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Chapter 5: Problem 87

The explicit finite difference formulation of a general interior node for transient heat conduction in a plane wall is given by $$ T_{m-1}^{i}-2 T_{m}^{i}+T_{m+1}^{i}+\frac{\dot{e}_{m}^{i} \Delta x^{2}}{k}=\frac{T_{m}^{i+1}-T_{m}^{i}}{\tau} $$ Obtain the finite difference formulation for the steady case by simplifying the relation above.

Short Answer

Expert verified
Answer: The simplified explicit finite difference equation for steady-state heat conduction in a plane wall is: \(T_{m-1}^{i}-2 T_{m}^{i}+T_{m+1}^{i}+\frac{\dot{e}_{m}^{i} \Delta x^{2}}{k}=0\).

Step by step solution

01

Identify steady-state condition

For a steady-state case, the temperature at any point in the wall does not change with time, which means the time derivative term in the given equation will be zero: \(\frac{T_{m}^{i+1}-T_{m}^{i}}{\tau}=0\).
02

Simplify the equation for steady-state

Using the result from Step 1, we can now simplify the given explicit finite difference equation: $$ T_{m-1}^{i}-2 T_{m}^{i}+T_{m+1}^{i}+\frac{\dot{e}_{m}^{i} \Delta x^{2}}{k}=\frac{T_{m}^{i+1}-T_{m}^{i}}{\tau} \Rightarrow T_{m-1}^{i}-2 T_{m}^{i}+T_{m+1}^{i}+\frac{\dot{e}_{m}^{i} \Delta x^{2}}{k}=0 $$
03

Final finite difference formulation for steady-state

The final finite difference formulation for steady-state heat conduction in a plane wall is: $$ T_{m-1}^{i}-2 T_{m}^{i}+T_{m+1}^{i}+\frac{\dot{e}_{m}^{i} \Delta x^{2}}{k}=0 $$ This equation can now be used to solve for temperature distribution in the wall under steady-state conditions.

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Most popular questions from this chapter

How can a node on an insulated boundary be treated as an interior node in the finite difference formulation of a plane wall? Explain.

Define these terms used in the finite difference formulation: node, nodal network, volume element, nodal spacing, and difference equation.

Consider steady one-dimensional heat conduction in a composite plane wall consisting of two layers \(A\) and \(B\) in perfect contact at the interface. The wall involves no heat generation. The nodal network of the medium consists of nodes 0,1 (at the interface), and 2 with a uniform nodal spacing of \(\Delta x\). Using the energy balance approach, obtain the finite difference formulation of this problem for the case of insulation at the left boundary (node 0 ) and radiation at the right boundary (node 2) with an emissivity of \(\varepsilon\) and surrounding temperature of \(T_{\text {surr }}\)

Consider steady two-dimensional heat transfer in a square cross section \((3 \mathrm{~cm} \times 3 \mathrm{~cm})\) with the prescribed temperatures at the top, right, bottom, and left surfaces to be \(100^{\circ} \mathrm{C}\), \(200^{\circ} \mathrm{C}, 300^{\circ} \mathrm{C}\), and \(500^{\circ} \mathrm{C}\), respectively. Using a uniform mesh size \(\Delta x=\Delta y\), determine \((a)\) the finite difference equations and \((b)\) the nodal temperatures with the Gauss-Seidel iterative method.

Starting with an energy balance on a volume element, obtain the steady two- dimensional finite difference equation for a general interior node in rectangular coordinates for \(T(x, y)\) for the case of variable thermal conductivity and uniform heat generation.
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