Question

Consider a large plane wall of thickness \(L=0.4 \mathrm{~m}\), thermal conductivity \(k=2.3 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and surface area \(A=30 \mathrm{~m}^{2}\). The left side of the wall is maintained at a constant temperature of \(T_{1}=90^{\circ} \mathrm{C}\), while the right side loses heat by convection to the surrounding air at $T_{\infty}=25^{\circ} \mathrm{C}\( with a heat transfer coefficient of \)h=24 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. Assuming constant thermal conductivity and no heat generation in the wall, \((a)\) express the differential equation and the boundary conditions for steady one-dimensional heat conduction through the wall, (b) obtain a relation for the variation of temperature in the wall by solving the differential equation, and (c) evaluate the rate of heat transfer through the wall.

Short answer

Answer: The rate of heat transfer through the wall is 32,400 W.

Step-by-step solution

Fourier's Law of Heat Conduction

In order to understand the temperature distribution and heat conduction through the wall, we will use Fourier's Law, which is given by: \(q=-kA\frac{dT}{dx}\), where \(q\) is the heat flux, \(k\) is the thermal conductivity, \(A\) is the surface area, and \(\frac{dT}{dx}\) is the temperature gradient.

Differential Equation for Steady One-Dimensional Heat Conduction

At steady-state conditions, the heat transfer through the wall is constant, which means the rate of heat transfer \(\frac{d^{2}T}{dx^{2}}\) is 0. So the differential equation is: \(\frac{d^{2}T}{dx^{2}}=0\).

Solving the Differential Equation

To solve the differential equation, we will perform two integrations: \(\int d(\frac{dT}{dx})= \int 0 dx \Rightarrow \frac{dT}{dx}=C_{1}\). To find the temperature distribution, perform the second integration: \(\int dT=C_{1}\int dx \Rightarrow T=C_{1}x+C_{2}\).

Applying Boundary Conditions

Given that the left surface is at \(T_{1}=90^{\circ} \mathrm{C}\) and \(x=0\), we can find \(C_{2}\): \(T_{1}=90=C_{2}\). Now, for the boundary condition on the right surface, we will be using Newton's law of cooling to account for the convective heat transfer: \(q=-kA\frac{dT}{dx}= hA(T_{2}-T_{\infty})\). Replace \(x=L\) and \(T_{2}\), and then substitute the known values: \((-k)(\frac{dT}{dx})= h(T_{2}-T_{\infty}) \Rightarrow C_{1}=-\frac{h}{k}(T_{2}-T_{\infty})\). Now we have the equation \(T=C_{1}x+90\) with \(C_{1}=-\frac{h}{k}(T_{2}-T_{\infty})\).

Finding the Rate of Heat Transfer

Using Fourier's Law, we have: \(q=-kA\frac{dT}{dx} = -kA \times C_{1} \). Replacing the values of \(k\), \(A\), and \(C_{1}\), we get: \(q= -\left( 2.3 \frac{W}{m\cdot K}\right) \left( 30 m^{2}\right) \left[-\frac{24 \frac{W}{m^{2}\cdot K}}{2.3 \frac{W}{m\cdot K}} (T_{2}-25^{\circ}C)\right]\), which simplifies to: \(q= 720[T_{2}-25^{\circ}C]\). Now, let's find the value of \(T_{2}\). To do this, we'll substitute \(x=L=0.4 m\) in the equation for \(T\), giving us: \(T_{2}=C_{1}\times 0.4+90\). Plugging in the value of \(C_{1}\), we have: \(T_{2}=-\frac{h}{k}(T_{2}-T_{\infty})\times 0.4+90\). Solving for \(T_{2}\), we get: \(T_{2}=70^{\circ}C\). Finally, substituting the value of \(T_{2}\) into the equation for \(q\), we find the rate of heat transfer: \(q=720(70-25)=720 \times 45 = 32,\!400 ~ W\). So, the rate of heat transfer through the wall is 32,400 W.