C H A P T E R CHAPTER 4

Two-Dimensional, Steady-State Conduction

For steady state with no heat generation, the Laplace equation applies. 𝜕2𝑇𝜕𝑥2

+ 𝜕2𝑇𝜕𝑦2

= 0 4.1 assuming constant thermal conductivity. The solution to this equation may be

obtained by analytical, numerical, or graphical techniques.

The objective of any heat-transfer analysis is usually to predict heat flow or the

temperature that results from a certain heat flow. The solution to Equation (4.1) will

give the temperature in a two-dimensional body as a function of the two independent

space coordinates x and y. Then the heat flow in the x and y directions may be

calculated from the Fourier equations

𝑞𝑥 = −𝑘𝐴𝑥

𝜕𝑇𝜕𝑥

4.2

𝑞𝑦 = −𝑘𝐴𝑦𝜕𝑇𝜕𝑦

4.3 These heat-flow quantities are directed either in the x direction or in the y direction.

The total heat flow at any point in the material is the resultant of the qx and qy at that

point. 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 Figure 4.1. So if the temperature

distribution in the material is known, we may easily establish the heat flow.

Figure 4.1 Sketch showing the heat flow in two dimensions.

4.1 MATHEMATICAL ANALYSIS OF TWO-DIMENSIONAL HEAT CONDUCTION We first consider an analytical approach to a two-dimensional problem and then indicate the numerical and graphical methods that may be used to advantage in many other problems.

It is worthwhile to mention here that analytical solutions are not always possible to obtain; indeed, in many instances they are very cumbersome and difficult to use. In these cases numerical techniques are frequently used to advantage. Consider the rectangular plate shown in Figure 4.2. 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. 4.1.1 Seperation of variable method: To solve Equation (4.1), the separation-of-variables method is used. The essential point of this method is that the solution to the differential equation is assumed to take a product form 𝑇 = 𝑋𝑌 4.4 Where: 𝑋 = 𝑋(𝑥) , 𝑌 = 𝑌(𝑦)

Figure 4.2 Isotherms and heat flow lines in a rectangular plate.

𝜕𝑇𝜕𝑥

= 𝑑𝑋𝑑𝑥𝑌

𝜕2𝑇𝜕𝑥2

= 𝑑2𝑋𝑑𝑥2

𝑌

Similarly,

𝜕2𝑇𝜕𝑦2

= 𝑑2𝑌𝑑𝑦2

𝑋

Substitute in equation 4.1

𝑑2𝑋𝑑𝑥2

𝑌 + 𝑑2𝑌𝑑𝑦2

𝑋 = 0 4.5

− 1𝑋𝑑2𝑋𝑑𝑥2

= 1𝑌𝑑2𝑌𝑑𝑦2

= 𝜆2 4.6

1𝑋𝑑2𝑋𝑑𝑥2

= −𝜆2 4.7

1𝑌𝑑2𝑌𝑑𝑦2

= 𝜆2 4.8

From equation 4.7

𝑑2𝑋𝑑𝑥2

+ 𝑋𝜆2 = 0 4.9

From equation 4.8

𝑑2𝑌𝑑𝑦2

− 𝑌𝜆2 = 0 4.10

Solution of equation 4.9

𝑋 = 𝐴 sin 𝜆𝑥 + 𝐵 cos 𝜆𝑥 4.11

Solution of equation 4.10

𝑌 = 𝐶 sinh 𝜆𝑦 + 𝐷 cosh 𝜆𝑦 4.12

General solution of equation 4.1

𝑇 = (𝐴 sin 𝜆𝑥 + 𝐵 cos 𝜆𝑥)(𝐶 sinh 𝜆𝑦 + 𝐷 cosh 𝜆𝑦) 4.13

Let 𝜃 = 𝑇 − 𝑇1

𝜃 = (𝐴 sin 𝜆𝑥 + 𝐵 cos 𝜆𝑥)(𝐶 sinh 𝜆𝑦 + 𝐷 cosh 𝜆𝑦)

𝜃 = 0

𝜃 = 0

𝜃 = 0

B.C

At 𝑥 = 0, 𝜃 = 0 for all y

𝐵 = 0

𝜃 = (𝐴 sin 𝜆𝑥)(𝐶 sinh 𝜆𝑦 + 𝐷 cosh 𝜆𝑦) 4.14

At 𝑦 = 0, 𝜃 = 0 for all 𝑥

𝐷 = 0

The general solution becomes:

𝜃 = (𝐴 ∗ 𝐶) sin 𝜆𝑥 sinh 𝜆𝑦 4.15

Let 𝐴 ∗ 𝐶 = 𝐸

At 𝑥 = 𝑊 , 𝜃 = 0 for all 𝑦

𝜃 = 𝐸 sin 𝜆𝑥 sinh 𝜆𝑦

0 = 𝐸 sin 𝜆𝑊 sinh 𝜆𝑦 4.16

sin 𝜆𝑊 = 0

Therefore,

𝜆 = 𝑛𝜋𝑊

where, 𝑛 = 0,1,2,3,4, … …

𝜃 = ∑ 𝐸𝑛 sin 𝑛𝜋𝑥𝑊

∞𝑛=1 ∗ sinh 𝑛𝜋𝑦

𝑊 4.17

To evaluate 𝐸𝑛, sibstitute the last B.C at 𝑦 = 𝐻

𝑇 − 𝑇1 = ∑ 𝐸𝑛 sin 𝑛𝜋𝑥𝑊

∞𝑛=1 ∗ sinh 𝑛𝜋𝐻

𝑊 4.18

How to evaluate 𝐸𝑛

This is a Fourier sine series, and the values of the 𝐸𝑛 may be determined by expanding the constant temperature difference 𝑇2 − 𝑇1 in a Fourier series over the interval 0<x<W. This series is 𝑇2 − 𝑇1 = 𝑇2 − 𝑇1

2𝜋∑ (−1)𝑛+1+1

𝑛∞𝑛=1 sin 𝑛𝜋𝑥

𝑊 4.19

𝐸𝑛 = 2

𝜋(𝑇2 − 𝑇1) 1

sinh𝑛𝜋𝐻𝑊

(−1)𝑛+1+1𝑛

The final solution:

𝑇−𝑇1𝑇2−𝑇1

= 2𝜋∑ (−1)𝑛+1+1

𝑛∞𝑛=1 sin 𝑛𝜋𝑥

𝑊

sinh𝑛𝜋𝑦𝑊sinh𝑛𝜋𝑊𝐻

4.20

4.2 Numerical Method of Analysis

In many practical situations the geometry or boundary conditions are such that an

analytical solution has not been obtained at all, or if the solution has been developed,

it involves such a complex series solution that numerical evaluation becomes

exceedingly difficult. For such situations the most fruitful approach to the problem is

one based on finite-difference techniques, the basic principles of which we shall

outline in this section.

4.2.1 Finite Difference Tichnique

Consider a two-dimensional body that is to be divided into equal increments in

both the x and y directions, as shown in Figure 4.3. The nodal points are designated as

shown, the m locations indicating the x increment and the n locations indicating the y

increment. We wish to establish the temperatures at any of these nodal points within

the body, using Equation (4.1) as a governing condition. Finite differences are used to

approximate differential increments in the temperature and space coordinates; and the

smaller we choose these finite increments, the more closely the true temperature

distribution will be approximated. Figure 4.3 Sketch illustrating nomenclature used in two-dimensional numerical analysis of heat conduction.

Thus the finite-difference approximation for Equation (4.1) becomes

𝑇𝑚+1,𝑛+𝑇𝑚−1,𝑛−2𝑇𝑚,𝑛

(∆𝑥)2+ 𝑇𝑚,𝑛+1+𝑇𝑚,𝑛−1−2𝑇𝑚,𝑛

(∆𝑦)2= 0

If ∆𝑥 = ∆𝑦, then

𝑇𝑚+1,𝑛 + 𝑇𝑚−1,𝑛 + 𝑇𝑚,𝑛+1 + 𝑇𝑚,𝑛−1 − 4𝑇𝑚,𝑛 = 0 4.21

Since we are considering the case of constant thermal conductivity, the heat flows

may all be expressed in terms of temperature differentials. Equation (4.21) states very

simply that the net heat flow into any node is zero at steady-state conditions. In effect,

the numerical finite-difference approach replaces the continuous temperature

distribution by fictitious heat-conducting rods connected between small nodal points

that do not generate heat. We can also devise a finite-difference scheme to take heat

generation into account.We merely add the term ˙q/k into the general equation and

obtain 𝑇𝑚+1,𝑛+𝑇𝑚−1,𝑛−2𝑇𝑚,𝑛

(∆𝑥)2+ 𝑇𝑚,𝑛+1+𝑇𝑚,𝑛−1−2𝑇𝑚,𝑛

(∆𝑦)2+ �̇�

𝑘= 0

Then for a square grid in which ∆𝑥 = ∆𝑦,

𝑇𝑚+1,𝑛 + 𝑇𝑚−1,𝑛 + 𝑇𝑚,𝑛+1 + 𝑇𝑚,𝑛−1 + �̇�∆𝑥2

𝑘− 4𝑇𝑚,𝑛 = 0 4.22

To utilize the numerical method, Equation (4.21) must be written for each node within

the material and the resultant system of equations solved for the temperatures at the

various nodes. A very simple example is shown in Figure 4.4, and the four equations

for nodes 1, 2, 3, and 4 would be

Figure 4.4 Four-node problem.

These equations have the solution

Of course, we could recognize from symmetry that T1 =T2 and T3 =T4 and would then only need two nodal equations,

Once the temperatures are determined, the heat flow may be calculated from

where the ∆𝑇 is taken at the boundaries. In the example the heat flow may be

calculated at either the 500◦C face or the three 100◦C faces. If a sufficiently fine grid

is used, the two values should be very nearly the same. As a matter of general

practice, it is usually best to take the arithmetic average of the two values for use in

the calculations. In the example, the two calculations yield:

500˚C face:

𝑞 = −𝑘 ∆𝑥∆𝑦

[(250 − 500) + (250 − 500)] = 500𝑘

100˚C face:

𝑞 = −𝑘 ∆𝑦∆𝑥

[(250 − 100) + (150 − 100) + (150 − 100) + (150 − 100) +(150 − 100) + (250 − 100)] = −500𝑘

4.2.2 Energy Balance Method

In many cases, it is desirable to develop the finite-difference equations by an

alternative method called the energy balance method. As will become evident, 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.

Consider a control volume about the interior node m,n of figure 4.5.

Figure 4.5: conduction to an interior node from its adjoining nodes.

−𝑘∆𝑦 𝑇𝑚,𝑛−𝑇𝑚+1,𝑛∆𝑥

− 𝑘∆𝑦 𝑇𝑚,𝑛−𝑇𝑚−1,𝑛∆𝑥

− 𝑘∆𝑥 𝑇𝑚,𝑛−𝑇𝑚,𝑛+1∆𝑦

− 𝑘∆𝑥 𝑇𝑚,𝑛−𝑇𝑚,𝑛−1∆𝑦

= 0

Taking ∆𝑥 = ∆𝑦, and simplify

𝑇𝑚+1,𝑛 + 𝑇𝑚−1,𝑛 + 𝑇𝑚,𝑛+1 + 𝑇𝑚,𝑛−1 − 4𝑇𝑚,𝑛 = 0

When the solid is exposed to some convection boundary condition, the temperatures

at the surface must be computed differently from the method given above. Consider

the boundary shown in Figure 4.5. The energy balance on node (m, n) is

−𝑘∆𝑦𝑇𝑚,𝑛 − 𝑇𝑚−1,𝑛

∆𝑥− 𝑘

∆𝑥2𝑇𝑚,𝑛 − 𝑇𝑚,𝑛+1

∆𝑦− 𝑘

∆𝑥2𝑇𝑚,𝑛 − 𝑇𝑚,𝑛−1

∆𝑦= ℎ∆𝑦(𝑇𝑚,𝑛 − 𝑇∞)

If ∆𝑥 = ∆𝑦, the boundary temperature is expresses in the equation

𝑇𝑚,𝑛 �ℎ∆𝑥𝑘

+ 2� − ℎ∆𝑥𝑘𝑇∞ − 1

2�2𝑇𝑚−1,𝑛 + 𝑇𝑚,𝑛+1 + 𝑇𝑚,𝑛−1� = 0 4.23

An equation of this type must be written for each node along the surface shown in

Figure 4.6. So when a convection boundary condition is present, an equation like

(4.23) is used at the boundary and an equation like (4.21) is used for the interior

points. Equation (4.23) applies to a plane surface exposed to a convection boundary

condition. It will not apply for other situations, such as an insulated wall or a corner

exposed to a convection boundary condition. Consider the corner section shown in

Figure 4.7. The energy balance for the corner section is

−𝑘 ∆𝑦2𝑇𝑚,𝑛−𝑇𝑚−1,𝑛

∆𝑥− 𝑘 ∆𝑥

2𝑇𝑚,𝑛−𝑇𝑚,𝑛−1

∆𝑦= ℎ ∆𝑥

2�𝑇𝑚,𝑛 − 𝑇∞� + ℎ ∆𝑦

2�𝑇𝑚,𝑛 − 𝑇∞�

If ∆𝑥 = ∆𝑦,

2𝑇𝑚,𝑛 �ℎ∆𝑥𝑘

+ 1� − 2 ℎ∆𝑥𝑘𝑇∞ − �𝑇𝑚−1,𝑛 + 𝑇𝑚,𝑛−1� = 0 4.24

EXAMPLE 4.1: Nine-Node Problem EXAMPLE 3-5 Consider the square of Figure Example 3-5. The left face is maintained at 100◦C and

the top face at 500◦C, while the other two faces are exposed to an environment at

Figure 4.7 Nomenclature for nodal equation with convection at a corner section.

Figure 4.6 Nomenclature for nodal equation with convective boundary condition.

100◦C: ℎ = 10 W/m2 ・ ◦C and k =10 W/m.◦C .The block is 1 m square. Compute the

temperature of the various nodes as indicated in Figure Example 4.1 and the heat

flows at the boundaries.

Solution The nodal equation for nodes 1, 2, 4, and 5 is

𝑇𝑚+1,𝑛 + 𝑇𝑚−1,𝑛 + 𝑇𝑚,𝑛+1 + 𝑇𝑚,𝑛−1 − 4𝑇𝑚,𝑛 = 0

Node 1:

𝑇2 + 𝑇4 + 500 + 100 − 4𝑇1 = 0

Node 2:

𝑇1 + 𝑇3 + 𝑇5 + 500 − 4𝑇2 = 0

Node 4:

𝑇1 + 𝑇5 + 𝑇7 + 100 − 4𝑇4 = 0

Node 5:

𝑇2 + 𝑇4 + 𝑇6 + 𝑇8 − 4𝑇5 = 0

ℎ∆𝑥𝑘

= (10)(1)(3)(10)

= 13

Node 3:

−𝑘∆𝑦 𝑇3−𝑇2∆𝑥

− 𝑘 ∆𝑥2𝑇3−500∆𝑦

− 𝑘 ∆𝑥2𝑇3−𝑇6∆𝑦

= ℎ∆𝑦(𝑇3 − 𝑇∞)

𝑇3 �ℎ∆𝑥𝑘

+ 2� − ℎ∆𝑥𝑘𝑇∞ − 1

2(2𝑇2 + 500 + 𝑇6) = 0

𝑇3 �13

+ 2� − 13𝑇∞ − 1

2(2𝑇2 + 500 + 𝑇6) = 0

2𝑇2 + 567 + 𝑇6 − 4.67𝑇3 = 0

Node 6:

−𝑘∆𝑦 𝑇6−𝑇5∆𝑥

− 𝑘 ∆𝑥2𝑇6−𝑇3∆𝑦

− 𝑘 ∆𝑥2𝑇6−𝑇9∆𝑦

= ℎ∆𝑦(𝑇3 − 𝑇∞)

𝑇6 �13

+ 2� − 13𝑇∞ − 1

2(2𝑇5 + 𝑇3 + 𝑇9) = 0

2𝑇5 + 𝑇3 + 𝑇9 + 67 − 4.67𝑇6 = 0

Node 7:

−𝑘∆𝑥 𝑇7−𝑇4∆𝑦

− 𝑘 ∆𝑦2𝑇7−𝑇8∆𝑥

− 𝑘 ∆𝑦2𝑇7−100∆𝑥

= ℎ∆𝑥(𝑇7 − 𝑇∞)

𝑇7 �13

+ 2� − 13𝑇∞ − 1

2(2𝑇4 + 100 + 𝑇8) = 0

2𝑇4 + 𝑇8 + 167 − 4.67𝑇7 = 0

Node 8:

−𝑘∆𝑥 𝑇8−𝑇5∆𝑦

− 𝑘 ∆𝑦2𝑇8−𝑇7∆𝑥

− 𝑘 ∆𝑦2𝑇8−𝑇9∆𝑥

= ℎ∆𝑥(𝑇8 − 𝑇∞)

𝑇8 �13

+ 2� − 13𝑇∞ − 1

2(2𝑇5 + 𝑇9 + 𝑇7) = 0

2𝑇5 + 𝑇7 + 𝑇9 + 67 − 4.67𝑇8 = 0

Figure Example 4.1: Numenucliature for Example 4.1.

Node 9:

−𝑘 ∆𝑥2𝑇9−𝑇6∆𝑦

− 𝑘 ∆𝑦2𝑇9−𝑇8∆𝑥

= ℎ ∆𝑥2

(𝑇9 − 𝑇∞) + ℎ ∆𝑦2

(𝑇9 − 𝑇∞)

2𝑇9 �ℎ∆𝑥𝑘

+ 1� − 2 ℎ∆𝑥𝑘𝑇∞ − (𝑇6 + 𝑇8) = 0

2𝑇9 �13

+ 1� − 23𝑇∞ − (𝑇6 + 𝑇8) = 0

𝑇6 + 𝑇8 + 67 − 2.67𝑇9 = 0

We thus have nine equations and nine unknown nodal temperatures. We shall discuss solution techniques shortly, but for now we just list the answers: Node Temperature, ◦C 1 280.67 2 330.30 3 309.38 4 192.38 5 231.15 6 217.19 7 157.70 8 184.71

9 175.62 The heat flows at the boundaries are computed in two ways: as conduction flows for

the 100 and 500◦C faces and as convection flows for the other two faces.

For the 500◦C face,

the heat flow into the face is

𝑞 = ∑𝑘∆𝑥 ∆𝑇∆𝑦

= (10) �(500 − 280.67) + (500 − 330.30) + (500 − 309.38) �12��

𝑞 = 4843.4 W/m.

The heat flow out of the 100◦C face is

𝑞 = ∑𝑘∆𝑦 ∆𝑇∆𝑥

= (10) �(280.67 − 100) + (192.38 − 100) + (157.7 − 100) �12��

𝑞 = 3019 W/m.

The convection heat flow out the right face is given by the convection relation

𝑞 = ∑ℎ∆𝑦(𝑇 − 𝑇∞) 𝑞 = (10) �1

3� �309.38 − 100 + 217.19 − 100 + (175.62 − 100) �1

2��

𝑞 = 1214.6 W/m.

Finally, the convection heat flow out the bottom face is

𝑞 = ∑ℎ∆𝑥(𝑇 − 𝑇∞) 𝑞 = (10) �1

3� �(100 − 100) �1

2� + (157.7 − 100) + (184.71 − 100) + (175.62 − 100) �1

2��

𝑞 = 600.7 W/m. The total heat flow out is 𝑞𝑜𝑢𝑡 = 3019 + 1214.6 + 600.7 = 4834.3 W/m.

Example 4.2: Irregular Boundaries

Node 1. On the inner boundary, subjected to convection, −𝑘 ∆𝑦

2(𝑇1−𝑇2)∆𝑥

− 𝑘 ∆𝑥2

(𝑇1−𝑇3)∆𝑦

= ℎ𝑖∆𝑥2

(𝑇1 − 𝑇𝑖) Taking ∆𝑥 = ∆𝑦, it simplifies to −𝑘 (𝑇1−𝑇2)

∆𝑥− 𝑘 (𝑇1−𝑇3)

∆𝑥= ℎ𝑖(𝑇1 − 𝑇𝑖)

𝑇1(∆𝑥ℎ𝑖𝑘

+ 2) = ∆𝑥ℎ𝑖𝑘𝑇𝑖 + (𝑇2 + 𝑇3)

𝑇1 =(∆𝑥ℎ𝑖𝑘 )𝑇𝑖+(𝑇2+𝑇3)

�∆𝑥ℎ𝑖𝑘 +2�

Node 2. On the inner boundary, subjected to convection, Figure −𝑘 ∆𝑦

2(𝑇2−𝑇1)∆𝑥

− 𝑘∆𝑥 (𝑇2−𝑇4)∆𝑦

= ℎ𝑖 ∆𝑥2 (𝑇2−𝑇𝑖) Taking∆𝑥 = ∆𝑦, it simplifies to 𝑇2(∆𝑥ℎ𝑖

2𝑘+ 3

2) = ∆𝑥ℎ𝑖

2𝑘𝑇𝑖 + (𝑇4 + 1

2𝑇1)

𝑇2(∆𝑥ℎ𝑖

𝑘+ 3) = ∆𝑥ℎ𝑖

𝑘𝑇𝑖 + (2𝑇4 + 𝑇1)

𝑇2 =�∆𝑥ℎ𝑖𝑘 �𝑇𝑖+(2𝑇4+𝑇1)

�∆𝑥ℎ𝑖𝑘 +3�

Nodes 3,

−𝑘 ∆𝑥2

(𝑇3−𝑇1)∆𝑦

− 𝑘∆𝑦 (𝑇3−𝑇4)∆𝑥

− 𝑘 ∆𝑥2

(𝑇3−𝑇6)∆𝑦

= 0

−2𝑇3 + 12𝑇1 + 𝑇4 + 1

2𝑇6 = 0

𝑇1 + 2𝑇4 + 𝑇6 − 4𝑇3 = 0

𝑇3 = 𝑇1+2𝑇4+𝑇64

Node 4,

−𝑘∆𝑥 (𝑇4−𝑇2)∆𝑦

− 𝑘∆𝑦 (𝑇4−𝑇3)∆𝑥

− 𝑘∆𝑦 (𝑇4−𝑇5)∆𝑥

− 𝑘∆𝑥 (𝑇4−𝑇7)∆𝑦

= 0

𝑇2 + 𝑇3 + 𝑇5 + 𝑇7 − 4𝑇4 = 0

𝑇4 = 𝑇2+𝑇3+𝑇5+𝑇74

Node 5,

−𝑘∆𝑦 (𝑇5−𝑇4)∆𝑥

− 𝑘∆𝑥 (𝑇5−𝑇8)∆𝑦

= 0

𝑇4 + 𝑇8 − 2𝑇5 = 0

𝑇5 = 𝑇4+𝑇82

Node 6. (On the outer boundary, subjected to convection and radiation) −𝑘 ∆𝑥

2(𝑇6−𝑇3)∆𝑦

− 𝑘 ∆𝑦2

(𝑇6−𝑇7)∆𝑥

= ℎ𝑜∆𝑥2

(𝑇6 − 𝑇𝑜) + 𝜀𝜎 ∆𝑥2�𝑇64 − 𝑇𝑠𝑘𝑦4�

Taking∆𝑥 = ∆𝑦, it simplifies to 𝑇6 �2 + ℎ𝑜∆𝑥

𝑘� = 𝑇3 + 𝑇7 + ℎ𝑜∆𝑥

𝑘𝑇𝑜 + 𝜀𝜎 ∆𝑥

𝑘�𝑇64 − 𝑇𝑠𝑘𝑦4�

𝑇6 =𝑇3+𝑇7+

ℎ𝑜∆𝑥𝑘 𝑇𝑜+𝜀𝜎

∆𝑥𝑘 �𝑇6

4−𝑇𝑠𝑘𝑦4�

�2+ℎ𝑜∆𝑥𝑘 �

Node 7. (On the outer boundary, subjected to convection and radiation, Fig −𝑘∆𝑥 (𝑇7−𝑇4)

∆𝑦− 𝑘 ∆𝑦

2(𝑇7−𝑇6)∆𝑥

− 𝑘 ∆𝑦2

(𝑇7−𝑇8)∆𝑥

= ℎ𝑜∆𝑥(𝑇7 − 𝑇𝑜) + 𝜀𝜎∆𝑥�𝑇74 − 𝑇𝑠𝑘𝑦4� Taking∆𝑥 = ∆𝑦, it simplifies to 𝑇7 �4 + 2ℎ𝑜∆𝑥

𝑘� = 2𝑇4 + 𝑇6 + 𝑇8 + 2ℎ𝑜∆𝑥

𝑘𝑇𝑜 + 2𝜀𝜎 ∆𝑥

𝑘�𝑇74 − 𝑇𝑠𝑘𝑦4�

𝑇7 =2𝑇4+𝑇6+𝑇8+

2ℎ𝑜∆𝑥𝑘 𝑇𝑜+2𝜀𝜎

∆𝑥𝑘 �𝑇7

4−𝑇𝑠𝑘𝑦4�

�4+2ℎ𝑜∆𝑥𝑘 �

Node 8. 𝑇8 �4 + 2ℎ𝑜∆𝑥

𝑘� = 2𝑇5 + 𝑇7 + 𝑇9 + 2ℎ𝑜∆𝑥

𝑘𝑇𝑜 + 2𝜀𝜎 ∆𝑥

𝑘�𝑇84 − 𝑇𝑠𝑘𝑦4�

𝑇8 =2𝑇5+𝑇7+𝑇9+

2ℎ𝑜∆𝑥𝑘 𝑇𝑜+2𝜀𝜎

∆𝑥𝑘 �𝑇8

4−𝑇𝑠𝑘𝑦4�

�4+2ℎ𝑜∆𝑥𝑘 �

Node 9. (On the outer boundary, subjected to convection and radiation, Fig. −𝑘 ∆𝑦

2(𝑇9−𝑇8)∆𝑥

= ℎ𝑜∆𝑥2

(𝑇9 − 𝑇𝑜) + 𝜀𝜎 ∆𝑥2�𝑇94 − 𝑇𝑠𝑘𝑦4�

Taking∆𝑥 = ∆𝑦, it simplifies to 𝑇9 �1 + ℎ𝑜∆𝑥

𝑘� = 𝑇8 + ℎ𝑜

∆𝑥𝑘𝑇𝑜 + 𝜀𝜎 ∆𝑥

𝑘�𝑇94 − 𝑇𝑠𝑘𝑦4�

𝑇9 =𝑇8+ℎ𝑜

∆𝑥𝑘 𝑇𝑜+𝜀𝜎

∆𝑥𝑘 �𝑇9

4−𝑇𝑠𝑘𝑦4�

�1+ℎ𝑜∆𝑥𝑘 �

This problem involves radiation, which requires the use of absolute temperature, and

thus all temperatures should be expressed in Kelvin. Alternately, we could use °C for

all temperatures provided that the four temperatures in the radiation terms are

expressed in the form (T + 273)4. Substituting the given quantities, the system of nine

equations for the determination of nine unknown nodal temperatures in a form

suitable for use with the Gauss-Seidel iteration method becomes

Solution Techniques The Matrix Inversion Method From the foregoing discussion we have seen that the numerical method is simply a

means of approximating a continuous temperature distribution with the finite nodal

elements. The more nodes taken, the closer the approximation; but, of course, more

equations mean more cumbersome solutions. Fortunately, computers and even

programmable calculators have the capability to obtain these solutions very quickly.

In practical problems the selection of a large number of nodes may be unnecessary

because of uncertainties in boundary conditions. For example, it is not uncommon to

have uncertainties in h, the convection coefficient, of ±15 to 20 percent. The nodal

equations may be written as

where T1, T2, . . . , Tn are the unknown nodal temperatures. By using the matrix

notation

Equation (4.25) can be expressed as

and the problem is to find the inverse of [A] such that

Designating [𝐴]−1 by

the final solutions for the unknown temperatures are written in expanded form as

4.25

4.27

4.26

Clearly, the larger the number of nodes, the more complex and time-consuming the

solution, even with a high-speed computer. For most conduction problems the matrix

contains a large number of zero elements so that some simplification in the procedure

is afforded. For example, the matrix notation for the system of Example 4.1 would be

Example 4.3:

Consider steady two-dimensional heat transfer in a long solid body whose cross

section is given in the figure. The temperatures at the selected nodes and the thermal

conditions at the boundaries are as shown. The thermal conductivity of the body is k =

214 W/m · °C, and heat is generated in the body uniformly at a rate of �̇� =6*106

W/m3. With a mesh size of ∆x=∆y = 5.0 cm, determine(a) the temperatures at nodes 1,

2, and 3 and (b) the rate of heat loss from the bottom surface through a 1-m-long

section of the body.

4.28

a:

Node 1:

−𝑘 ∆𝑦2

(𝑇1−325)∆𝑥

− 𝑘 ∆𝑦2

(𝑇1−240)∆𝑥

− 𝑘∆𝑥 (𝑇1−290)∆𝑦

+ �̇�(∆𝑥∆𝑦) = ℎ∆𝑥(𝑇1 − 𝑇∞)

𝑇1 �4 + 2 ℎ∆𝑥𝑘� = 325 + 240 + 2(290) + 2�̇�(∆𝑥∆𝑦)

𝑘+ 2ℎ∆𝑥

𝑘𝑇∞

𝑇1 =325+240+2(290)+2�̇�(∆𝑥∆𝑦)

𝑘 +2ℎ∆𝑥𝑘 𝑇∞

�4+2ℎ∆𝑥𝑘 �

𝑇1 = 319 ˚C.

Node 2:

−𝑘 ∆𝑥2

(𝑇2−350)∆𝑦

− 𝑘 ∆𝑥2

(𝑇2−325)∆𝑦

− 𝑘∆𝑦 (𝑇2−290)∆𝑥

+ �̇�(∆𝑥∆𝑦) = 0

4𝑇2 = 350 + 325 + 2(290) + 2�̇�(∆𝑥∆𝑦)𝑘

𝑇2 =350+325+2(290)+2�̇�(∆𝑥∆𝑦)

𝑘4

𝑇2 = 348.8 ˚C.

Node 3:

290 + 200 + 260 + 240 − 4𝑇3 + �̇�(∆𝑥∆𝑦)𝑘

= 0

𝑇3 =290+200+260+240+�̇�(∆𝑥∆𝑦)

𝑘4

𝑇3 = 265 ˚C. b:

𝑞 = ℎ ∆𝑥2

(200− 𝑇∞) + ℎ∆𝑥(240− 𝑇∞) + ℎ∆𝑥(𝑇1 − 𝑇∞) + ℎ ∆𝑥2

(325 − 𝑇∞)

𝑞 = 50 0.052

(200− 20) + 50 ∗ 0.05(240− 20) + 50 ∗ 0.05(319− 20) + 50 0.052

(325− 20)

𝑞 = 1903.75 W

5–52 Consider steady two-dimensional heat transfer in a long solid bar whose cross

section is given in the figure. The measured temperatures at selected points on the

outer surfaces are as shown. The thermal conductivity of the body is k = 20 W/m · °C,

and there is no heat generation. Using the finite difference method with a mesh size

of ∆𝑥 = ∆𝑦 =1.0 cm, determine the temperatures at the indicated points in the

medium. Answers: T1 = T4 = 143°C, T2 = T3 = 136°C.

solution:

node 1:

−𝑘∆𝑦 𝑇1−𝑇2∆𝑥

− 𝑘 ∆𝑥2𝑇1−100∆𝑦

− 𝑘 ∆𝑥2𝑇1−200∆𝑦

= 0

12

(200 + 100) + 𝑇2 − 2𝑇1 = 0

𝑇1 = 150+𝑇22

(1)

Node 2:

−𝑘∆𝑦 𝑇2−𝑇1∆𝑥

− 𝑘∆𝑦 𝑇2−100∆𝑥

− 𝑘∆𝑥 𝑇2−100∆𝑦

− 𝑘∆𝑥 𝑇2−200∆𝑦

= 0

𝑇1 + 100 + 100 + 200 − 4𝑇2 = 0

𝑇2 = 𝑇1+100+100+2004

(2)

Substitute equation (2) in equation (1):

𝑇1 =150+𝑇1+100+100+2004

2

𝑇1 = 75 + 50 + 𝑇18

𝑇1 = 1251−18

= 143 ˚C

𝑇1 = 𝑇4 = 143 ˚C

𝑇2 = 𝑇3 = 143+100+100+2004

= 136 ˚C

5–61 Consider a long solid bar whose thermal conductivity is k = 12 W/m · °C and

whose cross section is given in the figure. The top surface of the bar is maintained at

50°C while the bottom surface is maintained at 120°C. The left surface is insulated

and the remaining three surfaces are subjected to convection with ambient air at T∞ =

25°C with a heat transfer coefficient of h = 30 W/m2 · °C. Using the finite difference

method with a mesh size of ∆x = ∆y = 10 cm, (a) obtain the finite difference

formulation of this problem for steady two dimensional heat transfer and (b)

determine the unknown nodal temperatures by solving those equations. Answers: (b)

85.7°C, 86.4°C, 87.6°C

Solution:

Node 1:

−𝑘∆𝑦 (𝑇1−𝑇2)∆𝑥

− 𝑘 ∆𝑥2

(𝑇1−50)∆𝑦

− 𝑘 ∆𝑥2

(𝑇1−120)∆𝑦

= 0

𝑇2 + 502

+ 1202− 2𝑇1 = 0

−2𝑇1 + 𝑇2 = −85 (1)

Node 2:

−𝑘 ∆𝑦2

(𝑇2−𝑇3)∆𝑥

− 𝑘∆𝑦 (𝑇2−𝑇1)∆𝑥

− 𝑘 ∆𝑥2

(𝑇2−50)∆𝑦

− 𝑘∆𝑥 (𝑇2−120)∆𝑦

= ℎ ∆𝑥2

(𝑇2 − 𝑇∞) + ℎ ∆𝑦2

(𝑇2 − 𝑇∞)

𝑇1 − �3 + ℎ ∆𝑥𝑘�𝑇2 + 1

2𝑇3 = −145− ℎ ∆𝑥

𝑘𝑇∞

𝑇1 − �3 + 30 0.112�𝑇2 + 1

2𝑇3 = −145− 30 0.1

12(25)

𝑇1 − 3.25𝑇2 + 0.5𝑇3 = −151.25 (2)

Node 3:

−𝑘 ∆𝑦2

(𝑇3−𝑇2)∆𝑥

− 𝑘 ∆𝑥2

(𝑇3−120)∆𝑦

= ℎ ∆𝑥2

(𝑇3 − 𝑇∞) + ℎ ∆𝑦2

(𝑇3 − 𝑇∞)

0.5𝑇2 − �1 + ℎ ∆𝑥𝑘�𝑇3 = −60− ℎ ∆𝑥

𝑘𝑇∞

0.5𝑇2 − �1 + 30 0.112�𝑇3 = −60− 30 0.1

12(25)

0.5𝑇2 − 1.25𝑇3 = −66.25 (3)

Three equations with three anouns can be solved using matrix invers method.

−2𝑇1 + 𝑇2 = 85

𝑇1 − 3.25𝑇2 + 0.5𝑇3 = −151.25

0.5𝑇2 − 1.25𝑇3 = −66.25

[A]=�−2 1 01 −3.25 0.50 0.5 −1.25

�

[C]=�−85

−151.25−66.25

�

Det[A]= �−2 1 01 −3.25 0.50 0.5 −1.25

�−2 11 −3.250 0.5

=[(-2*-3.25*-1.25)+(1*0.5*0)+(0*1*0.5)]-[(0*-3.25*0)+(0.5*0.5*-2)+(-1.25*1*1)]

=[-8.125-(-0.5)-(-1.25)]=-6.375

Adj[A]=

⎣⎢⎢⎢⎢⎡�−3.25 0.5

0.5 −1.25� − � 1 00.5 −1.25� � 1 0

−3.25 0.5�

− �1 0.50 −1.25� �−2 0

0 −1.25� − �−2 01 0.5�

�1 −3.250 0.5 � − �−2 1

0 0.5� �−2 11 −3.25�⎦

⎥⎥⎥⎥⎤

�−3.25 0.50.5 −1.25�=[(-3.25*-1.25)-(0.5*0.5)]=3.8125

�1 0.50 −1.25�=[(1*-1.25)-(0.5*0)]=-1.25

�1 −3.250 0.5 � =[(1*0.5)-(0*-3.25)]=0.5

� 1 00.5 −1.25� =[(1*-1.25)-(0*0.5)]=-1.25

�−2 00 −1.25� =[(-2*-1.25)-(0)]=2.5

�−2 10 0.5� = [(−2 ∗ 0.5) − (0 ∗ 1)] = −1

� 1 0−3.25 0.5� =[(1*0.5)-(0*-3.25)]=0.5

�−2 01 0.5� =[(-2*0.5)-(0*1)]=-1

�−2 11 −3.25� =[(-2*-3.25)-(1*1)]=5.5

Adj[A]= �3.8125 1.25 0.5

1.25 2.5 10.5 1 5.5

�

[A]-1=Adj[A]Det[A]

= �−0.598 −0.196 −0.0784−0.196 −0.392 −0.157−0.0784 −0.157 −0.862

�

�𝑇1𝑇2𝑇3�=�

−0.598 −0.196 −0.0784−0.196 −0.392 −0.157−0.0784 −0.157 −0.862

� ∗ �−85

−151.25−66.25

�

�𝑇1𝑇2𝑇3� = �

85.786.3587.5

�