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Heat Exchanger Effectiveness and Outlet Temperature Question

  1. Aug 18, 2013 #1
    Dear All,

    I am having trouble getting the effectiveness and outlet temperature for counter flow heat exchanger which is described in the attachment below.

    I already have the answers but do not know what equations to use to get them. I have equations for the effectives and for the outlet temperatures, but there arn't enough variables from the question to put into them to get the correct answers.

    Any advice on where to start would be great, thank you.

    Q5 Heat Exchangers Tutorial.PNG

    Answers to the question are shown on bottom right corner of attached picture. thanks for looking
     
  2. jcsd
  3. Aug 19, 2013 #2

    SteamKing

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    Well, would you mind sharing these equations?
     
  4. Aug 19, 2013 #3
    Not at all.
    When I said 'I have equations' i meant i have a hand out of derived equations which might not be useful in their curent form but i will try my best to give the ones i think are 'meant' to be used.
    I will give the ones I wrote down on my note pad in class when i was clearly lost as to what was going on in that class:

    Q=mc*Cpc*(Tc2-Tc1);

    LMTD= (Outlet temp - inlet temp)/ (In(outlet temp/inlet temp));

    mh*Cph > mc*Cpc;

    NTU= (UA*A)/(mc*Cpc);

    E = E.PNG


    where:

    mc = 0.3 kg/s (mass flow rate of cold fluid)
    Cpc = specific heat of cold fluid?
    Tc1= 15 degs C (Cold fluid inlet temp)
    Tc2 = cold fluid outlet temp?

    mh= 0.5 kg/s (mass flow rate hot fluid: oil)
    Th1= 130 degs C (hot fluid inlet temp)
    UA= 1.53 KW/ m^2 (overal heat transfer coefficient)
    A= 2.4 m^2 (heat transfer surf. area)
    Cp mean = 2.22 KJ/Kg K


    Hope i haven't confused you more because i think i am.
    (The handout was wrote by my teacher!!! The handout from the Heat Exchanger lecture is attached; Question 5 can be found on the last pg.)
     

    Attached Files:

    Last edited: Aug 19, 2013
  5. Aug 19, 2013 #4

    SteamKing

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  6. Aug 19, 2013 #5
    Thank you very much. That has moved me a lot closer to solving.

    I have all required variables to solve using NTU method except Cph and Cpc.

    Is it possible to calculate this from the 'Cp mean' value of 2.22 KJ/Kg K?
     
  7. Aug 19, 2013 #6
    If i can somehow get Cph and Cpc, I must be able to complete the other equations as these are the missing values.
     
  8. Aug 19, 2013 #7

    SteamKing

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    Cph is going to be the spec. heat of the oil (Cph = spec. heat of the hot fluid)

    Cpc is going to be the spec. heat of the cooling water (Cpc = spec.heat of the cold fluid)
     
  9. Aug 19, 2013 #8
    Thank you but I don't have the values for Cpc and Cph.

    I have a 'Cp mean' value.

    Can I get the unknown values of Cpc and Cph by using the given 'Cp mean' value?
     
  10. Aug 19, 2013 #9

    SteamKing

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    Re-read Post #7 very carefully.

    The mean Cp value for the oil is all you have for the 'hot' fluid.

    You know that the 'cold' fluid is water. Don't you think the Cp of water can be obtained from a table of specific heats for various fluids?
     
  11. Aug 20, 2013 #10
    Thank you very much, I can solve the problem now.

    I am still confused about terminology:
    In the question it says to find 'fluid outlet T' (Tc2), and at first I obtained the oil outlet T (Th2), but realised that 'Fluid temperature' refers to the water T.
    Is this just a convention in these questions? As oil is obviously a fluid as well.

    Also, where it says 'mean specific heat of the fluid', how do I know it only refers to the Cp of the oil and not the water? Cph oppose to Cpc?
     
  12. Aug 20, 2013 #11

    SteamKing

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    Different oils can have different spec. heats. Water, on the other hand, is the most common fluid on earth. There are tables of the properties of water readily available. IMO, you are supposed to determine the outlet temps. of both the water and the oil, since both are unknown.
     
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