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?
(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:
Answer:
a, 71.8° C, 51° C
b, 191.8° C
Explanation:
Given the data:
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 heat generation formula can be articulated as follows:
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
Applying the energy balance equation,
Energy In = Energy Out
This equates to q, which is 754L
From the initial analysis, the temperature at the interface between the rod and sleeve is found to be 71.8° C
Additionally, the outer surface temperature records as 51° C
Furthermore, based on the second analysis, the calculated temperature at the center of the rod is determined to be 191.8° C
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