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condition. However, this backward difference formula for the boundary condition is just first-order accurate, whereas we will be using second-order accurate formulae everywhere else in the domain. It is possible to improve the accuracy by cleverly introducing a “ghost” point outside of the domain. You can see below how this saves the day while retaining the simplicity of the method. Tricks such as this are a staple of the numerical community, and commercial packages have many such clever hacks thrown in but hidden from the user.

3. Discretize the governing equation and boundary conditions: The discretized form of the governing ODE and boundary conditions are as follows

•Governing equation.

FIGURE 9.2

One-dimensional conduction in the fin for Example 9.1.

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numerical solution is inaccurate and yet to converge to the correct solution. Therefore, we need more numbers of iterations for the correct solution. The high value of the stopping criterion at the end of 8 iterations confirms this.

We now apply the same process for two-dimensional steady state heat transfer without any heat generation.

Example 9.2: Consider steady two-dimensional heat transfer in a square metal plate, as shown in the figure below. The temperature at the boundaries is prescribed to be 300 K, all sides except at the top where the temperature is maintained at 400 K. The thermal conductivity of the body is k = 180 W/mK. There is no heat generation in the slab. Using the finite difference method with a mesh size of ∆x = ∆ y = 0.1 m, determine the temperature at nodes 1, 2, 3, and 4. Perform iterations with the Gauss-

Seidel method until ∑i4=1 (Ti j+1 − Ti j )2 ≤ 1×10−4 , where j is the iteration number, and

find the heat flux at a point P3(0.1,0.1). Start with an initial guess of 325 K for all the temperatures.

Solution:

1.Mathematical modeling: The governing equation for two-dimensional steady state heat conduction without heat generation is

∂2 T + ∂2 T = 0

∂x2 ∂y2

2.Create a mesh or grid: The two-dimensional grid is discretized into 4 × 4 uniformly spaced discrete points as shown in Fig. 9.3.

3.Discretize the governing equations: Using the central difference method, we

discretize the governing equation as follows.

FIGURE 9.3

Two-dimensional conduction without heat generation for Example 9.2.

9.5 Steady conduction 309

Table 9.2  Progress of Gauss-Seidel iterations for Example 9.2.

T1 = 337.5K ,T2 = 337.5K ,T3 = 312.5K ,T4 = 312.5K

In this problem, ∆x = ∆y

4.Solve the set of equations: Using the Gauss-Seidel method of iteration, we get Table 9.2.

5.Postprocess to find desired quantities: The heat flux vector at P3 is given by

at this point is 35.576 kW/m2K.

We now consider two-dimensional, steady state conduction with heat generation in the next example.

Example 9.3: Consider steady two-dimensional heat transfer in a rectangular metal plate with constant thermophysical properties, as shown in Fig. 9.4. The temperature at the boundaries are as prescribed in the figure. The thermal conductivity of the

310 CHAPTER 9  Numerical heat transfer

FIGURE 9.4

Two-dimensional conduction with heat generation for Example 9.3.

body is k = 150 W/m K. Assume that heat is generated in the plate uniformly at a rate of qv = 1 × 107 W/m3, as shown in the figure. Write the finite difference formulation for the given two-dimensional heat conduction problem with internal heat generation. Using the finite difference method with a mesh size of ∆x = 0.1 and ∆y = 0.08 m, determine the temperature at nodes 1, 2, 3, and 4. Perform iterations with the Gauss-

Seidel method until ∑i4=1 (Ti j+1 − Ti j )2 ≤ 1×10−4 , where j is the iteration number. Start

with an initial guess of 310 K for all the temperatures.

Solution:

1. Mathematical modeling: The governing equation for two–dimensional, steady state heat conduction with heat generation is

∂2 T + ∂2 T + qv = 0

∂x2 ∂y2 k

2.Create a mesh or grid: The two-dimensional grid is discretized into 4 × 4 uniformly spaced discrete points as shown in Fig. 9.4.

3.Discretize the governing equations: Using the central difference method, we

have

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heat generation as well. If you try to solve these problems analytically, you will notice that each of these problems requires vastly different analytical methods and tricks. Contrast this with the numerical process. Notice that, throughout, our finite difference process hardly changed, and we were able to easily adapt it to the problem even as the equations changed. Indeed, it is this easy adaptability of the numerical approximation approach that makes it such a powerful tool for practical problems.

9.6  Unsteady conduction

The previous section looked at steady state conduction cases, usually governed by2T = 0. However, if we are interested in either the transient (i.e., how we got to the steady state), or a generally unsteady problem, we need to add one more wrinkle— time dependence—into the problem and see how to handle it numerically. Note that the steady state conduction problem is the steady state of a general, unsteady equation of the form

We will start with an extremely simple case and then proceed to a more complex example. The example has been a little bit open-ended deliberately to encourage exploration and discussion and foster some intuition about discretization and its effect on the numerical solution.

Example 9.4: Consider a slab of thickness L = 1.6 cm with walls maintained at temperatures TL = 25 and TR = 100, on the left and right respectively. The initial temperature in the slab is uniformly Tint = 25 everywhere. The slab’s diffusivity is α = 1 × 105 cm2/s. Answer the following questions:

1.Write the governing equation, boundary conditions, and initial condition for this problem.

2.Discretize the slab with uniformly spaced points. Write approximate expressions for the governing equation. Comment on the order of accuracy of your approximation.

3.Are there alternate versions of the discretization that you can think of? Comment on the order of accuracy of these alternates.

4.Rearrange your discretization to write an expression for how the temperature of any location evolves in time. Does any nondimensional parameter naturally arise from this? What does it remind you of?

5.Calculate the temperatures in the center of the slab at Fo = 0.125,0.25,0.5,1,2

using an explicit scheme. Including the boundary points, try with 3,5 points in the computational domain. What are the effects of changing ∆t in your discretizations? What can you conclude from this?

6.Repeat the above for the implicit scheme.

9.6 Unsteady conduction 313

Solution:

1. The governing equation for the temperature T(x,t) is the unsteady conduction equation and is given by

The boundary conditions are Dirichlet and are given by

The initial condition is given by

2. If there are just three points in the domain, there is just one unknown temperature,

Tmid. Let us say the three temperatures are T1 = TL, T2 = Tmid, and T3 = TR. Note that we have to discretize in time too. This is very much like the two-dimensional grid we had

used purely in space earlier. The only difference is that since the direction y is a spatial direction, the customary space-time notation is to use the index n as a superscript for stepping in time and i for stepping in space. That is, Tin denotes T (xi , tn ), where the superscript n denotes the index in time.

Fig. 9.5 shows a typical space-time stencil for unsteady problems. We can now discretize each of the derivatives in the governing equation as follows.

FIGURE 9.5

Space-time stencil for a one-dimensional, unsteady problem.