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Heat exchanger

  1. Jan 28, 2018 #1
    Fluid comes in at ##T_1## and comes out at ##T_2## and I know that heat is exchanged with steam at constant ##T_{steam}##.

    If I want to know the Surface of the exchanger then

    ##US(T_{steam}-T_{fluid}) = \dot m_{fluid} cp_{fluid} (T_2-T_1)##

    My question is, since ##T_{fluid}## changes, which value of ##T_{fluid} ##should I use in the above formula?

    PS: Since I don't know the surface, I don't know how the exchanges is designed so I have really know clue on what to do
     
  2. jcsd
  3. Jan 28, 2018 #2

    BvU

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  4. Jan 28, 2018 #3
    But how do I use LMTD if the steam remains at constant T? Does it make any difference? Or should I just do ##LMTD = \frac {(T_2 - T_{steam})-(T_1 - T_{steam})} {ln \frac {T_2-T_{steam}} {T_1 - T_{steam}}}##?
     
  5. Jan 28, 2018 #4

    BvU

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    Yep
     
  6. Jan 28, 2018 #5
    Thank you! :)
     
  7. Jan 31, 2018 #6
    Excuse me, what if ##T_2 ≈ T_{steam}##? The above formula breaks because ## ln \frac {{T_2} - {T_{steam}}} {T_1 - T_{steam}} = -\infty##

    Btw, I check how LMTD is derived for countercurrent heat exchanger... I don't know if I can apply it if the steam remains at the same T during the exchange.
     
  8. Jan 31, 2018 #7

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    If temperatures are equal, no heat transfer takes place. Just like the expression says. Not very realistic: in reality there is always a non-zero approach.
     
  9. Jan 31, 2018 #8
    The final temperatures are equal, but the fluid comes in at a lower temperature. The heat transfer is maximum at the beginning and zero at the end. I need a mean value, so I need a mean value of temperature. But I can't use the above formula.

    Say the fluid comes in at 100 °C and comes out at nearly 200°C (let's say 199°C). To heat it up I use steam (at equilibrium) at 200°C so it condenses and it remains at 200°C. How am I suppose to find the surface of the exchanger ?
     
  10. Jan 31, 2018 #9

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    You can ask someone to dig out the equipment drawings :smile:

    From the temperatures it appears that the surface area is at least $$Q\over U\; {\rm LMTD}$$ with lmtd = 21.5 for 1 degree and 8.7 for .001 degree approach.
     
  11. Jan 31, 2018 #10
    yeah, LMTD is very sensitive in this very case, so I was thinking that maybe it is not the best approach here.

    I have a test in two days, so I can't think to much about it now, later I will try to derive the dependence of Q from x:
    ##dQ = h(T_{steam} - T(x))\pi Ddx## and see if it leads to something... but I'm not to sure it will be of any help
     
  12. Feb 1, 2018 #11
    It will lead to the LMTD. Why are you saying that there is sensitivity of the use of LMTD when the steam temperature is constant? In the example you gave, you are not going to let the fluid temperature approach the steam temperature to 1C since that would require too large a heat transfer area.
     
  13. Feb 2, 2018 #12

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    I hope it went well -- and I guess there's no connection between the test and this thread...

    Is this a theoretical exercise ? If not, is your expedition UTCWAP (up the creek without a paddle) because it is based on plant measurements ? Could you post a sketch of the actual setup ?
     
  14. Feb 3, 2018 #13
    Thanks, I'm waiting for the results

    It's just something I was thinking about.

    Yeah, but I had to check myself :)

    No reason. Sorry for the trouble. I tend to be anxious before a test and I don't actually think straight. I've checked with my mind clear today and I got it immidiately.

    Sorry, and thanks for the replies!
     
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