A long bar of rectangular cross section 03 m × 06 m and having a thermal conductivity of k = 1 Wm-K is subjected to the boundary conditions shown Two of the sides are maintained at a uniform temperatu

We will use the finite-volume method to solve for the temperature distribution in the bar and then calculate the heat transfer rate at the convective boundary.

First, we divide the bar into a grid with a spacing of Δx = Δy = 0.1 m, as shown in the diagram below.

|-------------------------------------------|
|                                           |
| 200     T     T     T     T     T     200  |
|                                           |
|  T      T     T     T     T     T     T   |
|                                           |
|  T      T     T     T     T     T     T   |
|                                           |
|  T      T     T     T     T     T     T   |
|                                           |
|  T      T     T     T     T     T     T   |
|                                           |
|  T      T     T     T     T     T     T   |
|                                           |
|  T      T     T     T     T     T     T   |
|                                           |
|  T      T     T     T     T     T     T   |
|                                           |
|  T      T     T     T     T     T     T   |
|                                           |
|  T      T     T     T     T     T     T   |
|                                           |
|  T      T     T     T     T     T     T   |
|                                           |
|  T      T     T     T     T     T     T   |
|                                           |
|  T      T     T     T     T     T     T   |
|                                           |
|  T      T     T     T     T     T     T   |
|                                           |
|  T      T     T     T     T     T     T   |
|                                           |
|  T      T     T     T     T     T     T   |
|                                           |
|  T      T     T     T     T     T     T   |
|                                           |
|  T      T     T     T     T     T     T   |
|                                           |
|  T      T     T     T     T     T     T   |
|                                           |
|  T      T     T     T     T     T     T   |
|                                           |
|  T      T     T     T     T     T     T   |
|                                           |
|  T      T     T     T     T     T     T   |
|                                           |
|  T      T     T     T     T     T     T   |
|                                           |
|  T      T     T     T     T     T     T   |
|                                           |
| 20       T     T     T     T     T     20 |
|                                           |
|-------------------------------------------|

We can see that there are 5 nodes in the x-direction and 12 nodes in the y-direction, giving a total of 60 nodes. We label the nodes as shown in the diagram above, with the boundary nodes labeled with their known temperatures.

Next, we write the energy balance equation for each control volume. For the interior nodes (labeled with T), we have:

$$\frac{T_{i,j+1}-T_{i,j-1}}{2\Delta y}\cdot\Delta x+\frac{T_{i+1,j}-T_{i-1,j}}{2\Delta x}\cdot\Delta y=0$$

Simplifying this equation and rearranging, we get:

$$T_{i,j}=\frac{\Delta y^2(T_{i+1,j}+T_{i-1,j})+\Delta x^2(T_{i,j+1}+T_{i,j-1})}{2(\Delta x^2+\Delta y^2)}$$

For the boundary nodes, we have:

• For the adiabatic boundary (node 2), we have:

$$\frac{T_{1,j+1}-T_{1,j-1}}{2\Delta y}\cdot\Delta x+\frac{T_{2,j}-T_{1,j}}{\Delta x}=0$$

Simplifying and solving for T2,j, we get:

$$T_{2,j}=T_{1,j}$$

• For the convective boundary (node 4), we have:

$$\frac{T_{i,j+1}-T_{i,j-1}}{2\Delta y}\cdot\Delta x+\frac{T_{i+1,j}-T_{i-1,j}}{2\Delta x}\cdot\Delta y+h\Delta x(T_\infty-T_{i,j})=0$$

Simplifying and solving for T4,j, we get:

$$T_{4,j}=\frac{\Delta y^2(T_{3,j}+T_{5,j})+\Delta x^2(T_{4,j+1}+T_{4,j-1})+2h\Delta x\Delta yT_\infty}{2(\Delta x^2+\Delta y^2)+2h\Delta x\Delta y}$$

We can now use these equations to iteratively solve for the temperatures at each node until convergence is reached. We can set an initial guess for all the interior nodes as 100°C, and then use the boundary equations to update the temperatures at the boundary nodes. We can then use the interior equation to update the temperatures at the interior nodes, and repeat this process until the temperatures converge.

Once we have the temperature distribution, we can calculate the heat transfer rate at the convective boundary using the following equation:

$$q''=-h(T_{4,j}-T_\infty)$$

where q'' is the heat transfer rate per unit area, and T4,j is the temperature at the convective boundary node. We can then multiply q'' by the area of the boundary to get the total heat transfer rate per unit length of the bar.

Note: This problem assumes steady-state conditions. If the bar is initially at a different temperature than 200°C, then transient heat transfer analysis would be required to determine the temperature distribution and heat transfer rate at the convective boundary