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# CONDUCTION LESSON 3 ```MODULE II
HEAT TRANSFER BY CONDUCTION
Lesson 1:
Conduction Rate Equation
Lesson 2:
Systems
Lesson 3:
Conduction with Film Coefficient of
Convection
THELMA T. OBILLO, PME
FACULTY, MECHANICAL ENGINEERING
1ST SEMESTER SY 2021-2022
Module II
28
Lesson 3

Conduction with Convection Film Coefficient
When a moving fluid comes into contact with a stationary surface, a thin
boundary layer develops adjacent to the wall and in this layer, there is no relative
velocity with respect to surface. In a heat exchange process, this layer is called
stagnant film and heat flow in the layer is covered both by conduction and
convection processes. Since thermal conductivity of fluids is low, the heat flow from
the moving fluid of the wall is mainly due to convection.
The rate of convective heat transfer between a solid boundary and adjacent
fluid is given by the Newton-Rikhman law, q=hA(ts-tf), where tf is the temperature
of the hot moving fluid, ts is the temperature of the wall surface, and A is the area
exposed to heat transfer. The convective coefficient (film coefficient) h depends
upon the thermodynamic and transport properties (e.g., density, viscosity, specific
heat and thermal conductivity) of the fluid flow and the prevailing thermal
conditions. The resistance of the stagnant boundary layer is included in the
convection coefficient. The heat transfer through
A.
Plane Wall (Film Considered)
Equivalent electrical analogy
Figure 2-3
Based on the figure above, a plane wall with fluids A and B on both sides are
considered where the fluid A is hotter and fluid B is colder. The temperature profile
and equivalent thermal network is also shown. As discussed in the previous lesson, a
convection film coefficient, h, measures how effectively a fluid transfers heat by
convection. It is measured in W/m2-K, and is determined by factors such as the fluid
density, viscosity, and velocity. Thus, for the given illustration above, a combination
of conduction and convection are considered: conduction heat transfer across the
plane wall and convection heat transfer for fluids A and B. Therefore, the
corresponding resistance and total heat transfer will be:
Module II Heat Transfer by Conduction
THELMA T. OBILLO, PME
Faculty, Mechanical Engineering
29
qwall = qFluid A = qFluid B
qFluid A = h1 A (TA − T1 )
q = h1 A (TA − T1 ) =
q=
qwall =
kA
(T1 − T2 )
x
qFluid B = h2 A (T2 − TB )
kA
(T1 − T2 ) =h2 A (T2 − TB )
x
Overall Temperature Drop
Total Resistance
q=
TA − TB
1 x 1
+
+
h1 A kA h2 A
B. Composite Walls with Convection Film Coefficient
Considering Figure 2-4 with fluids A and B on both sides of the composite
(series-parallel) walls and hA and hB are the convection coefficients of the
two fluids, respectively, the following are the corresponding equations for
the resistances and total heat transfer across the composite wall:
Figure 2-4
thermal heat transfer:
q=
Overall Temperature Drop
T
=
Total Resistance
 Rthoverall
Module II Heat Transfer by Conduction
THELMA T. OBILLO, PME
Faculty, Mechanical Engineering
30
thermal resistances:
For the composite wall :
x A
xB
RA =
RB =
k A AA
k B AB
RE =
xE
k E AE
RF =
xF
k F AF
RC =
xC
kC AC
RG =
xG
kG AG
RD =
xD
For the two fluids :
1
1
RFluidA =
RFluidB =
hA A
hB A
The total thermal resistances will be:
 Rth
overall
= RFluidA + R12 + R23 + R34 + R35 + RFluidB
Where:
R12 = thermal resistance across 1 to 2 or across material A
R23 = thermal resistance across 2 to 3 or across materials B, C and D
R34 = thermal resistance across 3 to 4 or across material E
R45 = thermal resistance across 4 to 5 or across materials F and G
RFluidA = thermal resistance of Fluid A (hot fluid)
RFluidB = thermal resistance of Fluid B (cold fluid)
NOTE:
Insulation and R-Value
In classifying the performance of insulation, it is a common practice in the
building industry to use a term called the R-value, which is defined as
R=
T
q
A
---------Equation 2.6
The units for R are ◦C&middot;m2/W or ◦F&middot;ft2&middot;h/Btu. Note that this differs from the thermalresistance concept discussed above in that a heat flow per unit area is used. At this
point it is worthwhile to classify insulation materials in terms of their application
and allowable temperature ranges. Table 2-1 (Heat Transfer, J.P. Holman 10th Ed.)
furnishes such information and may be used as a guide for the selection of insulating
materials.
Module II Heat Transfer by Conduction
THELMA T. OBILLO, PME
Faculty, Mechanical Engineering
31
The Overall Heat Transfer Coefficient, U
Equivalent electrical analogy
Figure 2-3
q=
Overall Temperature Drop
Total Resistance
q=
TA − TB
1 x 1
+
+
h1 A kA h2 A
Observe that the value 1/hA is used to represent the convection resistance. The
overall heat transfer by combined conduction and convection is frequently expressed
in terms of an overall heat-transfer coefficient U, defined by the relation
q = UAToverall
---------Equation 2.7
where A is some suitable area for the heat flow.
The overall heat transfer coefficient represents the intensity of heat transfer
from one fluid to another through a wall separating them. It is numerically equal to
the quantity of heat passing through unit area of wall surface in a unit time at a
temperature difference of unit degree.
Referring to the figure (Figure 2-3), the overall heat-transfer coefficient
would be
U=
1
1 x 1
+
+
h1 k h2
 1 x 1 
U = +
+ 
 h1 k h2 
Module II Heat Transfer by Conduction
−1
THELMA T. OBILLO, PME
Faculty, Mechanical Engineering
32
The overall heat-transfer coefficient is also related to the R value
U=
1
RValue
=
U = ( RValue )
1
RTotal
−1
---------Equation 2.8
Heat Transfer by Conduction through Walls (with convection film)
Solved Problems:
Problem 10:
A composite wall is made up of an external thickness of brickwork 110 mm
thick inside which is a layer of fiberglass 75 mm thick. The fiberglass is faced
internally by an insulating board 25 mm thick. The coefficient of thermal
conductivities are as follows:
Brickwork = 1.5 W/m-K
Fiberglass = 0.04 W/m-K
Insulating board = 0.06 W/m-K
The surface transfer coefficient of the inside wall is 3.1 W/m 2-K while that
of the outside wall is 2.5 W/m2-K. Determine the (a) overall coefficient of
heat transfer and using this coefficient, (b) find the heat lost through such
wall 6 m high and 10 m long. Take the internal ambient temperature as 10 oC
and external temperature as 27oC. (c) Find also the interface temperature
between the brickwork and the fiberglass.
Given:
Module II Heat Transfer by Conduction
THELMA T. OBILLO, PME
Faculty, Mechanical Engineering
33
Required:
a. Overall heat transfer coefficient, U
b. Heat lost, Q
c. Interface temperature between brickwork and fiberglass
Solution:
a.
Overall heat transfer coefficient, U
U=
1
x
x
x
1
1
+ 1 + 2 + 3 +
ho k BW k FG k IB hi
1
1
0.11m
0.075m
0.025m
1
+
+
+
+
2
2.5W / m − K 1.5W / m − K 0.04W / m − K 0.06W / m − K 3.1W / m 2 − K
U = 0.3239W / m2 − K
U=
b.
Heat lost, Q
Q = UAT
Q = ( 0.3239W / m 2 − K ) (10m  6m )( 27 − 10 ) C o
Q = 330.38W
c.
Interface temperature between brickwork and fiberglass
Q1− 2 =
A ( t1 − t2 )
x1
k BW
------equation 1
Solving for t1 ,
Qo −1 =
A ( to − t1 )
1
ho
330.38W =
(10m  6m ) ( 27o C − t1 )
1
2.5W / m 2 − K
Substituting t1 in equation 1:
330.38W =
(10m  6m ) ( 24.8o C − t2 )
0.11m
1.5W / m − K
t2 = 24.4o C
Module II Heat Transfer by Conduction
THELMA T. OBILLO, PME
Faculty, Mechanical Engineering
34
Problem 11:
Hot gases at 980oC flow past the upper surface of the blade of a gas turbine
and the lower surface is cooled by air bled off the compressor. The
convective heat transfer coefficient at the upper and lower surfaces are
estimated to be 2830 and 1415 W/m 2-oC respectively. The blade material has
a thermal conductivity of 11.6 W/m-oC. If metallurgical considerations limit
the blade temperature at 870oC, workout the temperature of the cooling air.
Consider the blade as a flat plate 0.115 cm thick and presume that steady
state conditions have been reached.
Given:
Required:
Temperature of the coolant air
Solution:
Heat transfer per unit area through the coolant
q = hc A ( tl − tc )
q
= q &quot; = hc ( tl − tc )
A
Heat flow rate per unit area, from the hot gases to the upper surface of the blade is
q = hu A ( t g − tu )
(
)
q
= 2830W 2 o ( 980 − 870 ) oC
m − C
A
q
= q &quot; = 311,300W 2
m
A
Module II Heat Transfer by Conduction
THELMA T. OBILLO, PME
Faculty, Mechanical Engineering
35
Since steady state conditions prevail, this heat would be conducted across the blade.
Then from Fourier’s law of heat conduction,
kA ( tu − tl )
x
k ( tu − tl )
q
= q&quot; =
A
x
q=
11.6W
(870 C − t )
(
m − C)
=
o
311,300W
o
m
2
l
0.00115m
tl = 839.138 C
o
The heat conducted across the blade would finally be transferred to the coolant by
convection, then
q = hc A ( tl − tc )
q
= q &quot; = hc ( tl − tc )
A
311,300W 2 = 1415W 2 o (839.138 oC − tc )
m
m − C
tc = 619.138 oC
(
Module II Heat Transfer by Conduction
)
THELMA T. OBILLO, PME
Faculty, Mechanical Engineering
36
Conduction with Film Coefficient of Convection
Cylinders:
➢ The figure below shows conduction of heat through a composite cylindrical
wall having three layers of different materials and with inside and outside
convection films considered.
➢ There is a perfect contact between the layers and so an equal interface
temperature for any two neighboring layers.
➢ For example, steam pipe used for conveying high pressure steam in a steam
power plant may have cylindrical metal wall, a layer of insulation material
and then a layer of protecting plaster.
➢ The arrangement is called lagging of the pipe system.
Figure 3.9
Module II Heat Transfer by Conduction
THELMA T. OBILLO, PME
Faculty, Mechanical Engineering
37
➢ If the internal and external heat transfer coefficients for the composite
cylinder as shown in Figure 3.9 are ho and ho, respectively, then for a steady
state conduction, the heat flow through each layer is same and it can be
described by the following set of equations:
qi1 = hi Ai ( ti − t1 )
q23 =
q=
Where:
t2 − t3
D 
ln  3 
 D2 
2 k2 L
q4 o = ho Ao ( t4 − to )
q34 =
q12 =
t1 − t2
D 
ln  2 
 D1 
2 k1 L
t3 − t4
D 
ln  4 
 D3 
2 k3 L
2 L ( ti − to )
r
r
r
ln  2  ln  3  ln  4 
r
r
r
1
1
+  1+  2+  3+
hi Ai
k1
k2
k3
ho Ao
Ai = 2 ri L =  Di L
Ao = 2 ro L =  Do L
ri = r1
Di = D1
ti  t1
ro = r4
Do = D4
to  t 4
OVERALL HEAT TRANSFER COEFFICIENT, U
➢ The heat transfer equation can be written as
q = UAt
➢ Since the flow area varies for a cylindrical pipe, it becomes necessary to
specify the area on which U is based.
➢ Thus, depending upon whether the inner or outer area is specified, two
different values are defined for U.
qo = U o Ao t
qi = U i Ai t
Uo= Overall heat transfer coefficient based on the outside surface area
Ui= Overall heat transfer coefficient based on the inside surface area
Module II Heat Transfer by Conduction
THELMA T. OBILLO, PME
Faculty, Mechanical Engineering
38
Derivation of Uo:
Equating equations of heat transfer,
t
 Rt
t
U o Ao t =
=
 Rt
U o Ao t =
U o ( D4 L ) =
t
D
D
D
ln 4
ln 2
ln 3
D3
D1
D2
1
1
+
+
+
+
hi Ai 2 k1 L 2 k2 L 2 k3 L ho Ao
1
D
D
D2
ln 4
ln
ln 3
D3
D1
D2
1
1
+
+
+
+
hi ( D1 L ) 2 k1 L 2 k2 L 2 k3 L ho ( D4 L )
divide both sides by  D4 L :
Uo =
1
D 
D 
D 
D4 ln  2  D4 ln  3  D4 ln  4 
D4
 D1  +
 D2  +
 D3  + 1
+
hi D1
2k1
2k 2
2 k3
ho


D 
D 
D 
D4 ln  2  D4 ln  3  D4 ln  4 


D3  1 
D1 
D2 
D4




Uo =
+
+
+
+
 hi D1
2k1
2k 2
2 k3
ho 




−1
Do the same procedure for Ui:
Ui =
1
D 
D 
D 
D1 ln  2  D1 ln  3  D1 ln  4 
1
 D1  +
 D2  +
 D3  + D1
+
hi
2k1
2k 2
2 k3
ho D4


D 
D 
D 
D1 ln  2  D1 ln  3  D1 ln  4 


D3 
D1 
D2 
D1 
1




Ui =
+
+
+
+
 hi
2k1
2k 2
2 k3
ho D4 




Module II Heat Transfer by Conduction
−1
THELMA T. OBILLO, PME
Faculty, Mechanical Engineering
39
Hollow Sphere:
Heat conduction through composite sphere can be obtained similar to heat
conduction through composite cylinder. If the convective heat transfer is considered
with hi and ho as the inner and outer convection film coefficients, respectively, then
the heat conduction through composite sphere will be:
q=
ti − to
1 1 1 1
 −   − 
r r
r r
1
1
+ 1 2+ 2 3+
hi Ai
4 k1
4 k2
ho Ao
where :
Ai = 4 ri 2 = 4 r12
Ao = 4 ro 2 = 4 r32
q=
q=
ti − to
 r2 − r1   r3 − r2 


 
r1r2   r2 r3 
1
1

+
+
+
2
4 k1
4 k2
hi ( 4 r1 )
ho ( 4 r32 )
4 ( ti − to )
 r2 − r1   r3 − r2 


 
r1r2   r2 r3 
1
1

+
+
+
2
hi r1
k1
k2
ho r32
Module II Heat Transfer by Conduction
THELMA T. OBILLO, PME
Faculty, Mechanical Engineering
40
Critical Thickness of Insulation
on a surface always brings about a decrease in the heat transfer rate.
− But addition of insulating material to the outside surfaces of cylindrical or
spherical walls (geometries which have non-constant cross-sectional areas) may
increase the heat transfer rate rather than decrease under the certain
circumstances.
− To establish this fact, consider a thin walled metallic cylinder of length l,
radius ri and transporting a fluid at temperature ti which is higher than the ambient
temperature.
− Insulation of thickness (r - ri) and conductivity k is provided on the surface
of the cylinder.
− With assumption
b. One-dimensional heat flow only in radial direction
c. Negligible thermal resistance due to cylinder wall
d. Negligible radiation exchange between outer surface of insulation and
surrounding
e. The heat transfer can be expressed as
➢ Where hi and ho are the convection coefficients at inner and outer surface
respectively.
Module II Heat Transfer by Conduction
THELMA T. OBILLO, PME
Faculty, Mechanical Engineering
41
➢ The denominator represents the sum of thermal resistance to heat flow.
➢ The value of k, ri, hi and ho constant; therefore, the total thermal resistance
will depend upon thickness of insulation which depends upon the outer radius
of the arrangement.
➢ It is clear from the equation above that with increase of radius r (i.e.
thickness of insulation), the conduction resistance of insulation increases but
the convection resistance of the outer surface decreases.
➢ Therefore, addition of insulation can either increase or decrease the rate of
heat flow depending upon a change in total resistance with outer radius r.
➢ To determine the effect of insulation on total heat flow, differentiate the
total resistance Rt with respect to r and equating to zero.
➢ To determine whether the foregoing result maximizes or minimizes the total
resistance, the second derivative need to be calculated
➢ which is positive, so r = k ⁄ ho represent the condition for minimum resistance
and consequently maximum heat flow rate.
➢ The insulation radius at which resistance to heat flow is minimum is called
➢ The critical radius, designated by rc is dependent only on thermal quantities
k and ho.
Module II Heat Transfer by Conduction
THELMA T. OBILLO, PME
Faculty, Mechanical Engineering
42
➢ From the above equation it is clear that with increase of radius of insulation
heat transfer rate increases and reaches the maximum at r=rc and then it will
decrease.
➢ Two cases of practical interest are:
When ri &lt; rc:
It is clear from equation 2.14a below, that with addition of insulation to bare
pipe increases the heat transfer rate until the outer radius of insulation
becomes equal to the critical radius.
Because with addition of insulation decrease the convection resistance of
surface of insulation which is greater than increase in conduction resistance
of insulation.
➢ Any further increase in insulation thickness decreases the heat transfer from
the peak value but it is still greater than that of for the bare pipe until a
certain amount of insulation r*.
➢ So, insulation greater than (r* - ri) must be added to reduce the heat loss
below the bare pipe.
➢ This may happen when insulating material of poor quality is applied to pipes
➢ This condition is used for electric wire to increase the heat dissipation from
the wire which helps to increase the current carrying capacity of the cable.
Module II Heat Transfer by Conduction
THELMA T. OBILLO, PME
Faculty, Mechanical Engineering
43
When ri &lt; rc
➢ It is clear from the figure 2.14b that increase in insulation thickness always
decrease the heat loss from the pipe.
➢ This condition is used to decrease the heat loss from steam and refrigeration
pipes.
➢ Critical radius of insulation for the sphere can be obtain in the similar way:
Problem 13:
A hot fluid is being conveyed through a long pipe of 4 cm outer dia. And covered
with 2 cm thick insulation. It is proposed to reduce the conduction heat loss to the
surroundings to one-third of the present rate by further covering with some
insulation. Calculate the additional thickness of insulation.
Module II Heat Transfer by Conduction
THELMA T. OBILLO, PME
Faculty, Mechanical Engineering
44
Module II Heat Transfer by Conduction
THELMA T. OBILLO, PME
Faculty, Mechanical Engineering
45
Problem 14:
A pipe carrying the liquid at -20oC is 10 mm in outer diameter and is exposed
to ambient at 25oC with convective heat transfer coefficient of 50 W/m-K. It
is proposed to apply the insulation of material having thermal conductivity of
0.50 W/m-K. Determine the thickness of insulation beyond which the heat
gain will be reduced. Also calculate the heat loss for 2.5 mm, 7.5 mm and 15
mm thickness of insulation over 1m length. Which one is more effective
thickness of insulation?
Module II Heat Transfer by Conduction
THELMA T. OBILLO, PME
Faculty, Mechanical Engineering
46
Module II Heat Transfer by Conduction
THELMA T. OBILLO, PME
Faculty, Mechanical Engineering
47
Problem 15:
An insulated steam pipe located where the ambient temperature is 32 oC has
an inside diameter of 50 mm with 10 mm thick wall. The outside diameter of
the corrugated asbestos insulation is 125 mm and the surface coefficient of
still air, ho=12 W/m2-oC. Inside the pipe is steam having a temperature of 150
W/m2-oC with film coefficient, hi=6000 W/m2-oC. Thermal conductivity of
pipe and asbestos insulation are 45 and 0.12 W/m-K respectively. Determine
the heat loss per unit length of pipe and the interface temperature.
Given:
Required:
a) heat loss per unit length of pipe, q/L
b) interface temperature, t2
Solution:
a) heat loss per unit length of pipe, q/L
q=
ti − to
D
D2
ln 3
D1
D2
1
1
+
+
+
hi ( D1 L ) 2 k1L 2 k2 L ho ( D3 L )
ln
Module II Heat Transfer by Conduction
THELMA T. OBILLO, PME
Faculty, Mechanical Engineering
48
q
=
L
ti − to
D
D
ln 2 ln 3
D1
D2
1
1
+
+
+
hi ( D1 ) 2 k1 2 k2 ho ( D3 )
150 − 32 ) oC
(
q
=
L 

 70 
 125 
ln   ln 



1
1
1
50
70
 +


+  + 
  ( 6000 )( 0.05) 2 ( 45) 2 ( 0.12 )  (12 )( 0.125)  W
o

 m − C
q
= ________ W
m
L
b) interface temperature, t2
q=
ti − t2
D2
D1
1
+
hi ( D1 L ) 2 k1 L
=
ln
substitute q
L
t 2 − to
D3
D2
1
+
2 k2 L ho ( D3 L )
ln
from (a)
150 oC − t2
q
=
L 
 70  
ln   

1
1
50

+  
  ( 6000 )( 0.05 ) 2 ( 45 )  W
o

 m − C
t2 = ________ oC
Module II Heat Transfer by Conduction
THELMA T. OBILLO, PME
Faculty, Mechanical Engineering
49

MODULE SUMMARY
The rate of heat conduction in a specified direction is proportional to the
temperature gradient, which is the rate of change in temperature with distance in
that direction. One dimensional steady state heat conduction through homogenous
material is given by Fourier Law of heat conduction.
The Fourier law is essentially based on the following assumptions:
1. Steady state heat conduction, i.e. temperature at fixed point does
not change with respect to time.
2. One dimensional heat flow.
3. Material is homogenous and isotropic, i.e. thermal conductivity has
a constant value in all the directions.
4. Constant temperature gradient and a linear temperature profile.
5. No internal heat generation.
The Fourier law helps to define thermal conductivity of the material. Hence
thermal conductivity may be defined as the amount of heat conducted per unit time
across unit area and through unit thickness, when a temperature difference of unit
degree is maintained across the bounding surface.
When a moving fluid comes into contact with a stationary surface, a thin
boundary layer develops adjacent to the wall and in this layer, there is no relative
velocity with respect to surface. In a heat exchange process, this layer is called
stagnant film and heat flow in the layer is covered both by conduction and convection
processes. Since thermal conductivity of fluids is low, the heat flow from the moving
fluid of the wall is mainly due to convection. The rate of convective heat transfer
between a solid boundary and adjacent fluid is given by the Newton-Rikhman law
or otherwise known as Newton’s law of cooling.
The overall heat transfer coefficient represents the intensity of heat transfer
from one fluid to another through a wall separating them. It is numerically equal to
the quantity of heat passing through unit area of wall surface in a unit time at a
temperature difference of unit degree.
Congratulations! You have just studied Module 2.
Now you are ready to evaluate how much you have
summative test. Good Luck!!!
Module II Heat Transfer by Conduction
THELMA T. OBILLO, PME
Faculty, Mechanical Engineering
50
Prepared by:
Recommending Approval:
Approved:
THELMA T. OBILLO
Professor
ROY N. LAQUIDAN
Program Chair
LORENZO L. BACANI
Dean
Module II Heat Transfer by Conduction
THELMA T. OBILLO, PME
Faculty, Mechanical Engineering
51
August 3, 2021
August 10, 2021
Module II Heat Transfer by Conduction
August 10, 2021
THELMA T. OBILLO, PME
Faculty, Mechanical Engineering
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