CHAPTER TWO
Steady-State Conduction—One Dimension
Dr. Haider Ali Hussein
CHAPTER TWO
2.1 Introduction
We now wish to examine the applications of Fourier’s law of heat conduction to
calculation of heat flow in some simple one-dimensional systems. Several
different physical shapes may fall in the category of one-dimensional systems:
cylindrical and spherical systems are one-dimensional when the temperature in
the body is a function only of radial distance and is independent of azimuth angle
or axial distance. The heat transfer through the wall is in the normal direction to
the wall surface, and no significant heat transfer takes place in the wall in other
directions (Fig. 2–1).
FIGURE 2–1 Heat flow through a wall is one dimensional when the
temperature of the wall varies in one direction only.
2-2 THE PLANE WALL
2.2.1 Thermal conductivity (function of temperature)
First consider the plane wall where a direct application of Fourier’s law may be
made. Integration yields:
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CHAPTER TWO
Steady-State Conduction—One Dimension
q = KA
Dr. Haider Ali Hussein
T2 − T1
ΔX
when the thermal conductivity is considered constant. The wall
thickness is Δx, and T1 and T2 are the wall-face temperatures. If the
thermal conductivity varies with temperature according to some linear
relation k =K 0 (1+βT)
2.2.2 The Thermal Resistance Concept
At this point we note that, for the special case of one-dimensional heat transfer
with no internal energy generation and with constant properties. Defining
resistance as the ratio of a driving potential to the corresponding transfer rate, it
follows from Fourier’s law that the thermal resistance for conduction in a plane
wall is:
q = KA
T2 − T1
ΔX
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Steady-State Conduction—One Dimension
𝑅𝑅𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 =
Dr. Haider Ali Hussein
𝑇𝑇2 −𝑇𝑇1
𝑞𝑞
=
𝐿𝐿
𝐾𝐾𝐾𝐾
A thermal resistance may also be associated with heat transfer by convection at
a surface. From Newton’s law of cooling,
q= hA(𝑻𝑻𝑺𝑺 - 𝑻𝑻∞ )
The Composite Wall
𝑅𝑅𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 =
𝑇𝑇𝑆𝑆 − 𝑇𝑇∞
𝑞𝑞
=
1
ℎ𝐴𝐴
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
thermal resistances due to layers of different materials. Consider the series
composite wall of Figure 3.2. The one-dimensional heat transfer rate for this
system may be expressed as
where 𝑇𝑇∞1 , 𝑇𝑇∞4 is the overall temperature difference and the summation
includes all thermal resistances. Hence
Alternatively, the heat transfer rate can be related to the temperature difference
and resistance associated with each element. For example,
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CHAPTER TWO
Steady-State Conduction—One Dimension
Dr. Haider Ali Hussein
FIGURE 2.2 Equivalent thermal circuit for a series composite wall.
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
Q
=UA ΔT
where ΔT is the overall temperature difference. The overall heat transfer
coefficient is related to the total thermal resistance
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CHAPTER TWO
Steady-State Conduction—One Dimension
Dr. Haider Ali Hussein
EXAMPLE 2.1
Consider a 0.8-m-high and 1.5-m-wide glass window with a thickness of 8 mm
and a thermal conductivity of k = 0.78 W/m · °C. Determine the steady rate of
heat transfer through this glass window and the temperature of its inner
surface for a day during which the room is maintained at 20°C while the
temperature of the outdoors is 10°C. Take the heat transfer coefficients on the
inner and outer surfaces of the window to be h1 = 10 W/m2 · °C and h2 = 40
W/m2 · °C .
SOL
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CHAPTER TWO
Steady-State Conduction—One Dimension
Dr. Haider Ali Hussein
EXAMPLE 2.2
Consider a 0.8-m-high and 1.5-m-wide double-pane window consisting of two 4mm-thick layers of glass (k = 0.78 W/m · °C) separated by a 10-mm-wide
stagnant air space (k= 0.026 W/m · °C). Determine the steady rate of heat
transfer through this double-pane window and the temperature of its inner
surface for a day during which the room is maintained at 20°C while the
temperature of the outdoors is 10°C. Take the convection heat transfer
coefficients on the inner and outer surfaces of the window to be h1 = 10 W/m2
· °C and h2 = 40 W/m2 · °C.
SOL
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CHAPTER TWO
Steady-State Conduction—One Dimension
Dr. Haider Ali Hussein
GENERALIZED THERMAL RESISTANCE NETWORKS
The thermal resistance concept or the electrical analogy can also be used to
solve steady heat transfer problems that involve parallel layers or combined
series-parallel arrangements. Consider the composite wall shown in Fig. 2.3,
which consists of two parallel layers. The thermal resistance network, which
consists of two parallel resistances, can be represented as shown in the figure.
Noting that the total heat transfer is the sum of the heat transfers through
each layer, we have.
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CHAPTER TWO
Steady-State Conduction—One Dimension
Dr. Haider Ali Hussein
FIGURE 3.2 Thermal resistance network for two parallel layers
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CHAPTER TWO
Steady-State Conduction—One Dimension
Dr. Haider Ali Hussein
EXAMPLE 2.3
A 3-m-high and 5-m-wide wall consists of long 16-cm 22-cm cross section
horizontal bricks (k = 0.72 W/m · °C) separated by 3-cm-thick plaster layers (k =
0.22 W/m · °C). 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/m · °C) on the inner side of the wall.
The indoor and the outdoor temperatures are 20°C and 10°C, and the
convection heat transfer coefficients on the inner and the outer sides are h1 =
10 W/m2 · °C and h2 =25 W/m2 · °C, respectively. Assuming one-dimensional
heat transfer and disregarding radiation, determine the rate of heat transfer
through the wall.
SOL
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CHAPTER TWO
Steady-State Conduction—One Dimension
Dr. Haider Ali Hussein
HEAT CONDUCTION IN CYLINDERS AND SPHERES
Consider a long cylinder of inside radius ri, outside radius ro, and length L, such
as the one shown in Figure 2.4.We expose this cylinder to a temperature
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CHAPTER TWO
Steady-State Conduction—One Dimension
Dr. Haider Ali Hussein
differential Ti −To 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. Again, Fourier’s law is used by inserting the proper area relation. The area for
heat flow in the cylindrical system is
Figure 2.4 One-dimensional heat flow through a hollow cylinder and electrical
analog
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CHAPTER TWO
Steady-State Conduction—One Dimension
Dr. Haider Ali Hussein
Spheres
Spherical systems may also be treated as one-dimensional when the temperature
is a function of radius only. The heat flow is then
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CHAPTER TWO
Steady-State Conduction—One Dimension
Dr. Haider Ali Hussein
EXAMPLE 2.4
A thick-walled tube of stainless steel [18% Cr, 8% Ni, k =19 W/m・ ◦C] with
2-cm inner diameter (ID) and 4-cm outer diameter (OD) is covered with a 3-cm
layer of asbestos insulation [k =0.2 W/m・ ◦C]. If the inside wall temperature
of the pipe is maintained at 600◦C, calculate the heat loss per meter of length.
Also calculate the tube–insulation interface temperature
SOL
where Ta is the interface temperature, which may be obtained as
Ta =595.8◦C
Case 1
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CHAPTER TWO
Steady-State Conduction—One Dimension
Dr. Haider Ali Hussein
Case 2
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
for wall
for cylinder
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CHAPTER TWO
Steady-State Conduction—One Dimension
Dr. Haider Ali Hussein
Case 3 Critical insulation thickness
EXAMPLE
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/𝑚𝑚2 ◦C.
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CHAPTER TWO
Steady-State Conduction—One Dimension
Dr. Haider Ali Hussein
Calculate the heat loss from a 200◦C, 5.0-cm-diameter pipe when covered with
the critical radius of insulation and without insulation
Sol
HW
2-1 A wall 2 cm thick is to be constructed from material that has an average
thermal conductivity of 1.3 W/m・ ◦C. The wall is to be insulated with material
having an average thermal conductivity of 0.35W/m・ ◦C, so that the heat loss
per square meter will not exceed 1830W. Assuming that the inner and outer
surface temperatures of the insulated wall are 1300 and 30◦C, calculate the
thickness of insulation required.
2-2 A certain material 2.5 cm thick, with a cross-sectional area of 0.1 m2, has one
side maintained at 35◦C and the other at 95◦C. The temperature at the center
plane of the material is 62◦C, and the heat flow through the material is 1 kW.
Obtain an expression for the thermal conductivity of the material as a function of
temperature.
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CHAPTER TWO
Steady-State Conduction—One Dimension
Dr. Haider Ali Hussein
2.3 Find the heat transfer per unit area through the composite wall in Figure .
Assume one-dimensional heat flow.
2.4 A steel tube having k =46W/m・ ◦C has an inside diameter of 3.0 cm and
a tube wall thickness of 2 mm. A fluid flows on the inside of the tube producing
a convection coefficient of 1500W/m2 ・ ◦Con the inside surface, while a
second fluid flows across the outside of the tube producing a convection
coefficient of 197 W/m2 ・ ◦C on the outside tube surface. The inside fluid
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CHAPTER TWO
Steady-State Conduction—One Dimension
Dr. Haider Ali Hussein
temperature is 223◦C while the outside fluid temperature is 57◦C. Calculate
the heat lost by the tube per meter of length.
2.5 A spherical tank, 1 m in diameter, is maintained at a temperature of 120◦C
and exposed to a convection environment.With h=25W/m2 ・ ◦C and T∞ =15
◦C, what thickness of urethane foam should be added to ensure that the outer
temperature of the insulation does not exceed 40◦C? What percentage reduction
in heat loss results from installing this insulation?
2.6 Derive an expression for the thermal resistance through a hollow spherical
shell of inside radius ri and outside radius ro having a thermal conductivity k.
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