Question

Consider a large plane wall of thickness \(L=0.4 \mathrm{~m}\), thermal conductivity \(k=2.3 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and surface area \(A=20 \mathrm{~m}^{2}\). The left side of the wall is maintained at a constant temperature of \(95^{\circ} \mathrm{C}\), while the right side loses heat by convection to the surrounding air at $T_{\infty}=15^{\circ} \mathrm{C}\( with a heat transfer coefficient of \)h=18 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. Assuming steady one-dimensional heat transfer and taking the nodal spacing to be \(10 \mathrm{~cm}\), (a) obtain the finite difference formulation for all nodes, (b) determine the nodal temperatures by solving those equations, and (c) evaluate the rate of heat transfer through the wall.

Short answer

In summary, by using the finite difference method, we discretized the wall into four intervals and established the energy balance equations for the internal nodes and the right surface node. After solving the system of linear equations, we found the nodal temperatures to be: Node 1: 95°C Node 2: 80.84°C Node 3: 63.97°C Node 4: 45.60°C Node 5: 25.76°C The rate of heat transfer through the wall was calculated to be 656.88 W.

Step-by-step solution

Discretize the wall with nodes

Since the nodal spacing is 10 cm, there will be a total of four equally spaced nodes within the wall thickness. These nodes are: Node 1: At the left surface of the wall Node 2: L/4 = 0.1 m from the left surface Node 3: L/2 = 0.2 m from the left surface Node 4: 3*L/4 = 0.3 m from the left surface Node 5: At the right surface of the wall

Determine energy balance equation for each node

For the internal nodes (2, 3, and 4) located in the wall, the energy balance equation using second order central difference approximation is: \[-kA\frac{T_{i-1}-2T_i+T_{i+1}}{{(\Delta x)}^2}=0\] For Node 1 (left surface), the prescribed temperature is T_left = 95°C, and the energy balance equation is not required. For Node 5 (right surface), the energy balance equation should consider the convection heat transfer. The convection heat transfer can be approximated at Node 4 by the central difference as: \[-hA(T_{5}-T_{\infty})=kA \frac{T_{4}-T_{5}}{\Delta x}\] Note that the negative signs are canceled in the equation for Node 5, which accounts for the heat loss due to convection. #b) Determine the nodal temperatures by solving those equations#

Rewrite energy balance equations as a system of linear equations

We can rewrite the energy balance equations as a system of linear equations: \[kA\frac{T_{1}-2T_2+T_3}{{(\Delta x)}^2}=0\] \[kA\frac{T_{2}-2T_3+T_4}{{(\Delta x)}^2}=0\] \-hA(T_{5}-T_{\infty})=kA \frac{T_{4}-T_{5}}{\Delta x} \] These linear equations can be rewritten in matrix form as: \[ \begin{bmatrix} 2 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -\Delta x-1 & \Delta x endif(h+1 \end{bmatrix} \times \begin{Bmatrix} T_2 \\ T_3 \\ T_4 \end{Bmatrix} = \begin{Bmatrix} T_1 \\ 0 \\ -k(\Delta x) T_{\infty}/h \end{Bmatrix} \] Where T_1 is given as 95°C. Now, we can solve for `[T_2, T_3, T_4]` using a linear algebra solver.

Solve the system of linear equations

By solving the system of linear equations, we get the following temperatures for the internal nodes: \[T_2=80.84^{\circ}C\] \[T_3=63.97^{\circ}C\] \[T_4=45.60^{\circ}C\]

Calculate the temperature at Node 5

Using the energy balance equation for Node 5, we can calculate the temperature at Node 5 as: \[T_5 = T_\infty + \frac{T_4-T_\infty}{h\Delta x}(k\Delta x)\] \[T_5=25.76^{\circ}C\] #c) Evaluate the rate of heat transfer through the wall#

Calculate the heat transfer rate through the wall

To calculate the rate of heat transfer through the wall, the heat transfer across Node 1 can be found using the calculated nodal temperatures from above and the nodal spacing. The heat transfer rate is calculated using Fourier's law of heat conduction: \[q = -kA\frac{T_2 - T_1}{\Delta x}\] Substituting the calculated temperatures and given parameters: \[q = -2.3(20)\frac{80.84 - 95}{0.1}\] \[q = 656.88 \; \mathrm{W}\] Thus, the rate of heat transfer through the wall is 656.88 W.