Chapter (2/2)
Heat and Mass Transfer lectures at BUE
Dr. Ahmed Abdel-Azim Ahmed
By the end of today’s lecture, you should be able to:
Learn how to deal with combined modes and composite walls for the
three applications, plan wall, cylinder and sphere.
You will be familiar with the concept of contact resistances and critical
radios.
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4
Ts ,1 − Ts , 2 Ts ,1 − Ts , 2
qx =
=
Rth
L / kA
Ts ,1
Ts,2
qx
x=L
x=0
Ts,2
Ts,1
R=L/k
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You must consider the following:
T,1
– Convection from hot fluid to wall
– Conduction through wall
T,2 , h2
Ts ,1
Cold fluid
– Convection from wall to cold fluid
Ts,2
T ,1 − T , 2
qx =
Rtot
For a set of resistors in series,
the total thermal resistance is:
1
L
1
Rtot =
+ +
h1 A kA h2 A
qx
T,1 , h1
T , 2
Hot fluid x=0
T,1
x=L
R=1/h1
T,2
Ts,2
Ts,1
R=L/k
R=1/h2
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Cold fluid
T,1
T,2 , h2
Ts ,1
T ,1 − T , 2
qx =
1
L
1
+
+
h1 A kA h2 A
Ts,2
qx
T,1 , h1
T , 2
Hot fluid x=0
T,1
x=L
R=1/h1
T,2
Ts,2
Ts,1
R=L/k
R=1/h2
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Cold fluid
T,1 − Ts ,1
qx =
1 / h1 A
Ts ,1 − Ts , 2
=
L / kA
Ts , 2 − T, 2
=
1 / h2 A
T,1
(solve for Ts,1)
T,2 , h2
Ts ,1
Ts,2
(solve for Ts,2)
qx
T,1 , h1
T , 2
Hot fluid x=0
T,1
x=L
R=1/h1
T,2
Ts,2
Ts,1
R=L/k
R=1/h2
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Layers of different materials may comprise the wall of the object. These layers
may involve any number of series or parallel thermal resistances. We can use
“equivalent thermal circuits” to solve these problems.
Typically we use this method to determine
the overall heat transfer coefficient, U.
Consider the following example:
Express the following geometry in terms
of an equivalent thermal circuit ?
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For engineers, problems like these are solved on a day-by-day basis.
For convenience of daily use, we simplify the problem into a overall
heat transfer equation.
Rtot =
LC
1 1
LA
LB
1 
(
)
+
(
)
+
(
)
+
(
)
+
(
)

A  h1
kA
kB
kC
h4 
T ,1 − T , 4
qx =
Rtot
The problem given above is an example of
a simple circuit. More complex circuits can
arise when drawing up equivalent thermal
circuits for a series-parallel composite wall.
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12
Rtot = RE + Req + RH
➢ For resistances in series:
➢ For resistances in parallel:
Rtot=R1+R2+…+Rn
1/Rtot=1/R1+1/R2+…+1/Rn
−1




LE 
1
1
LH

Rtot =
+
+
+

LG
k E A  LF
kH A
 k ( A / 2) k ( A / 2) 
G
 F

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15
(a) Ideal (perfect)
thermal contact
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(b) Actual (imperfect)
thermal contact
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Heat transfer through the interface is the sum of the heat transfers
through the solid contact spots and the gaps in the noncontact
areas and can be expressed as:
Q = Qcontact + Qgap
Q = hc A Tinterface
Where:
A; is the apparent interface area (which is the same as the
cross-sectional area of the rods)
Tinterface ; is the effective temperature difference at the
interface. hc; is called the thermal contact conductance which
corresponds to the convection heat transfer coefficient.
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Heat and Mass Transfer lectures at BUE
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the thermal contact conductance is related to thermal contact resistance by:
1
Rc =
hc
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Heat and Mass Transfer lectures at BUE
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Consider a hollow
cylinder, whose inner
and outer surfaces are
exposed to fluids at
different temperatures
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Based on the previous solution, the heat transfer rate can be calculated In
terms of equivalent thermal circuit :
T,1 − T, 2
qr =
Rtot
1
ln( r2 / r1 )
1
Rtot =
+
+
h1 (2r1L)
2kL
h2 (2r2 L)
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Express the following
geometry in terms of a an
equivalent thermal circuit?
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Thermal insulations consist of low thermal conductivity materials combined
to achieve an even lower system thermal conductivity.
An exterior house wall
containing (a) an air space
and (b) insulation.
Trapped air makes a good insulator
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Heat and Mass Transfer lectures at BUE
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Thermal insulations consist of low thermal conductivity materials combined
to achieve an even lower system thermal conductivity.
For small diameter tubes it is sometimes possible to increase heat loss by
adding insulation on the outer surface!
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Heat and Mass Transfer lectures at BUE
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Heat loss from a pipe:
q = hA(Ts − T )
➢ If A, is increased, q will increase. When insulation is added to a pipe, the
outside surface area of the pipe will increase. This would indicate an
increased rate of heat transfer
➢ The insulation material has a low thermal conductivity,
· it reduces the conductive heat transfer
· lowers the temperature difference between the outer surface
temperature of the insulation and the surrounding bulk fluid
temperature.
• This contradiction indicates that there must be a critical thickness of
insulation.
· The thickness of insulation must be greater than the critical thickness, so
that the rate of heat loss is reduced as desired.
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Heat and Mass Transfer lectures at BUE
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For small diameter tubes covered by an insulation, the heat transfer
through the insulation to outside is obtained by:
Ti − T
qr =
ln(ro / ri )
1
+
2kL
ho (2ro L)
The numerator is a given CONSTANT in the heat flow
equation above. To maximize ‘q’, we must minimize the
denominator by changing ro:
d  ln(ro / ri )
1 
+

=0
dro  k
ho (ro ) 
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Heat and Mass Transfer lectures at BUE
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The numerator is a given CONSTANT in the heat flow equation above. To
maximize ‘q’, we must minimize the denominator by changing ro:
d  ln(ro / ri )
1 
+

=0
dro  k
ho (ro ) 
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Heat and Mass Transfer lectures at BUE
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Minimum resistance and maximum heat flow
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Heat and Mass Transfer lectures at BUE
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Summary
 We obtained temperature distributions, thermal resistances, heat flow,
for problems involving steady-state, one-dimensional conduction in
Cartesian, cylindrical and spherical coordinates, without energy
generation and for different applications
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