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3 years ago

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A cylindrical shell of inner and outer radii, ri and ro, respectively, is filled with a heat-generating material that provides a

uniform volumetric generation rate (W/m3) of Qdot. The inner surface is insulated, while the outer surfaced is exposed to a fuild at Tinfinity and a convection coefficient h.
a) Obtain an expression for the steady-state temperature distribution T(r) in the shell, expressing your result in terms of ri, ro, Qdot, h, Tinfinity, and the thermal conductivity k of the shell material
b) Determine an expression for the heat flux, q''(ro), at the outer radius of the shell in terms of Qdot and shell dimensions.

1 answer:

mestny [16]3 years ago

8 0

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)

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