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Homework Help: Heat Exchanger - Boiling Rate - HT Coefficient

  1. Apr 21, 2013 #1
    1. The problem statement, all variables and given/known data

    In Attachment

    2. Relevant equations

    3. The attempt at a solution
    Second Attachment.
    - I have only attempted part (i) so far, because I am not sure if I am doing it correctly. I was thinking since the value of the latent heat of pentane is given, surely it has to be used when calculating the boiling rate of pentane (I did not use it) Maybe its used in the second part, not sure. Can someone, just confirm if what I am doing is correct, because I have a good idea on how to tackle (ii) and (iii)
    - The only thing bugging me in part (ii), is how to decide whether the method of main resistance can be used to calc. the length. I have no idea, what this is meant to explain

    Attached Files:

  2. jcsd
  3. Apr 22, 2013 #2
    Hey man. I'm back for a short time. I'm still waiting for your answer to my question on the other thread.

    Here are my comments on this problem.

    I didn't check your numbers, but you had the right idea on how to calculate the cooling rate.

    Your diagram is not right. The pentane is saturated, and its temperature stays 35 C throughout the heat exchanger. All you need to do to get the pentane boiling rate is to divide the cooling load by the heat of vaporization of pentane.

    Once you calculate the heat transfer coefficient on the water side, you can compare it with the heat transfer coefficient on the pentane side. If the heat transfer coefficient on the water side is much lower than that on the pentane side, then the overall heat transfer coefficient will be very close to the water heat transfer coefficient, and the heat transfer coefficient on the pentane side can be neglected. The main resistance to heat transfer will reside on the water side. This would be an example of the "main resistance method."

  4. Apr 26, 2013 #3
    Okay, so your comment were obviously helpful it helped me finish the question. I have attached screenshots of my working, solely for the benefit of other user if they want to know how to work it out. I will explain the main steps, if anyone needs any clarifications I will type out the detailed answer.

    So, as you said my diagram was wrong. Pentane's temp. stays constant. So I calculated the boiling rate by dividing the cooling load by the latent heat vap (pentant) giving a value of 43759 kg/s (does that seem reasonable)

    Next I calculated the delta temp (LM) = 23.27, as this will be later used. To get the h(water), first calculate the Re of water in inner tube, gives 10,612. This is Turbulent, so use the 1st heat transfer coeff. correlation given in the question. This gives a Nu = 70.99. We can thus calculate 'h' from it, giving 2271.68 W/Km^2.

    Now for the discussion of main resistance, using your comments. I said since for water it is quite less than pentane, the h(pentane) can be neglected and the main resistance to heat transfer will be on water side.

    I then, calculated the overall heat transfer coefficient, which was basically that of water (that is correct, right?)

    Using equation: Q = UA delta.temp(LM), I get a value for length as 7.37m
    Just a few questions:
    1. In the question, there are 3 h.t coeff. correlations, the 1st is Turbulent, 2nd is Laminar, what is the 3rd for?
    2. The discussion for main resistance does that seem reasonable? And is there some boundary, I mean when can main resitance not be used, how much smaller should a heat transfer coefficient be, for the rule to be applicable.
    3. When, calculating the overall heat transfer coefficient, I did not take into account Pentane, does that seem reasonable?

    Thanks man

    Attached Files:

  5. Apr 26, 2013 #4
    Does the pentane evaporation rate seem reasonable to you? After all, you have this tiny little heat exchanger processing only about 10 kg/min of water, and it is causing 20 tons/second of pentane to evaporate? Does that make sense? For one thing, 560 kJ/kg is equal to 560000 J/kg, not 0.56 J/kg.

    My calculation of the heat load is:

    Q = 10.0 (4200) (80 - 45) = 1,470,000 J/min = 24500 J/s
    If the heat of vaporization of the pentane is 560000 J/kg, the pentane evaporation rate is 2.625 kg/min.

    You calculated a heat transfer coefficient on the water side of 2272 (I'm assuming this is correct). The heat transfer coefficient on the pentane side is 7000. The overall heat transfer coefficient is given (see your textbook) by

    1/U = (1/7000) + (1/2272)

    From this equation, you get U = 1715 W m-2K-1

    So the conclusion here is that, if the water side heat transfer coefficient is about 1/3 of the pentane side heat transfer coefficient, assuming that the water heat transfer coefficient is dominant is not that good an approximation. In order to make this approximation, the ratio should be more like 1/10.

    The heat transfer coefficient 5.6 applies to laminar flow, but only at long distances from the pipe inlet. At short distances, the heat transfer coefficient is a function of the Re, Pr, and L/D.

    The last equation for the heat transfer coefficient, if I remember correctly, applies to heat transfer to a sphere.
  6. Apr 27, 2013 #5
    Okay thanks, so basically the method of main resistance cannot be used since the difference is not that low. Although, say if the water side heat transfer coefficient was 700 (1/10 of pentane), I can use that as the overall heat transfer coefficient (and this would mean using the method of main resistance?)
  7. Apr 27, 2013 #6
    In practice, you should try to determine the heat transfer coefficient as accurately as possible. That would mean not using the method of main resistance. So under what circumstances would you use the method of main resistance?...Well, if you are confident that one of the heat transfer resistances dominates, and you don't need a very accurate answer, you can use it. Or suppose it is very costly to determine the heat transfer coefficient experimentally, but you are sure that it is going to be very high.
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