Question

A plane wall of thickness \(L\) is subjected to convection at both surfaces with ambient temperature \(T_{\infty \text { el }}\) and heat transfer coefficient \(h_{1}\) at the inner surface and corresponding \(T_{\infty 2}\) and \(h_{2}\) values at the outer surface. Taking the positive direction of \(x\) to be from the inner surface to the outer surface, the correct expression for the convection boundary condition is (a) \(k \frac{d T(0)}{d x}=h_{1}\left[T(0)-T_{\infty 1}\right]\) (b) $k \frac{d T(L)}{d x}=h_{2}\left[T(L)-T_{\infty 22}\right]$ (c) $-k \frac{d T(0)}{d x}=h_{1}\left(T_{\infty 1}-T_{\infty 22}\right)(d)-k \frac{d T(L)}{d x}=h_{2}\left(T_{\infty \infty 1}-T_{\infty 22}\right)$ (e) None of them

Short answer

Question: Determine the correct expressions for the convection boundary condition at the inner and outer surfaces of a plane wall. Answer: The correct expressions are (a) for the inner surface: \(-k \frac{dT(0)}{dx} = h_1(T(0) - T_{\infty1})\) and (b) for the outer surface: \(-k \frac{dT(L)}{dx} = h_2(T(L) - T_{\infty2})\).

Step-by-step solution

Write down the general heat conduction equation for this problem

The differential equation governing the steady-state one-dimensional thermal conduction through the wall is given by Fourier's law: \(\frac{d^2T}{dx^2} = 0\)

Apply the inner surface boundary condition for convection

At \(x=0\) (inner surface), the conduction heat transfer rate is equal to the convection heat transfer rate. According to Fourier's law, the conduction heat transfer rate is given by \(q_x = -k \frac{dT}{dx}\) The convection heat transfer rate can be found using Newton's law of cooling: \(q_{conv} = h_1(T(0) - T_{\infty1})\) By equating these two, we have: \(-k \frac{dT(0)}{dx} = h_1(T(0) - T_{\infty1})\)

Apply the outer surface boundary condition for convection

At \(x=L\) (outer surface), the conduction heat transfer rate is equal to the convection heat transfer rate. According to Fourier's law, the conduction heat transfer rate is given by \(q_x = -k \frac{dT}{dx}\) The convection heat transfer rate can be found using Newton's law of cooling: \(q_{conv} = h_2(T(L) - T_{\infty2})\) By equating these two, we have: \(-k \frac{dT(L)}{dx} = h_2(T(L) - T_{\infty2})\)

Verify which option matches with our derived expressions

We have derived the following expressions for the inner and outer surface boundary conditions: At inner surface: \(-k \frac{dT(0)}{dx} = h_1(T(0) - T_{\infty1})\) At outer surface: \(-k \frac{dT(L)}{dx} = h_2(T(L) - T_{\infty2})\) Comparing with the options given in the exercise, we can clearly see that the correct expressions are: (a) for the inner surface and (b) for the outer surface.