR 7 7R → 7 7R
4
S /N OQ U U
4
S U /KR
Adding the three equations on the right column which eliminates the wall temperatures gives:
4
4
/
S/ 7L 7R
OQ U U
KL U
N
KR U
S OQ u u : P 15
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Heat Transfer: Exercises
Conduction
Insulated pipe
4
/
S 7L 7R
OQ U U OQ U U
KL U
N
N LQV
KR U
4
/
S OQ OQ u u : P 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.
16
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Heat Transfer: Exercises
Conduction
Solution
The equation for heat flow through a pipe per unit length was developed in Example 2.3:
Q=
2πL(Ti − To )
ln (r2 / r1 )
1
1
+
+
hi r1
k
ho r2
Hence substituting into this equation:
4
S u OQ u u u :
For the case with insulation, we also use the equation from Example 2.3
4
4
S/ 7L 7R
OQ U U OQ U U
KL U
N
N LQV
KR U
S u OQ OQ u u u :
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Heat Transfer: Exercises
Conduction
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:
4
S/ 7L 7R
OQ U U
N LQV
S u OQ u : 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, U
NJ P ) 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 ofI 7E
q& , when the external
air temperature Tf is 20 C. Under these conditions, the external heat transfer coefficient h is 12 W/m2 K.
o
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:
7 7I
7E 7 I FRVK P / [
VLQK P/
where P K3
and Ac is the cross sectional area.
N$F
Determine the total convective heat transfer from the heat sink, the fin effectiveness and the
fin efficiency.
18
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Heat Transfer: Exercises
Conduction
Solution
Total heat fluxed is that from the un-finned surface plus the heat flux from the fins.
Q = Qu + Q f
Qu = Au h (Tb − T f ) = w × s ( N − 1) )h (Tb − T f
4X
u u )
:
For a single fin:
4I
§ G7 ·
N$F ¨
¸ © G[ ¹ [ Where Ac is the cross sectional area of each fin
Since
7 7I
7E 7 I FRVK P / [
VLQK P/
678'<)25<2850$67(5©6'(*5((
&KDOPHUV8QLYHUVLW\RI7HFKQRORJ\FRQGXFWVUHVHDUFKDQGHGXFDWLRQLQHQJLQHHU
LQJDQGQDWXUDOVFLHQFHVDUFKLWHFWXUHWHFKQRORJ\UHODWHGPDWKHPDWLFDOVFLHQFHV
DQGQDXWLFDOVFLHQFHV%HKLQGDOOWKDW&KDOPHUVDFFRPSOLVKHVWKHDLPSHUVLVWV
IRUFRQWULEXWLQJWRDVXVWDLQDEOHIXWXUH¤ERWKQDWLRQDOO\DQGJOREDOO\
9LVLWXVRQ&KDOPHUVVHRU1H[W6WRS&KDOPHUVRQIDFHERRN
19
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Heat Transfer: Exercises
Conduction
Then
G7
G[
VLQK P / [
P 7E 7 I FRVK P/
4I
§ G7 ·
N$F ¨
¸
© G[ ¹ [
Thus
4I
§ VLQK P/ ·
N$F ¨
¸ P 7E 7 I © FRVK P/ ¹
N$F P 7E 7 I WDQK P/
KSN$F
7E 7 I WDQK P/
Since
P
§ K3
¨¨
© N$F
·
¸¸ ¹
3
ZW
$F
ZuW
u u P
u P P
§ u · ¨
¸
© u u ¹
P/ u WDQK P/
4I
P WDQK u u u u u u So total heat flow:
4
4X 4 I
u :
20
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: ILQ Heat Transfer: Exercises
Conduction
Finn effectiveness
H ILQ
)LQKHDWWUDQVIHUUDWH
+HDWWUDQVIHUUDWHWKDWZRXOGRFFXULQWKHDEVHQFHRI WKHILQ
H ILQ
u u 4I
K$F 7E 7 I
Fin efficiency:
$FWXDOKHDWWUDQVIHUWKURXJKWKHILQ
+HDWWKDWZRXOGEHWUDQVIHUUHGLI DOOWKHILQDUHDZHUHDWWKHEDVHWHPSHUDWXUH
K ILQ
4I
K ILQ
K$V 7E 7 I
$V
Z/ Z/ /W /W
$V
u K ILQ
/ Z W
u P u u Example 2.6
For the fin of example 4.5, a fan was used to improve the thermal performance, and as a result, the heat
transfer coefficient is increased to 40 W/m2 K. Justify the use of the lumped mass approximation to
predict the rate of change of temperature with time. Using the lumped mass approximation given below,
calculate the time taken, τ, for the heat sink to cool from 60oC to 30oC.
7 7I
§ K$ W ·
7L 7 I H[S ¨ V ¸ © P& ¹
21
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Heat Transfer: Exercises
Conduction
Solution
Consider a single fin (the length scale L for the Biot number is half the thickness t/2)
K/
N
%L
KuW N
u | Since Bi < 1, we can use he “lumped mass” model approximation.
7 7I
7L 7 I
W
§ K$ W ·
H[S ¨ V ¸ © P& ¹
P& §¨ 7 7 I
OQ
K$V ¨© 7L 7 I
·
¸
¸
¹
m = ρAs t / 2
W
§ 7 7I
OQ¨
K ¨© 7L 7 I
U&W
·
¸
¸
¹
u u § ·
OQ¨
¸
u © ¹
VHFRQGV
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Heat Transfer: Exercises
Conduction
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
Solution
Consider a single fin:
3
ZW
$F
ZuW
P
§ K3
¨¨
© N$F
P/
4I
P u u P ·
¸¸
¹
§ u ·
¨
¸
© u u ¹
P u WDQK P/
4I
u K3N$F
7E 7 I WDQK P/ u u u u (From example 2.5)
u u : 23
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Heat Transfer: Exercises
Conduction
So for 1 kW, the total number of fins required:
1
Example 2.8
An air temperature probe may be analysed as a fin. Calculate the temperature recorded by a probe of
length L = 20 mm, k = 19 W/m K, D = 3 mm, when there is an external heat transfer coefficient of h = 50
W/m2K, an actual air temperature of 50oC and the surface temperature at the base of the probe is 60oC.
Solution
The error should be zero when Ttip = T∞. The temperature distribution along the length of the probe
(from the full fin equation) is given by:
7[ 7f
7E 7f
P
§ K3 ·
¨
¸
© N$ ¹
FRVK P / [ FRVK P/ KWLS
PN
KWLS
PN
VLQK P / [
VLQK P/
A = π D 2 / 4,
P =π D
At the tip, x = L, the temperature is given by ( cosh(0) = 1 , sinh(0) = 0 ):
7WLS 7f
7E 7f
FRVK P/ KWLS
PN
I
VLQK P/
24
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Heat Transfer: Exercises
Conduction
Where is the dimensionless error:
φ = 0, Ttip = T∞ (no error)
φ = 1, TL = Tb (large error)
For /
7f
PP N
q& $ S ' P
P/
§ K3 ·
¸
¨
© N$ ¹
7E
: P . '
PP
K
KWLS
: P . q& 3
S'
§ KS ' u ·
¸
¨¨
¸
© NS ' ¹
§ K ·
¨
¸
© N' ¹
§ u ·
¨
¸
© u ¹
P u h
50
=
= 0.0444
km 59.235 × 19
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Heat Transfer: Exercises
Conduction
Tx − T∞
1
=
= 0.539
Tb − T∞ cosh (1.185) + 0.0444 × sinh (1.185)
Ttip = 0.539(Tb − T∞ ) + T∞
7WLS
q& Hence error
q& 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 $H
P , perimetre P = 0.6 m (The
actual floor of the balcony in this case may be considered to be insulated from the wall
c) No balcony.
Solution
26
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Heat Transfer: Exercises
Conduction
a) For the concrete balcony
Treat the solid balcony as a fin Bi =
%L
ho t / 2
kb
u Not that Bi is not << 1, thus 2D analysis would be more accurate. However, treating it as a fin will give
an acceptable result for the purpose of a quick calculation.
P = 2( H + t ) = 2(4 + 0.2) = 8.4 m
Ac = H × t = 4 × 0.2 = 0.8 m 2
To decide if the fin is infinite, we need to evaluate mL (which is in fact in the notation used here is mW)
P:
§ K3 ·
¨
¸
© N$ ¹
§ u ·
¨
¸
© u ¹
:
u
This is large enough to justify the use of the fin infinite equation.
4E
KR 3N E $F
7 7R

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 ) (2)
qb =
kb
(T1 − T2 ) (3)
L
Adding equations 1, 2 and 3 and rearranging:
TE
7R 7L
/ § $F
¨
KL N E ¨© KR 3N E
·
¸¸
¹
(4)
This assumes 1D heat flow through the wall, the concrete balcony having a larger k than the wall may
introduce some 2-D effects.
27
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Heat Transfer: Exercises
Conduction
From (4)
TE
§
·
¨
¸
© u u ¹
: P Compared with no balcony:
TE
7R 7L
/
KL N Z KR
: P The difference for one balcony is $F u : For 350 apartments, the difference is 6891 W.
For the steel supported balcony where $F
P and 3
P .
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Heat Transfer: Exercises
Conduction
As before, however, in this case Bi << 1 because k s > k b
§ K3 ·
¨
¸
© N$ ¹
P:
Z
§ u ·
¨
¸
© u ¹
u | P: !! , so we can use the infinite fin approximation as before
TE
7R 7L
/ § $F
¨
KL N V ¨© KR 3N V
·
¸¸
¹
§ ·
¨
¸
© u u ¹
: P Qb = Ac qb = 0.01 × 182 = 1.82 W / beam
For 350 apartments, Qb = 1915 W
Example 2.10
In free convection, the heat transfer coefficient varies with the surface to fluid temperature difference
(T
− T f ) . Using the low Biot number approximation and assuming this variation to be of the form
n
h = G (Ts − T f ) Where G and n are constants, show that the variation of the dimensionless temperature
s
ratio with time will be given by
T Q
QKLQLW O W
Where
θ=
(T
s
− Tf
Tinit − T f
)
, λ=
Area
Mass × Specific Heat Capacity
and hinit = the heat transfer coefficient at t = 0. Use this expression to determine the time taken for an
aluminium motorcycle fin ( U
NJ P &
- NJ. ) of effective area 0.04 m2 and thickness
2mm to cool from 120oC to 40oC in surrounding air at 20oC when the initial external heat transfer
coefficient due to laminar free convection is 16 W/m2 K. Compare this with the time estimated from
the equation ( θ = e − hλt ) which assumes a constant value of heat transfer coefficient.
29
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Heat Transfer: Exercises
Conduction
Solution
Low Biot number approximation for free convection for Bi < 1
Heat transfer by convection = rate of change of internal energy
K$ 7V 7 I
P&
G 7V 7 I
(1)
GW
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:
1X
*U 3U
in turbulent flow or 1X
*U 3U
for laminar flow)
Equation 1 then becomes:
* 7V 7 I
Q
W
*$
³W P& GW
*Q$W
P&
7V 7 I
³
Q
s
Q 7V 7 I
W (T
P& G 7V 7 I
$
GW
G 7V 7 I
W
7V 7 I
At t = 0,
7V 7 I
Q
W − T f ) = (Ts ,i − T f
(
(2)
)
)
(T − T )
And use the definition θ =
(T − T )
−n
If we divide equation 2 by Ts ,i − T f
s
s ,i
f
f
We obtain
*Q$W
P& 7V L 7 I
Q
T Q *Q$W
7V L 7 I
P&
Q
30
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Heat Transfer: Exercises
(
Conduction
)
Since G Ts ,i − T f = h i , the heat transfer coefficient at time t = 0, then
T Q
KL $W
P&
Or T Q
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Heat Transfer: Exercises
For aluminium U
Conduction
NJ P &
- NJ . For laminar free convection, n = ¼
P
O
T Q
When
U $;
$
P&
u u u NJ u P . -
QKL O W which gives
W
T Q QKL O
7
q&
W
u u u T
Then
V For the equation θ = e − hλt
which assumes that the heat transfer coefficient is independent of surface-to-fluid temperature difference.
W
OQ T
KO
OQ u u Percentage error =
V u 32
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Heat Transfer: Exercises
Conduction
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?
Solution
Spherical bead: ܽ‫ ܽ݁ݎ‬ൌ ߨ݀ଶ ଷ
‫ ݁݉ݑ݈݋ݒ‬ൌ ߨ݀ Τ͸
Assume this behaves as a lumped mass, then
ܶ௕ െ ܶஶ
= 0.995
ൌ
‡š’ሺെߣ‫ݐ‬ሻ ൌ ͲǤͻͻͷ
ܶஶ െ ܶ௜
(given)
݄‫ܣ‬
ǡ‫ ݐ‬ൌ ͲǤͲͳ‫ݏ‬
ߣൌ
݉‫ܥ‬
ߣ‫ ݐ‬ൌ ͲǤͲͲͷ
ߣ ൌ ͲǤͷ
33
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Heat Transfer: Exercises
Conduction
For lumped mass on cooling from temperature Ti
ܶ௕ െ ܶஶ
ൌ ‡š’ሺെߣ‫ݐ‬ሻ ൌ ͲǤͻͻͷ
ܶஶ െ ܶ௜
݄‫ܣ‬
ǡ‫ ݐ‬ൌ ͲǤͲͳ‫ݏ‬
ߣൌ
݉‫ܥ‬
ߣ‫ ݐ‬ൌ ͲǤͲͲͷ
ߣ ൌ ͲǤͷ
Which gives the required value of heat transfer coefficient
݄‫ܣ‬
ൌ ͲǤͷ
ߩܸ‫ܥ‬
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Heat Transfer: Exercises
Conduction
So
݄ ൌ ͲǤͷ
݄ൌ
ߨ݀ ଷ ‫ߩܥ‬
ͲǤͷ݀‫ߩܥ‬
ൌ
ଶ
͸
͸ ߨ݀
ͲǤͷ ൈ ͳͲିଷ ൈ ͶͲͲ ൈ ͹ͺͲͲ
ൌ ʹ͸Ͳ ܹ Τ݉ଶ ‫ ܭ‬
͸
ܰ‫ݑ‬஽ ൌ
݄‫ʹ ܦ‬͸Ͳ ൈ ͳͲିଷ
ൌ
ൌ ͻǤͻ
݇
ͲǤͲʹ͸ʹ
For a sphere
ଵȀଶ
ܰ‫ݑ‬஽ ൌ ʹ ൅ ቄͲǤͶܴ݁஽
ଶȀଷ
൅ ͲǤͲ͸ܴ݁஽ ቅ ܲ‫ ݎ‬଴Ǥସ From which with Pr = 0.707
ଵȀଶ
ଶȀଷ
݂ ൌ ͲǤͶܴ݁஽ ൅ ͲǤͲ͸ܴ݁஽ െ ͻǤͶ ൌ Ͳ
ିଵȀଶ
ିଵȀଷ
݂ሖ ൌ ͲǤʹܴ݁஽
൅ ͲǤͲͶܴ݁஽ Using Newton iteration
‫ ݔ‬ሺ௡ାଵሻ ൌ ‫ ݔ‬௡ െ
݂ሺ‫ݔ‬ሻ
ሖ
݂ሺ‫ݔ‬ሻ
Starting with ReD = 300
ሺଵሻ
ܴ݁஽ ൌ ͵ͲͲ െ
ൣͲǤͶξ͵ͲͲ ൅ ͲǤͲ͸ሺ͵ͲͲሻଶȀଷ െ ͻǤͶ൧
ͲǤʹʹʹ
ൌ ͵ͲͲ െ
ͲǤʹ
ͲǤͲͶ
ͲǤͲͳ͹ͺʹ
൨
൤
൅
ξ͵ͲͲ ͵ͲͲଵȀଷ
Which is close enough to 300
From which
‫ݑ‬ஶ ൌ
ܴ݁ߤ
ൌ ͶǤͷ݉Ȁ‫ݏ‬
‫ߩܦ‬
35
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Heat Transfer: Exercises
Convection
3 Convection
Example 3.1
Calculate the Prandtl number (Pr = µCp/k) for the following
a) Water at 20°C: µ = 1.002 × 10−3 kg/m s, Cp = 4.183 kJ/kg K and k = 0.603 W/m K
b) Water at 90°C: ρ = 965 kg/m3, ν = 3.22 × 10−7 m2/s, Cp = 4208 J/kg K and k = 0.676 W/m K
c) Air at 20°C and 1 bar: R = 287 J/kg K, ν = 1.563 × 10−5 m2/s, Cp = 1005 J/kg K and
k = 0.02624 W/m K
u 7 kg/m s
7
d) Air at 100°C: P
&S
u 7 u 7 N- NJ . (Where T is the absolute temperature
in K) and k = 0.03186 W/m K.
e) Mercury at 20°C: µ = 1520 × 10−6 kg/m s, Cp = 0.139 kJ/kg K and k = 0.0081 kW/m K
f) Liquid Sodium at 400 K: µ = 420 × 10−6 kg/m s, Cp = 1369 J/kg K and k = 86 W/m K
g) Engine Oil at 60°C: µ = 8.36 × 10−2 kg/m s, Cp = 2035 J/kg K and k = 0.141 W/m K
Solution
a) 3U
b) 3U
c) 3U
U
P &S
u u N
P &S
UQ & S
N
N
u u u UQ & S
N
3
57
u NJ P 3U
u u u 36
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Heat Transfer: Exercises
d) P
&S
Convection
u 7 7
u u u 7 u 7 u NJ P V
u u u u - NJ .
3U
e) 3U
f) 3U
g) 3U
u u P &S
N
P &S
N
P &S
N
u u u u u u u Click here
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Heat Transfer: Exercises
Convection
Comments:
• Large temperature dependence for water as in a) and b);
• small temperature dependence for air as in c) and d);
• use of Sutherland’s law for viscosity as in part d);
• difference between liquid metal and oil as in e), f) and g);
• units of kW/m K for thermal conductivity;
• use of temperature dependence of cp as in part a).
Example 3.2
Calculate the appropriate Reynolds numbers and state if the flow is laminar or turbulent for
the following:
a) A 10 m (water line length) long yacht sailing at 13 km/h in seawater ρ = 1000 kg/m3 and
µ = 1.3 × 10−3 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
P
u 7 kg/m s
7
c) 0.05 kg/s of carbon dioxide gas at 400 K flowing in a 20 mm diameter pipe. For the viscosity
u 7 P
kg/m s
take
7
d) The roof of a coach 6 m long, travelling at 100 km/hr in air (ρ = 1.2 kg/m3 and µ = 1.8 ×
10−5 kg/m s)
e) The flow of exhaust gas (p = 1.1 bar, T = 500ºC, R = 287 J/kg K and µ = 3.56 × 10−5 kg/m s)
over 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)
Solution
a)
5H
b) 7
UX/
P
u u u u
u (turbulent)
.
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Heat Transfer: Exercises
P
Convection
u u u NJ P V :
u S
X
:U
UDG V P V u U
3
57
u NJ P Characteristic length is r not D
5H
UX'
P
c) P
U X$
X
P
US' u u u (turbulent)
u UX u
S' 5H
UX'
P
U u P '
US' P
P
S'P
P
5H
d) X
5H
u u u S u u u u UX/
P
u NJ P V
u (turbulent)
P V
u u u u (turbulent)
39
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Heat Transfer: Exercises
Convection
 be the mass flow through the exhaust port
e) Let m
m = inlet density X volume of air used in each cylinder per second
u u
u
P u
X
5H G
5H
NJ V P
S 'U
UXG
P
u u S u u u (laminar)
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Heat Transfer: Exercises
Convection
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 5H | plate
5H | u )
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
(ρ = 1.2 kg/m3, Pr = 0.72 and µ = 1.8 × 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 (ρ = 854 kg/m3, Pr = 546, β = 0.7 × 10−3 K−1 and
µ = 3.56 × 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 (ρ = 1000 kg/m3, Pr = 6.95, β = 0.227 × 10−3K−1 and
µ = 1.00 × 10-3kg/m s)
d) Air at 20ºC (ρ = 1.2 kg/m3, Pr = 0.72 and µ = 1.8 × 10−5 kg/m s) adjacent to a 60 mm
diameter vertical, light bulb with a surface temperature of 90°C
Solution
U J E '7 /
P
a) *U
'7
E
7
*U
*U 3U
. . u u u u u u u u u (mostly laminar)
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Heat Transfer: Exercises
b) /
$UHD
3HULPHWHU
Convection
u u '7
*U
U J E '7 /
P
*U 3U
P . u u u u u u u u u u Heated surface facing downward results in stable laminar flow for all Gr Pr
c)
'7
*U
U J E '7 /
P
*U 3U
.
u u u u u u u u u u (laminar)
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Heat Transfer: Exercises
d) /
$UHD
3HULPHWHU
'7
E
7
*U
*U 3U
Convection
S' S'
'
.
U J E '7 /
P
. u u u u u u u u u (laminar)
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Heat Transfer: Exercises
Convection
Comments:
• Note evaluation of β for a gas is given by β = 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, µ = 4.63 × 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 σ = 56.7 × 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.
Solution
a) 3U
N
1X
P &S
N
P &S
3U
K/
N
u u u : P .
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Heat Transfer: Exercises
b) 1X
T
'7
/
1X
c) 1X
K/
N
4
$
Convection
T /
'7 N
u : P q& $UHD
3HULPHWHU
K/
N
PP
u u P
TF /
'7 N
In this case, q must be the convective heat flux – radiative heat flux
7V
. 7f
.
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Heat Transfer: Exercises
Convection
45
V$ 7V 7f
'7
u u u :
.
Qc = Q − QR = 1000 − 416 = 584 W
qc =
Qc
584
=
= 649 W / m 2
A 1.5 × 0.6
1X
TF /
'7 N
u
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Heat Transfer: Exercises
d) '7
T
Convection
.
N E 7 7
:
q& (assuming 1-D conduction)
T
1X
TF /
'7 N
: P u
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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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 1X [ & 5H Q[ . In free convection from a vertical surface the local Nusselt number
is represented by 1X [ & *U[P , 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) 1X [
5H [
K[
N
UX[
P
For forced convection: 1X [
k ρux

Hence h = C1 
x  µ 
& 5H Q[
n
This shows that the heat transfer coefficient for forced does not depend on temperature difference.
For free convection 1X [
*U[
& *U[P U J E '7 [ P
m
k  ρ 2 g β ∆T x 3 
 (1)
Hence h = C 2 
x 
µ2

So for free convection, heat transfer coefficient depends on ∆T m
b) From (1), with m = 1/3 for turbulent free convection:
k  ρ 2 g β ∆T x 3 

h = C 2 
x 
µ2

1/ 3
k  ρ 2 g β ∆T
h = C 2 
x 
µ2



§ U J E '7
N& ¨¨
P
©
·
¸¸
¹
K
1/ 3
x
Hence the convective heat transfer coefficient does not depend on x
48
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Heat Transfer: Exercises
Convection
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 7f q& & S N- NJ . Q u P V U NJ P and 3U .
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 7f
q& and Q = 0.5 W .
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Heat Transfer: Exercises
Convection
Solution
Firstly, we need to estimate if the flow is laminar or turbulent.
Assuming a critical (transition) Reynolds number of 5H
X WXUE
u P
U/
u Q
/
u the velocity required would be:
u u u u P V Wind speed is very unlikely to reach this critical velocity, so the flow can be assumed to be laminar.
1X [
1X DY
5H/ T DY
5H/ 5H /
Xf
5H [ 3U 5H/ 3U T DY /
7V 7f N
T DY /
7V 7f N u 3U u : P u u u u u 5H / Q
/
u u u u P V 50
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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 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 × 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
3U
P &S
N
gives: N
P &S
3U
u u : P .
First we need to determine if these flows are laminar or turbulent.
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Heat Transfer: Exercises
Convection
For the inside (Free convection):
E
7
*U
U J E '7 /
P
*U
u '7 . u u '7 u u u (Flow will be turbulent over most of the surface for all reasonable values of ∆T )
For the outside (Forced convection)
5H /
U Xf /
P
u u u u (Flow will be turbulent for most of the surface apart from the first 0.3 m)
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Heat Transfer: Exercises
Convection
Hence we use the following correlations:
*U On the inside surface: 1X [
On the outside surface: 1X [
5H [ 3U For the inside:
1X [
K[
N
§ U J E 7L 7V [ ·
¸¸
¨¨
P
©
¹
[
K FRQVWDQWu
[
FRQVWDQW
Hence heat transfer coefficient is not a function of x
KDY
K[ / (1)
For the outside:
1X [
K[
N
§ U J E 7L 7V [ ·
¸¸
¨¨
P
©
¹
h = constant ×
KDY (x )0.8
[ /
K G[
/ [³
x
&
/
= C x −0.2
[ /
³[
[ G[
K[ /
(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)
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Heat Transfer: Exercises
Convection
From equation 1
§ U J 7L 7V + ·
¸
¨¨
¸
P u 7L
©
¹
KL +
N
KL
§ u u 7 7V
¨
¨ u u ©
KL
7L 7V
·
¸ u ¸
¹
(4)
From equation 2:
KR :
N
§ U X :
¨
¨© P
·
¸¸
¹
3U KR
§ u u ·
u
¨
¸
© u ¹
KR
: P . (5)
u From (3) with (4) and (5)
7L 7V
7V
7V
7V 7R 7V 7V
(6)
To solve this equation for Ts an iterative approach can be used
First guess: 7V
q&
Substitute this on the right hand side of equation 6:
7V
q& 54
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Heat Transfer: Exercises
Convection
For the second iteration we use the result of the first iteration:
7V
q& The difference between the last two iterations is 0.1°C , so we can consider this converged.
7V | q& From which:
T
KR 7V 7R
4
T$ u u :
: P N:
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.
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Heat Transfer: Exercises
Convection
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.
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 × 105
Nux = 0.02 Rex0.8 Pr1/3
Rex ≥ 3 × 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 × 10-5 kg /m s and ρ = 1.0 kg / m3
Solution
56
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Heat Transfer: Exercises
Convection
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
*U
U J E '7 /
P
'7
.
E
. u u u *U
u u (Laminar)
u (L here is height because it is in the direction of the free convection boundary layer)
So we use:
1X [
*U[ 3U
/
K G[
/ ³
KDY
/
FRQVWDQW³ [ G[
K[ /
KDY
1X DY
*U/ 3U
1X DY
u u KDY
1X DY N
/
T DY
KDY '7 4
T DY $
4
:
u KDY '7$
: P .
u u u u (Last factor of 2 is for both sides)
57
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Heat Transfer: Exercises
Convection
For forced convection, we need to evaluate Re to see if flow is laminar or turbulent
5H
UX/
P
u u u u (Laminar)
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Heat Transfer: Exercises
Convection
(L here is the width because flow is along that direction)
1X [
5H[ 3U KDY
KG[
/ ³
1X DY
KDY
/
K[ /
5H/ 3U 1X DY N
/
4
T DY $
4
:
u u u KDY '7$
u : P .
u u u u '7
q& 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
Solution
1X DY
KDY N
/
59
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Heat Transfer: Exercises
Convection
Where for a constant surface-to-fluid temperature:
KDY
[
/
½°
­° /
K
G[
K
G[
® ODPLQDU
³ WXUEXOHQW ¾° / °̄ ³
[/
¿
Since for laminar flow ( 5H [ ):
1X [
5H[ 3U KODP
N § U Xf
u ¨¨
[© P
·
¸¸
¹
KODP
§ U Xf
u N ¨¨
© P
·
¸¸
¹
[ 3U 3U [ & ODP [ Where C lam does not depend on x
Similarly:
hturb = C turb x −0.2
Where
&WXUE
§ U Xf
u N ¨¨
© P
·
¸¸
¹
3U Hence
KDY
[
/
½°
­° /
® ³ &ODP [ G[ ³ & WXUE [ G[ ¾
/ °̄ °¿
[/
KDY
ª [ º
ª [ º
­°
&
&
® ODP «
WXUE «
»
»
/ °̄
¬ ¼ ¬ ¼ [
[/
½°
¾
°
/ ¿
/
60
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Heat Transfer: Exercises
But
Convection
1X DY
KDY N
/
1X DY
&ODP & WXUE / [ / [/ N
N
1X DY
§ U Xf
¨¨
© P
U Xf [/
P
>
·
¸¸
¹
@
[
/
3U
ª§ U X / · § U X [
f
f /
¸¸ ¨¨
«¨¨
«¬© P ¹
© P
·
¸¸
¹
º » 3U
»¼
(The transition Reynolds number)
So
>
1X DY
3U u 5H / u 1X DY
5H / 3U @
61
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Heat Transfer: Exercises
Convection
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 × 10-5 kg/m s
62
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Heat Transfer: Exercises
Convection
Solution
T DY
3U
5H /
U
5H /
4
$
u P& 3
: P u u N
U Xf /
P
3
57
NJ P u u u u u Using the formula for Nusselt Number obtained in Example 3.9:
1X DY
5H / 3U 1X DY
u u 1X DY
KDY N
/
'7
T DY /
1X DY N
^
`
u T DY /
'7N
u u q&
Ts = T∞ + ∆T
7V
q&
63
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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:
7
§ $ ) 7 $ )7
¨¨ © $ ) $ )
·
¸¸ ¹
where F13 and F23 are the appropriate view factors.
b) Show that the total radiative heat transfer to the tubes, Q2, is given by:
4 §
$) $ ) ·
¨¨ $ ) ¸¸V 7 7 $ ) $ ) ¹
©
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Heat Transfer: Exercises
Radiation
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)
V $ ) 7 7
7
7
V $ ) 7 7 $ ) 7 $ ) 7
$ ) $ )
§ $ ) 7 $ ) 7 ·
¨
¸
¨
¸
$ ) $ )
©
¹
since $L )LM
$ M ) ML 65
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Heat Transfer: Exercises
Radiation
b) From (1)
4 V $ ) 7 7 V $ ) 7 7 4 V $ ) 7 7 V $ ) 7 7 4 V $ ) 7 7 V $ ) ¨¨
4 V $ ) 7 7
4 V $ ) 7 7 V $ ) ¨¨
4 V $ ) 7 7 V $ ) 7 7 ¨¨
4 V $ ) 7 7 V 7 7 ¨¨
4 V 7 7 ¨¨ $ ) c) 7
$ ) 7 $ ) 7
$ ) $ )
§ $ ) 7 $ ) 7
·
7 ¸¸
$ ) $ )
©
¹
§ $ ) 7 $ ) 7 $ )7 $ )7 ·
¸¸ V $ ) ¨¨
$ ) $ )
©
¹
§ $ ) 7 $ )7
© $ ) $ )
·
¸¸
¹
§
·
$ )
¸¸ © $ ) $ ) ¹
§ $ ) $ ) ·
¸¸ $
)
$
)
¹
© §
©
$ ) $ ) ·
¸
$ ) $ ) ¸¹
$ ) 7 $ ) 7
$ ) $ )
7
u u u u u u 4 u ·
§
u ¨ ¸
¹
©
.
u :
66
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Heat Transfer: Exercises
Radiation
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 e1 = 0.4, T1 = 800 K, e2 = 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
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− ε i
j =1
εi
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Heat Transfer: Exercises
Radiation
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
)
(Although this is given in the question, it can be obtained from appropriate tables with the
above parameters)
)
(As surface 1 is flat, it cannot see itself)
)
)
(Symmetry)
)
)
(From the relation
¦)
LM
in an enclosure)
68
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Heat Transfer: Exercises
Radiation
S u u u S u u )
$
)
$
)
(Symmetry)
)
n
E b ,i − J i
= ∑ (J i − J j )Fij
b) 1 − ε i
j =1
εi
Apply to surface 1, (i = 1)
Let
1 − ε1
ε1
= φ1
I >) - - ) - - @ (E - ( E - ^ I ) I ) ` I ) - I ) - Eb ,1 = σ T14
J 3 = σ T34 (Radiatively black surface)
φ1 =
1 − ε1
ε1
=
1 − 0.4
= 1.5
0.4
σ T14 = 2.5 J 1 − 0.9 J 2 − 0.6 σ T34
u u u - u - u u u 2.5 J 1 − 0.9 J 2 = 45545 W / m 2 (1)
Applying to surface 2 (i = 2)
(E - ^ I ) I ) ` I ) - I ) - Eb , 2 = σ T24
69
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Heat Transfer: Exercises
φ2 =
1− ε2
ε2
=
Radiation
1 − 0.3
= 2.333
0.3
σ T24 = 3.333 J 2 − 1.4 J 1 − 0.9333σ T34
3.333 J 2 − 1.4 J 1 = 48334 W / m 2 (2)
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 Wake
the only emission we want to leave behind
.QYURGGF'PIKPGU/GFKWOURGGF'PIKPGU6WTDQEJCTIGTU2TQRGNNGTU2TQRWNUKQP2CEMCIGU2TKOG5GTX
6JGFGUKIPQHGEQHTKGPFN[OCTKPGRQYGTCPFRTQRWNUKQPUQNWVKQPUKUETWEKCNHQT/#0&KGUGN6WTDQ
2QYGTEQORGVGPEKGUCTGQHHGTGFYKVJVJGYQTNFoUNCTIGUVGPIKPGRTQITCOOGsJCXKPIQWVRWVUURCPPKPI
HTQOVQM9RGTGPIKPG)GVWRHTQPV
(KPFQWVOQTGCVYYYOCPFKGUGNVWTDQEQO
70
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Heat Transfer: Exercises
Radiation
The net radiative flux to surface 2 is given by
T
(E - H
H
u u u : P The minus sign indicates a net influx of radiative transfer as would be expected from
consideration of surface temperatures.
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:
-
V 7 )7 )7 )7
) ) )
The following equation may be used without proof:
( E L - L
HL HL
1
¦)
M LM
-L - M 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
71
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Heat Transfer: Exercises
Radiation
Solution
a) A1 = 2 π r1 h
A2 = 2 π r2 h
(
A3 = A4 = π r22 − r12
)
)
)
)
) ) ) )
but
$ )
)
$ ) $
)
$
U
)
U
u Thus
)
)
) ) ) )
72
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Heat Transfer: Exercises
)
)
Radiation
) )
) ) ) )
)
$ )
)
$ )
$ ) $
)
$
S U K
)
S U U
u u u S U K
)
S U U
u u u $ )
)
$
)
$
)
I joined MITAS because
I wanted real responsibili�
I joined MITAS because
I wanted real responsibili�
Real work
International
Internationa
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�ree wo
work
or placements
�e Graduate Programme
for Engineers and Geoscientists
Maersk.com/Mitas
www.discovermitas.com
�e G
for Engine
Ma
Month 16
I was a construction
Mo
supervisor
ina const
I was
the North Sea super
advising and the No
he
helping
foremen advis
ssolve
problems
Real work
he
helping
fo
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�ree wo
work
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ssolve pr
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Heat Transfer: Exercises
Radiation
Similarly (using symmetry)
)
)
)
)
)
)
)
( E L - L
b) H L
HL
Q
¦M L
- M )LM
For surface 2, i = 2, j = 1, 3, 4
(E - H
) - - ) - - ) - - H
ε 2 = 0.5 ,
1 − 0.5
=1
0.5
J 1 = Eb ,1 , J 3 = Eb ,3 , J 4 = Eb , 4 (1, 3, 4 are black)
(E - ) - ( E ) - ( E ) - ( E - ) ) ) -
c) - -
V 7 V 7 ) V 7 ) V 7 )
V 7 7 ) 7 ) 7 )
) ) ) u 7 u u u u 7 On the outside of surface 2:
(
− q 2 = σ ε 2, 0 T24 − T44
)
74
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Heat Transfer: Exercises
Radiation
Also
T
(E - H
V 7 u 7 H
u 7 u u 7 T2 = 1029 K
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 H . 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:
(EL - L
HL HL
1
¦)
M LM
-L - M 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.
75
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Heat Transfer: Exercises
Radiation
d) Briefly explain how this analysis could be extended to make it more realistic
Solution
a) $
$
S u S
PP S '/ S u u S
PP $
S U
www.job.oticon.dk
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Heat Transfer: Exercises
Radiation
)
(Flat surface)
)
(Given)
)
)
By Symmetry:
)
)
)
)
)
)
$
)
$
)
(By symmetry)
)
) )
Since A1 = A3
b) For surface 2, i = 2
(E - H
) - - ) - - H
J 1 = σ T14 (Black body)
H
,
Eb , 2 = σ T24
V 7 - ) - V 7 ) - - -
)V 7 ) - ) )
V 7 77
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Heat Transfer: Exercises
-
-
Radiation
u u 7 u u u - u 7 - We are also given that
J 3 = 107210 + 0.222 J 2
0.1 J 3 = 10721 + 0.0222 J 2
Hence
-
u u - 0.97778 J 2 = 5508 + 81756
J 2 = 89247 W / m 2
Also
T
(E - H
H
u u u : P 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.
78
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Heat Transfer: Exercises
Radiation
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: F1,2, 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:
(EL - L
HL HL
1
¦)
M LM
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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 × 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
From the figure: )
$ )
$ ) 80
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Heat Transfer: Exercises
Radiation
$
)
$
)
Z
)
K
) ) )
)
)
)
u ) ) )
)
)
)
For a grey body radiative heat transfer in an enclosure (n surfaces)
( E L - L
HL
Q
¦M HL
L
- M )LM
Applying for surface 1, i = 1 (the casing)
( E - H
) - - ) - - H
Eb ,1 = σ T14
J 3 = σ T34
1 − ε1
ε1
=
1 − 0.5
= 1.0
0.5
81
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Heat Transfer: Exercises
Radiation
So
-
V 7 ) - )V 7
) )
-
u 7 - u u u -
u 7 - : P (1)
c)
Given: - -
u 7 - u u - : P
: P J 2 = 5796 + 0.0972 J 1 (2)
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Heat Transfer: Exercises
Radiation
Substituting from equation 2 into equation 1:
-
u 7 - : P Which gives:
-
u 7 : P Applying a heat balance to surface 1
TLQ
TRXW TLQ
ª
º
«( - »
E »
«
« H »
« H
»
¬
¼
T LQ
u 7 T RXW
H V 7 7f
u 7 u 7 u u 7 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 Transfer: Exercises
Heat Exchangers
5 Heat Exchangers
Example 5.1
A heat exchanger consists of numerous rectangular channels, each 18 mm wide and 2.25 mm high. In
an adjacent pair of channels, there are two streams: water k = 0.625 W/m K and air k = 0.0371 W/m
K, separated by a 18 mm wide and 0.5 mm thick stainless steel plate of k = 16 W/m K. The fouling
resistances for air and water are 2 × 10−4 m2 K/W and 5 × 10−4 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.
b) Calculate the overall heat transfer coefficient with these resistances.
c) Which is the controlling heat transfer coefficient?
Solution:
Hydraulic Diameter = 4 x Area / Wetted perimetre
u u u u
u 'K
K
1X ' N
'K
a) KZDWHU
u u KDLU
u u 8
º
ª «¬ »¼
b) 8
u : P .
: P .
: P . º
ª u u u »
«
¼
¬ : P .
c) The controlling heat transfer coefficient is the air heat transfer coefficient.
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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, ρ = 1000 kg/m3, µ = 0.651 × 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:
1X '
5H ' 3U Solution:
P
U9$
V =
m
ρA
5H '
U9'
P
P
S'P
u S u u u 678'<)25<2850$67(5©6'(*5((
&KDOPHUV8QLYHUVLW\RI7HFKQRORJ\FRQGXFWVUHVHDUFKDQGHGXFDWLRQLQHQJLQHHU
LQJDQGQDWXUDOVFLHQFHVDUFKLWHFWXUHWHFKQRORJ\UHODWHGPDWKHPDWLFDOVFLHQFHV
DQGQDXWLFDOVFLHQFHV%HKLQGDOOWKDW&KDOPHUVDFFRPSOLVKHVWKHDLPSHUVLVWV
IRUFRQWULEXWLQJWRDVXVWDLQDEOHIXWXUH¤ERWKQDWLRQDOO\DQGJOREDOO\
9LVLWXVRQ&KDOPHUVVHRU1H[W6WRS&KDOPHUVRQIDFHERRN
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Heat Transfer: Exercises
Heat Exchangers
1X '
u u 1X '
K'
N
K
1X ' N
'
8
º
ª «¬ »¼
u : P .
: P .
Example 5.3
a) Show that the overall heat transfer coefficient for a concentric tube heat exchanger is given
by the relation:
8R
ª UR § UR · UR
º
»
« OQ¨¨ ¸¸ ¬ N © UL ¹ KL UL KR ¼ 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:
&S
-NJ.P
&S
-NJ.P
u NJPVN
:P.
Water:
u NJPVN
:P.
Using the relations:
1X '
5H ' 1X '
5H ' 3U 5H ' ! 86
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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 )
4
S UL /KL 7L 7 (1)
For the convection outside
Q = Ao ho (To − T1 )
4
S UR /KR 7R 7 (2)
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Heat Transfer: Exercises
Heat Exchangers
For conduction through the pipe material
4
G7
S U N
G7
GU
§ 4 · GU
¸¸ (3)
¨¨
© S U / ¹ U
Integrating between 1 and 2:
7 7
§ 4 · § UR ·
¨¨
¸¸ OQ¨¨ ¸¸ (4)
© S U / ¹ © UL ¹
From 1 and 2
7L 7
§ 4 ·
¸¸ (5)
¨¨
© S UL /KL ¹
7 7R
·
§
4
¸¸ (6)
¨¨
© S UR /KR ¹
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Heat Transfer: Exercises
Heat Exchangers
Adding 4, 5 and 6
4 § OQ UR UL
·
¸
¨¨
S/ ©
N
KL UL KR UR ¸¹
7L 7R
Rearranging
4
S/UR
7L 7R
§ UR § UR · UR
·¸
¨ OQ¨ ¸ ¨ N ¨ U ¸ KU K ¸
R ¹
© L¹ LL
©
8 R 7L 7R
Therefore, overall heat transfer coefficient is
8R
§ UR § UR · UR
·¸
¨ OQ¨ ¸ ¨ N ¨ U ¸ KU K ¸
R ¹
© L¹ LL
©
b)
i. To calculate the overall heat transfer coefficient, we need to evaluate the convection heat
transfer coefficient both inside and outside.
5H
U9P 'K
P
Vm =
m
,
ρA
5H
P
S'P
u S u u u
P &S
u u For water:
3U
N
A=
πD 2
4
Re > 2300 (turbulent flow)
Therefore: 1X '
5H ' 3U From which: KL
1X ' N
'
u u u : P .
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Heat Transfer: Exercises
Heat Exchangers
For oil:
Dh =
4π (rb2 − ra2 )
4 Area
=
= 2(rb − ra ) = 2(0.045 − 0.025) = 0.034 m
Perimeter 2π (rb + ra )
5H
U9P 'K
P
P UE UD
S UE UD P
P
S UE UD P
u S u u u Re < 2300 (Laminar flow)
Therefore: 1X '
KR
1X ' N
'K
8R
§ § ·
·
OQ¨ ¸ ¨¨
¸¸
© ¹ u ¹
© u W/m2 K
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. 7KL
q& , Tci = 100C, 7KR
q&
7FR is unknown. This can be computed from an energy balance
For the oil side:
4
P K& SK 7KL 7KR
u W
4
P F& SF 7FR 7FL
u 7FR : W
Therefore 7FR
q& Evaluate LMTD
'7
q& '7
q& '7OP
'7 '7
OQ '7 '7
OQ q&
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Heat Transfer: Exercises
4 8$'7OP
/
Heat Exchangers
8 R u S UR /'7OP
4
8 R u S UR '7OP
u S u u P
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
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Heat Transfer: Exercises
Heat Exchangers
Use the following relations:
Using the relations:
1X '
5H ' 1X '
5H ' 3U 5H ' ! 92
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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:
U9/
P 5H
L = Dh (Hydraulic diameter)
uFURVVVHFWLRQDODUHD
SHULPHQWHU
'K
u Zu +
Z +
u u For a single passage:
9
P + uZ U
5H
u u u u u u P V
5H (laminar flow)
1X '
K
1X ' N
'K
u u : P .
93
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PP
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.
8
ª
º
W « 5 I K 5 I F »
N KF
¬ KK
¼
ª º
u »
«
¬
¼
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 )
7F R
7F L 7KL 7K R
R &
Also
4 8$'7OP Tlm is constant in a balanced flow heat exchanger
7OP
4
q&
u P & S 7K L 7K R
Z SDVVDJH
Area of passage:
$
4
8'7OP
u P And since: A = w × L
L=
0.211
= 0.704 m
0.3
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