Consider a steam pipe of length \(L\), inner radius \(r_{1}\), outer radius \(r_{2}\), and constant thermal conductivity \(k\). Steam flows inside the pipe at an average temperature of \(T_{i}\) with a convection heat transfer coefficient of \(h_{i}\). The outer surface of the pipe is exposed to convection to the surrounding air at a temperature of \(T_{0}\) with a heat transfer coefficient of \(h_{o^{*}}\) Assuming steady one-dimensional heat conduction through the pipe, \((a)\) express the differential equation and the boundary conditions for heat conduction through the pipe material, \((b)\) obtain a relation for the variation of temperature in the pipe material by solving the differential equation, and (c) obtain a relation for the temperature of the outer surface of the pipe.

To find the governing equation for steady-state heat conduction through the pipe's material, we will use the heat equation in cylindrical coordinates, assuming one-dimensional heat conduction in the radial direction, which is given by: \(\frac{1}{r} \frac{d}{d r}(r k \frac{dT}{dr}) = 0\) To solve the problem, we need boundary conditions at the inner and outer surfaces of the pipe. At the inner surface of the pipe, the heat transfer rate is given by convection between the steam and the pipe wall: \(h_i A (T_i - T(r_1)) = 2 \pi r_1 L (T_i - T(r_1))\) At the outer surface of the pipe, the heat transfer rate is given as: \(h_o^* A (T(r_2) - T_0) = 2 \pi r_2 L (T(r_2) - T_0)\)

Now we'll solve the differential equation for the temperature distribution in the pipe wall: Step 1: Separating variables \(\frac{1}{r} \frac{d}{dr}(r k \frac{dT}{dr}) = 0\) \(\frac{d}{dr}(r k \frac{dT}{dr}) = 0\) Step 2: Integrate the equation \(\int\frac{d}{dr}(r k \frac{dT}{dr}) dr = \int 0 dr\) \(r k \frac{dT}{dr} = C_1\) Step 3: Integrate once more \(k \frac{dT}{dr} = \frac{C_1}{r}\) \(\int\frac{dT}{ dr = \int\frac{C_1}{r k}dr\) \(T(r) = \frac{C_1}{k} \ln\vert r\vert + C_2\)

Now we'll use the boundary conditions to find the temperature at the outer surface of the pipe: Step 1: Use the first boundary condition at \(r = r_1\) \(h_i (T_i - T(r_1)) = \frac{k}{r_1}(T(r_1) - T(r_2))\) Using the temperature distribution function we derived: \(h_i (T_i - (\frac{C_1}{k} \ln\vert r_1\vert + C_2)) = \frac{k}{r_1}(T(r_1) - T(r_2))\) Now, we can solve for \(C_1\) and \(C_2\): \(C_1 = -\frac{h_i k^2 (T_i - C_2)}{r_1}\) Step 2: Use the second boundary condition at \(r = r_2\) \(h_o^* (T(r_2) - T_0) = \frac{k}{r_2}(T(r_1) - T(r_2))\) Using the temperature distribution function and \(C_1\) expression: \(h_o^* (\frac{-h_i k^2 (T_i - C_2)}{r_1 r_2} + C_2 - T_0) = \frac{k}{r_2}(T(r_1) - T(r_2))\) Finally, solving for the temperature at the outer surface of the pipe: \(T(r_2) = \frac{-h_i k^2 (T_i - C_2)}{r_1 r_2} + C_2\) In conclusion, we obtained a relation for the temperature distribution in the pipe wall under steady-state heat conduction conditions. The temperature at the outer surface of the pipe is given by \(T(r_2) = \frac{-h_i k^2 (T_i - C_2)}{r_1 r_2} + C_2\).