Chris Long Naser Sayma
Heat Transfer: Exercises
Heat Transfer: Exercises © 2010 Chris Long, Naser Sayma & Ventus Publishing ApS ISBN 978-87-7681-433-5
Contents
Preface
1. Introduction
2. Conduction
3. Convection
4. Radiation
5. Heat Exchangers
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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
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 16 o C and that on the outside is 6
C. 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
dx L
16 6 20 W / m
Q qA 20 6 7 840 W
The minus sign indicates heat flux from inside to outside.
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 2 h 6 W / m K , find the heat loss by convection
per metre length of the pipe when the external surface temperature is 80 o
C and the surroundings are at
20 o C. Assuming black body radiation what is the heat loss by radiation?
Solution
conv h T s T f 6 80 20 360 W / m
For 1 metre length of the pipe:
Q conv q conv A q conv 2 r 360 2 0 . 01 22 . 6 W / m
For radiation, assuming black body behaviour:
rad T s T f
rad 5 . 67 10 353 293
q 2 rad 462 W / m
For 1 metre length of the pipe
rad q rad A 462 2 0 . 01 29 . 1 W / m
A value of h = 6 W/m 2 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.
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 o 16 W / m K ), the top surface is exposed to an airstream of temperature 20
C. 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 100 o
C. 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 2
1666 . 7 W / m
This will equal the conducted heat through the plate:
t qt
T 1 T 2 100
98 . 75 C (371.75 K)
The conducted heat will be transferred by convection and radiation at the surface:
2 12 . 7 W / m K T 1 T f
Example 1.4
An electronic component dissipates 0.38 Watts through a heat sink by convection and radiation (black body) into surrounds at 20 o
C. What is the surface temperature of the heat sink if the convective heat
transfer coefficient is 6 W/m 2 K, and the heat sink has an effective area of 0.001 m ?
Solution
6 T s 293 5 . 67 10 T s 293
8 5 4 . 67 10 T
s 6 T s 2555 . 9 0
This equation needs to be solved numerically. Newton-Raphson’s method will be used here: 8 f 4 5 . 67 10 T
s 6 T s 2555 . 9
df 8 3
22 . 68 10 T s 6
dT s
f n 5 . 67 10 T s 6 T s 2555 . 9
df
22 . 68 T s 6
dT s
Start iterations with 0 T
300 K
T s 300
3 324 . 46 K
T s 324 . 46
3 323 K
The difference between the last two iterations is small, so:
323 K 50 C
The value of 300 K as a temperature to begin the iteration has no particular significance other than being above the ambient temperature.
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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:
k Were is the thermal diffusivity.
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
Substituting in equation 2.1:
( mcT )
Fourier’s law in the normal direction of the outward normal n:
Q r kA
Q z kA
Equation 2.1 becomes
(2.3) r
k 2 r z r k 2 r r z mc
Noting that the mass of the control volume is given by:
Equation 2.3 becomes
k r r k r z cr r r
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:
Using the definition of the thermal diffusivity: and expanding the first term using the product rule:
which gives the required outcome:
Example 2.2
An industrial freezer is designed to operate with an internal air temperature of -20 o
C when the external
air temperature is 25 2 C and the internal and external heat transfer coefficients are 12 W/m K and 8 W/m 2 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/m 2 .
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Solution
q U T where U is the overall heat transfer coefficient given by:
0 . 333 W / m K
T 25 ( 20 )
h i k p k i k s h o 0 . 333
1 1 0 . 003 0 . 001 1 L i k i
h 0 . 333 12 1 16 8 i p s o
0 . 333 h k
L i 0 . 195 m
(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.
when the water temperature is 15 o C, the outside air temperature is -10
C, the water side heat
transfer coefficient is 30 kW/m 2 K and the outside heat transfer coefficient is 20 W/m 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 2 r 1 Lh i T i T 1
2 r 1 Lh i
2 Lk T 1 T 2
ln r 2 / r 1
2 Lk / ln( r 2 / r 1 )
Q 2 r 2 Lh o T 2 T o
2 r 2 Lh o
Adding the three equations on the right column which eliminates the wall temperatures gives:
1 ln r 2 / r 1 1
1 ln 0 . 052 / 0 . 05
163 . 3 W / m
Insulated pipe
1 ln r 2 / r 1 ln( r 3 / r 2 ) 1
k ins
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1 ln 0 . 052 / 0 . 05 ln( 0 . 15 / 0 . 052 ) 1
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 80 o
C 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/m 2 K. The outer
surface of the pipe loses heat by convection to air at 20 2 C and the heat transfer coefficient is 200 W/m 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:
1 ln r 2 / r 1 1
Hence substituting into this equation:
1 ln 50 / 47
Q 6 0 . 329 10 W
For the case with insulation, we also use the equation from Example 2.3
1 ln r 2 / r 1 ln( r 3 / r 2 ) 1
k ins
1 ln 50 / 47 ln( 100 / 50 )
Q 3 5 . 39 10 W
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:
2 L T i T o 2 100 80 20
Q 3 5 . 44 10 W
ln( r 3 / r 2 )
ln( 100 / 50 )
k ins
This has less than 1% error compared with the full thermal resistance.
Example 2.5
A diagram of a heat sink to be used in an electronic application is shown below. There are a total of 9 aluminium fins (k = 175 W/m K, C = 900 J/kg K, 3 2700 kg / m ) 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 T b 60 C , when the external
C. Under these conditions, the external heat transfer coefficient h is 12 W/m K. The fin may be assumed to be sufficiently thin so that the heat transfer from the tip can be
air temperature 2 T
f is 20
neglected. The surface temperature T, at a distance, x, from the base of the fin is given by:
T b T f cosh m ( L x )
where m
2 hP
and A c is the cross sectional area. sinh mL
kA c
Determine the total convective heat transfer from the heat sink, the fin effectiveness and the fin efficiency.
Solution
Total heat fluxed is that from the un-finned surface plus the heat flux from the fins.
Q u 0 . 04 0 . 003 9 1 ) 12 60 20 0 . 461 W
For a single fin:
dT Q f kA c
dx x 0
Where A c is the cross sectional area of each fin
Since
T b T f cosh m ( L x )
T T f sinh mL
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Then dT sinh m ( L x )
dx cosh mL
Thus
dT
sinh mL Q f kA c kA c
dx x 0 cosh mL
f kA c m T b T f tanh( mL ) hpkA c T b T f tanh mL
Since
1 2 hP m kA
P 2 ( w t ) 2 ( 0 . 04 0 . 001 ) 0 . 082 m
c w t 0 . 04 0 . 0001 40 10 m
6 11 . 856 m
mL 11 . 856 0 . 06 0 . 7113
tanh mL tanh 0 . 7113 0 . 6115
f 12 0 . 082 175 40 10 60 20 0 . 6115 2 . 03 W / fin
So total heat flow:
Q Q u Q f 0 . 461 9 2 . 03 18 . 7 W
Finn effectiveness
Q f fin
Fin heat trans fer rate
Heat trans fer rate that woul d occur in the absence of the fin hA c T b T f
2 . 03 fin
Fin efficiency:
Actual heat trans fer throug h the fin fin
Heat that would be transferr ed if all the fin area were at the base temperatu re
Q f fin
hA s T b T f
A s wL wL Lt Lt 2 L ( w t )
s 2 0 . 06 ( 0 . 04 0 . 001 ) 4 . 92 10 m
2 . 03 fin
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/m 2 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, o
, for the heat sink to cool from 60
C to 30 C.
hA s T T f T i T f exp
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
kk
Since B i 1 , we can use he “lumped mass” model approximation.
hA s
exp
mC
mC T T f
ln
hA s
Ct T T f 2700 900 0 . 001 30 20
ln
ln
42 seconds
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
20 o C. 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/m 2 K acts over the remaining surfaces.
Estimate the number of fins required to ensure the base temperature does not exceed 120 o C
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Solution
Consider a single fin:
P 2 ( w t ) 2 ( 0 .. 005 0 . 03 ) 0 . 07 m
c w t 0 . 005 0 . 03 150 10 m
hP 15 0 . 07 2
m 1
6 6 . 2361 m
kA c 180 150 10
mL 6 . 2361 0 . 1 0 . 62361
tanh mL 0 . 5536
f hPkA c T b T f tanh( mL )
(From example 2.5)
f 15 0 . 07 180 150 10 120 20 0 . 5536 9 . 32 W
So for 1 kW, the total number of fins required:
N 1000 108
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/m 2 K,
an actual air temperature of 50 o C and the surface temperature at the base of the probe is 60 C.
Solution
The error should be zero when T tip T . The temperature distribution along the length of the probe (from the full fin equation) is given by:
h tip
cosh m ( L x )
sinh m ( L x )
mk
cosh mL
h tip
sinh mL
mk
1 / 2 hP
kA
At the tip, x L , the temperature is given by ( cosh( 0 ) 1 , sinh( 0 ) 0 ):
T tip T 1
cosh mL
h tip
sinh mL
mk
Where is the dimensionless error:,
0 , T tip T
(no error)
T L T b (large error)
For 2 L 20 mm , k 19 W / m K , D 3 mm , h h
tip 50 W / m K
T 50 C ,
T b 60 C
1 / 2 1 / 2 1 / 2 1 / 2 hP
m 1
59 . 235 m
kA
kD
mL 59 . 235 0 . 02 1 . 185
h 50
mk 59 . 235 19
T b T cosh 1 . 185 0 . 0444 sinh 1 . 185
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T tip 0 . 539 T b T T
T tip 0 . 539 60 50 50 55 . 39 C
Hence error 39 5 . C
Example 2.9
A design of an apartment block at a ski resort requires a balcony projecting from each of the 350 separate apartments. The walls of the building are 0.3 m wide and made from a material with k = 1 W/m K. Use the fin approximation to examine the implications on the heat transfer for two separate suggestions for this design. In each case, the balcony projects 2 m from the building and has a length
(parallel to the wall) of 4 m. Assume an inside temperature of 20 o
C and an outside temperature of -
5 o C; the overall (convective + radiative) heat transfer coefficient on the inside of the building is 8
2 W/m 2 K and on that on the outside of the building is 20 W/m 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 2 A
e 0 . 01 m , perimetre P 0 . 6 m (The actual floor of the balcony in this case may be considered to be insulated from the wall
c) No balcony.
Solution
a) For the concrete balcony
Treat the solid balcony as a fin B i
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.
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
kA
This is large enough to justify the use of the fin infinite equation.
b h o Pk b A c T 2 T o
h o Pk b A c T 2 T o
1 / 2 h o Pk b
T 2 T o
Also assuming 1-D conduction through the wall:
Adding equations 1, 2 and 3 and rearranging:
h i k b h o 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.
From (4)
1 / 2 77 . 2 W / m
Compared with no balcony:
52 . 6 W / m
h i k w h o 8 1 20
The difference for one balcony is A c ( 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 2 A
c 0 . 01 m and P 0 . 6 m
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As before, however, in this case Bi << 1 because k s k b
1 / 2 1 / 2 hP 20 6
mW w
kA
mW 2 , so we can use the infinite fin approximation as before
1 / 2 182 W / m
h i k s h o Pk s
Q b A c q b 0 . 01 182 1 . 82 W / beam
For 350 apartments, Q b 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
T s T f Where G and n are constants, show that the variation of the dimensionless
temperature ratio with time will be given by
n 1 nh init t
Where
Area ,
T init T f Mass Specific Heat Capacity
and h init = the heat transfer coefficient at t = 0. Use this expression to determine the time taken for an
3 aluminium motorcycle fin ( 2 2750 kg / m , C 870 J / kgK ) of effective area 0.04 m and
thickness 2mm to cool from 120 o C to 40 C in surrounding air at 20
C when the initial external heat
transfer coefficient due to laminar free convection is 16 W/m 2 K. Compare this with the time estimated
h from the equation ( t e ) which assumes a constant value of heat transfer coefficient.
Solution
Low Biot number approximation for free convection for B i 1
Heat transfer by convection = rate of change of internal energy
dt
We know that n h G ( T
Where G is a constant.
(Note that this relation arises from the usual Nusselt/Grashof relationship in free convection; for
1 / 3 1 / example: 4 Nu 0 . 1 Gr Pr in turbulent flow or Nu 0 . 54 Gr Pr for laminar flow)
Equation 1 then becomes:
mC d ( T s T f )
A dt
t GA d ( T
dt
t 0 mC
GnAt
T s T f T s T f
t 0 (2)
mC
T s T f T s , i T f
At t 0 ,
If we divide equation 2 by T s , i T f
And use the definition
We obtain
GnAt
GnAt
mC T s , i T f
mC
Since G T s , i T f h i , the heat transfer coefficient at time t = 0, then
h i At
1 mC
Or n nh
For aluminium 3 2750 kg / m ,
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 4 2
2 . 1 10 m K / J
mC 0 . 22 870 n nh
i t 1 which gives
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When
T 40 C
Then
4 590 s
h For the equation t e
which assumes that the heat transfer coefficient is independent of surface-to-fluid temperature difference.
ln
ln 0 . 2
4 479 s
h 16 2 . 1 10
590 479 Percentage error =
Example 2.11
A 1 mm diameter spherical thermocouple bead (C = 400 J/kg K, � � 7800 �� � � ) is required to respond to 99.5% change of the surrounding air
� � 1.2 �� � �� � � � 1.8 � 10 ��� � � , � � 0.0262 � � � ⁄ and Pr = 0.77) temperature in 10 ms. What is the minimum air speed at which this
will occur?
Solution
Spherical bead: ���� � �� �
Assume this behaves as a lumped mass, then
For lumped mass on cooling from temperature T i
Which gives the required value of heat transfer coefficient
So
For a sphere
From which with Pr = 0.707
Using Newton iteration
Starting with Re D = 300
Which is close enough to 300
From which
3. Convection
Example 3.1
Calculate the Prandtl number (Pr = C p /k) for the following
a) Water at 20 C:
3 = 1.002 x 10 kg/m s, C
p = 4.183 kJ/kg K and k = 0.603 W/m K
b) 2 Water at 90 C: 3 = 965 kg/m , = 3.22 x 10 m /s, C
p = 4208 J/kg K and k = 0.676 W/m K
c) 2 Air at 20 C and 1 bar: R = 287 J/kg K, = 1.563 x 10 5 m /s, C p = 1005 J/kg K and k = 0.02624 W/m K
6 3 1 2 . 46 10 T
d) Air at 100 C:
110 T
kg/m s
C p 0 . 917 2 . 58 10 T 3 . 98 10 T kJ / kg K (Where T is the absolute temperature in
K) and k = 0.03186 W/m K.
e) Mercury at 20 C: = 1520 x 10 6 kg/m s, C p = 0.139 kJ/kg K and k = 0.0081 kW/m K
f) Liquid Sodium at 400 K: = 420 x 10 6 kg/m s, C p = 1369 J/kg K and k = 86 W/m K
g) 2 Engine Oil at 60 C: = 8.36 x 10 kg/m s, C p = 2035 J/kg K and k = 0.141 W/m K
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Solution
C p 3 1 . 002 10 4183
a) Pr
b) Pr
C p
b) Pr
c) Pr
C
c) p Pr k
Pr P
3 1 . 19 kg / m
RT 287 293
RT 5 1 . 19
Pr
d) Pr
1 . 46 10 T
d)
5 2 . 18 10 kg / m s
110 T
p 0 . 917
2 . 58 10 T 3 . 98 10 T 0 . 917 2 . 58 10 373 3 . 98 10 373
1007 . 7 J / kg K
Pr
e)
Pr
e) Pr
Pr
C p 6 420 10 1369
f)
Pr
g) Pr
C p 2 8 . 36 10 2035
g)
Pr
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 c p 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) 3 A 10 m (water line length) long yacht sailing at 13 km/h in seawater = 1000 kg/m and 3 = 1.3 x 10 kg/m s,
b)
A compressor disc of radius 0.3 m rotating at 15000 rev/min in air at 5 bar and 400 C and
1 . 46 10 T
kg/m s 110 T
c)
0.05 kg/s of carbon dioxide gas at 400 K flowing in a 20 mm diameter pipe. For the viscosity
kg/m s 233 T
6 3 2 1 . 56 10 T
take
d) The roof of a coach 6 m long, travelling at 100 km/hr in air (
3 5 = 1.2 kg/m and = 1.8 x 10 kg/m s)
e) The flow of exhaust gas (p = 1.1 bar, T = 500ºC, R = 287 J/kg K and = 3.56 x 10 5 kg/m s) over
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/m 3 and an exhaust
port diameter of 25 mm)
Solution
uL
a) Re
3 2 . 78 10 (turbulent)
b) T 400 273 673 K
5 3 . 26 10 kg / m s
2 1571 rad / s
u r 1571 0 . 3 471 . 3 m / s
P 100000
3 2 . 59 kg / m
RT 287 673
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Characteristic length is r not D
uD 2 . 59 471 . 3 3 7
Re 5 1 . 12 10 (turbulent)
c) m uA u
4 m u 2 D
uD 4 m D 4 m
Re
6 3 1 2 . 56 10 400 5 1 . 97 10 kg / m s
Re 5 1 . 6 10 (turbulent)
d) u
27 . 8 m / s
uL 1 . 2 27 . 8 6 7
Re 5 11 . 1 10 (turbulent)
e) Let m be the mass flow through the exhaust port
m = inlet density X volume of air used in each cylinder per second
0 . 012 kg / s
ud Re d
Re
(laminar)
Comments:
Note the use of D to obtain the mass flow rate from continuity, but the use of d for the characteristic length Note the different criteria for transition from laminar flow (e.g. for a pipe Re 2300 plate
Re 5 3 10 )
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 18C ( =
3 1.2 kg/m 5 , Pr = 0.72 and = 1.8 x 10 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 3 C ( = 854 kg/m , Pr = 546, = 0.7 x 10 3 K 1 and = 3.56 x 10 2 kg/m s)
c) The external surface of a heating coil, 30 mm diameter, having a surface temperature of 80 C in
water at 20 -3 C ( = 1000 kg/m , Pr = 6.95, = 0.227 x 10 K and = 1.00 x 10 kg/m s)
3 d) Air at 20ºC ( 5 = 1.2 kg/m , Pr = 0.72 and = 1.8 x 10 kg/m s) adjacent to a 60 mm diameter
vertical, light bulb with a surface temperature of 90 C
Solution
a) Gr
T 75 18 57 K
T 18 273 291
Gr 9
9 Gr 9 Pr 1 . 84 10 0 . 72 1 . 3 10 (mostly laminar)
b) L
Area
Perimeter 2 0 . 4 0 . 2
0 . 0667 m
T 75 40 35 K
2 3 2 3 3 g T L 854 9 . 81 0 . 7 10 35 0 . 0667
Gr 4
4 Gr 7 Pr 4 . 1 10 546 2 . 24 10
Heated surface facing downward results in stable laminar flow for all Gr Pr
c)
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T 80 20 60 K
2 3 2 3 3 g T L 1000 9 . 81 0 . 227 10 60 0 . 03
Gr 6
6 Gr 6 Pr 3 . 6 10 6 . 95 25 10 (laminar)
Area 2 D D
d) L
Perimeter 4 D 4
T 90 20 70 K
T 20 273 293
2 3 2 3 g T L 1 . 2 9 . 8 70 0 . 015
Gr 4
4 Gr 4 Pr 3 . 5 10 0 . 72 2 . 5 10 (laminar)
Comments:
Note evaluation of for a gas is given by 1 / T For a horizontal surface L A / p
Example 3.4
Calculate the Nusselt numbers for the following:
a) A flow of gas (Pr = 0.71, = 4.63 x 10 5 kg/m s and C p = 1175 J/kg K) over a turbine blade of chord length 20 mm, where the average heat transfer coefficient is 1000 W/m 2 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 4 = 56.7 x 10 W/m K )
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 12C
Solution
a) Pr k
0 . 0766 W / m K
Pr
h L 1000 0 . 02
Nu
b) Nu k
q 2 2000 W / m
A 0 . 01 0 . 005
T 35 20 15 C
Area
mm 0 . 001667 m
Perimeter 30 3
h L 2000 0 . 001667
Nu
c) Nu T k
In this case, q must be the convective heat flux – radiative heat flux
T s 80 273 353 K
T 20 273 293 K
s T 56 . 7 10
1 . 5 0 . 6 353 353 293 293 416 416 W W
T 80 20 60 K
Q c Q Q R 1000 416 584 W
Q c 584 q 2
649 W / m
Nu
q c L 649 0 . 6
T k 60 0 . 026
d) T 12 4 8 K
60 C
(assuming 1-D conduction)
q 2 12 W / m
Nu
q c L 12 3
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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Example 3.5
In forced convection for flow over a flat plate, the local Nusselt number can be represented by the general expression n Nu
x C 1 Re x . In free convection from a vertical surface the local Nusselt number is represented by m Nu
x C 2 Gr x , where C 1 ,C 2 , 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 (T m
s T )
b) In turbulent free convection, it is generally recognised that m = 1/3. Show that the local heat transfer coefficient does not vary with coordinate x.
Solution
a) Nu x k
x u Re x
For forced convection: n Nu
x C 1 Re x
Hence
h C 1
This shows that the heat transfer coefficient for forced does not depend on temperature difference.
For free convection m Nu
x C 2 Gr x
2 3 g T x Gr x
Hence
h C 2
So for free convection, heat transfer coefficient depends on m T
b) From (1), with m = 1/3 for turbulent free convection:
h C 2
x
h kC 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 5 2 air with 3 T
20 C , C p 1 . 005 kJ / kg K , 1 . 522 10 m / s 1 . 19 kg / m and Pr 0 . 72 . The surface temperature, T of the foil is to be measured at the trailing edge – but can be assumed to
be constant. Estimate the wind speed when T 32 C and Q 0 . 5 W .
Solution
Firstly, we need to estimate if the flow
is laminar
or turbulent.
Assuming a critical (transition) Reynolds number of 5 Re 3 10 the velocity required would be:
u turb
3 304 m / s
Wind speed is very unlikely to reach this critical velocity, so the flow can be assumed to be laminar.
1 / 2 1 / Nu 3 x 0 . 331 Re x Pr
Nu av 0 . 662 Re L Pr
1 / 2 1 / 3 q av L
1 / 2 q av L Re L
T s T k 0 . 662 Pr
q av
1250 W / m
Re L
Re 4 L 3 10
4 Re 5
L 3 10 1 . 522 10
3 18 . 3 m / s
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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 -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
3 numbers to estimate the average heat transfer coefficients. For air take: -5 ρ= 1.2 kg / m , μ = 1.8 x 10 kg / m s, C p = 1 kJ / kg K and Pr = 0.7.
9 Free convection in air, laminar (Gr 1/4
x < 10 ): Nu x = 0.3 Gr x
9 Free convection in air, turbulent (Gr 1/3
x > 10 ): Nu x = 0.09 Gr x
5 0.5 Forced convection, laminar (Re 1/3
x < 10 ): Nu x = 0.33 Re x Pr
5 0.8 Forced convection, turbulent (Re 1/3
x > 10 ): Nu x = 0.029 Re x Pr
Solution
C p 1 . 8 10 1000
Pr
k gives:
0 . 026 W / m K
Pr
First we need to determine if these flows are laminar or turbulent.
For the inside (Free convection):
T 20 273 293
2 3 2 3 g T L 1 . 2 9 . 81 T 7
Gr
Gr 10 5 . 1 10 T
(Flow will be turbulent over most of the surface for all reasonable values of T )
For the outside (Forced convection)
u Re
L 1 . 2 15 30 7
L 5 3 10
(Flow will be turbulent for most of the surface apart from the first 0.3 m)
Hence we use the following correlations:
1 / On the inside surface: 3 Nu
x 0 . 09 Gr
0 . 8 1 / On the outside surface: 3 Nu
x 0 . 029 Re x Pr
For the inside:
Nu x
k
x
h constant
constant
Hence heat transfer coefficient is not a function of x
h av h x L
For the outside:
0 . h 8 x x u
1 / Nu 3 x 0 . 029 k
Pr
0 . h 2 constant C x
x dx
Write a heat balance:
Assuming one-dimensional heat flow and neglecting the thermal resistance of the glass
From equation 1
k 2 T
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1 . 2 9 . 81 T 1 T s h i 0 . 09
i 1 . 24 T i T s
From equation 2:
0 . h 8 o W 0 . 029 W u
1 / 3 Pr
h 2 o 26 . 7 W / m K
From (3) with (4) and (5)
T i T s 26 . 7 T s T o
20 T s 26 . 7 T s 15
s 0 . 0464 20 T s 15 (6)
To solve this equation for T s an iterative approach can be used
First guess: T s 10 C
Substitute this on the right hand side of equation 6:
s 0 . 0464 20 10 15 10 . 7 C
For the second iteration we use the result of the first iteration:
s 0 . 0464 20 10 . 7 15 10 . 6 C
The difference between the last two iterations is 0 . 1 C , so we can consider this converged.
T s 10 . 6 C
From which:
o T s T o 26 . 7 10 . 6 15 117 W / m
Q qA 117 30 7 24600 W 24 . 6 kW
Example 3.8
The figure below shows part of a heat exchanger tube. Hot water flows through the 20 mm diameter tube and is cooled by fins which are positioned with their longest side vertical. The fins exchange heat by convection to the surrounds that are at 27 C.
Estimate the convective heat loss per fin for the following conditions. You may ignore the contribution and effect of the cut-out for the tube on the flow and heat transfer.
a) natural convection, with an average fin surface temperature of 47 C;
b) forced convection with an air flow of 15 m / s blowing parallel to the shortest side of the fin and
with an average fin surface temperature of 37 C.
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The following correlations may be used without proof, although you must give reasons in support of your choice in the answer.
Nu 5 x = 0.02 Re x Pr
For air at these conditions, take: Pr = 0.7, k = 0.02 W / m K, 3 μ = 1.8 x 10 kg /m s and ρ = 1.0 kg / m
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 3 g T L Gr
T 47 27 20 K
Gr 6
(L here is height because it is in the direction of the free convection boundary layer)
So we use:
1 / Nu 4
x 0 . 5 Gr x Pr
av
h dx constant x dx
h av
Nu av Gr L Pr
Nu av 2 10 0 . 7 23
Nu av k 23 0 . 02 2
h av
q av h av T
Q q av A h av 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
L u 1 15 0 . 05 4
Re 5 4 . 17 10 (Laminar)
(L here is the width because flow is along that direction)
1 / 2 1 / Nu 3 x 0 . 3 Re x Pr
hdx
av
Nu av 0 . 6 Re L Pr 0 . 6 4 . 17 10 0 . 7 109
Nu av k 109 0 . 02 2
h av
43 . 5 W / m K
Q q av A h av TA 43 . 5 10 0 . 1 0 . 05 2 T 10 C
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:
x = 0.04 Re x Pr (Re x ≥ 10 )
Show that for a plate of length, L, the average Nusselt number is:
0.8 Nu 1/3 av = (0.05 Re L - 310) Pr
Solution
h av k Nu av L
Where for a constant surface-to-fluid temperature:
h av h laminar dx h turbulent dx
0 x L
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Since for laminar flow ( 5 Re
x 10 ):
1 / 2 1 / Nu 3 x 0 . 3 Re x Pr
h lam 0 . 3 x
x Pr
1 / 3 1 / 2 1 / h 2 lam 0 . 3 k
Pr x
C lam x
Where C lam does not depend on x
Similarly:
0 . h 2 turb C turb x
Where
0 . 8 u
1 / C 3 turb 0 . 04 k
Pr
Hence
C x lam dx turb dx
h av C x
L 0 x L
h av C lam
C turb
1 / 2 0 0 . 8 x L
h av k Nu av L
Nu av
C lam 1 / 2 C turb 0 . 8 0 . 8
Nu av 0 . 6
x L Pr 0 . 05
Pr
But 5 10 (The transition Reynolds number)
So
av Pr 0 . 6 10 0 . 05 Re L 0 . 05 10
1 / 3 5 1 / 2 0 . 8 5 0 . Nu 8
av 0 . 05 Re L 310 Pr
0 . 8 1 / Nu 3
Example 3.10
A printed circuit board dissipates 100 W from one side over an area 0.3m by 0.2m. A fan is used to cool this board with a flow speed of 12 m / s parallel to the longest dimension of the board. Using the average Nusselt number relationship given in Example 3.9 to this question, calculate the surface temperature of the board for an air temperature of 30 ºC.
Take an ambient pressure of 1 bar, R = 287 J / kg K,
Cp = 1 kJ / kg K, k = 0.03 W / m K and -5 μ = 2 x 10 kg/m s
Solution
Q 100 q 2
av
1666 . 7 W / m
Pr
u L Re L
P 5 10
3 1 . 15 kg / m
RT 287 303
Re L
Using the formula for Nusselt Number obtained in Example 3.9:
av 0 . 05 Re L 310 Pr
0 . 8 1 / Nu 3
av 0 . 05 2 . 07 10 310 0 . 667 511
5 0 . 8 1 / Nu 3
h av k q av L Nu av L
Tk
q av L 1666 . 7 0 . 3
32 . 6 C
Nu av k
T s 30 32 . 6 62 . 6 C
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 T 1 and T 2 respectively and their areas are A 1 and A 2 .
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:
A 1 F 13 T 1 A 2 F 23 T 2
T 3
A 2 F 23 A 1 F 13
where F 13 and F 23 are the appropriate view factors.
b) Show that the total radiative heat transfer to the tubes, Q 2 , is given by:
A 1 F 13 A 2 F 23
2 A 1 F 12
A 2 F 23 A 1 F 13
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1 =A 2 = 12m and the view factors are each 0.5?
a) Q 2 Q 1 2 Q 3 2 (1)
Since the walls are adiabatic
Q 3 2 Q 1 3
From (2)
3 F 32 T 3 T 2 A 1 F 13 T 1 T 3
4 A 1 F 13 T 1 A 3 F 32 T 2
A 3 F 32 A 1 F 13
1 F 13 T 1 A 2 F 23 T 2
A 2 F 23 A 1 F 13
since A i F ij A j F ji
b) From (1)