Answer:
q = 12.95 W
Step-by-step explanation:
Given:
- The thermal conductivity of the rod k = 377 W / m-K
- The diameter of the rod D = 10 mm
- The base temperature T_b = 150°C
- The temperature of the environment T_air = 22°C
- The heat transfer coefficient is h_o = 11 W/m^2K
Find:
Calculate the rate of heat loss from the rod q.
Solution:
- We can apply the fin equations with respect to boundary conditions:
- The heat loss from a cylindrical rod is given by:
q = sqrt ( h_o * P * k * A ) * ( T_base - T_air )
Where P is the perimeter circular surface = pi*D
A is the cross sectional area = pi*D^2 / 4
- Plug in values:
q = sqrt ( 11 * pi*D * 377 * pi*D^2 / 4 ) * ( T_base - T_air )
q = sqrt ( 11 * pi*0.01 * 377 * pi*0.01^2 / 4 ) * ( 150 - 22 )
q = sqrt (0.01023231)*128
q = 12.95 W
- The heat loss from the cylindrical rod is 12.95 W.