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Steady State Ideal Gas Where Do I start

  1. May 9, 2009 #1
    Uncle Sam claims that he invented a black-box device which operates at steady state, and has two inlet streams and one outlet stream as shown in the figure below. Inlet 1 allows air to enter at a pressure of 200 kPa, temperature of -50 oC, and mass flow rate of 30 kg/min. Inlet 2 allows air in at 100 kPa, temperature of 150 oC, and mass flow rate of 20 kg/min. The outlet is at a pressure of 190 kPa. He further claims that this device is insulated and requires no work input. After a thermodynamic analysis, one can show that a work input to the device is actually needed. In your analysis treat the air as an ideal gas with constant specific heats. Also R=0.287 kJ/kgK and cp=1 kJ/kgK. Determine the minimum amount of work that must be done on the system to make the device possible. Assume that the outlet pressure remains unchanged

    So this is what I was thinking:
    1) change deg. C to deg K
    2)interpolate for h

    However: I think I should find the entropy change 1st but, I'm unsure how to
  2. jcsd
  3. May 9, 2009 #2
    I believe T3= (m1T1+m2T2)/(m1+m2)=303.15K
    howver the answer my prof gave me is 307.4K .... why isn't my equation correct?
  4. May 10, 2009 #3


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    Not sure offhand how to solve this, but to answer your question:

    Your equation would be correct if no energy is added to the air. If energy is added to the air, that makes the output temperature higher.
  5. May 10, 2009 #4
    steady state means m1=m2 but in this case can you say m1+m2=m3?
  6. May 10, 2009 #5
    how do I use this to find Q?
    I know:
    T1, T2, and I found T3
    P1, P2, P3
    cp=1 kJ/kgK
    if m1+m2=m3 then m3=50kg/min ... right?
  7. May 10, 2009 #6


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    Yes, steady state means

    minlets = moutlets
  8. May 10, 2009 #7
    I ment "how do I find Wcv"?
  9. May 10, 2009 #8
    also how do I relate d S = δq/T to my problem and my knowns or can/should I?
  10. May 11, 2009 #9


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    Hi CRich,
    Normally, the first law is written:

    dU = dQ + dW

    For a control volume with flow in and/or out, heat in and/or out, work in and/or out, and the change in internal energy being only that change internal to the control volume, the first law can be rewritten:

    dU = dQ + dW + dH

    Where dH = Hin - Hout
    and H=mh (mass flow rate times specific enthalpy)
    This assumes energy entering the volume is positive and energy leaving the volume is negative.

    For you problem then, the first law can be rewritten:

    dU = Win + H1in + H2in - Hout

    For a steady state process, dU = 0 so you can now solve for work. Note that to do this, you need a way of determining h.

    Also, you can neglect velocity on the inlet and outlet. Although it is more correct to include it, there isn't any information that allows it to be calculated and even if there was, the error imposed is generally less than the error in the values for enthalpy in your database.
  11. May 11, 2009 #10
    okay I was thinking along those same lines...but my equation was a little off... I'll give this one a try. Thanks
  12. May 11, 2009 #11
    Final Is Thursday Pls Help ASAP!!! Steady State Ideal Gas Where Do I start

    okay so my T equations is correct except for the addition of work...
    In order to calculate work I tried :
    inserting my values

    30*1*(223.15-303.15)+20*1*(423.15-303.15)=0 .... But according to my prof W=-3.67kJ
    So what am I doing wrong???????????

    Also how do I find δ? What equation do I use? How do I use the 2nd Law to find δ?
  13. May 11, 2009 #12


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    Hi CRich. That's a good shot at it. What you've done however is to simply calculate this as an isenthalpic process. There's no calculation to see what pressure should be assuming the low pressure stream is isentropically compressed.

    I've been playing around with this for a while now and I still haven't nailed it down. However, here's what I'd suggest.

    The best that one could hope for is to take the work that comes from the isentropic expansion of flowstream 1 from 200 kPa to outlet pressure (190 kPa), and put that work into an isentropic compression of flowstream 2.* The pressure in flowstream 2 must then come up to some higher pressure. If that's not high enough to bring that flow stream up to the outlet pressure (190 kPa), then additional work must be put in. You can assume that work is put in at 100% isentropic efficiency as eluded to in the problem statement.

    Once you've brought flowstream 2 up to outlet pressure, the rest is just the mixing of two flow streams which you've already done (enthalpy in = enthalpy out). This gives you the final outlet temperature.

    I appologize for not being more helpful. I'll have another look tomorrow and see if I can finish solving this but I've run out of time tonight. Good luck on your exam.

    *Edit: Actually, if you were to transfer as much heat as possible from flowstream 2 to flowstream 1 prior to expanding to the lower pressure, you could get more work out of flowstream 1 and it would take less work to compress flowstream 2. That could be done using a perfect counterflow heat exchanger. The part I struggle with is how to determine that this is theoretically, the best possible process. I suspect it is but can't prove it.
    Last edited: May 12, 2009
  14. May 12, 2009 #13
    How do I do/show that?

    Also how do I find δ? What equation do I use? How do I use the 2nd Law to find δ?
  15. May 13, 2009 #14
    okay so Ive tried several other things and I still cant calculate entropy production ...
    I tried
    -entropy production=m1(cp*ln(T3/T1)-R*ln(P3/P1)+m2(cp*ln(T3/T2)-R*ln(P3/P2)
    but I don't get the answer my prof gave ... and even if I did how would I use my 'knowns' to calculate work?
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