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?

Question

Johnson

Engineering

31 January, 04:50

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?

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Paityn Mayo · Answer 1

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