Question

A hollow aluminum sphere, 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 0.15 and 0.18 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 0.12 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/m2·K. If 70W are dissipated by the heater under steady-state conditions, what is the thermal conductivity of the insulation?

Answer 1

Answer:

Explanation:

This is a questions that can be answered using electrical resistance analogy. The expression for the

Answer 2

Answer:

$k_{ins} = 0.0541 W/m. K$

Explanation:

Given data:

inner radii of sphere is 0.15

outer radii of sphere is 0.18 m

thickness of given insulation is 0.12m

outer radius of given insulation is 0.18 + 0.12 = 0.30 m

inner temperature of surface os 250 degree C  = 523 k

$T_{\infity} = 293 K$

convective coffcient is h= 30 W/m^2 -K

For given condition, rate of heat transfer is given as

$q = \frac{T_s -T_{\infty}}{R}$

where R is thermal resistance

$R = R_{cond} + R_{cond, ins} + R_{conv}$

$R =\frac{1}{4\pi K_a} [\frac{1}{r_i} - \frac{1}{r_o}] + \frac{1}{4\pi K_{ins}} [\frac{1}{r_0} - \frac{1}{r}] + \frac{1}{4\pi r^2 h}$

fro aluminum at T = 523 K, Thermal conductivity is$K_a = 230 W/m. K$

$R =\frac{1}{4\pi \times 230} [\frac{1}{0.15} - \frac{1}{0.18}] + \frac{1}{4\pi K_{ins}} [\frac{1}{.18} - \frac{1}{0.30}] + \frac{1}{4\pi 0.30^2 30}$

$R = 0.02985 + \frac{0.1763}{k_{ins}} K/W$

FROM

$q = \frac{T_s -T_{\infty}}{R}$

$70 = \frac{523 - 293}{0.02985 + \frac{0.1763}{k_{ins}}}$

solving for$R_{ins}$

$0.02985 + \frac{0.1763}{k_{ins}} = 3.285$

$\frac{0.1763}{k_{ins}} = 3.255$

$k_{ins} = 0.0541 W/m. K$