Consider a large plane wall of thickness L = 0.3 m, thermal conductivity k = 2.5 W/m · °C, and surface area A = 12 m2. The left side of the wall at x = 0 is subjected to a net heat flux of q0 = 700 W/m2 while the temperature at that surface is measured to be T1 = 80°C. Assuming constant thermal conductivity and no heat generation in the wall, a. Express the differential equation and the boundary conditions for steady onedimensional heat conduction through the wall. b. Obtain a relation for the variation of temperature in the wall by solving the differential equation. c. Evaluate the temperature of the right surface of the wall at x = L.

a) - k * dT / dx = q_o

b) T (x) = - 280*x + 80

c) T (L) = - 4 C

Explanation:

Given:

- large plane wall of thickness L = 0.3 m

- thermal conductivity k = 2.5 W/m · °C

- surface area A = 12 m2.

- left side of the wall at net heat flux q_o = 700 W/m2 @ x = 0

- temperature at that surface is measured to be T1 = 80°C.

Find:

- Express the differential equation and the boundary conditions for steady one dimensional heat conduction through the wall.

- Obtain a relation for the variation of temperature in the wall by solving the differential equation

- Evaluate the temperature of the right surface of the wall at x = L.

Solution:

- The mathematical formulation of Rate of change of temperature is as follows:

d^2T / dx^2 = 0

- Using energy balance:

E_out = E_in

-k * dT / dx = q_o

- Integrate the ODE with respect to x:

T (x) = - (q_o / k) * x + C

- Use the boundary conditions, T (0) = T_1 = 80C

80 = - (q_o / k) * 0 + C

C = 80 C

-Hence the Temperature distribution in the wall along the thickness is:

T (x) = - (q_o / k) * x + 80

T (x) = - (700/2.5) * x + 80

T (x) = - 280*x + 80

- Use the above relation and compute T (L):

T (L) = - 280*0.3 + 80

T (L) = - 84 + 80 = - 4 C