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Is this even possible? (heat exchanger problem)

  1. Aug 27, 2007 #1
    I am currently dealing with a problem involving the heat transfer between two gases. Would it be possible to take JP-8 Fuel Exhaust (990C) and heat up water (starting at 140C, 9.8MPa) to produce 900 KW? The flow-rate of the water is 0.9kg/sec.

    There is an extreme size-restriction on this heat exchanger, which is limited to 1.5m x 0.6m x 0.6m.

    I read through this "Guide to Compact Heat Exchangers" (link below), and found that the Module 2.5 was the closest to the one I am trying to design, however the size-constraints and input-ranges specified are not within my range of desired inputs.

    Thanks for all your help.


    link to "Guide to Compact Heat Exchangers"
    http://www.ept.ntnu.no/fag/tep32/Pensum/CompactHeatExchangers.pdf [Broken]
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Aug 27, 2007 #2


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    Two issues I see right off the bat:

    1) Have you done a basic energy balance between the two streams? By that I mean a basic first law energy balance? Forget about losses and such. Can you even, theoretically get to where you need to go?

    [tex]\Sigma \dot{m}h_{in} = \Sigma \dot{m}h_{out}[/tex]

    2) What process/cycle will you be employing to produce the 900 kW?
  4. Aug 27, 2007 #3
    1) Yes. I calculated around 1500HP, or 1200 kW, using the values below.

    2) I'm hoping to use a plates-in-shell heat exchanger, as described by "Module 2.5" in the provided link. This 900 kW (~1200HP) will be powering a turbine.


    Perhaps I am approaching this problem from the wrong angle.

    Here is the basic outline of the problem:

    "H20, 1":

    T= 272F
    P= 1433 psia
    h= 245.23 Btu/lbm

    "JP-8 Diesel Fuel Exhaust":
    P= TBD
    h= TBD

    "H20, 1" enters heat exchanger and is heated by the JP-8 Fuel Exhaust, it is now "H20, 2"

    "H20, 2":
    T = 1500F
    P = 1040 psia
    h = 1786.62 Btu/lbm

    "H20, 2" enters turbine then becomes "H20, 3"

    "H20, 3" (isentropic):
    T = 160F
    P = 4.72 psia
    h = 1109.9 Btu/lbm

    "H20, 3" (actual):
    T = 483 F
    P = 4.72 psia
    1280.00 Btu/lbm


    The problem is basically getting "H20, 1" into "H20, 2" within a 5' x 2' x 2' container. I believe that it would require a tremendous amount of surface area. I'm now wondering if the Module 2.5 is the best choice (as seen in the below link, page 58)

    link to "Guide to Compact Heat Exchangers"
    http://www.ept.ntnu.no/fag/tep32/Pensum/CompactHeatExchangers.pdf [Broken]
    Last edited by a moderator: May 3, 2017
  5. Aug 28, 2007 #4


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    Not to sluff you off, but have you contacted a heat exchanger manufacturer? All of the ones I have ever dealt with could tell me exactly what I needed, in about two minutes, as far as overall HT coefficient as well as expected required HT area. You give them the two fluid streams and desired conditions and they can size you a heat exchanger very efficiently. Of course, if this is for a school project, then that changes things.

    The other thing you need to be concerned with, that is very difficult for us lay people, is the structural aspects of the heat exchanger. This is especially true when you have a wide delta between conditions. The thermal stresses alone get to be very harsh.
  6. Aug 31, 2007 #5
    Doing just that. Thanks for your help & suggestions, Fred.
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