Chapter two
2.1 Study state conduction in one dimension:
2.1.1 The Plane Wall
a. temperature destribution
FIGURE 2.1 Heat transfer through a plane wall. (a) Temperature distribution. (b)
Equivalent thermal circuit.
Assumption:
Temperature varies with x-direction only no heat generation. So, the general equation
is reduces to:
𝜕2 𝑇
𝜕𝑥 2
=0
First integration:
𝜕𝑇
𝜕𝑥
= 𝐶1
Second integration:
𝑇(𝑥) = 𝐶1 𝑥 + 𝐶2
2.1
this is the general solution
𝐶1 and 𝐶2 are constant of integration.
Baundary condition:
Substitute in general solution:
2.2
the heat transfer rate:
2.3
b. Thermal resistance
The thermal resistance for conduction is:
2.4
The thermal resistance for convection is:
2.5
Circuit representations provide a useful tool for both conceptualizing and quantifying
heat transfer problems. The equivalent thermal circuit for the plane wall with
convection surface conditions is shown in Figure 2.1b. The heat transfer rate may be
determined from separate consideration of each element in the network. Since qx is
constant throughout the network, it follows that
2.1.2 The Composite Wall:
a. Material in series:
Equivalent thermal circuits may also be used for more complex systems, such as
composite walls. Such walls may involve any number of series and parallel
thermalresistances due to layers of different materials. Consider the series composite
wall of Figure 2.2. The one-dimensional heat transfer rate for this system may be
expressed as
FIGURE 2.2 Equivalent thermal circuit for a series composite wall.
*Overall heat transfer coefficient
With composite systems it is often convenient to work with an overall heat transfer
coefficient, U, which is defined by an expression analogous to Newton’s law of
cooling. Accordingly,
Or
𝒒𝒙 = 𝑼𝑨(𝑻∞𝟏 − 𝑻∞𝟒 )
That means:
b. Material in parallel:
𝑞𝑥 = 𝑞1 + 𝑞2
𝑞1 =
𝑇1 −𝑇2
∆𝑥
𝑘1 𝐴1
𝑇1
𝑞
1
2
∆𝑥
𝑇2
𝑞2 =
𝑇1 −𝑇2
∆𝑥
𝑘2 𝐴2
𝑞𝑥 = (𝑇1 − 𝑇2 ) �
1
𝑅𝑒
1
1
=𝑅 +𝑅
1
1
∆𝑥
𝑘1 𝐴1
+� + �
1
∆𝑥
𝑘2 𝐴2
�=
∆𝑇
𝑅𝑒
2
Where:
𝑅𝑒 : Equavelant resistance.
Example 2.1: Find the heat transfer per unit area through the composite wall in
Figure below. Assume one-dimensional heat flow.
Solusion:
∆𝑇
𝑞 = ∑𝑅
∆𝑥
𝑅 = 𝑘𝐴
0.025
𝑅𝐴 = (150)(0.1) = 1.667 ∗ 10−3
0.075
𝑅𝐵 = (30)(0.05) = 0.05
0.05
𝑅𝐶 = (50)(0.1) = 0.01
0.075
𝑅𝐷 = (70)(0.05) = 0.02143
𝑅 = 𝑅𝐴 + 𝑅𝐶 +
1
1
1
+
𝑅𝐵 𝑅𝐷
= 2.667 ∗ 10−2
𝑞=
∆𝑇
𝑅
370−66
= 2.667∗10−2 = 11400 W
H.W CH2:1,2,3,4
1.1.2 Radial system
a. Cylindrical
Consider a long cylinder of inside radius r i , outside radius ro, and length L, such as
the one shown in Figure 2-3. We expose this cylinder to a temperature differential
T i − T o and ask what the heat flow will be. For a cylinder with length very large
compared to diameter, it may be assumed that the heat flows only in a radial direction,
so that the only space coordinate needed to specify the system is r.
The general conduction equation in cylindrical coordinate:
2.6
Assumptions:
 steady state.
 One dimension with radias only.
 No heat generation.
Equation 2.6 will reduced to:
1 𝜕
𝜕𝑇
�𝑘𝑟 𝜕𝑟 � = 0
𝑟 𝜕𝑟
2.7
We may determine the temperature distribution in the cylinder by solving Equation
2.7 and applying appropriate boundary conditions. Assuming the value of k to be
constant, Equation 2.7 may be integrated twice to obtain the general solution
𝑟
𝜕𝑇
𝜕𝑟
= 𝐶1
FIGURE 2-3: Hollow cylinder with convective surface conditions.
To obtain the constants of integration C 1 and C 2 , we introduce the following
boundary conditions:
Applying these conditions to the general solution, we then obtain
Solving for C 1 and C 2
𝐶1 =
𝑇𝑠,1 −𝑇𝑠,2
𝑟
ln( 1�𝑟2 )
𝐶2 = 𝑇𝑠,2 −
𝑇𝑠,1 −𝑇𝑠,2
𝑟
𝑙𝑛� 1�𝑟2 �
∗ 𝑙𝑛(𝑟2 )
and substituting into the general solution, we then obtain
2.8
The rate at which energy is conducted across any cylindrical surface in the solid may
be expressed as
𝜕𝑇
𝜕𝑇
𝑞𝑟 = −𝑘𝐴 𝜕𝑟 = −𝑘(2𝜋𝑟𝐿) 𝜕𝑟
2.9
2.10
From this result it is evident that, for radial conduction in a cylindrical wall, the
thermal resistance is of the form
K/W, ˚C/W
2.11
*Composite Cylinder Wall
FIGURE 2-4: Temperature distribution for a composite cylindrical wall.
Consider now the composite system of Figure 2.4. Recalling how we treated the
composite plane wall and neglecting the interfacial contact resistances, the heat
transfer rate may be expressed as
2.12
*Overall Heat Transfer Coefficient
The foregoing result may also be expressed in terms of an overall heat transfer
coefficient. That is,
2.13
If U is defined in terms of the inside area, 𝐴1 = 𝟐𝝅𝒓𝟏 𝑳, Equations 2.12 and 2.13
may be equated to yield
2.14
This definition is arbitrary, and the overall coefficient may also be defined in
terms of A 4 or any of the intermediate areas. Note that
2.15
Example 2.2:
Water flows at 50◦C inside a 2.5-cm-inside-diameter tube such that h 1 = 3500
W/m2 · ˚C. The tube has a wall thickness of 0.8 mm with a thermal conductivity
of 16 W/m · ˚C. The outside of the tube loses heat by free convection with h 2 =
7.6 W/m2 · ˚C. Calculate the overall heat-transfer coefficient and heat loss per unit
length to surrounding air at 20◦C.
Solution
There are three resistances in series for this problem, with L = 1.0 m, d 1 = 0.025
m, and d 2 = 0.025 + (2)(0.0008) = 0.0266 m, the resistances may be calculated as:
𝑅1 =
1
2𝜋𝑟1 𝐿ℎ1
𝑅𝑡,𝑐𝑜𝑛𝑑 =
𝑅2 =
1
=
𝑟
𝑙𝑛( 2 )
𝑟1
2𝜋𝑘𝐿
2𝜋𝑟2 𝐿ℎ2
=
1
(3500)𝜋(0.025)(1)
=
0.0266
)
0.025
𝑙𝑛(
2𝜋(16)(1)
1
= 0.00364 ˚C/W.
= 0.00062 ˚C/W.
(7.6)𝜋(0.0266)(1)
= 1.575 ˚C/W.
𝑞𝑟 = 𝑈1 𝐴1 �𝑇∞,1 − 𝑇∞,2 � = 𝑈2 𝐴2 �𝑇∞,1 − 𝑇∞,2 �
𝑈1 =
𝑈1 =
1
𝐴1 𝑅𝑡𝑜𝑡𝑎𝑙
=
1
(𝜋𝑑1 𝐿)𝑅𝑡𝑜𝑡𝑎𝑙
1
𝜋(0.025)(1)(0.00364+0.00062+1.575)
= 8.064
𝑈2 =
𝑈2 =
1
𝐴2 𝑅𝑡𝑜𝑡𝑎𝑙
=
1
(𝜋𝑑2 𝐿)𝑅𝑡𝑜𝑡𝑎𝑙
1
𝜋(0.0266)(1)(0.00364+0.00062+1.575)
= 7.577
𝑞𝑟 = 8.064(𝜋)(0.025)(1)(50 − 20) = 19 W
𝑞𝑟 = 7.577(𝜋)(0.0266)(1)(50 − 20) = 19 W
Example 2.3:
The heat flow is given by
*Critical Thickness of Insulation
Let us consider a layer of insulation which might be installed around a circular
pipe, as shown in Figure 2-5. The inner temperature of the insulation is fixed at T i ,
and the outer surface is exposed to a convection environment at T∞. From the
thermal network the heat transfer is
Figure 2-5 Critical insulation thickness.
2.16
Now let us manipulate this expression to determine the outer radius of insulation
ro, which will maximize the heat transfer. The maximization condition is
which gives the result
2.17
Equation (2.17) expresses the critical-radius-of-insulation concept. If the outer
radius is less than the value given by this equation, then the heat transfer will be
increased by adding more insulation. For outer radii greater than the critical value
an increase in insulation thickness will cause a decrease in heat transfer.
Example 2.4:
Calculate the critical radius of insulation for asbestos [k = 0.17 W/m · ◦C]
surrounding a pipe and exposed to room air at 20◦C with h = 3.0 W/m2 · ◦C.
Calculate the heat loss from a 200◦C, 5.0-cm-diameter pipe when covered with the
critical radius of insulation and without insulation.
Solution:
From Equation (2-17) we calculate r o as
The inside radius of the insulation is 5.0/2 = 2.5 cm, so the heat transfer is
calculated from Equation (2.16) as
Without insulation the convection from the outer surface of the pipe is
So, the addition of 3.17 cm (5.67 − 2.5) of insulation actually increases the heat
transfer by 25 percent. As an alternative, fiberglass having a thermal conductivity
of 0.04 W/m · ◦C might be employed as the insulation material. Then, the critical
radius would be
Now, the value of the critical radius is less than the outside radius of the pipe (2.5
cm), so addition of any fiberglass insulation would cause a decrease in the heat
transfer.
b. Spherial
Figure 2.6 show a hollow sphere of radius R1 at Ts,1 and R2 at Ts,2. For sphere
the heat equation is gives by equation 1.9c
Assumption:
 One dimension heat flow r-direction.
 Steady state.
 No heat generation.
For the control volume in the figure energy conservation required that
The approperiate form of Fouries low is
The general heat equation reduced to
1 𝜕
𝑟 2 𝜕𝑟
𝜕
𝜕𝑇
�𝑘𝑟 2 𝜕𝑟 � = 0
𝜕𝑇
�𝑘𝑟 2 𝜕𝑟 � = 0
𝜕𝑟
First integration:
𝑟2
𝜕𝑇
𝜕𝑟
𝜕𝑇
𝜕𝑟
=
= 𝐶1
𝐶1
Figure 2.6: conduction in spherical shell
𝑟2
Second integration:
𝐶1
𝑻(𝒓) = −
𝒓
+ 𝐶2
..........a
B.C: at 𝑟 = 𝑟1 , 𝑇 = 𝑇𝑠,1
at 𝑟 = 𝑟2 , 𝑇 = 𝑇𝑠,2
apply B.C in equation a
𝐶
..........b
𝐶
.............c
𝑇𝑠,1 = − 𝒓 1 + 𝐶2
𝟏
𝑇𝑠,2 = − 𝒓 1 + 𝐶2
𝟐
Solve equation b and c we get:
𝐶1 =
(𝑇𝑠,1 −𝑇𝑠,2 )
1
1
−
𝑟2 𝑟1
𝐶2 = 𝑇𝑠,1 −
1 (𝑇𝑠,1 −𝑇𝑠,2 )
𝑟1
1
1
−
𝑟1 𝑟2
Sub. 𝐶1 & 𝐶2 in equation a, for temperature distribution:
𝑇(𝑟)−𝑇𝑠,1
=
𝜕𝑇
, 𝐴 = 4𝜋𝑟 2
𝑇𝑠,2 −𝑇𝑠,1
𝜕𝑟
=
𝐶1
𝑟2
𝑟2
𝑟
𝑞𝑟 = −4𝜋𝑘𝑟 2
𝑞𝑟 = 4𝜋𝑘
𝑅𝑡,𝑐𝑜𝑛𝑑 =
𝑟−𝑟
�𝑟 −𝑟1 �
2
1
(𝑻𝒔,𝟏 −𝑻𝒔,𝟐 )
𝟏 𝟏
𝒓𝟐 𝒓𝟏
𝒓𝟐 ( − )
(𝑻𝒔,𝟏 −𝑻𝒔,𝟐 )
𝟏 𝟏
𝒓𝟏 𝒓𝟐
( − )
1
1
−
𝑟1 𝑟2
4𝜋𝑘
2.2 Conduction with thermal energy generation:
2.2.1 Plane wall with heat generation:
To investigate the effect of uniform internal heat generation in one dimension
temperature distribution within the plane wall. Figure 2.7 shows the plane wall with
internal uniform heat generation, 𝑞̇ .
Figure 2.7 Conduction in a plane wall with uniform heat generation with Symmetrical
boundary conditions.
The general conduction equation is:
For study, one dimension:
𝜕2 𝑇
𝜕𝑥 2
𝑞̇
+ =0
𝑘
First integration gives:
𝜕𝑇
𝜕𝑥
=−
𝑞𝑥̇
+ 𝐶1
𝑘
Second integration gives:
𝑞̇
𝑇(𝑥) = −
2𝑘
𝑥 2 + 𝐶1 𝑥 + 𝐶2 ................a
B.C: at 𝑥 = 𝐿, 𝑇 = 𝑇𝑠
at 𝑥 = −𝐿, 𝑇 = 𝑇𝑠
:
Apply B.C to equation a:
𝑇𝑠 = −
𝑇𝑠 = −
So:
𝑞̇
2𝑘
𝑞̇
2𝑘
𝐿2 + 𝐶1 𝐿 + 𝐶2 .........b
𝐿2 − 𝐶1 𝐿 + 𝐶2
𝐶2 = 𝑇𝑠 +
𝑞̇
2𝑘
...........c
𝐿2 and 𝐶1 = 0
Sub. 𝐶1 &𝐶2 in equation a gives:
𝑞̇
𝑇(𝑥) − 𝑇𝑠 =
2𝑘
Thus at x=0,
𝜕𝑇
𝐿2 �1 −
𝑥2
𝐿2
�
..........d
This is a parabolic temperature distribution that is symmetry about x=0
𝜕𝑥
=0
And the maximum temperature axist at x=0
𝑇𝑚𝑎𝑥 − 𝑇𝑠 =
𝑞̇
2𝑘
𝐿2
From equation d
𝑞̇
2𝑘
𝐿2 =
𝑇(𝑥)−𝑇𝑠
𝑥2
�1− 2 �
𝐿
.........e
Sub. In equation e we get:
𝑇(𝑥)−𝑇𝑠
𝑇𝑚𝑎𝑥 −𝑇𝑠
=1−
𝑥2
𝐿2
in case the temperature distribution is asymmetrical as shown in Figure 2.8 the
temperature distribution is as below:
Figure 2.8: Conduction in a plane wall with uniform heat generation with
Asymmetrical boundary conditions.
2.2.2 Cylinder with heat generation:
Heat generation may occur in a variety of radial geometries. Consider the long, solid
cylinder of Figure 2.9, which could represent a current-carrying wire or a fuel
element in a nuclear reactor. For steady-state conditions the rate at which heat is
generated within the cylinder must equal the rate at which heat is convected from the
surface of the cylinder to a moving fluid. This condition allows the surface
temperature to be maintained at a fixed value of T s .
FIGURE 2.9 Conduction in a solid cylinder with uniform heat generation.
To determine the temperature distribution in the cylinder, we begin with the
appropriate form of the heat equation.
For constant thermal conductivity k, the heat conduction Equation reduces to:
1 𝑑
𝑟 𝑑𝑟
𝑑𝑇
𝑞̇
�𝑟 𝑑𝑟 � + 𝑘 = 0
Separating variables and assuming uniform generation, this expression may be
integrated to obtain
𝑟
𝑑𝑇
𝑑𝑟
=−
𝑞̇
2𝑘
𝑟 2 + 𝐶1
Repeating the procedure, the general solution for the temperature distribution
becomes
𝑇(𝑟) = −
𝑞̇
4𝑘
𝑟 2 + 𝐶1 𝑙𝑛𝑟 + 𝐶2
.........a
To obtain the constants of integration C 1 and C 2 , we apply the boundary conditions
𝑑𝑇
𝑑𝑟
= 0 𝑎𝑡 𝑟 = 0 , 𝑇 = 𝑇𝑠 𝑎𝑡 𝑟 = 𝑟𝑜
Therefore:
𝐶1 = 0
𝐶2 = 𝑇𝑠 +
𝑞̇
𝑟2
4𝑘 𝑜
The temperature distribution is therefore
𝑇(𝑟) − 𝑇𝑠 =
𝑟2
𝑞̇ 𝑟𝑜2
�1 − 𝑟 2�
4𝑘
𝑇 = 𝑇𝑚𝑎𝑥 𝑎𝑡 𝑟 = 0
𝑜
...........b
Sub in equation
𝑇𝑚𝑎𝑥 − 𝑇𝑠 =
𝑞̇ 𝑟𝑜2
..............c
4𝑘
Devided equation (b) to equation (c) we obtained
𝑇(𝑟)−𝑇𝑠
𝑟 2
𝑇𝑚𝑎𝑥 −𝑇𝑠
=1−� �
𝑟
𝑑𝑇
𝑞̇
From equation (b)
𝑑𝑟
=−
2𝑘
𝑟
𝑜
𝑑𝑇
𝑞(𝑟) = −𝑘2𝜋𝑟𝑙 𝑑𝑟
𝑞̇
𝑞(𝑟) = −𝑘2𝜋𝑟𝐿(− 2𝑘 𝑟)
𝑞(𝑟) = 𝜋𝐿𝑞̇ 𝑟 2
(heat conduction in the radius direction)
.........d
𝑞(𝑟𝑜 ) = 𝜋𝐿𝑞̇ 𝑟𝑜2
To relate the surface temperature, 𝑇𝑠 , to the temperature of the cold fluid, 𝑇 , either a
surface energy balance or an overall energy balance may be used.
𝑞𝑐𝑜𝑛𝑑 = 𝑞𝑐𝑜𝑛𝑣
𝜋𝐿𝑞̇ 𝑟𝑜2 = ℎ(2𝜋𝑟𝑜 𝐿)(𝑇𝑠 − 𝑇∞ )
𝑇𝑠 = 𝑇∞ +
𝑞̇ 𝑟𝑜
2ℎ
H.W a similar analysis applies to a hollow cylinder as shown in the figure below:
Baundary condition: at 𝑟 = 𝑟2 𝑇(𝑟) = 𝑇𝑠,2 ∶ 𝑟 = 𝑟2
Example:
𝑑𝑇
𝑑𝑟
=0
A current of 200 A is passed through a stainless-steel wire [k = 19 W/m · ◦C] 3 mm in
diameter. The resistivity of the steel may be taken as 70 μΩ · cm, and the length of
the wire is 1 m. The wire is submerged in a liquid at 110 ◦C and experiences a
convection heat-transfer coefficient of 4kW/m2 · ◦C. Calculate the center temperature
of the wire.
Solution:
All the power generated in the wire must be dissipated by convection to the liquid:
𝑃 = 𝐼 2 𝑅 = 𝑞 = ℎ𝐴(𝑇𝑠 − 𝑇∞ )
..........a
The resistance of the wire is calculated from
𝐿
𝑅 = 𝜌𝐴 =
𝑐
(70𝑥10−6 )(100)
𝜋(0.15)2
= 0.099Ω
where ρ is the resistivity of the wire. The surface area of the wire is πdL, so from
Equation (a)
(200)2 (0.099) = 4000𝜋(3𝑥10−3 )(1)(𝑇𝑠 − 110)
𝑇𝑠 =215˚C.
The heat generated per unit volume 𝑞̇ is calculated from
𝑃 = 𝑞̇ 𝑉 = 𝑞̇ 𝜋𝑟𝑜2 𝐿
So that,
𝑞̇ =
3960
𝜋(1.5𝑥10−3 )2 (1)
= 560.2 𝑀𝑊/𝑚3
Finally, the center temperature of the wire is calculated from Equation below:
𝑇𝑚𝑎𝑥 − 𝑇𝑠 =
𝑇𝑚𝑎𝑥 =
𝑞̇ 𝑟𝑜2
4𝑘
(5.605𝑥108 )(1.5𝑥10−3 )2
(4)(19)
+ 215 = 231.6 ℃
2.2.3 sphere with heat generation
The general heat conduction equation
For one dimension (r direction only) study state the general equation has been reduced
to:
Baundary condition: at 𝑟 = 0
𝑑
𝑑𝑟
𝑑𝑇
𝑞̇
�𝑟 2 𝑑𝑟 � = − 𝑘 𝑟 2
1 𝑑 2 𝑑𝑇
𝑞̇
�𝑟
�+ =0
2
𝑟 𝑑𝑟
𝑑𝑟
𝑘
𝑑𝑇
𝑑𝑟
= 0 & 𝑎𝑡 𝑟 = 𝑟𝑜 𝑇 = 𝑇𝑠
First integration gives:
𝑑𝑇
𝑞̇
𝑟 2 𝑑𝑟 = − 3𝑘 𝑟 3 + 𝐶1
𝑑𝑇
𝑑𝑟
𝑞̇
𝐶
= − 3𝑘 𝑟 + 𝑟 12
Second integration gives:
𝑇(𝑟) = −
𝑞̇ 𝑟 2
6𝑘
−
𝐶1
𝑟
+ 𝐶2
H.W find the temperature distribution and heat transfer.
3.3 Contact Resistance
Although neglected until now, it is important to recognize that, in composite systems,
the temperature drop across the interface between materials may be appreciable. This
temperature change is attributed to what is known as the thermal contact resistance,
R t,c . The effect is shown in Figure 2.10, and for a unit area of the interface, the
resistance is defined as
FIGURE 2.10 Temperature drop due to thermal contact resistance.