Question

Consider a 20 -cm-thick concrete plane wall \((k=0.77\) $\mathrm{W} / \mathrm{m} \cdot \mathrm{K})\( subjected to convection on both sides with \)T_{\infty 1}=27^{\circ} \mathrm{C}\( and \)h_{1}=5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\( on the inside and \)T_{\infty 22}=8^{\circ} \mathrm{C}$ and \(\mathrm{h}_{2}=12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) on the outside. Assuming constant thermal conductivity with no heat generation and negligible radiation, (a) express the differential 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 temperatures at the inner and outer surfaces of the wall.

Short answer

(a) Express the differential equation and boundary conditions for steady one-dimensional heat conduction through the wall. The differential equation for steady one-dimensional heat conduction is: $$ \frac{d^2T}{dx^2} = 0 $$ The boundary conditions at the inner and outer surfaces of the wall are: $$ -k\frac{dT}{dx}\Bigg|_{x=0} = h_1(T_1 - T_{\infty 1}) $$ $$ k\frac{dT}{dx}\Bigg|_{x=L} = h_2(T_{\infty 2}-T_2) $$ (b) Obtain a relation for the temperature variation in the wall by solving the differential equation. The temperature variation in the wall is given by the linear function: $$ T(x) = Ax + B $$ where the constants A and B are: $$ A = \frac{h_1 T_{\infty 1} + h_2 T_{\infty 2}}{h_1 + h_2 + k L^{-1}(h_1 h_2)} $$ $$ B= \frac{k(h_1+h_2) T_{\infty 1} +h_1 h_2 L(T_{\infty 2}-T_{\infty 1})}{(h_1+h_2)(h_1 + h_2 + k L^{-1}(h_1 h_2))} $$ (c) Evaluate the temperatures at the inner and outer surfaces of the wall. The temperatures at the inner and outer surfaces of the wall are: $$ T_1 = T(0) = B = \frac{k(h_1+h_2) T_{\infty 1} +h_1 h_2 L(T_{\infty 2}-T_{\infty 1})}{(h_1+h_2)(h_1 + h_2 + k L^{-1}(h_1 h_2))} $$ $$ T_2 = T(L) = AL + B = \frac{h_2 T_{\infty 1} + h_1 T_{\infty 2}}{h_1 + h_2 + k L^{-1}(h_1 h_2)} $$

Step-by-step solution

Express the differential equation for steady one-dimensional heat conduction

In the absence of heat generation and with constant thermal conductivity, the differential equation for steady one-dimensional heat conduction through the wall is given by: $$ \frac{d^2T}{dx^2} = 0 $$ where \(T\) is the temperature and \(x\) is the distance along the wall.

Express the boundary conditions

The boundary conditions at the inner and outer surfaces of the wall can be expressed using Fourier's law of heat conduction and Newton's law of cooling for convection. At the inner surface, we have $$ -k\frac{dT}{dx}\Bigg|_{x=0} = h_1(T_1 - T_{\infty 1}) $$ At the outer surface, we have $$ k\frac{dT}{dx}\Bigg|_{x=L} = h_2(T_{\infty 2}-T_2) $$ where \(k\) is the thermal conductivity, \(T_1\) and \(T_2\) are temperatures at the inner and outer surfaces of the wall, \(L\) is the thickness of the wall, and \(h_1\) and \(h_2\) are the inside and outside convection coefficients, respectively.

Solve the differential equation

Given that the second derivative of temperature over distance is zero, the solution to this differential equation is a linear temperature profile. Thus, we can express the temperature as a linear function of distance \(x\): $$ T(x) = Ax + B $$ where \(A\) and \(B\) are constants to be determined using the boundary conditions.

Apply the boundary conditions

Using the boundary conditions in Step 2, we can write two equations to solve for the constants \(A\) and \(B\): At \(x = 0\), we have $$ -kA = h_1(B - T_{\infty 1}) $$ At \(x = L\), we have $$ kA = h_2(T_{\infty 2} - (AL + B)) $$ Solving these equations simultaneously, we get the values of \(A\) and \(B\) as: $$ A = \frac{h_1 T_{\infty 1} + h_2 T_{\infty 2}}{h_1 + h_2 + k L^{-1}(h_1 h_2)} $$ and $$ B= \frac{k(h_1+h_2) T_{\infty 1} +h_1 h_2 L(T_{\infty 2}-T_{\infty 1})}{(h_1+h_2)(h_1 + h_2 + k L^{-1}(h_1 h_2))} $$

Evaluate the temperatures at the inner and outer surfaces

We can now use the temperature expression from Step 3 and the constants from Step 4 to find the temperatures at the inner and outer surfaces: At \(x=0\) (inner surface): $$ T_1 = T(0) = B = \frac{k(h_1+h_2) T_{\infty 1} +h_1 h_2 L(T_{\infty 2}-T_{\infty 1})}{(h_1+h_2)(h_1 + h_2 + k L^{-1}(h_1 h_2))} $$ At \(x=L\) (outer surface): $$ T_2 = T(L) = AL + B = \frac{h_2 T_{\infty 1} + h_1 T_{\infty 2}}{h_1 + h_2 + k L^{-1}(h_1 h_2)} $$ Given the values from the problem statement, substitute them into the expressions above to evaluate the temperatures at the inner and outer surfaces.