Answer:
Explanation:
This is a questions that can be answered using electrical resistance analogy. The expression for the
Answer:
[tex]k_{ins} = 0.0541 W/m. K[/tex]
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
[tex]T_{\infity} = 293 K[/tex]
convective coffcient is h= 30 W/m^2 -K
For given condition, rate of heat transfer is given as
[tex]q = \frac{T_s -T_{\infty}}{R}[/tex]
where R is thermal resistance
[tex]R = R_{cond} + R_{cond, ins} + R_{conv}[/tex]
[tex]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}[/tex]
fro aluminum at T = 523 K, Thermal conductivity is[tex] K_a = 230 W/m. K[/tex]
[tex]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}[/tex]
[tex]R = 0.02985 + \frac{0.1763}{k_{ins}} K/W[/tex]
FROM
[tex]q = \frac{T_s -T_{\infty}}{R}[/tex]
[tex]70 = \frac{523 - 293}{0.02985 + \frac{0.1763}{k_{ins}}}[/tex]
solving for[tex] R_{ins}[/tex]
[tex]0.02985 + \frac{0.1763}{k_{ins}} = 3.285[/tex]
[tex]\frac{0.1763}{k_{ins}} = 3.255[/tex]
[tex]k_{ins} = 0.0541 W/m. K[/tex]
