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Heat exchanger question

  1. Feb 4, 2017 #1
    1. The problem statement, all variables and given/known data
    flue gases = 50,000kg h^-1
    surface area = 113.57m^2
    Do = 25 + (3*2) = 31mm

    The tubes within a boiler are 25mm inside diameter with a wall thickness of 3mm. The flue gas velocity through the tubes is to maintain the overall heat transfer coefficient value and minimise pressure losses to be more than 22ms^-1 and less than 28ms^-1

    Assuming average density of the flue gases is 1.108kgm^-3

    (i) calculate the minimum and maximum number of tubes in each pass
    (ii) overall length of tubes at each of these numbers of tubes
    (iii) the minimum number of tube passes in each case if the boiler tube is to be less than 5M

    2. Relevant equations
    number of tubes = (4 * qmc(s)) / (density * π * Di * V)
    A= π * Do * L * N
    L = A / π * Do * N
    A= π * D * L * N
    rearranged
    N= (A/L) / (π*Do)


    3. The attempt at a solution

    (i)

    qmc= 50,000/3600
    =13.88kg s^-1

    minimum velocity @ 22ms^1

    number of tubes = (4 * qmc(s)) / (density * π * Di * V)
    = (4 x 13.88) / (1.108 * π * (25*10^-3)*22)
    =1160 tubes

    minimum velocity @ 28ms^1

    number of tubes = (4 * qmc(s)) / (density * π * Di * V)
    = (4 x 13.88) / (1.108 * π * (25*10^-3)*28)
    =912 tubes

    (ii)

    A= π * Do * L * N

    rearranged

    L = A / π * Do * N

    1160 tubes
    L = 113.57 / π * (31*10^-3) * 1160
    L= 1.005M

    912 tubes
    L = 113.57 / π * (31*10^-3) * 912
    L= 1.279M

    (iii) COMPLETELY LOST!!! been barking up the following tree but cant make it work!! some help would be greatly appreciated!!

    calculate number of tubes required in each case based on a tube length of 5M

    A= π * D * L * N
    rearranged
    N= (A/L) / (π*Do)

    22 ms^-1

    =(113.57/5)/(π*(31*10^-3)
    = 233.2
    = 234 tubes @ 5m

    26 ms^-1

    =(113.57/5)/(π*(31*10^-3)
    = 233.2
    = 234 tubes @ 5m

    I was thinking of working out the area for that number of tubes (in each case) ie 912 and 1160 based on the tube lengths calculated in (ii) then plugging the new area in to (iii) using L=5m and leaving N as the variable?? SOS
     
  2. jcsd
  3. Feb 4, 2017 #2
    It looks like only one pass is required. There must be something wrong with the problem statement.
     
  4. Feb 4, 2017 #3
    http://imgur.com/a/D5qqN

    That is the original question paper ^^^
     
    Last edited: Feb 4, 2017
  5. Feb 4, 2017 #4
    My calculated heat transfer area doesn't match yours.
     
  6. Feb 4, 2017 #5
    I have just checked it. I got ;

    heat transfer duty = qmH * Hfg
    = (3598.5 * 2015) / 3600
    = 2014.1KW

    Area m^2 = heat transfer duty / U * LMTD
    = (2014.1*10^3) / (54 * 328.4)
    = 113.57m^2
     
  7. Feb 4, 2017 #6
    Heat load = (50000/3600)(1150)(1600-200)=22,400,000 W
     
  8. Feb 4, 2017 #7
    Im not entirely sure where that calc comes from (im not doubting it BTW), if that was the case the area would be;

    Area m^2 = heat transfer duty / U * LMTD
    = (22400*10^3) / (54 * 328.4)
    = 1263.1m^2

    seems a bit excessive but I could be wrong?
     
  9. Feb 4, 2017 #8
    That's the rate of heat loss from the flue gas, which gets transferred to the liquid water to make steam.
     
  10. Feb 4, 2017 #9
    ok, so if you follow it through what surface area do you get?
     
  11. Feb 4, 2017 #10
    I get the value you got.
     
  12. Feb 4, 2017 #11
    ok great, so if you follow the question through to where I was struggling;

    (iii) the minimum number of tube passes in each case if the boiler tube is to be less than 5M

    Do you have any thoughts on how to tackle this?
     
  13. Feb 4, 2017 #12
    The total area is ##\pi D L n##, where n is the number of tubes and L is the length of each tube. So, first find the length of each tube. The tube bank has to fit into a space no longer than 5 meters. If L is greater than 5 meters, you need to "fold" the tubes once or more so that there are enough tube passes to accommodate the total tube length within the allowable space.
     
  14. Feb 4, 2017 #13
    Ok great, ill have a look in the morning its very late here now (UK), thanks for your help Chester! Hopefully I will have some sensible answers tomorrow!!
     
  15. Feb 5, 2017 #14
    3ci)

    Di = 25mm
    D0= 31mm
    Min V = 22ms^-1
    Max V= 28ms^-1

    qMc = 50000/3600

    Heat load = 50000/3600 * 1150 * (1600-200)
    =24000

    Area = 22400*10^3 / (54*328.4)
    = 1263.1M^2

    Min number of tubes in each

    22 ms^-1

    N = 4*13.88 / 1.108 * π * (25*10^-3)^2 * 22
    N = 1160 Tubes @ 22 m s ^-1

    28 ms^-1

    N = 4*13.88 / 1.108 * π * (25*10^-3)^2 * 28
    N = 912 Tubes @ 22 m s ^-1

    3cii) Overall length of tubes in each case

    A = π * D * L * N

    Re arranged

    L = A / π * D * N
    = 1263 / π * (31*10^-3) * 1160
    = 11.18M @ 1160 Tubes

    L = A / π * D * N
    = 1263 / π * (31*10^-3) * 912
    = 14.22M @ 1160 Tubes

    3ciii) Minimum number of passes in each case

    V = 22 ms^-1

    L = 11.18M @ 1160
    = 11.18 / 5
    = 2.236 > 3 Passes required

    V = 28 ms^-1

    L = 14.22M @ 912
    = 14.22 / 5
    = 2.844 > 3 Passes required

    Hopefully this is somewhere near!!!???
     
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