Question

Consider a 20-cm-thick large concrete plane wall $(k=0.77 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})$ subjected to convection on both sides with \(T_{\infty 1}=22^{\circ} \mathrm{C}\) and $h_{1}=8 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\( on the inside and \)T_{\infty 22}=8^{\circ} \mathrm{C}$ and \(h_{2}=12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) on the outside. Assuming constant thermal conductivity with no heat generation and negligible radiation, \((a)\) express the differential equations 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 temperatures at the inner and outer surfaces of the wall.

Short answer

Question: Express the differential equations and boundary conditions for steady one-dimensional heat conduction through a concrete wall, the relation for the temperature variation, and find the temperatures at the inner and outer surfaces. Solution: 1. The heat conduction equation: \(\frac{d^2T}{dx^2} = 0\) 2. The temperature profile: \(T(x) = C_1 x + C_2\) 3. Boundary conditions: - At x = 0: \(-k \frac{dT}{dx} \big|_{x=0} = h_1 (T(0) - T_{\infty1})\) - At x = L: \(-k \frac{dT}{dx} \big|_{x=L} = h_2 (T(L) - T_{\infty2})\) 4. Relationship for the temperature variation in the wall: \(T(x) = -\frac{h_1 h_2 (T(L) - T_{\infty1} + T_{\infty2})}{k (h_1 + h_2)} x + \frac{h_1 T_{\infty1} + h_2 T(L)}{h_1 + h_2}\) 5. The temperatures at the inner and outer surfaces: - \(T(0) = \frac{h_1 T_{\infty1} + h_2 T(L)}{h_1 + h_2}\) - \(T(L) = -\frac{h_1 h_2 (T(L) - T_{\infty1} + T_{\infty2})}{k (h_1 + h_2)} L + \frac{h_1 T_{\infty1} + h_2 T(L)}{h_1 + h_2}\)

Step-by-step solution

Write the heat conduction equation for the wall

We will use the general heat conduction equation for steady-state and one-dimensional heat conduction, without heat generation: \(\frac{d^2T}{dx^2} = 0\)

Integrate the equation to obtain the temperature profile

We will integrate the equation twice to obtain the temperature profile T(x): First integration: \(\frac{dT}{dx} = C_1\) Second integration: \(T(x) = C_1 x + C_2\)

Write the boundary conditions

We are given convective boundary conditions at both sides of the wall, so we will have two boundary conditions: At x = 0: \(-k \frac{dT}{dx} \big|_{x=0} = h_1 (T(0) - T_{\infty1})\) At x = L (20 cm = 0.2 m): \(-k \frac{dT}{dx} \big|_{x=L} = h_2 (T(L) - T_{\infty2})\)

Apply the boundary conditions to obtain the constants \(C_1\) and \(C_2\) in the temperature profile

First, let's apply the boundary condition at x = 0: \(-k C_1 = h_1 (C_2 - T_{\infty1})\) Second, apply the boundary condition at x = L: \(k C_1 = h_2 (T(L) - T_{\infty2})\) Now, solve the system of two linear equations to obtain \(C_1\) and \(C_2\).

Solve for the constants \(C_1\) and \(C_2\)

Let's summarize the equations: 1) \(k C_1 = -h_1 (C_2 - T_{\infty1})\) 2) \(k C_1 = h_2 (T(L) - T_{\infty2})\) Divide the first equation by k and isolate \(C_1\): \(C_1 = -\frac{h_1}{k} (C_2 - T_{\infty1})\) Plug this expression of \(C_1\) into the second equation: \(-\frac{h_1}{k} (C_2 - T_{\infty1}) = \frac{h_2}{k} (T(L) - T_{\infty2})\) Now, multiply both sides by k and solve for \(C_2\): \(C_2 = \frac{h_1 T_{\infty1} + h_2 T(L)}{h_1 + h_2}\) Finally, substitute the value of \(C_2\) back into the equation for \(C_1\) to find \(C_1\): \(C_1 = -\frac{h_1 h_2 (T(L) - T_{\infty1} + T_{\infty2})}{k (h_1 + h_2)}\)

Substitute the constants \(C_1\) and \(C_2\) into the temperature profile equation to get the temperature distribution in the wall

Now, we can write the temperature profile as follows: \(T(x) = -\frac{h_1 h_2 (T(L) - T_{\infty1} + T_{\infty2})}{k (h_1 + h_2)} x + \frac{h_1 T_{\infty1} + h_2 T(L)}{h_1 + h_2}\)

Calculate the temperatures at the inner and outer surfaces of the wall

To find the temperature at the inner surface (x=0), substitute x=0 into the temperature profile equation: \(T(0) = \frac{h_1 T_{\infty1} + h_2 T(L)}{h_1 + h_2}\) To find the temperature at the outer surface (x=L), substitute x=L into the temperature profile equation: \(T(L) = -\frac{h_1 h_2 (T(L) - T_{\infty1} + T_{\infty2})}{k (h_1 + h_2)} L + \frac{h_1 T_{\infty1} + h_2 T(L)}{h_1 + h_2}\) Now, we can plug in the given values of \(k\), \(h_1\), \(h_2\), \(T_{\infty1}\), \(T_{\infty2}\), and \(L\) to find \(T(0)\) and \(T(L)\).