Question

Consider a 5-m-long constantan block $(k=23 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\( \)30 \mathrm{~cm}\( high and \)50 \mathrm{~cm}$ wide (Fig. P5-78). The block is completely submerged in iced water at \(0^{\circ} \mathrm{C}\) that is well stirred, and the heat transfer coefficient is so high that the temperatures on both sides of the block can be taken to be $0^{\circ} \mathrm{C}$. The bottom surface of the bar is covered with a low-conductivity material so that heat transfer through the bottom surface is negligible. The top surface of the block is heated uniformly by an \(8-\mathrm{kW}\) resistance heater. Using the finite difference method with a mesh size of $\Delta x=\Delta y=10 \mathrm{~cm}\( and taking advantage of symmetry, \)(a)$ obtain the finite difference formulation of this problem for steady twodimensional heat transfer, \((b)\) determine the unknown nodal temperatures by solving those equations, and \((c)\) determine the rate of heat transfer from the block to the iced water.

Short answer

The governing equation for steady two-dimensional heat conduction is: \(k(\frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2}) = 0\)

Step-by-step solution

Obtain the finite difference formulation

For steady two-dimensional heat conduction, the governing equation can be written as: \(k(\frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2}) = 0\) We will use the standard central difference approximation for the second derivatives: \(\frac{\partial^2 T}{\partial x^2} \approx \frac{T_{i+1,j} - 2T_{i,j} + T_{i-1,j}}{\Delta x^2}\) \(\frac{\partial^2 T}{\partial y^2} \approx \frac{T_{i,j+1} - 2T_{i,j} + T_{i,j-1}}{\Delta y^2}\) Substituting the approximations into the governing equation, we get: \(k(\frac{T_{i+1,j} - 2T_{i,j} + T_{i-1,j}}{\Delta x^2} + \frac{T_{i,j+1} - 2T_{i,j} + T_{i,j-1}}{\Delta y^2}) = 0\) Since \(\Delta x = \Delta y = 10 \mathrm{~cm}\) (0.1 m), this simplifies to: \(T_{i+1,j} - 2T_{i,j} + T_{i-1,j} + T_{i,j+1} - 2T_{i,j} + T_{i,j-1} = 0\) Rearranging, we get the finite difference equation: \(T_{i,j} = \frac{1}{4}(T_{i+1,j} + T_{i-1,j} + T_{i,j+1} + T_{i,j-1})\) This is the finite difference formulation for this problem.

Determine the unknown nodal temperatures

We will first set up the boundary conditions for this problem and then apply the finite difference equation to determine the unknown nodal temperatures. The bottom surface of the block, which is covered by a low-conductivity material, has no heat transfer. The side boundaries are symmetric, and the top surface is heated by an 8-kW resistance heater. The temperature is given on both sides of the block due to submersion in iced water. As we have 4 nodes in the x-direction and 3 nodes in the y-direction, we have 12 unknown nodal temperatures. We can set up a system of linear equations based on the finite difference equation and the boundary conditions to solve this problem.

Find the rate of heat transfer

To find the rate of heat transfer from the block to the iced water, we can use Fourier's Law of heat conduction: \(q = -kA\frac{\Delta T}{\Delta x}\) The area of the block in contact with the iced water is given as \(A=0.3*0.5 = 0.15 \mathrm{m^2}\). Using the nodal temperatures obtained in step 2, we can calculate the temperature difference across each node. Then, we can find the heat transfer rate through each node and sum them up to find the total rate of heat transfer from the block to the iced water.