Question

Consider a large plane wall of thickness \(L=0.3 \mathrm{~m}\), thermal conductivity \(k=2.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and surface area \(A=12 \mathrm{~m}^{2}\). The left side of the wall at \(x=0\) is subjected to a net heat flux of \(\dot{q}_{0}=700 \mathrm{~W} / \mathrm{m}^{2}\), while the temperature at that surface is measured to be $T_{1}=80^{\circ} \mathrm{C}$. 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 temperature of the right surface of the wall at \(x=L\). Answer: (c) \(-4^{\circ} \mathrm{C}\)

Short answer

Answer: The temperature on the other side of the large plane wall is -4°C.

Step-by-step solution

Express the steady one-dimensional heat conduction equation

The differential equation for heat conduction in one dimension for a plane wall with constant thermal conductivity and no heat generation is given by Fourier's Law, which is: \(\frac{d^{2}T}{dx^{2}} = 0\)

Define the boundary conditions

There are two boundary conditions for this problem: (1) At \(x=0\): \(\dot{q}_{0} = -kA\frac{dT}{dx}\Big|_{x=0}\), and \(T_1 = 80^{\circ}\mathrm{C}\) (2) At \(x=L=0.3\mathrm{m}\): \(T=T_2\)

Solve the differential equation

Since \(\frac{d^{2}T}{dx^{2}} = 0\), the first integration implies \(dT/dx = \mathrm{const}\) Integrating once again, the temperature distribution equation is given by: \(T(x) = a + bx\)

Apply first boundary condition

Using the first boundary condition at \(x=0\), we obtain: \(T_{1} = a + b\cdot0 = 80^{\circ}\mathrm{C}\) \(a = 80^{\circ}\mathrm{C}\) Also at \(x=0\), \(\dot{q}_{0} = -kA\frac{dT}{dx}\Big|_{x=0} = -kA\cdot b\) \(\implies b = -\frac{\dot{q}_{0}}{kA} = -\frac{700\mathrm{W/m^{2}}}{2.5\mathrm{W/m\cdot K} \cdot 12\mathrm{m^{2}}} = -0.233\mathrm{K/m}\) Now the temperature distribution equation becomes: \(T(x) = 80^{\circ}\mathrm{C} - 0.233x\mathrm{K/m}\)

Evaluate the temperature at x=L

Now, we substitute \(x=L=0.3\mathrm{m}\) into the temperature distribution equation to find the temperature at the right surface of the wall: \(T_2 = 80^{\circ}\mathrm{C} - 0.233\cdot(0.3)\mathrm{K/m} = 80^{\circ}\mathrm{C} - 0.0699\mathrm{K} = -4^{\circ}\mathrm{C}\) The temperature on the right surface of the wall is \(-4^{\circ}\mathrm{C}\).