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2 years ago

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A long cylindrical rod of diameter 200 mm with thermal conductivity of 0.5 W/m⋅K experiences uniform volumetric heat generation

of 24,000 W/m3. The rod is encapsulated by a circular seeve having an outer diameter of 400 mm and a thermal conductivity of 4 W/m⋅K. The outer surface of the sleeve is exposed to cross flow air at 27∘C with a convection coefficient of 25 W/m2⋅K.
(a) Find the temperature at the interface between the rod and sleeve and on the outer surface.
(b) What is the temperature at the center of the rod?

1 answer:

LuckyWell [14K]2 years ago

6 0

Answer:

a, 71.8° C, 51° C

b, 191.8° C

Explanation:

Given that

D(i) = 200 mm

D(o) = 400 mm

q' = 24000 W/m³

k(r) = 0.5 W/m.K

k(s) = 4 W/m.K

k(h) = 25 W/m².K

The expression for heat generation is given by

q = πr²Lq'

q = π . 0.1² . L . 24000

q = 754L W/m

Thermal conduction resistance, R(cond) = 0.0276/L

Thermal conduction resistance, R(conv) = 0.0318/L

Using energy balance equation,

Energy going in = Energy coming out

Which is = q, which is 754L

From the attachment, we deduce that the temperature between the rod and the sleeve is 71.8° C

At the same time, we find out that the temperature on the outer surface is 51° C

Also, from the second attachment, the temperature at the center of the rod was calculated to be, 191.8° C

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<u>Electrostatic Force</u>

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