Ask Question
14 October, 23:07

A hollow aluminum sphere (kAl = 234 W/mK), with an electrical heater in the center, is used in tests to determine the thermal conductivity of insulating materials. The inner and outer radii of the sphere are 0.18 and 0.21 m, respectively, and testing is done under steady-state conditions with the inner surface of the aluminum maintained at 250°C. In a particular test, a spherical shell of insulation is cast on the outer surface of the sphere to a thickness of 0.15 m. The system is in a room for which the air temperature is 20°C and the convection coefficient at the outer surface of the insulation is 30 W/m2K. If 80W are dissipated by the heater under steady-state conditions, what is the thermal conductivity of the insulation?

+2
Answers (1)
  1. 15 October, 02:44
    0
    thermal conductivity of the insulation; k_ins = 0.062 W/m. k

    Explanation:

    We are given;

    inner radii of sphere; r_i = 0.15 m

    outer radii of sphere; r_o = 0.18 m

    Thickness of given insulation is 0.12m

    outer radius of given insulation; r = 0.18 + 0.12 = 0.30 m

    convective coefficient; h = 30 W/m². K

    inner temperature of surface; T_s = 250 °C = 250 + 273K = 523 k

    T (∞) = 20°C = 20 + 273 = 293K

    Thermal conductivity of aluminium; k_Al = 234 W/m. K

    For this problem, the formula for the rate of heat transfer is given as;

    q = [T_s - T (∞) ]/R

    R is thermal resistance and is given by the formula;

    R = R_cond + R_cond, ins + R_conv

    When expanded, it results in;

    R = (1 / (4•π•234)) [1/r_i - 1/r_o] + (1 / (4•π•k_ins)) [1/r_o - 1/r] + (1 / (4•π•r²•h))

    Where k_ins is thermal conductivity of insulator

    Plugging in the relevant values to obtain;

    R = (1 / (4•π•234)) [1/0.15 - 1/0.18] + (1 / (4•π•k_ins)) [1/0.18 - 1/0.3] + (1 / (4•π•0.3²•30))

    R = 0.00037786 + 0.1768/k_ins + 0.02947

    R = 0.029848 + 0.1768/k_ins

    We recall that;

    q = [T_s - T (∞) ]/R

    We are given; q = 80W

    Thus;

    80 = 523 - 293 / (0.029848 + 0.1768/k_ins)

    Thus;

    80 (0.029848 + 0.1768/k_ins) = 230

    Divide both sides by 80;

    (0.029848 + 0.1768/k_ins) = 230/80

    0.1768/k_ins = 230/80 - 0.029848

    0.1768/k_ins = 2.8452

    Thus;

    k_ins = 0.1768/2.8452

    k_ins = 0.062 W/m. k
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “A hollow aluminum sphere (kAl = 234 W/mK), with an electrical heater in the center, is used in tests to determine the thermal conductivity ...” in 📘 Engineering if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers