In this chapter, Heat diffusion equation in cartesian, cylindrical and spherical coordinate system is covered. Representation of plane wall, cylindrical wall using thermal resistance and Composite material steady state heat analysis is also covered.

1. UNIT TWO
One Dimensional Steady State
Conduction

2. THE HEAT DIFFUSION EQUATION
Consider the Control Volume given below
The control volume
 Is homogeneous
 There is no bulk motion
within it.

3. As a control volume we can apply energy conservation
principle on it.
𝐸𝑖𝑛 − 𝐸𝑜𝑢𝑡 + 𝐸𝑔𝑒𝑛 = ∆𝐸𝑠𝑡𝑜𝑟𝑒𝑑
The control volume is in no motion
at all. There is no
 Mechanical Energy
 Work being done.
Therefore only Thermal Energy is
going to be considered.

4. Taking infinitesimal (differential) Control Volume we can show the
temperature distribution in each dimension on a Cartesian
coordinate plane.

5.  Remember Taylor Series Approximation
 The conduction heat rates at the opposite surfaces
can then be expressed as a Taylor series
expansion where, neglecting higher-order terms.
 𝑞𝑥+𝑑𝑥 = 𝑞𝑥 +
𝜕𝑞𝑥
𝜕𝑥
𝑑𝑥
 𝑞𝑦+𝑑𝑦 = 𝑞𝑦 +
𝜕𝑞𝑦
𝜕𝑦
𝑑𝑦
 𝑞𝑧+𝑑𝑧 = 𝑞𝑧 +
𝜕𝑞𝑧
𝜕𝑧
𝑑𝑧 qx

6.  The energy generated system is represented by
𝐸𝑔 = 𝑞𝑑𝑥𝑑𝑦𝑑𝑧
Where
𝑞 is rate at which energy is generated per unit
volume (W/m3).
 For a solid body that is not experiencing any phase change the
stored energy can be represented by
𝐸𝑠𝑡 = 𝑑𝑚𝑐𝑝
𝜕𝑇
𝜕𝑡
𝐸𝑠𝑡 = 𝜌𝑐𝑝
𝜕𝑇
𝜕𝑡
𝑑𝑥𝑑𝑦𝑑𝑧,
where
d𝑚 = 𝜌𝑑𝑣 = 𝜌𝑑𝑥𝑑𝑦𝑑𝑧

7.  Applying Rate of Energy Balance on the infinitesimal
control Volume 𝑑𝑥 ∙ 𝑑𝑦 ∙ 𝑑𝑧
𝐸𝑖𝑛 + 𝐸𝑔𝑒𝑛 − 𝐸𝑜𝑢𝑡 = 𝐸𝑠𝑡
𝐸𝑖𝑛 𝐸𝑔𝑒𝑛 𝐸𝑜𝑢𝑡 𝐸𝑠𝑡
𝑞𝑥 = −𝑘𝑑𝐴
𝜕𝑇
𝜕𝑥
𝑞𝑦 = −𝑘𝑑𝐴
𝜕𝑇
𝜕𝑦
𝑞𝑧 = −𝑘𝑑𝐴
𝜕𝑇
𝜕𝑧
𝐸𝑔 = 𝑞𝑑𝑥𝑑𝑦𝑑𝑧
𝑞𝑥+𝑑𝑥 = 𝑞𝑥 +
𝜕𝑞𝑥
𝜕𝑥
𝑑𝑥
𝑞𝑦+𝑑𝑦 = 𝑞𝑦 +
𝜕𝑞𝑦
𝜕𝑦
𝑑𝑦
𝑞𝑧+𝑑𝑧 = 𝑞𝑧 +
𝜕𝑞𝑧
𝜕𝑧
𝑑𝑧
𝐸𝑠𝑡 = 𝜌𝑐𝑝
𝜕𝑇
𝜕𝑡
𝑑𝑥𝑑𝑦𝑑𝑧

8. 𝐸𝑖𝑛 + 𝐸𝑔𝑒𝑛 − 𝐸𝑜𝑢𝑡 = 𝐸𝑠𝑡
𝑞𝑥 + 𝑞𝑦 + 𝑞𝑧 + 𝑞𝑑𝑥𝑑𝑦𝑑𝑧 − 𝑞𝑥 +
𝜕𝑞𝑥
𝜕𝑥
𝑑𝑥 + 𝑞𝑦 +
𝜕𝑞𝑦
𝜕𝑦
𝑑𝑦 + 𝑞𝑧 +
𝜕𝑞𝑧
𝜕𝑧
𝑑𝑧 = 𝜌𝑐𝑝
𝜕𝑇
𝜕𝑡
𝑑𝑥𝑑𝑦𝑑𝑧
Simplified to
𝑞𝑑𝑥𝑑𝑦𝑑𝑧 −
𝜕𝑞𝑥
𝜕𝑥
𝑑𝑥 +
𝜕𝑞𝑦
𝜕𝑦
𝑑𝑦 +
𝜕𝑞𝑧
𝜕𝑧
𝑑𝑧 = 𝜌𝑐𝑝
𝜕𝑇
𝜕𝑡
𝑑𝑥𝑑𝑦𝑑𝑧
Lets substitute the value of 𝑞𝑥, 𝑞𝑦, 𝑎𝑛𝑑 𝑞𝑧 in the upper equation.
𝑞𝑥 = −𝑘𝑑𝐴
𝜕𝑇
𝜕𝑥
= −𝑘𝑑𝑦𝑑𝑧
𝜕𝑇
𝜕𝑥
𝑞𝑦 = −𝑘𝑑𝐴
𝜕𝑇
𝜕𝑦
= −𝑘𝑑𝑥𝑑𝑧
𝜕𝑇
𝜕𝑦
𝑞𝑧 = −𝑘𝑑𝐴
𝜕𝑇
𝜕𝑧
= −𝑘𝑑𝑥𝑑𝑦
𝜕𝑇
𝜕𝑧

9. 𝑞𝑑𝑥𝑑𝑦𝑑𝑧 −
𝜕 −𝑘𝑑𝑦𝑑𝑧
𝜕𝑇
𝜕𝑥
𝜕𝑥
𝑑𝑥 +
𝜕 −𝑘𝑑𝑥𝑑𝑧
𝜕𝑇
𝜕𝑦
𝜕𝑦
𝑑𝑦 +
𝜕 −𝑘𝑑𝑥𝑑𝑦
𝜕𝑇
𝜕𝑧
𝜕𝑧
𝑑𝑧 = 𝜌𝑐𝑝
𝜕𝑇
𝜕𝑡
𝑑𝑥𝑑𝑦𝑑𝑧
taking the value of thermal conductivity as uniform
𝑞𝑑𝑥𝑑𝑦𝑑𝑧 + 𝑘
𝜕 𝑑𝑦𝑑𝑧
𝜕𝑇
𝜕𝑥
𝜕𝑥
𝑑𝑥 +
𝜕 𝑑𝑥𝑑𝑧
𝜕𝑇
𝜕𝑦
𝜕𝑦
𝑑𝑦 +
𝜕 𝑑𝑥𝑑𝑦
𝜕𝑇
𝜕𝑧
𝜕𝑧
𝑑𝑧 = 𝜌𝑐𝑝
𝜕𝑇
𝜕𝑡
𝑑𝑥𝑑𝑦𝑑𝑧
Rearranging
𝑞𝑑𝑥𝑑𝑦𝑑𝑧 + 𝑘
𝜕
𝜕𝑇
𝜕𝑥
𝜕𝑥
𝑑𝑥𝑑𝑦𝑑𝑧 +
𝜕
𝜕𝑇
𝜕𝑦
𝜕𝑦
𝑑𝑦𝑑𝑥𝑑𝑧 +
𝜕
𝜕𝑇
𝜕𝑧
𝜕𝑧
𝑑𝑥𝑑𝑦𝑑𝑧 = 𝜌𝑐𝑝
𝜕𝑇
𝜕𝑡
𝑑𝑥𝑑𝑦𝑑𝑧
Simplified
𝜕2
𝑇
𝜕𝑥2
+
𝜕2
𝑇
𝜕𝑦2
+
𝜕2
𝑇
𝜕𝑧2
+
𝑞
𝑘
=
𝝆𝒄𝒑
𝒌
𝜕𝑇
𝜕𝑡
𝜶 =
𝒌
𝝆𝑪𝒑

10. 𝜶
𝝏𝟐𝑻
𝝏𝒙𝟐
+
𝝏𝟐𝑻
𝝏𝒚𝟐
+
𝝏𝟐𝑻
𝝏𝒛𝟐
+
𝒒
𝒌
=
𝝏𝑻
𝝏𝒕
Heat diffusion equation
Can be simplified to
 When there is no energy generation
𝜶
𝝏𝟐𝑻
𝝏𝒙𝟐 +
𝝏𝟐𝑻
𝝏𝒚𝟐 +
𝝏𝟐𝑻
𝝏𝒛𝟐 =
𝝏𝑻
𝝏𝒕
 When the system is in steady state
𝝏𝟐
𝑻
𝝏𝒙𝟐 +
𝝏𝟐
𝑻
𝝏𝒚𝟐 +
𝝏𝟐
𝑻
𝝏𝒛𝟐 +
𝒒
𝒌
= 𝟎
 When the system is in steady state and no heat generation
𝝏𝟐𝑻
𝝏𝒙𝟐 +
𝝏𝟐𝑻
𝝏𝒚𝟐 +
𝝏𝟐𝑻
𝝏𝒛𝟐 = 𝟎

11. CYLINDRICAL COORDINATES

12.  When the del operator 𝛻 of the heat equation in
Cartesian coordinate system is expresses in
cylindrical coordinates, the general form of heat flux
vector
𝑞′′ = −𝑘𝛻𝑇 = −𝑘 𝑖
𝜕𝑇
𝜕𝑟
+ 𝑗
1
𝑟
𝜕𝑇
𝜕∅
+ 𝑘
𝜕𝑇
𝜕𝑧
Where
𝑞𝑟
′′ = −𝑘
𝜕𝑇
𝜕𝑟
𝑞∅
′′
= −
𝑘
𝑟
𝜕𝑇
𝜕∅
𝑞𝑧
′′ = −𝑘
𝜕𝑇
𝜕𝑧
are heat flux components in radial circumferential and
axial directions, respectively.
 Applying energy balance to the differential control
volume
𝟏
𝒓
𝝏
𝝏𝒓
𝒌𝒓
𝝏𝑻
𝝏𝒓
+
𝟏
𝒓𝟐
𝝏
𝝏∅
𝒌
𝝏𝑻
𝝏∅
+
𝝏
𝝏𝒛
𝒌
𝝏𝑻
𝝏𝒛
+ 𝒒 = 𝝆𝒄𝒑
𝝏𝑻
𝝏𝒕

13. SPHERICAL COORDINATES

14.  In spherical coordinates, the general form of heat flux vector
𝑞′′
= −𝑘𝛻𝑇 = −𝑘 𝑖
𝜕𝑇
𝜕𝑟
+ 𝑗
1
𝑟
𝜕𝑇
𝜕𝜃
+ 𝑘
1
𝑟𝑠𝑖𝑛𝜃
𝜕𝑇
𝜕∅
Where
𝑞𝑟
′′
= −𝑘
𝜕𝑇
𝜕𝑟
𝑞𝜃
′′
= −
𝑘
𝑟
𝜕𝑇
𝜕𝜃
𝑞∅
′′
= −
𝑘
𝑟𝑠𝑖𝑛𝜃
𝜕𝑇
𝜕∅
are heat flux components in radial, polar, and azimuthal directions,
respectively.
 Applying energy balance to the differential control volume
𝟏
𝒓𝟐
𝝏
𝝏𝒓
𝒌𝒓𝟐
𝝏𝑻
𝝏𝒓
+
𝟏
𝒓𝟐𝒔𝒊𝒏𝟐𝜽
𝝏
𝝏∅
𝒌
𝝏𝑻
𝝏∅
+
𝟏
𝒓𝟐𝒔𝒊𝒏𝜽
𝝏
𝝏𝜽
𝒌𝒔𝒊𝒏𝜽
𝝏𝑻
𝝏𝜽
+ 𝒒 = 𝝆𝒄𝒑
𝝏𝑻
𝝏𝒕

15. 𝜶
𝝏𝟐𝑻
𝝏𝒙𝟐
+
𝝏𝟐𝑻
𝝏𝒚𝟐
+
𝝏𝟐𝑻
𝝏𝒛𝟐
+
𝒒
𝒌
=
𝝏𝑻
𝝏𝒕
Heat diffusion equation in Cartesian Coordinate
To determine the temperature distribution in a medium, it
is necessary to solve the appropriate form of heat
equation.
They are dependent on physical condition or time.
 Boundary Condition
 Initial Condition

16. Boundary Conditions

17. 1. The first condition corresponds to a situation for which the surface is maintained at a
fixed temperature Ts.
 termed as a Dirichlet condition, or a boundary condition of the first kind.
 It is closely approximated, for example, when the surface is in contact with a melting solid or a
boiling liquid.
 In both cases there is heat transfer at the surface, while the surface remains at the
temperature of the phase change process.
2. The second condition corresponds to the existence of a fixed or constant heat flux at
the surface. This heat flux is related to the temperature gradient at the surface by Fourier’s
law.
 Termed as a Neumann condition, or a boundary condition of the second kind, and may
be realized by bonding a thin film electric heater to the surface.
3. The boundary condition of the third kind corresponds to the existence of convection
heating (or cooling) at the surface and is obtained from the surface energy balance

18. THE PLANE WALL
 For one dimensional conduction in a plane wall,
temperature is a function of x only and heat is
transferred in this direction only.
 Consider a plane wall separating two fluids of different
temperatures.

19.  Heat transfer occurs
 By convection from hot fluid, 𝑇∞,1 to one surface 𝑇𝑆,1
 By conduction through the wall
 By convection from the other surface 𝑇𝑆,2 to the cold fluid,
𝑇∞,2.
 First lets determine temperature distribution T(x) and
then rate of heat transfer.
 T(x) can be determined by solving heat equation with
proper boundary conditions.

20.  For steady state conditions with no distributed source or
sink of energy within the wall, the appropriate heat
equation is
𝝏𝟐𝑻
𝝏𝒙𝟐
= 𝟎
 And within this conditions, the heat flux is constant ,
independent of x.
 The heat equation above is integrated twice and gives
𝑇 𝑥 = 𝐶1𝑥 + 𝐶2
To obtain the constants of integration, 𝐶1 & 𝐶2, boundary
conditions must be introduced.

21. Conditions at 𝑋 = 0 𝑎𝑛𝑑 𝑋 = 𝐿
T 0 = TS,1 and T L = TS,2
Applying the condition at x=0 to the general solution.
T 0 = C1 0 + C2 = Ts,1
C2 = Ts,1
Similarly for X=L
T L = C1 L + C2 = Ts,2
C1 =
Ts,2 − Ts,1
L
Substituting the values of C1 and C2 in to the general solution
T x =
Ts,2 − Ts,1
L
x + Ts,1
In steady state condition with no generation and constant
property, the temperature varies linearly with x.

22. 𝑇 𝑥 =
𝑇𝑠,2 − 𝑇𝑠,1
𝐿
𝑥 + 𝑇𝑠,1
 Since we have determined temperature distribution,
then lets determine the rate of heat transfer using
Fourier’s Law of Conduction.
𝑞𝑥 = −𝐾𝐴
𝑑𝑇
𝑑𝑥
=
𝐾𝐴
𝐿
𝑇𝑆,1 − 𝑇𝑆,2

23. THERMAL RESISTANCE
 For the special case of one-dimensional heat transfer
with no internal energy generation and with constant
properties, a very important concept is suggested
𝑞𝑥 =
𝑇𝑆,1 − 𝑇𝑆,2
𝐿 𝐾𝐴
& 𝐼 =
𝐸𝑠,1 − 𝐸𝑠,2
𝐿 𝜎𝐴
In particular, an analogy exists between the diffusion
of heat and electrical charge.
 Defining resistance as the ratio of a driving potential to
the corresponding transfer rate, the thermal
resistance for conduction in a plane wall is
𝑹𝒕𝒉,𝑪𝒐𝒏𝒅 =
𝑻𝑺,𝟏−𝑻𝑺,𝟐
𝒒𝒙
=
𝑳
𝑲𝑨
& 𝑹𝒆 =
𝑬𝒔,𝟏−𝑬𝒔,𝟐
𝑰
= 𝑳 𝝈𝑨

24.  A thermal resistance may also be associated with heat
transfer by convection at a surface. From Newton’s law of
cooling
𝑞 = ℎ𝐴 𝑇𝑆 − 𝑇∞ =
𝑻𝑺 − 𝑻∞
𝒉𝑨
 the thermal resistance for convection
𝑹𝒕𝒉,𝑪𝒐𝒏𝒗 =
𝑻𝑺 − 𝑻∞
𝒒
=
𝟏
𝒉𝑨
 Circuit representations provide a useful tool for both
conceptualizing and quantifying heat transfer problems

25.  When the wall is surrounded by a gas, the radiation effects,
can be significant and may need to be considered. The rate
of radiation heat transfer between a surface of emissivity Ɛ and
area As at temperature Ts and the surrounding surfaces at
some average temperature Tsurr can be expressed as
Where
hrad is the radiation heat transfer coefficient. Note that both Ts
and Tsurr must be in K in the evaluation of hrad.
The definition of the radiation heat transfer coefficient enables us
to express radiation conveniently in an analogous manner to
convection in terms of a temperature difference. But hrad
depends strongly on temperature while hconv usually does not.

26. A surface exposed to the surrounding air involves
convection and radiation simultaneously, and the total heat
transfer at the surface is determined by
adding (or subtracting, if in the opposite direction) the
radiation and convection components. The convection and
radiation resistances are parallel to each other, as shown
in Fig. below, and may cause some complication in the
thermal resistance network.
When Tsurr =T, the radiation effect can properly be
accounted for by replacing h in the convection resistance
relation by
where hcombined is the combined heat
transfer coefficient.

27. T V
I q
Re R th

28.  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
𝑞𝑥 =
𝑻∞,𝟏 − 𝑻𝒔,𝟏
𝟏 𝒉𝟏𝑨
=
𝑇𝑆,1 − 𝑇𝑆,2
𝐿 𝑘𝐴
=
𝑻𝒔,𝟐 − 𝑻∞,𝟐
𝟏 𝒉𝟐𝑨
 In terms of the overall temperature difference, 𝐓∞,𝟏 − 𝐓∞,𝟐, and the total
thermal resistance, 𝑹𝒕𝒐𝒕, the heat transfer rate may also be expressed as
𝑞𝑥 =
𝐓∞,𝟏 − 𝐓∞,𝟐
𝑹𝒕𝒐𝒕
 Since the conduction and convection resistances are in series they may be
summed
𝑹𝒕𝒐𝒕 =
𝟏
𝒉𝟏𝑨
+
𝑳
𝒌𝑨
+
𝟏
𝒉𝟐𝑨

37. THE COMPOSITE WALL
Composite walls may involve any
number of series and parallel thermal
resistances due to layers of different
materials.
Equivalent thermal circuits may also be
used.

38. Equivalent circuit may also be used
Considering the overall temperature difference, all thermal resistance
will also be considered.
𝒒𝒙 =
𝑻∞,𝟏 − 𝑻∞,𝟒
𝟏 𝒉𝟏𝑨 + 𝑳𝑨 𝒌𝑨𝑨 + 𝑳𝑩 𝒌𝑩𝑨 + 𝑳𝑪 𝒌𝑪𝑨 + 𝟏 𝒉𝟒𝑨
Rate of heat transfer 𝒒𝒙 can also be determined on a separate element
basis
𝒒𝒙 =
𝑻∞,𝟏 − 𝑻𝒔,𝟏
𝟏 𝒉𝟏𝑨
=
𝑻𝒔,𝟏 − 𝑻𝟐
𝑳𝑨 𝒌𝑨𝑨
= ⋯

42. The thermal resistance concept can be used to solve steady state heat transfer problem in
parallel layers or combined series‐parallel arrangements.
It should be noted that these problems are often two‐ or three dimensional, but approximate
solutions can be obtained by assuming one dimensional heat transfer (using thermal
resistance network).
𝑞 = 𝑞1 + 𝑞2 =
𝑇1 − 𝑇2
𝑅1
+
𝑇1 − 𝑇2
𝑅2
= 𝑇1 − 𝑇2
1
𝑅1
+
1
𝑅2
𝑞 =
𝑇1 − 𝑇2
𝑅𝑡𝑜𝑡𝑎𝑙
1
𝑅𝑡𝑜𝑡𝑎𝑙
=
1
𝑅1
+
1
𝑅2
→
1
𝑅𝑡𝑜𝑡𝑎𝑙
=
𝑅1𝑅2
𝑅1 + 𝑅2

43. Consider the combined series‐parallel arrangement shown in figure below. Assuming one –
dimensional heat transfer, determine the rate of heat transfer.
𝑞 =
𝑇1 − 𝑇∞
𝑅𝑡𝑜𝑡𝑎𝑙
𝑅𝑡𝑜𝑡𝑎𝑙 = 𝑅12 + 𝑅3 + 𝑅𝐶𝑜𝑛𝑣 =
𝑅1𝑅2
𝑅1 + 𝑅2
+ 𝑅3 + 𝑅𝐶𝑜𝑛𝑣
Two approximations commonly used in solving complex multi‐dimensional heat transfer problems by
transfer problems by treating them as one dimensional, using the thermal resistance network:
1‐ Assume any plane wall normal to the x‐axis to be isothermal, i.e. temperature to vary in one
direction only T = T(x)
2‐ Assume any plane parallel to the x‐axis to be adiabatic, i.e. heat transfer occurs in the x‐ direction
only.
These two assumptions result in different networks (different results). The actual result lies between
these two results.

44. The thermal resistance concept can be used to solve steady state heat transfer
problem in parallel layers or combined series‐parallel arrangements.
It should be noted that these problems are often two‐ or three dimensional, but
approximate solutions can be obtained by assuming one dimensional heat
transfer (using thermal resistance network).

45.  Composite walls may also be characterized by series–parallel
configurations
 Although the heat flow is now multidimensional, it is often reasonable to
assume one-dimensional conditions.
 (a) it is presumed that surfaces normal to the x-direction are isothermal,
whereas for case
 (b) it is assumed that surfaces parallel to the x-direction are adiabatic.
 Different results are obtained for Rtot, and the corresponding values of q
bracket the actual heat transfer rate.
 These differences increase with increasing (kF – kG), as multidimensional
effects become more significant

52. Rf,top Rp,top
Rf,cen Rp,cen1 RBrick Rp,cen2
Rf,bot Rp,bot

53. OVERALL HEAT TRANSFER COEFFICIENT
 With composite system it is often convenient to
work with an overall heat transfer coefficient, U.
 It is defined by an expression analogous to
Newton’s law of Cooling.
𝑞𝑥 = 𝑈𝐴∆𝑇 =
∆𝑇
𝑅𝑡𝑜𝑡
Where ∆𝑇 is the overall temperature difference.
The overall heat transfer coefficient is related to the
total thermal resistance
𝑈𝐴 =
1
𝑅𝑡𝑜𝑡
=⇒ 𝑈 =
1
𝑅𝑡𝑜𝑡𝐴
𝑈 =
1
𝟏 𝒉𝟏 + 𝑳𝑨 𝒌𝑨 + 𝑳𝑩 𝒌𝑩 + 𝑳𝑪 𝒌𝑪 + 𝟏 𝒉𝟒

54. RADIAL SYSTEMS
 Cylindrical and spherical systems often experience
temperature gradients in the radial direction only and
may therefore be treated as one-dimensional.
 Moreover, under steady-state conditions with no
heat generation, such systems may be analysed by
using the standard method, which begins with the
appropriate form of the heat equation, or the alternative
method, which begins with the appropriate form of
Fourier’s law.

55. THE CYLINDER
A common example is the hollow cylinder whose inner
and outer surfaces are exposed to fluids at different
temperatures.

56. For steady-state conditions with no heat generation the heat
diffusion equation
𝟏
𝒓
𝒅
𝒅𝒓
𝒌𝒓
𝒅𝑻
𝒅𝒓
= 𝟎 −−−∗
where, for the moment, k is treated as a variable. The physical
significance of this result becomes evident if we also consider
the appropriate form of Fourier’s law. The rate at which energy
is conducted across any cylindrical surface in the solid may be
expressed as
𝑞𝑟 = −𝑘𝐴
𝑑𝑇
𝑑𝑟
= −𝑘 2𝜋𝑟𝐿
𝑑𝑇
𝑑𝑟
−−∗∗
where 2𝝅𝒓𝑳 is the area normal to the direction of heat transfer.
Since the equation * dictates that the quantity 𝒌𝒓
𝒅𝑻
𝒅𝒓
is
independent of r, it follows from equation ** that the conduction
heat transfer rate qr (not the heat flux qr’’) is a constant in the
radial direction.

57.  We may determine the temperature distribution in the cylinder by
solving Equation * and applying appropriate boundary conditions.
 Assuming the value of k to be constant, Equation * may be integrated
twice to obtain the general solution .
𝑇 𝑟 = 𝐶1 ln 𝑟1 + 𝐶2
To obtain the constants of integration, 𝐶1 & 𝐶2, boundary conditions must
be introduced.
T 𝑟1 = TS,1 and T 𝑟2 = TS,2
Applying these conditions to the general solution, we then obtain
TS,1 = 𝐶1 ln 𝑟1 + 𝐶2 and TS,2 = 𝐶1 ln 𝑟2 + 𝐶2
Solving for 𝐶1&𝐶2
𝑎𝑛𝑑 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑔 𝑖𝑛 𝑡𝑜 𝑡ℎ𝑒 𝑔𝑒𝑛𝑒𝑟𝑎𝑙 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛, 𝑤𝑒 𝑡ℎ𝑒𝑛 𝑜𝑏𝑡𝑎𝑖𝑛
𝑇 𝑟 =
(𝑇𝑠,1 − 𝑇𝑠,2)
ln 𝑟2 𝑟1
ln
𝑟
𝑟2
+ 𝑇𝑠,2

58.  If the temperature distribution above, is now used with
Fourier’s law, Equation **, we obtain the following
expression for the heat transfer rate:
𝑞𝑟 =
2𝜋𝐿𝑘(𝑇𝑠,1 − 𝑇𝑠,2)
ln 𝑟2 𝑟1
 From this result it is evident that, for radial conduction in
a cylindrical wall, the thermal resistance is of the form
𝑅𝑡,𝑐𝑜𝑛𝑑 =
ln 𝑟2 𝑟1
2𝜋𝐿𝑘

59. Consider now the composite system of Figure below Recalling how we treated
the composite plane wall and neglecting the interfacial contact resistances,

60. The heat transfer rate may be expressed as
𝑞𝑟 =
𝑇∞,1 − 𝑇∞,4
1
2𝜋𝑟1𝐿ℎ1
+
ln 𝑟2 𝑟1
2𝜋𝑘𝐴𝐿
+
ln 𝑟3 𝑟2
2𝜋𝑘𝐵𝐿
+
ln 𝑟4 𝑟3
2𝜋𝑘𝐶𝐿
+
1
2𝜋𝑟4𝐿ℎ4
The foregoing result may also be expressed in terms of an overall heat
transfer coefficient.
𝑞𝑟 =
𝑇∞,1 − 𝑇∞,4
𝑅𝑡𝑜𝑡
= 𝑈𝐴 𝑇∞,1 − 𝑇∞,4

61. THE SPHERE
 Now consider applying the alternative method to analysing
conduction in the hollow sphere of Figure below
 For the differential control volume of the figure, energy
conservation requires that 𝑞𝑟 = 𝑞𝑟+𝑑𝑟 for steady-state, one-
dimensional conditions with no heat generation. The appropriate
form of Fourier’s law is
𝑞𝑟 = −𝑘𝐴
𝑑𝑇
𝑑𝑟
= −𝑘(4𝜋𝑟2)
𝑑𝑇
𝑑𝑟
Where A = 4𝜋𝑟2 is the area normal to the direction of heat flow.

62. Acknowledging that qr is a constant, independent of r,
from heat Equation may be expressed in the integral
form
𝑞𝑟
4𝜋 𝑟1
𝑟2 𝑑𝑟
𝑟2
= −
𝑇𝑠,1
𝑇𝑠,2
𝑘 𝑇 𝑑𝑇
Assuming constant k,
𝑞𝑟 =
4𝜋𝑘 𝑇𝑠,1 − 𝑇𝑠,2
1 𝑟1 − 1 𝑟2
Remembering that the thermal resistance is defined as
the temperature difference divided by the heat transfer
rate, we obtain
𝑅𝑡.𝑐𝑜𝑛𝑑 =
1
4𝜋𝑘
1
𝑟1
−
1
𝑟2

63.  Spherical composites may be treated in much the
same way as composite walls and cylinders, where
appropriate forms of the total resistance and overall
heat transfer coefficient may be determined.