Respuesta :
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
