Question

A large plane wall, with a thickness \(L\) and a thermal conductivity \(k\), has its left surface \((x=0)\) exposed to a uniform heat flux \(\dot{q}_{0}\). On the right surface \((x=L)\), convection and radiation heat transfer occur in a surrounding temperature of \(T_{\infty}\). The emissivity and the convection heat transfer coefficient on the right surface are \(\varepsilon\) and \(h\), respectively. Express the boundary conditions and the differential equation of this heat conduction problem during steady operation.

Short answer

Answer: The boundary conditions and differential equation for a plane wall during steady-state operation are: 1. Boundary condition at x=0: $$\frac{dT}{dx}\Big|_{x=0} = -\frac{\dot{q}_0}{k}$$ 2. Boundary condition at x=L: $$\frac{dT}{dx}\Big|_{x=L} = -\frac{1}{k}[h(T(L) - T_{\infty}) + \varepsilon \sigma (T(L)^4 - T_{\infty}^4)]$$ 3. Differential equation for steady-state heat conduction: $$\frac{d^2T}{dx^2} = 0$$

Step-by-step solution

Boundary condition at x=0

At the left surface of the wall where \(x=0\), the wall is exposed to a uniform heat flux \(\dot{q}_{0}\). The heat flux at \(x=0\) is given by Fourier's law as \(-\dot{q}_{0} = -k\frac{dT}{dx}|_{x=0}\). Therefore, the boundary condition at \(x=0\) is given by: $$\frac{dT}{dx}\Big|_{x=0} = -\frac{\dot{q}_0}{k}$$

Boundary condition at x=L

At the right surface of the wall where \(x=L\), convection, and radiation heat transfer occur. The convection heat transfer rate is given by Newton's law of cooling: \(q_{conv} = h(T(L) - T_{\infty})\). The radiation heat transfer rate is given by the Stefan-Boltzmann law for radiation: \(q_{rad} = \varepsilon \sigma (T(L)^4 - T_{\infty}^4)\), where \(\sigma\) is the Stefan-Boltzmann constant. The overall heat transfer rate at \(x=L\) is given by the sum of convection and radiation heat transfer rates: $$\dot{q}(L) = q_{conv} + q_{rad} = h(T(L) - T_{\infty}) + \varepsilon \sigma (T(L)^4 - T_{\infty}^4)$$ Since the heat flux at \(x=L\) is given by Fourier's law as \(-\dot{q}(L) = -k\frac{dT}{dx}|_{x=L}\), the boundary condition at \(x=L\) is given by: $$\frac{dT}{dx}\Big|_{x=L} = -\frac{1}{k}[h(T(L) - T_{\infty}) + \varepsilon \sigma (T(L)^4 - T_{\infty}^4)]$$ Now, let's derive the differential equation for heat conduction during steady-state operation.

Differential equation for steady-state heat conduction

During steady-state operation, the heat conduction problem is governed by the steady-state one-dimensional heat conduction equation, which is given by: $$\frac{d^2T}{dx^2} = 0$$ In summary, we have derived the following boundary conditions and differential equation for the heat conduction problem: 1. Boundary condition at \(x=0\): $$\frac{dT}{dx}\Big|_{x=0} = -\frac{\dot{q}_0}{k}$$ 2. Boundary condition at \(x=L\): $$\frac{dT}{dx}\Big|_{x=L} = -\frac{1}{k}[h(T(L) - T_{\infty}) + \varepsilon \sigma (T(L)^4 - T_{\infty}^4)]$$ 3. Differential equation for steady-state heat conduction: $$\frac{d^2T}{dx^2} = 0$$