Question

Water flows through a pipe at an average temperature of $T_{\infty}=70^{\circ} \mathrm{C}\(. The inner and outer radii of the pipe are \)r_{1}=6 \mathrm{~cm}$ and \(r_{2}=6.5 \mathrm{~cm}\), respectively. The outer surface of the pipe is wrapped with a thin electric heater that consumes \(300 \mathrm{~W}\) per \(\mathrm{m}\) length of the pipe. The exposed surface of the heater is heavily insulated so that all heat generated in the heater is transferred to the pipe. Heat is transferred from the inner surface of the pipe to the water by convection with a heat transfer coefficient of $h=85 \mathrm{~W} / \mathrm{m}^{2}$.K. Assuming constant thermal conductivity and one-dimensional heat transfer, express the mathematical formulation (the differential equation and the boundary conditions) of the heat conduction in the pipe during steady operation. Do not solve.

Short answer

Answer: The differential equation for steady-state heat conduction in the cylindrical pipe is given by: \( T(r) = C_{1} \ ln(r) + C_{2} \) The boundary conditions are: 1. \(k\frac{dT}{dr}|_{r=r_1} = h(T_{\infty}-T(r_1))\) 2. \(k\frac{dT}{dr}|_{r=r_2} = \frac{300\mathrm{~W}}{2\pi r_2}\)

Step-by-step solution

Write the general heat conduction equation

The general heat conduction equation for a cylindrical pipe is given by: \(\frac{1}{r} \frac{\partial}{\partial r}(r \frac{\partial T}{\partial r})=0\)

Integrate the equation twice

Integrating the equation once, with respect to \(r\), we get: \(r \frac{\partial T}{\partial r}=C_{1}\) Integrating again with respect to \(r\) gives us: \(T(r) = C_{1} \ ln(r) + C_{2}\)

Derive the boundary conditions

We have 2 boundary conditions to apply: 1. At the inner or \(r=r_1\): Heat is transferred by convection. \(q_{in}=h(T(r_1)-T_{\infty})\) We also know that \(q_{in}=-k\frac{dT}{dr}|_{r=r_1}\), where \(k\) is the constant thermal conductivity. So, \(k\frac{dT}{dr}|_{r=r_1} = h(T_{\infty}-T(r_1))\) 2. At the outer surface or \(r=r_2\): The electric heater consumes \(300 \mathrm{~W}\) per \(\mathrm{m}\) length of the pipe. Therefore, the heat transferred is equal to the power consumption: \(q_{out}=\frac{300\mathrm{~W}}{2\pi r_2}\) We also know that \(q_{out}=-k\frac{dT}{dr}|_{r=r_2}\). So, \(k\frac{dT}{dr}|_{r=r_2} = \frac{300\mathrm{~W}}{2\pi r_2}\)

Express the mathematical formulation of the problem

The steady-state heat conduction equation for the pipe is given by: \(T(r) = C_{1} \ ln(r) + C_{2}\) With boundary conditions: \(k\frac{dT}{dr}|_{r=r_1} = h(T_{\infty}-T(r_1))\) \(k\frac{dT}{dr}|_{r=r_2} = \frac{300\mathrm{~W}}{2\pi r_2}\) The mathematical formulation (the differential equation and the boundary conditions) of the heat conduction in the pipe during steady operation has been presented, as requested. The next step would be to solve the problem and analyze the temperature distribution, but that was not requested in this exercise.