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18 November, 06:36

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.

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  1. 18 November, 06:56
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    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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