Question

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.

Answer 1

Answer:

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