Answer:
A)The expression for the steady-state temperature distribution T(r) in the shell is;
T(r) = [(q'r²/4k)(ro² - r²)] + [(q'ri/2k)In(r/ro)] + (q'ro/2h)[1 - (ri/ro)²] + T∞
B) The expression for the heat flux, q''(ro), at the outer radius of the shell is; q'(ro) = q'π(ro² - ri²)
Explanation:
A) we want to find an expression for the steady-state temperature distribution T(r) in the shell.
First of all, one dimensional radial heat for cylindrical shell with uniform heat generation is;
(1/r)(d/dr)[rdT/dr] + (q'/k) = 0
Subtract (q'/k)from both sides to give;
(1/r)(d/dr)[rdT/dr] = - (q'/k)
Multiply both sides by r to give;
(d/dr)[rdT/dr] = - (q'r/k)
So, [rdT/dr] = - ∫(q'/k)
So [rdT/dr] = -(q'r²/2k) +C1
And;
[dT/dr] = - (q'r²/2k) +(C1)/r
Thus, T(r) = - (q'r²/4k) +(C1)In(r) +C2
Now, we apply the boundary condition at r = ri for dT/dr =0
Thus; (dT/dr)| at r=ri, is zero.
Thus, at r=ri
d/dr[(q'r²/2k) +(C1)In(r) +C2] = 0
Thus;
- (q'ri/2k) +(C1)/ri + 0 = 0
So, making C1 the subject of the formula,
C1 = (q'ri/2k)
Now, let's apply the boundary condition at r=ro for
q''(conduction) = q''(convection)
So, at r=ro, - kdT = h[T(ro - T∞)]
So, using previously gotten equation above, we obtain,
-k[(q'ro²/2k) + (C1)/ro] = h[-(q'ro²/4k) +(C1)In(ro) + C2- T∞)]
So making C2 the subject, we have;
C2 = (q'ro/2h)[1 - (ri/ro)²] + (q'ro²/2k)[(1/2) - (ri/ro)²In(ro)] + T∞
So putting the formulas for C1 and C2 in the equation earlier derived for T(r) to obtain;
T(r) = - (q'r²/4k) + (q'r²/2k)In(r) + (q'ro/2h)[1 - (ri/ro)²] + (q'ro²/2k)[(1/2) - (ri/ro)²In(ro)] + T∞
Thus;
T(r) = [(q'r²/4k)(ro² - r²)] + [(q'ri/2k)In(r/ro)] + (q'ro/2h)[1 - (ri/ro)²] + T∞
B) We want to find out heat rate at outer radius of the shell. So at r=ro, Formula is;
q'(ro) = - k(2πro) dT/dr
= - k(2πro) [- (q'ro²/2k) +(C1)/ro]
C1 = (q'ri/2k) from equation earlier. Thus;
q'(ro) = -k(2πro) [- (q'ro²/2k) +(q'ri/2k)/ro]
When we expand this, we obtain;
q'(ro) = q'π(ro² - ri²)
