TWO DIMENSIONAL STEADY STATE HEAT CONDUCTION

12/19/2017 Heat Transfer 1
HEAT TRANSFER
(MEng 3121)
TWO-DIMENSIONAL STEADY STATE
HEAT CONDUCTION
Chapter 3
Debre Markos University
Mechanical Engineering
Department
Prepared and presented by:
Tariku Negash
E-mail: thismuch2015@gmail.com
Lecturer at Mechanical Engineering
Department Institute of Technology, Debre
Markos University, Debre Markos, Ethiopia

12/19/2017Heat Transfer
2
For two dimensional steady state, with no heat generation, the Laplace
equation can be applies. Assuming: constant thermal conductivity (K).
3.1 Introduction
Then the heat ﬂow in the x and y directions may be calculated from the
Fourier equations
The total heat flow at
any point in the
materials is the
resultant of 𝒒 𝒙 𝑎𝑛𝑑 𝒒 𝒚
𝑎𝑡 𝑎𝑛𝑦 𝑝𝑜𝑖𝑛𝑡.

3
 The solution to this equation may be obtained by analytical, graphical
techniques. or numerical (finite-difference, finite-element, or
boundary-element)
Thus the total heat-flow vector is directed so that it is perpendicular to
the lines of constant temperature in the material, as shown in Fig. 3-1.
So if the temperature distribution in the material is known, we may
easily establish the heat flow.
 In contrast to the analytical methods, which provide exact results at any
point, graphi-cal and numerical methods can provide only approximate
results at discrete points.

4
3.1 Mathematical Analysis of 2D Heat
Conduction
 Consider the rectangular plate shown in Fig. 3-2.
 Three sides of the plate are maintained at the
constant temperature 𝑇1, and the upper side has
some temperature distribution impressed upon it.
 This distribution could be simply a constant
temperature or something more complex, such as a
sine-wave distribution. We shall consider both
cases.
 After some mathematical manipulation (see
Reference 3 @ p232-234)
(3.1)
 The temperature distribution T(x, y), but to simplify the solution we
introduce the transformation
T2

5
A two-dimensional rectangular plate is subjected to prescribed boundary
conditions. Using the results of the exact solution for the heat equation
(3.1) ,
(a). calculate the temperature at the midpoint (1, 0.5) by considering the
first five nonzero terms of the infinite series that must be evaluated.
(b). assess the error resulting from using only the first three terms of the
infinite series.
(c). Plot the temperature distributions T(x,0.5) and T(1.0, y).
Examples 3.1
Fig 3.1

6
The Conduction Shape Factor and the Dimensionless Conduction
Heat Rate
In general, finding analytical solutions to the
two- or three-dimensional heat equation is
time-consuming and, in many cases, not
possible.
Therefore, a different approach is often
taken.
For example, in many instances, two- or
three-dimensional conduction problems may
be rapidly solved by utilizing existing
solutions to the heat diffusion equation.
These solutions are reported in terms of a shape factor S or a steady-
state dimensionless conduction heat rate, q*ss.
The shape factor is defined such that
3.1 ∆𝑻 𝟏−𝟐 is the temperature difference between boundaries

7
Two-dimensional conduction resistance
3.2
Shape factors have been obtained analytically for numerous two- and
three-dimensional systems, and results are summarized in Table 2.1 for
some common configurations.

8
Conduction shape factors and dimensionless conduction heat rates for
selected systems.

12
In cases 1 through 8 and case 11, two dimensional conduction is presumed
to occur between the boundaries that are maintained at uniform
temperature, with ∆𝑻 𝟏−𝟐 = 𝑻 𝟏 − 𝑻 𝟐
In case 9, three-dimensional conduction exists in the corner region, while
in case 10 conduction occurs between an isothermal disk 𝑻 𝟏) and a
semi−infinite medium of uniform temperature (𝑻 𝟐) at locations well
removed from the disk.
Shape factors may also be defined for one−dimensional geometries, and
from the disk.
Cases 12 through 15 are associated with conduction from objects held at an
isothermal temperature (𝑻 𝟏) that are embedded within an infinite medium
of uniform temperature (𝑻 𝟐) at locations removed from the object.
For these infinite medium cases, useful results may be obtained by defining
a characteristic length
3.3

13
where 𝑨 𝑺 is the surface area of the object.
Conduction heat transfer rates from the object to the infinite medium
may then be reported in terms of a dimensionless conduction heat rate

14
The Energy Balance Method
In many cases, it is desirable to develop the finite-difference equations by
an alternative method called the energy balance method.
this approach enables one to analyze many different phenomena such as
problems
 involving multiple materials,
 embedded heat sources, or
 exposed surfaces that do not align with an axis of the coordinate
system.
In the energy balance method, the finite-difference equation for a node is
obtained by applying conservation of energy to a control volume about the
nodal region.
Since the actual direction of heat flow (into or out of the node) is often
unknown, it is convenient to formulate the energy balance by assuming
that all the heat flow is into the node.

15
3.1
 Consider applying Equation 3.1 to a control volume about the interior
node (m, n) of Figure 3.3.
 For two-dimensional conditions, energy exchange is influenced by
conduction between (m, n) and its four adjoining nodes, as well as by
generation. Hence Equation 3.1 reduces to
Fig 3.3 Conduction to an interior node from its adjoining nodes.
where i refers to the neighboring nodes,
𝑞 𝑖 →(𝑚, 𝑛) is the conduction rate between
nodes, and unit depth is assumed.
To evaluate the conduction rate terms, we
assume that conduction transfer occurs
exclusively through lanes that are oriented
in either the x- or y-direction.
3.2

16
Simplified forms of Fourier's law may therefore be used.
For example, the rate at which energy is transferred by conduction from
node (m-1, n) to (m, n) may be expressed as
The quantity (∆𝑦. 1) is the rate heat transfer area, and
the term
The remaining conduction rates may be expressed as
is the finite difference approximation to
the temperature gradient at the boundary between the two nodes.
3.3
3.4
3.5
3.6

17
Substituting Equations 3.3 through 3.6 into the energy balance (3.2) and
remembering that ∆𝑦 = ∆𝑥, it follows that the finite-difference equation
for an interior node with generation is
If there is no internally distributed source of energy (q=0) , this expression
reduces to
3.7
3.8
Finite difference for convective/insulated the surface
For example, the temperature may be unknown at an insulated surface or at
a surface that is exposed to convective conditions.
For points on such surfaces, the finite-difference equation must be obtained
by applying the energy balance method.

18
 The node represents the three-quarter
shaded section and exchanges energy by
convection with an adjoining fluid at 𝑇∞.
 Conduction to the nodal region (m, n)
occurs along four different lanes from
neighboring nodes in the solid.
 The conduction heat rates 𝑞 𝑐𝑜𝑛𝑑 may be
expressed as
3.9
3.10
3.11
3.12
Note that the areas for
conduction from nodal regions
(m-1, n) and (m, n+1) are
proportional to ∆𝑦 𝑎𝑛𝑑 ∆𝑥,
respectively, whereas conduction
from (m+1, n) and (m, n -1)
occurs along lanes of width
∆𝑦 /2 and ∆𝑥 /2, respectively.
Fig 3.4

19
Conditions in the nodal region (m, n) are also influenced by convective
exchange with the fluid, and this exchange may be viewed as
occurring along half-lanes in the x- and y- directions.
The total convection rate 𝑞 𝑐𝑜𝑛𝑣 may be expressed as
In the absence of transient, three-dimensional, and generation effects,
conservation of energy, Equation 3.1, requires that the sum of Equations
3.9 - 3.12 be zero. Summing these equations and rearranging, we therefore
obtain
3.13
3.14
where again the mesh is such that ∆𝑦 = ∆𝑥

20
Table 3.1 Summary of nodal ﬁnite-difference equations with 𝒒 𝒗 = 𝟎
i

22
Corresponding of thermal resistances for two dimensional heat rate
As shown from the fig 3.4 or using Eqn. 3.9 the rate of heat transfer by
conduction from node (m-1, n) to (m, n) may be expressed as
Similarly, the rate of heat transfer by convection to (m,n) may be expressed
as
Which is similar to equation 3.13
where, for a unit depth,
Fig 3.5
3.15
3.16
3.17

23
a. Using the energy balance method, derive the finite-difference
equation for the (m, n) nodal point located on a plane, insulated surface
of a medium with uniform heat generation.
Example 3.2
Assumptions:
 Steady state conditions.
 Two-dimensional conduction.
 Constant properties.
 Uniform internal heat generation.

24
a. As an application of the foregoing finite-difference
equation, consider the following two-dimensional
system within which thermal energy is uniformly
generated at an unknown rate . The thermal
conductivity of the solid is known, as are convection
conditions at one of the surfaces. In addition,
temperatures have been measured at
locations corresponding to the nodal points of a finite-
difference mesh.
Example 3.3

25
Verifying the Accuracy of the Solution
 Even when the finite-difference equations have been properly formulated
and solved, the results may still represent a coarse approximation to the
actual temperature field.
 This behavior is a consequence of the finite spacing (∆𝑦, ∆𝑥) between
nodes and of finite-difference approximations, such as
k(∆𝑦.1)(𝑇 𝑚−1,𝑛 − 𝑇 𝑚,𝑛)∆𝑥), to Fourier's law of conduction,
−𝑘(𝑑𝑦. 1)𝜕𝑇/𝜕𝑥.
 The finite-difference approximations become more accurate as the nodal
network is refined (∆𝑦 𝑎𝑛𝑑 ∆𝑥 are reduced).
 Hence, if accurate results are desired, grid studies should be performed,
whereby results obtained for a fine grid are compared with those
obtained for a coarse grid.
Two method for verifying Accuracy of the Solution
Method 1

26
 One could, for example, reduce ∆𝑦 𝑎𝑛𝑑 ∆𝑥 by a factor of 2, thereby
increasing the number of nodes and finite-difference equations by a
factor of 4.
 If the agreement is unsatisfactory, further grid refinements could
be made until the computed temperatures no longer depend
significantly on the choice of ∆𝑦 𝑎𝑛𝑑 ∆𝑥 .
 Such grid-independent results would provide an accurate
solution to the physical problem.
Method 2
 Another option for validating a numerical solution involves comparing
results with those obtained from an exact solution.
 For example, a finite-difference solution of the physical problem
described in Figure 3.2 could be compared with the exact solution given
by Equation 3.1.
 However, this option is limited by the fact that we seldom seek
numerical solutions to problems for which there exist exact solutions.

27
 Nevertheless, if we seek a numerical solution to a complex problem
for which there is no exact solution, it is often useful to test our finite
difference procedures by applying them to a simpler version of the
problem.

28
Consider steady heat transfer in an L-shaped solid body whose cross section is
given below Fig. Heat transfer in the direction normal to the plane of the paper is
negligible, and thus heat transfer in the body is two-dimensional. The thermal
conductivity of the body is k = 15 W/m · °C, and heat is generated in the body at a
rate of 𝑔 = 2𝑥106 𝑊
𝑚3. The left surface of the body is insulated, and the bottom
surface is maintained at a uniform temperature of 90°C. The entire top surface is
subjected to convection to ambient air at 𝑇∞ = 25℃ with a convection coefficient
of ℎ = 80
𝑊
𝑚2℃
, and the right surface is subjected to heat flux at a uniform rate of
𝑞 𝑅 = 5000 W/𝑚2. The nodal network of the
Example 3.4 Case: Insulated, convective and heat flux
problem consists of 15 equally spaced
nodes with ∆𝑥 = ∆𝑦 = 1.2𝑐𝑚, as shown
in the figure. Six of the odes are at the
bottom surface, and thus their
temperatures are known. Obtain the finite
difference equations at the remaining nine
nodes and determine the nodal
temperatures by solving them.

29
Assumptions
1 Heat transfer is steady and two-dimensional, as stated.
2 Thermal conductivity is constant.
3 Heat generation is uniform.
4 Radiation heat transfer is negligible.

30
Hot combustion gases of a furnace are flowing through a concrete
chimney (k=1.4 W/m · °C) of rectangular cross section. The flow section
of the chimney is 20 cm X 40 cm, and the thickness of the wall is 10 cm.
The average temperature of the hot gases in the chimney is 𝑇𝑖 =280°C,
and the average convection heat transfer coefficient inside the chimney
is ℎ𝑖 = 75𝑊/𝑚2
°C. The chimney is losing heat from its outer surface to
the ambient air at 𝑇0 = 15°C by convection with a heat transfer
coefficient of ℎ0 = 18𝑊/𝑚2
°C and to the sky by radiation. The
emissivity of the outer surface of the wall is 𝜀 = 0.9, and the effective
sky temperature is estimated to be 250 K. Using the finite difference
method with ∆𝑥 = ∆𝑦 = 10 𝑐𝑚 and taking full advantage of symmetry,
(a) obtain the finite difference formulation of this problem for steady
two dimensional heat transfer,
(b) determine the temperatures at the nodal points of a cross section,
and
(c) evaluate the rate of heat loss for a 1-m-long section of the chimney.
Example 3.5

31
Assumptions
1 Heat transfer is given to be steady and
two-dimensional since the height of the
chimney is large relative to its cross-section,
and thus heat conduction through the
chimney in the axial direction is
negligible. It is tempting to simplify the
problem further by considering heat transfer
in each wall to be one dimensional which
would be the case if the walls were thin and
thus the corner effects were negligible.
This assumption cannot be justified in this case since the walls are very
thick and the corner sections constitute a considerable portion of the
chimney structure.
2 There is no heat generation in the chimney.
3 Thermal conductivity is constant.

32
Reference
This lecture power point adapted from
1. Yunus Cengel, Heat and Mass Transfer A Practical Approach,
3rd edition
2. Jack P. Holman, Heat Transfer, Tenth Edition.
3. Frank P. Incropera, Theodore l. Bergman, Adrienne S.
Lavine, and David P Dewitt, fundamental of Heat and Mass
Transfer, 7th edition
4. Lecture power point of heat transfer by Mehmet Kanoglu
University of Gaziantep
Note: You can download chapter and two by using below URL (click )
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TWO DIMENSIONAL STEADY STATE HEAT CONDUCTION

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