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