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.

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)

answer:

the formula for amps is watts divided by volts. to use the chart, cover up the a with your finger and use the remaining chart calculation of w divided by v. using our sample panel data, 60 watts divided by 12 volts equals 5 amps.

explanation:

answer:

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