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Homework Help: Steam produced heat transfer heat exchangers

  1. Nov 15, 2014 #1
    1. The problem statement, all variables and given/known data

    Dry saturated steam at a temperature of 180oC is to be produced in a fire tube boiler from the cooling of 50 000 kg h–1 of flue gases from a pressurised combustion process. The gases enter the tubes of the boiler at 1600oC and leave at 200oC. The feed water is externally preheated to 180oC before entering the boiler.

    The mean specific heat capacity of the flue gases is 1.15 kJ kg–1 K–1. The latent heat of vaporisation of the water at 180oC is 2015 kJ kg–1. Feed water temperature = 180oC.

    Determine the amount of steam produced per hour, if the total heat loss is 10% of the heat available for steam raising.

    2. Relevant equations
    Mass steam= heat available / heat required to produce 1kg steam
    Qmc=heat transfer rate/ Cpc(tc2-tc1)
    Heat transfer rate=Qmh hfg
    3. The attempt at a solution
    Heat required would be 2257 from steam table at 100C evaporation
    Heat available
    Qmh hfg
    50000*2015=10075x10^4 kW
    Ht rate=10% of 10075x10^4 = 90675x10^3
    Mass steam=heat available / heat req 1kg
    Mass steam = 90675x10^3/2257
    =40175 kg h-1
    Can anyone check over this as I am unsure if it is correct
  2. jcsd
  3. Nov 15, 2014 #2


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    Why? The feed water is being supplied already pre-heated to 180 C
    The units here are not kW. 1 W = 1 J/s, and 1 hour ≠ 1 second

    Show the units here and carry thru with the multiplication. That's how to avoid mistakes.
    As mentioned above, the feed water is already pre-heated to 180 C when it enters the heat exchanger. The problem statement tells you that the latent heat of vaporization is 2015 kJ/kg for water at 180 C.

    The enthalpy of steam at 100 C is irrelevant here.
  4. Nov 15, 2014 #3
    Thanks for your reply
    Ah so if I work out the correct units and carry it through I will divide it by 2015 which is at 180C ? Does the rest Seem ok?
  5. Nov 15, 2014 #4


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    The heat supplied by the flue gas seems to have been overlooked. The gas enters at a temp. of 1600 C and leaves at a temp. of 200 C. The specific heat capacity of the flue gas is 1.15 kJ/kg-K. You've got to work out how much heat is available to be transferred from the gas to the feed water to make steam.
  6. Nov 15, 2014 #5
    Hi again,
    To find out this is this the spec heat*temp difference
    I'm not that clued up on this subject
    Once this is found is this part of the heat available
  7. Nov 15, 2014 #6


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    Yes. You are transferring the heat from the flue gas to the feed water to turn it into steam. Remember to include the 10% heat loss in your calculations of steam production.
  8. Nov 15, 2014 #7
    So it would be 1400*1.15*.9 for the ten percent heat loss which gives 1449 kW I'm assuming?
    So is this the heat available? Then we divide it by 2015 the enthalpy at 180C ?
    I believe I am missing a step out as this seems too low of a value
  9. Nov 15, 2014 #8


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    Again, you need to show and carry thru the units in these calculations. You are just assuming what units appear in the results.

    You've also neglected to include the flow rate of the flue gases in your heat calculations, which is why the result appears so low.

    For the heat available:

    H = eff. * flow * cp * ΔT, which has units

    H = kg/hr * kJ/kg-K * K

    H = kJ/hr, which is not the same units as kW, so stop calling the heat rate kW
  10. Nov 17, 2014 #9
    So, (1600-200)*1.15*.9*50,000=72450000kj/hr
    7245x10^4/2015=35955kg/hr ?
  11. Nov 17, 2014 #10


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    This looks better.
  12. Nov 17, 2014 #11
    Thanks for your help!
  13. Oct 8, 2015 #12
    On the same question: The overall heat transfer coefficient based on outside area of the tubes is given as 54 Wm K .Determine the area of heat transfer required to perform this duty
  14. Oct 9, 2015 #13
    Are these units correct?
  15. Oct 9, 2015 #14
    Should be Wm^-2 K^-1
  16. Oct 9, 2015 #15
    Good. Now, what's the equation for the heat transfer in a heat exchanger?
  17. Oct 10, 2015 #16
    I'm new to this forum and not sure yet how to write down the symbols, but heat transfer= Phi = U A (LMTD).So finding A would be phi/U(LMTD).Finding the LMTD is proving difficult as I'm not sure of the outlet temp of the feed water.
  18. Oct 10, 2015 #17


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    It's not clear what you are trying to analyze here.

    The original problem involved a fire tube boiler. The conditions at which the boiler produced steam were not disclosed in the OP. The only thing you know for certain is that the feed water entered the boiler at 180° C.
  19. Oct 10, 2015 #18
    If you refer to the first post from Mitch1 all the details of the first part are there.The second part of the question is asking for the area required to perform this duty when the overall heat coefficient is 54 Wm^-2 K^-1
  20. Oct 10, 2015 #19
    Good. You have a pool of water boiling at 180oC producing steam at 180oC. What might the water temperature be everywhere outside the tubes?
  21. Oct 10, 2015 #20
    Of course it's obvious when you look at it logically.Thanks for the heads up.I can know find the LMTD and ultimately find the required area
  22. Feb 17, 2016 #21

    I have a query to what lycee was trying to work out.

    When working out the LMTD I keep getting a negative number. My working is as follows:

    LMTD = (Tc1 -Tc2) / In[(Th-Tc1)/(Th-Tc2)]

    Where I have:

    Th = 180°C
    Tc1 = 1600°C
    Tc2 = 200°C

    Subbing those values in gives me an answer of -328.41°C.

    Do I have the values correct? Or even the equation?

    Thank you in advance for any response.
  23. Feb 17, 2016 #22
    Your LMTD equation is incorrect.
  24. Feb 17, 2016 #23


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    You can't calculate a valid LMTD for a boiler since there is latent heat involved.

  25. Feb 18, 2016 #24
    Is the equation I want:

    LMTD = (Tc1-Tc2) / [(In(Tc1) - In(Tc2))]


    1400 / 2.07 = 676.33°C

    *Source for that equation from Wiki page SteamKing posted.

    Edit: I've just looked again.. The equation on the Wiki page mentions temperature change at points A and B. Not set temperatures.

    Lemme have some time with this today and I'll report back. The only equation I know for the LMTD is the one I originally stated.
    Last edited: Feb 18, 2016
  26. Feb 18, 2016 #25
    Having worked out correctly what Mich1 was trying to work out, I now have the values for:

    qm = 35955 kgh-1
    Φ = 72.45 x 106kJh-1

    I am now trying to work out the area of heat transfer if the overall heat transfer coefficient based on the outside area of the tubes is given as:

    U = 54 Wm-2k-1

    I am now using the equation:

    Φ = UA(LMTD)

    Rearranged for A:

    A = Φ / (U x LMTD) ..

    So far so good?
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