Question

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 \(h_{\sigma}\). 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.

Short answer

Question: Determine the expression for the temperature profile within the pipe material and the temperature at the outer surface of a steam pipe. Short Answer: The temperature profile within the pipe material can be expressed as \(T(r) = C_1 \ln r + C_2\), where r is the radial distance from the center of the pipe, and \(C_1\) and \(C_2\) are constants determined by the boundary conditions. The temperature at the outer surface of the pipe is given by the expression \(T_0 = C_1 \ln r_2 + C_2\), where \(r_2\) is the outer radius of the pipe.

Step-by-step solution

Set up the differential equation for heat conduction through the pipe material

We will use the equation for steady one-dimensional heat conduction in cylindrical coordinates for a pipe: $$\frac{d}{dr}\left(r\frac{dT}{dr}\right) = 0$$

Integrate the differential equation to obtain the general solution

Integrate the equation above to obtain the general solution: $$r\frac{dT}{dr} = C_1$$ where \(C_1\) is a constant of integration. Next, let's rearrange the equation and then integrate again to obtain the variation of temperature in the pipe material. $$\frac{dT}{dr} = \frac{C_1}{r}$$ Integrate this equation to obtain the temperature profile in the pipe material: $$T(r) = C_1 \ln r + C_2$$ where \(C_2\) is another constant of integration.

Determine the boundary conditions

Since the temperature at the inner radius \(r_1\) is \(T_i\) and at the outer radius \(r_2\) is \(T_0\), we can set up the following boundary conditions: 1. \(T(r_1) = T_i\) 2. \(T(r_2) = T_0\)

Determine the constants of integration \(C_1\) and \(C_2\)

Apply the boundary conditions to the temperature profile equation T(r) to find the constants \(C_1\) and \(C_2\): From condition 1: \(T_i = C_1 \ln r_1 + C_2\) From condition 2: \(T_0 = C_1 \ln r_2 + C_2\) Solve these two linear equations for the constants \(C_1\) and \(C_2\). After obtaining these constants, plug them back into the temperature profile equation T(r).

Obtain the relation for the temperature at the outer surface of the pipe

Since the temperature at the outer surface of the pipe is \(T_0\), evaluate the temperature profile equation \(T(r)\) at \(r = r_2\): $$T_0 = T(r_2)$$ The final expression for the outer surface temperature is: $$T_0 = C_1 \ln r_2 + C_2$$ This step completes the exercise and now we have obtained the expressions for the temperature profile within the pipe material and the temperature at the outer surface of the pipe.