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Homework Help: Mixing chamber with multiple inlets

  1. Oct 29, 2011 #1
    Hi I was wondering if someone could help me with this question. Question 4 uploaded in picture.
    Ok, so far this is what I have found

    Pipe 1:
    m1=2kg/s
    P1=1000kpa
    T1=100 C
    h1=334.82kJ/kg

    Pipe 2:
    sat-liquid
    T2=60 C
    P2= 1000kpa
    h2=291.36KJ/kg

    Outlet:
    sat vapor
    P3=1000kpa
    V=20 m/s
    h3=267.97 kJ/kg
    v3= 0.0202 m^3/kg

    and I know that the mass in = mass out
    so m1 + m2 = mout
    My problem is that I don't know what to do with the velocity because I do not have area to use formula Vout=VA

    Also I believe to solve the problem the formula m1*h1 + m2*h2 = m3*h3
    but I have tried a variation of this and it did not work.
     

    Attached Files:

  2. jcsd
  3. Oct 30, 2011 #2

    rude man

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    Enthalpy is conserved for each input, right. That's because dH = dQ + Vdp, dQ = 0, dp = 0. So you have m1h1 + m2h2 = m3h3. You also have m1 + m2 = m3. In other words, you've already stated all the equations you need.

    Two equations, 2 unknowns (m2 ans m3). Outflow = m3*V.

    I'm not claiming total knowledge here - does anyone see any flaws in this?
     
  4. Oct 30, 2011 #3
    I have tied to use that method which would be
    m1h1 + m2h2 = (m1 + m2)h3
    but then m2 = -5.716 and this is wrong
    I don't now how to relate the velocity into the equation. You said that outflow = m3*V right? what would outflow be? wouldn't the units be like kgm/s^2 or something like that?
     
  5. Oct 30, 2011 #4

    rude man

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    My apologies. I did not think when I wrote that outflow = m3*V. Outflow should be just m3.

    Something bothers me. Your enthalpy at the outlet is less than the enthalpy at inlet 2. That sounds weird. At 2 we have saturated liquid, and at 3 we have saturated vapor , so h3 must be > h2, yet you have h3 < h2. h3 looks OK since its temp. is 100 vs. 60 for the other two, so its h should be the highest of all.

    I don't know what to do with V, if it's a red herring or needed. Must think more.

    EDIT:

    To conserve energy, looks like h3 should be reduced by ρV^2 where ρ = density at outlet.

    BTW your insert is really hard to read, at least on my display.
     
    Last edited: Oct 30, 2011
  6. Oct 30, 2011 #5
    Actually I believe the h3 value is correct because r-134a at 1000kpa the saturation temp is only 39.39 C and the sat vapor value in the table hg= 267.97. The numbers are much less intuitive because it is not water. Still confused by this velocity given. It feels like It is a necessary piece of info but without area of pipe or diameter I can't seem to do anything with it.
     
  7. Oct 30, 2011 #6

    rude man

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    According to my table, saturated liquid enthalpy of r134a at 1 MPa is 105.29 kg/kJ, not 291.36. The table agrees with your T = 39.39C and sat. vapor h of 267.97.

    http://energy.sdsu.edu/testhome/Test/solve/basics/tables/tablesPC/pSatR134a.html

    I don't see how inlet 2 can have T = 60C and p = 1 Mpa. There's something fishy about that. As you can see from the table, and as I believed a priori, there is one and only one temperature associated with saturated r134a at 1 MPa, and that is 39.39C. I wonder if there is something tricky about the statement "after the valve", though I sure can't see it.
     
  8. Oct 30, 2011 #7
    Sorry is this better?
     

    Attached Files:

  9. Oct 30, 2011 #8
    Oh my bad you are very correct. I read the table wrong. I believe the pressure at inlet 2 is so high due to the valve.
     
  10. Oct 30, 2011 #9
    hmm. I'm not sure. Like there shouldn't be. min to valve should = mout
     
  11. Oct 30, 2011 #10

    rude man

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    Much! Thanks.
     
  12. Oct 30, 2011 #11
    Ok I got it. So h3 will be 267.97 because it is sat vapor at 1Mpa and h2 will be 137.42 which is the sat-liquid value for 60 C at a pressure of 1681.3 Kpa. And because the hin = hout of a valve this can be used. Plug all this in the the formula we talked about and a got the right answer. Thanks for your help and for your time.
     
  13. Oct 30, 2011 #12

    rude man

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    You absolutely got it. I was just writing to point out that the valve made it a throttling process with h conserved. Good going!

    PS so V was never used?
     
  14. Oct 30, 2011 #13
    Nope. I guess it was to throw you off as well as the pressure at p2 after the valve was never used. I hate extra info that that. Thanks again
     
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