A DC motor delivers mechanical power to a rotating stainless steel shaft ( \(k=15.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) with a length of \(25 \mathrm{~cm}\) and a diameter of \(25 \mathrm{~mm}\). The DC motor is in a surrounding with ambient air temperature of \(20^{\circ} \mathrm{C}\) and convection heat transfer coefficient of \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and the base temperature of the motor shaft is \(90^{\circ} \mathrm{C}\). Using a uniform nodal spacing of \(5 \mathrm{~cm}\) along the motor shaft, determine the finite difference equations and the nodal temperatures by solving those equations.

The variables given in the problem are: - Thermal conductivity of the shaft (\(k\)): \(15.1\,\mathrm{W/(m\cdot K)}\) - Length of the shaft (\(L\)): \(25\,\mathrm{cm}\) or \(0.25\,\mathrm{m}\) - Diameter of the shaft (\(D\)): \(25\,\mathrm{mm}\) or \(0.025\,\mathrm{m}\) - Ambient air temperature (\(T_\infty\)): \(20^{\circ}\mathrm{C}\) - Convection heat transfer coefficient (\(h\)): \(25\,\mathrm{W/(m^2\cdot K)}\) - Base temperature of the motor shaft (\(T_0\)): \(90^{\circ}\mathrm{C}\) - The uniform nodal spacing (\(\Delta x\)): \(5\,\mathrm{cm}\) or \(0.05\,\mathrm{m}\)

We are given a convection heat transfer coefficient, which helps to determine the rate of heat transfer between the shaft and the surrounding air. We also need to determine the shaft area (\(A\)) for heat transfer in the radial direction, which can be calculated as: $$A = \pi D L$$ Using the given values for diameter and length, $$A = \pi (0.025\,\mathrm{m})(0.25\,\mathrm{m}) \approx 0.0196\,\mathrm{m^2}$$

The finite difference method requires dividing the shaft length into equal segments based on the uniform nodal spacing. We have four nodes with distances \(0.05\,\mathrm{m}\), \(0.10\,\mathrm{m}\), \(0.15\,\mathrm{m}\), and \(0.20\,\mathrm{m}\) from the base (where \(x=0\)). Let \(T_i\) represent the temperature at each node along the shaft. For instance, \(T_1\) is the temperature at the first node, located at \(0.05\,\mathrm{m}\) from the base of the shaft.

Using the heat transfer properties, we can create equations using the finite difference method for each node. The general equation for a node \(i\) on the shaft is: $$\frac{k}{\Delta x^2} (T_{i-1} - 2T_i + T_{i+1}) + h A (T_\infty - T_i) = 0$$ For the first node (node 1), the equation will be: $$\frac{k}{\Delta x^2} (T_{0} - 2T_1 + T_{2}) + h A (T_\infty - T_1) = 0$$ For the second node (node 2), the equation will be: $$\frac{k}{\Delta x^2} (T_{1} - 2T_2 + T_{3}) + h A (T_\infty - T_2) = 0$$ For the third node (node 3), the equation will be: $$\frac{k}{\Delta x^2} (T_{2} - 2T_3 + T_{4}) + h A (T_\infty - T_3) = 0$$ For the fourth node (node 4), the equation will be: $$\frac{k}{\Delta x^2} (T_{3} - 2T_4 + T_{5}) + h A (T_\infty - T_4) = 0$$ Since the shaft is divided into 5 segments, one would simplify the last equation by considering the effect of the boundary at \(x = L\). This could either be insulation or convection at the end of the shaft. The explanation to this should be provided to solve the problem completely.

Now that we have the finite difference equations for each node, we can solve them simultaneously to find the nodal temperatures, \(T_1\), \(T_2\), \(T_3\), and \(T_4\) for the given uniform nodal spacing of \(5\,\mathrm{cm}\). The system of equations can be solved using linear algebra methods, such as matrix inversion or Gaussian elimination.