0
$\begingroup$

Assume we have a cylindrical fin which has the effective length of L and its efficiency is given by the equation: $$η=exp(-0.32mL)$$ where $$m=\sqrt{\frac{hP}{kA}}$$ where P is circumference and A is the cross sectional area of the fin.

If the volume of the fin remains constant, which of the following statements is true?

By increasing the length of the fin ...

  1. Heat transfer increases.

  2. Heat transfer decreases.

  3. Heat transfer increases then decreases.

  4. Heat transfer remains constant because the volume is constant.

$\endgroup$
2
  • $\begingroup$ Are you testing us? $\endgroup$
    – Solar Mike
    May 3 '18 at 11:17
  • $\begingroup$ No its the test i've seen in my exam today! $\endgroup$
    – Moreza7
    May 3 '18 at 14:31
0
$\begingroup$

The math for this problem goes like this: The fin has a circular cross section and a constant volume. We can write a relationship between volume, length and radius: $V = L*pi*r^2$

Solving that for radius: $r = \sqrt{\frac{V}{L \pi}}$

Efficiency is the ratio of the effectiveness of a fin divided by how effective that fin would be if 100% of its area were at the temperature of the base. Total heat transfer is proportional to efficiency times area.

$Q = hA\eta = hL\pi r = hL\pi\sqrt{\frac{V}{L\pi}} = h\sqrt{L\pi V} \eta$

We can do similar substitutions for efficiency. $\eta = exp(-0.32\sqrt{\frac{hP}{kA}}) = exp(-0.32\sqrt{\frac{h\pi r}{k\pi r^2}}) = exp(-0.32\sqrt{\frac{h}{k r}}) = exp(-0.32\sqrt{\frac{h}{k \sqrt{\frac{V}{L \pi}}}}) = exp(-0.32\sqrt{\frac{h}{k \sqrt{\frac{V}{\pi}}}}L^{1/4})$

When we put these two terms together we can ignore everything that isn't L for the purpose of this question:

$Q = things*L^{1/2}*exp(-things*L^{1/4})$

The first term with L increases with L, the second decreases and definitely not at the same rate. Put in some dummy values for "things" and you can get this graph with L on the horizontal and Q on the vertical. It clearly goes up and then down.

L vs Q

All of this math was not really necessary; Common sense could suffice. The fin equation does not account for heat transfer from the fin tip. You need an additional term for that. If we have a very short fin, it will be $\sqrt{very}$ big around, and have very little area (very > $\sqrt{very}$). Even though it will have 100% efficiency, there will be very little heat transfer. If we have a very long fin, the efficiency will be terrible because most of the fin is very far from the heat source and is cold. At some point in the middle there is an optimum.

$\endgroup$
0
$\begingroup$

The solution for the problem you posed would depend on the relation between P and A. This relation would be specific for each type of cross sectional area(circle, rectangle etc.). So you need to specify the exact shape of cross sectional area for this to be answerable.

$\endgroup$
1
  • $\begingroup$ it has circular cross section $\endgroup$
    – Moreza7
    May 3 '18 at 14:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.