Answer:
q = 16.366 W/m^2
L = 0.646m
Explanation:
Given:
- The inside surface temperature T_i = 925 K
- The pipe surface temperature T_1 = 650 K
- The ambient temperature T_a = 300 K
- The outermost surface temperature = T_rw
- The thermal conductivity coefficient :
- Total thickness of the wall = 1.2 m
k = 0.0073*( 1 + 0.0054*T)
- The convection heat transfer coefficient h = 23 W / m^2K
Find:
How far from the hot surface should the pipe be located? What is the heat flux for the wall?
Solution:
- The conduction through the wall is given by:
[tex]q = -k*\frac{dT}{dx}\\\\ q.x\limits^1^.^2_0 = - 0.0073*\int\limits^T_L {( 1+ 0.0054*T)} \, dT\\\\ 1.2*q = - 0.0073*(T + 0.0027*T^2)\limits^T^r^w_9_2_5 \\\\q = 0.00608333*(T_r_w + 0.0027*T_r_w^2 - 3235.1875)\\\\q = h*( T_r_w - 300 )\\\\0.00608333*(T_r_w + 0.0027*T_r_w^2 - 3235.1875) = 23*( T_r_w - 300 )[/tex]
- Solve the above quadratic equation we get:
T_rw = 300.71 K
q = 16.366 W/m^2
- The pipe must be located where the surface temperature of 650 K can be maintained. Hence, we have:
[tex]q.x\limits^L_0 = - 0.0073*\int\limits^T_L {( 1+ 0.0054*T)} \, dT\\\\ L*q = - 0.0073*(T + 0.0027*T^2)\limits^6^5^0_9_2_5 \\\\L = 0.00044605*(650 + 0.0027*650^2 - 3235.1875)\\\\L = 0.646m[/tex]
