A long thin copper rod (K = 377 W/m K) 10 mm in diameter is exposed to an environment at 22 oC. The base temperature of the rod is maintained at 150 oC. The heat transfer coefficient between the rod and the environment is 11 W/m2 K. Calculate the rate of heat loss from the rod.

q = 12.95 W

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.