- #1

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Given:

Given:

Sh

_{D}≡h

_{m}(r)D/D

_{AB}=Sh

_{o}[1 + a (r/r

_{o})

^{n}] (1)

Sh

_{o}=h

_{m}(r=0)D/D

_{AB}=0.814Re

_{D}

^{1/2}Sc

^{0.36}(2)

**Relevant equations:**

Average nusselt number is defined as Nu

_{av}=h

_{av}D/k

where k is thermal conductivity, D is diameter of disk and h

_{av}is average convection coefficient.

The heat and mass transfer analogy states that

Nu

_{av}/Pr

^{n}=Sh

_{av}/Sc

^{n}(3)

Sh is Sherwood number, Nu is Nusselt number, Pr is Prandtl and Sc is Schmidt. In this case n=0.36, from given data.

h

_{av}is defined as: h

_{av}=(1/A

_{s})∫ h dA

_{s}(4)

where you integrate h, the convection coefficient, over the surface area, A

_{s}.

**Solution??**

If I solve for h

_{m}(r) in (1), and integrate over the surface area, I am still stuck with the constant (D

_{AB}/D), but this should NOT be in the final answer. This is obviously the wrong approach, but the rest of the answer is correct, so I am on to something, I am just not sure how to use the analogy correctly.

Somehow I need to use the analogy and combine it with the formula for h

_{av}to obtain the average Nusselt number. Apparently the solution is to integrate Sh

_{o}[1 + a (r/r

_{o})

^{n}] over the area A

_{s}, and just replace Sh

_{o}with 0.814Re

_{D}

^{1/2}Pr

^{0.36}. But why can I do this? I understand that it has something to do with the analogy, but I don't understand how or why. Can someone help me out here?