2.

ONE DIMENSIONAL STEADY STATE HEAT

CONDUCTION
2.1 Objectives of conduction analysis
The primary objective is to determine the temperature field, T(x,y,z,t), in a body (i.e. how
temperature varies with position within the body)
T(x,y,z,t) depends on:
- Boundary conditions
- Initial condition
- Material properties (k, cp, )
- Geometry of the body (shape, size)
Why do we need T (x, y, z, t)?
- To compute heat flux at any location (using Fourier’s eqn.)
- Compute thermal stresses, expansion, and deflection due to temp. Etc.
- Design insulation thickness
- Chip temperature calculation
- Heat treatment of metals
2.2 General Conduction Equation
Recognize that heat transfer involves an energy transfer across a system boundary. The
analysis for such process begins from the 1st Law of Thermodynamics for a closed system:
𝑑𝐸
| = 𝑄𝑖𝑛 − 𝑊𝑜𝑢𝑡
𝑑𝑡 𝑠𝑦𝑠𝑡𝑒𝑚

The above equation essentially represents Conservation of Energy. The sign convention on
work is such that negative work out is positive work in.
𝑑𝐸
| = 𝑄𝑖𝑛 + 𝑊𝑖𝑛
𝑑𝑡 𝑠𝑦𝑠𝑡𝑒𝑚

The work – in term could describe an electric current flow across the system boundary and
through a resistance inside the system. Alternatively it could describe a shaft turning across the
system boundary and overcoming friction within the system. The net effect in either case would
cause the internal energy of the system to rise. In heat transfer we generalize all such terms as
“heat sources”.
𝑑𝐸
| = 𝑄𝑖𝑛 + 𝑄𝑔𝑒𝑛
𝑑𝑡 𝑠𝑦𝑠𝑡𝑒𝑚

The energy of the system will in general include system’s ethalpy, H, potential energy, ½ mgz,
or kinetic energy, ½ mv2. In case of heat transfer problems, the latter two terms could often be
neglected. In this case,
𝐸 = 𝐻 = 𝑚. ℎ = 𝑚𝑐𝑝 (𝑇 − 𝑇𝑟𝑒𝑓 ) = 𝜌. 𝑉. 𝑐𝑝 (𝑇 − 𝑇𝑟𝑒𝑓 )
where Tref is the reference temperature at which the energy of the system is defined as zero.
When we differentiate the above expression with respect to time, the reference temperature,
being constant, disappears:
 𝑑𝑇
𝜌𝑐𝑝 . 𝑉. 𝑑𝑡 | = 𝑄𝑖𝑛 + 𝑄𝑔𝑒𝑛
𝑠𝑦𝑠𝑡𝑒𝑚

Consider the differential control element shown below. Heat is assumed to flow through the
element in the positive directions as shown by the 6 heat vectors.

In the equation above we substitute the 6 heat inflows/outflows using the appropriate sign:
𝑑𝑇
𝜌. 𝑐𝑝 . (∆𝑥. ∆𝑦. ∆𝑧). 𝑑𝑡 | = 𝑞𝑥 − 𝑞𝑥+∆𝑥 + 𝑞𝑦 − 𝑞𝑦+∆𝑦 + 𝑞𝑧 − 𝑞𝑧+∆𝑧 + 𝑄𝑔𝑒𝑛
𝑠𝑦𝑠𝑡𝑒𝑚

Substituting for each of the conduction terms using the Fourier Law gives:
𝜕𝑇
𝜌. 𝑐𝑝 . (∆𝑥. ∆𝑦. ∆𝑧). |
𝜕𝑡 𝑠𝑦𝑠𝑡𝑒𝑚
𝜕𝑇 𝜕𝑇 𝜕 𝜕𝑇
= {−𝑘(∆𝑦. ∆𝑧). − [−𝑘(∆𝑦. ∆𝑧). + (−𝑘. (∆𝑦. ∆𝑧). ) . ∆𝑥]}
𝜕𝑥 𝜕𝑥 𝜕𝑥 𝜕𝑥
𝜕𝑇 𝜕𝑇 𝜕 𝜕𝑇
+ {−𝑘(∆𝑥. ∆𝑧). − [−𝑘(∆𝑥. ∆𝑧). + (−𝑘. (∆𝑥. ∆𝑧). ) . ∆𝑦]}
𝜕𝑦 𝜕𝑦 𝜕𝑦 𝜕𝑦
𝜕𝑇 𝜕𝑇 𝜕 𝜕𝑇
+ {−𝑘(∆𝑥. ∆𝑦). − [−𝑘(∆𝑥. ∆𝑦). + (−𝑘. (∆𝑥. ∆𝑦). ) . ∆𝑧]}
𝜕𝑧 𝜕𝑧 𝜕𝑧 𝜕𝑧
+ 𝑞(∆𝑥. ∆𝑦. ∆𝑧)
Where, q, is defined as the internal heat generation per unit volume.
The above equation reduces to:
𝜕𝑇
𝜌. 𝑐𝑝 . (∆𝑥. ∆𝑦. ∆𝑧). |
𝜕𝑡 𝑠𝑦𝑠𝑡𝑒𝑚
𝜕 𝜕𝑇 𝜕 𝜕𝑇
= {− [ (−𝑘. (∆𝑦. ∆𝑧). )] . ∆𝑥} + {− [ (−𝑘(∆𝑥. ∆𝑧). )] . ∆𝑦}
𝜕𝑥 𝜕𝑥 𝜕𝑦 𝜕𝑦
𝜕 𝜕𝑇
+ {− [ (−𝑘(∆𝑥. ∆𝑦). )] . ∆𝑧} + 𝑞(∆𝑥. ∆𝑦. ∆𝑧)
𝜕𝑧 𝜕𝑧
Dividing by the volume (xyz),
 𝜕𝑇 𝜕 𝜕𝑇 𝜕 𝜕𝑇 𝜕 𝜕𝑇
𝜌. 𝑐𝑝 𝜕𝑡 | = − 𝜕𝑥 (−𝑘. 𝜕𝑥 ) − 𝜕𝑦 (−𝑘. 𝜕𝑦) − 𝜕𝑧 (−𝑘. 𝜕𝑧 ) + 𝑞
𝑠𝑦𝑠𝑡𝑒𝑚

which is the general conduction equation in three dimensions.

In the case where k is independent of x, y and z then;
𝜌𝑐𝑝 𝜕𝑇 𝜕2 𝑇 𝜕2 𝑇 𝜕2 𝑇 𝑞
. 𝜕𝑡 | = 𝜕𝑥 2 + 𝜕𝑦 2 + 𝜕𝑧 2 + 𝑘
𝑘 𝑠𝑦𝑠𝑡𝑒𝑚
𝑘
It is known that, 𝛼 = 𝜌𝑐 = a property known as the thermal diffusivity. Hence,
𝑝

1 𝑑𝑇 𝜕2 𝑇 𝜕2 𝑇 𝜕2 𝑇 𝑞
. | = 𝜕𝑥 2 + 𝜕𝑦 2 + 𝜕𝑧 2 + 𝑘
𝛼 𝑑𝑡 𝑠𝑦𝑠𝑡𝑒𝑚
1 𝑑𝑇 𝑞
Or: 𝛼 . 𝑑𝑡 | = ∇2 𝑇 + 𝑘
𝑠𝑦𝑠𝑡𝑒𝑚

2.3 Boundary and Initial Conditions

The objective of deriving the heat diffusion equation is to determine the temperature distribution
within the conducting body.
We have set up a differential equation, with T as the dependent variable. The solution will give
us, T(x,y,z). Solution depends on boundary conditions (BC) and initial conditions (IC).
How many BC’s and IC’s?
- Heat equation is second order in spatial coordinate. Hence, 2 BC’s needed for
each coordinate.
* 1D problem: 2 BC in x-direction
* 2D problem: 2 BC in x-direction, 2 in y-direction
* 3D problem: 2 in x-dir., 2 in y-dir., and 2 in z-dir.
- Heat equation is first order in time. Hence one IC needed.
2.4 Heat Diffusion Equation for a One Dimensional System
Consider the system shown above. The top, bottom, front and back of the cube are insulated, so
that heat can be conducted through the cube only in the x direction. The internal heat
generation per unit volume is q (W/m3).
Consider the heat flow through a differential element of the cube;

From the 1st Law we write for the element:

(𝐸𝑖𝑛 − 𝐸𝑜𝑢𝑡 ) + 𝐸𝑔𝑒𝑛 = 𝐸𝑠𝑡 …………………………………………………………………(2.1)
𝜕𝐸
𝑞𝑥 − 𝑞𝑥+∆𝑥 + 𝐴𝑥 (∆𝑥)𝑞 = ……………………………………………………………….(2.2)
𝜕𝑡
𝜕𝑇
𝑞𝑥 = −𝑘𝐴𝑥 . 𝜕𝑥 …………………………………………………………………(2.3)
𝜕𝑞𝑥
𝑞𝑥+∆𝑥 = 𝑞𝑥 + ∆𝑥 ………………………………………………………………….(2.4)
𝜕𝑥
𝜕𝑇 𝜕𝑇 𝜕 𝜕𝑇 𝜕𝑇
−𝑘𝐴 𝜕𝑥 + 𝑘𝐴 𝜕𝑥 + 𝐴 𝜕𝑥 (𝑘 𝜕𝑥 ) ∆𝑥 + 𝐴∆𝑥. 𝑞 = 𝜌𝐴𝑐∆𝑥 𝜕𝑡 ………………………………….(2.5)
Hence,
𝜕 𝜕𝑇 𝜕𝑇
(𝑘 ) + 𝑞 = 𝜌𝑐 …………………………………………………………...(2.6)
𝜕𝑥 𝜕𝑥 𝜕𝑡

The first term of equation (2.6), represents longitudinal conduction, the second term is for
internal heat generation and the one on the RHS is thermal inertia.
If, k, is constant, then;
𝜕2 𝑇 𝑞 𝜌𝑐 𝜕𝑇 1 𝜕𝑇
+𝑘 = = 𝛼 . 𝜕𝑡 …………………………………………………………………….(2.7)
𝜕𝑥 2 𝑘 𝜕𝑡
 For T to rise, LHS must be positive (heat input is positive)
 For a fixed heat input, T rises faster for higher 
 In this special case, heat flow is 1D. If sides were not insulated, heat flow could be
2D, 3D.
2.5 One Dimensional Steady State Heat Conduction
The plane wall:

𝑑 𝑑𝑇
The differential equation governing heat diffusion is: 𝑑𝑥 (𝑘 𝑑𝑥 ) = 0
With constant k, the above equation may be integrated twice to obtain the general solution:
𝑇(𝑥) = 𝐶1 𝑥 + 𝐶2
Where, C1 and C2 are constants of integration. To obtain the constants of integration, we apply
the boundary conditions at x = 0 and x = L, in which case;
T(0) = Ts,1 and T(L) = Ts,2
Once the constants of integration are substituted into the general equation, the temperature
distribution is obtained:

The heat flow rate across the wall is given by:

𝑑𝑇 𝑘𝐴 𝑇𝑠,1 −𝑇𝑠,2
𝑞𝑥 = −𝑘𝐴. 𝑑𝑥 = (𝑇𝑠,1 − 𝑇𝑠,2 ) =
𝐿 𝐿/𝑘𝐴
Thermal resistance (electrical analogy):
Physical systems are said to be analogous if they obey the same mathematical equation. The
above relations can be put into the form of Ohm’s law:

Using this terminology, it is common to speak of a thermal resistance:

∆𝑇 = 𝑞𝑅𝑡ℎ𝑒𝑟𝑚

A thermal resistance may also be associated with heat transfer by convection at a surface.
From Newton’s law of cooling,

the thermal resistance for convection is then;

Applying thermal resistance concept to the plane wall, the equivalent thermal circuit for the
plane wall with convection boundary conditions is shown in the figure below;

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;

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;

Since the resistance are in series, it follows that;

Composite walls:
Thermal Resistances in Series
Consider three blocks, A, B and C, as shown. They are insulated on top, bottom, front and back.
Since the energy will flow first through block A and then through blocks B and C, we say that
these blocks are thermally in a series arrangement.
The steady state heat flow rate through the walls is given by:

1
Where, 𝑈 = 𝑅 , is the overall heat transfer coefficient. In the above case, U is expressed as;
𝑡𝑜𝑡 .𝐴

Series-parallel arrangement

The following assumptions are made with regard to the above thermal resistance model:
1) Face between B and C is insulated.
2) Uniform temperature at any face normal to X.
1-D radial conduction through a cylinder
One frequently encountered problem is that of heat flow through the walls of a pipe or
through the insulation placed around a pipe. Consider the cylinder shown below. The pipe is
either insulated on the ends or is of sufficient length, L, that heat losses through the ends is
negligible. Assume no heat sources within the wall of the tube. If T1>T2, heat will flow
outward, radially, from the inside radius, R1, to the outside radius, R2. The process will be
described by the Fourier Law.
 1 𝑑 𝑑𝑇
The differential equation governing heat diffusion is: . (𝑟. )=0
𝑟 𝑑𝑟 𝑑𝑟

The solution of the differential equation can be obtained as;

𝑇 = 𝐶1 ln(𝑟) + 𝐶2 , and the constants are determined from the boundary conditions; i.e. if T = T1
at r = R1 and T = T2 at r = R2, then;
𝑇1 −𝑇 ln(𝑟/𝑟1 )
= {𝑙𝑛(𝑟 } ………………………………………………………………………..(2.8)
𝑇1 −𝑇2 2 /𝑟1 )

Similarly, since the heat flow, 𝑄 = −𝑘𝐴(𝑑𝑇/𝑑𝑟), then for a length, L, in the axial direction, then
the heat flow can be determined by differentiating equation (2.8) as;
2𝜋𝑘𝐿(𝑇1 −𝑇2 )
𝑄= 𝑟
𝑙𝑛( 2 )
𝑟1

𝑟
𝑙𝑛( 2 )
𝑟1
Hence, the thermal resistance in this case can be expressed as: 2𝜋𝑘𝐿
Composite cylindrical walls
 Critical Insulation Thickness

Objective: To decrease q, Rtot must be increased. This is done by varying, ro. But as ro increases,
the first term of the Rtot expression increases and the second term decreases.
This leads to a maximum – minimum problem.

The point of extrema can be found by setting;

In order to determine if it is a maxima or a minima, we make the second derivative zero:

1-D radial conduction in a sphere:

2.6 Summary of Electrical Analogy

2.7 One-Dimensional Steady State Conduction with Internal Heat Generation
Applications: Current carrying conductor, chemically reacting systems, nuclear reactors.
𝐸
Energy generated per unit volume is given by: 𝑞 = 𝑉
Plane wall with heat source
Assumptions: 1D conduction, steady state, constant k, and uniform, q.

𝑑2 𝑇 𝑞
The governing differential is; 𝑑𝑥 2 + 𝑘 = 0
Boundary conditions: x = -L, T = Ts,1
x = L, T = Ts,2
𝑞
The solution of the governing differential is; 𝑇 = − 2𝑘 𝑥 2 + 𝐶1 𝑥 + 𝐶2
The boundary conditions are used to find C1 and C2 and the final solution is:
𝑞𝐿2 𝑥2 𝑇𝑠,2 −𝑇𝑠,1 𝑥 𝑇𝑠,2 +𝑇𝑠,1
𝑇= (1 − 𝐿2 ) + .𝐿 +
2𝑘 2 2
𝑑𝑇
The heat flux (Heat conducted per unit cross sectional area), 𝑞𝑥′′ = −𝑘. 𝑑𝑥
Note
From the above expressions, it may be observed that the solution for temperature is no longer
linear. As an exercise, show that the expression for heat flux is no longer independent of x.
Hence thermal resistance concept is not correct to use when there is internal heat generation
Cylinder with heat source
Assumptions: 1D conduction, steady state, constant k, and uniform, q .


1 𝑑 𝑑𝑇 𝑞
The governing differential equation is; 𝑟 . 𝑑𝑟 (𝑟. 𝑑𝑟 ) + 𝑘 = 0
Boundary conditions are; r = ro, T = Ts
 r = 0, dT/dr = 0
The solution of the DE with the boundary conditions considered comes out as;
𝑞 𝑟2
𝑇(𝑟) = 4𝑘 𝑟𝑜2 (1 − 𝑟 2 ) + 𝑇𝑠
𝑜

Exercise
Ts may not be known. Instead, T∞ and h may be specified. Eliminate Ts, using T∞ and h.
Problems
1. The steady-state temperature distribution in a one–dimensional slab of thermal conductivity
50W/m.K and thickness 50 mm is found to be T= a+bx2, where a=200oC, b = -2000oC/ m2, T
is in degrees Celsius and x in meters.
(a) What is the heat generation rate in the slab? (Ans: 2×105
3
W/m )
(b) Determine the heat fluxes at the two wall faces. From the given temperature distribution
and the heat fluxes obtained, can you comment on the heat generation rate? (Ans: 0 W/m2
and 10,000 W/m2)
2. Consider a solar pond having three distinct layers of water-salt solution. The top and bottom
layers are well mixed with salt. These layers are subjected to natural convention, but the
middle layer is stationary. With this arrangement, the top and bottom surfaces of the middle
layer is maintained at uniform temperature T1 and T2, where T1 > T2. Solar radiation is
absorbed in the middle layer in the form q=Ae-mx, resulting in the following temperature
distribution in the central layer;

In the above equation, k is the thermal conductivity, and the constants A (W/m 3), a (1/m), B
(K/m) and C(K) are also known.
Obtain expressions for the interfacial heat flux from the bottom layer to the middle layer, and
from the middle layer to the top layer. Are the conditions are steady or transient? Next, obtain
an expression for the rate at which thermal energy is generated in the entire middle layer, per
unit surface area.
3. Consider 1D heat transfer across a slab with thermal conductivity k and thickness L. The
steady state temperature is of the form T=Ax3+Bx2+Cx+D. Find expressions for the heat
generation rate per unit volume in the slab and heat fluxes at the two wall faces (i.e. x = 0, L).
4. A 1–m-long metal plate with thermal conductivity k=50W/m.K is insulated on its sides. The
top surface is maintained at 100oC while the bottom surface is convectively cooled by a fluid
at 20oC. Under steady state conditions and with no volumetric heat generation, the
temperature at the midpoint of the plate is measured to be 85oC. Calculate the value of the
convection heat transfer coefficient at the bottom surface.
2
(Ans: 30 W.m K)
5. Consider a person standing in a room maintained at 20oC with an exposed surface area of 1.7
m2. The deep body temperature of a human body is 37oC, and the thermal conductivity of the
human tissue near the skin is about 0.3 W/mK. The body is losing heat at a rate of 150 W by
 natural convection and radiation to the surroundings. Taking the body temperature 0.5 cm
beneath the skin to be 37oC, determine the skin temperature of the person. (Ans: 35.5oC)
6. The wall of a building is a multi-layered composite consisting of brick (100-mm layer), a
100-mm layer of glass fiber (paper faced. 28kg/m2), a 10-mm layer of gypsum plaster
(vermiculite), and a 6-mm layer of pine panel. If hinside is 10W/m2.K and houtside is 70W/m2.K,
calculate the total thermal resistance and the overall coefficient for heat transfer.
(Ans: 2.93 m2.K/W, 0.341 W/m2K)
7. The wall of an oven is a composite of the following layers. Layers A has a thermal
conductivity kA=20W/m.K, and layer C has a thermal conductivity k C=50W/m.K. The
corresponding thicknesses are LA=0.30m and LC=0.15m, respectively. Layer B is sandwiched
between layers A and C, is of known thickness, LB = 0.15m, but unknown thermal
conductivity kB. Under steady-state operating conditions, the outer surface temperature is
measured to be Ts,0=200oC. Measurements also tell us that the inner surface temperature Ts,i
is 600oC and the oven air temperature is T =800oC. The inside convection coefficient h is
known to be 25W/m2.K. Find the value of kB. (Ans: 1.53 W/mK)
8. A 3 m high and 5 m wide wall consists of long 16 cm × 22 cm cross section horizontal bricks
(k = 0.72 W/mK) separated by 3 cm thick plaster layers (k = 0.22 W/mK. There are also 2
cm thick plaster layers on each side of the brick and a 3 cm thick rigid foam (k = 0.026
W/mK) on the inner side of the wall, as shown in Figure below. The indoor and outdoor
temperatures are 20oC and -10oC, and the convection heat transfer coefficients on the inner
and the outer sides are h1 = 10 W/m2K and h2 = 25 W/m2K respectively. Assuming one –
dimensional heat transfer and disregarding radiation, determine the rate of heat transfer
through the wall. (Ans: 263 W)

9. A steam pipe of 0.12 m outside diameter is insulated with a 20-mm-thick layer of calcium
silicate. If the inner and outer surfaces of the insulation are at temperatures of Ts,1=800 K and
Ts,2=490 K, respectively, what is the heat loss per unit length of the pipe? (Ans: 603 W/m)
10. A cylindrical nuclear fuel rod of 0.1m dia has a thermal conductivity k=0.0W/m.K and
generates uniformly 24,000W/m3. This rod is encapsulated within another cylinder having an
outer radius of 0.2m and a thermal conductivity of 4W/m.K. The outer surface is cooled by a
coolant fluid at 100oC, and the convection coefficient between the outer surface and the
coolant is 20W/m2.K. Find the temperatures at the interface between the two cylinders and at
the outer surface. (Ans: 150.8oC, 130oC)