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Rankine Cycle Efficiency

  1. Jul 27, 2017 #1
    1. The problem statement, all variables and given/known data

    Superheated steam at a pressure of 40 bar and a temperature of 500°C is supplied to the turbine of a Rankine cycle. If the condenser pressure is 0.03 bar.

    Find the thermal efficiency of the cycle. (Neglect feed pump work).

    I used steam tables found here

    2. Relevant equations
    S1 = Sg

    Sg = (1 - xg)Sf2 + xgSg2

    h2 = (1 - xg)hf2 + xghg2

    Specific work (W) = h1 - h2

    Specific Heat (Q) = h1 - hf2

    Eff (η) = W / Q

    3. The attempt at a solution
    From steam tables:

    At 40 Bar and 500°C (Before turbines):

    hg = 3445 kJ Kg-1
    Sg = 7.089 kJ Kg-1 K-1

    At 0.03 Bar (After turbines):

    hf2 = 101 kJ Kg-1
    hg2 = 2545 kJ Kg-1
    hfg2 = 2444 kJ Kg-1

    Sf2 = 0.354 kJ Kg-1 K-1
    Sg2 = 8.576 kJ Kg-1 K-1
    Sfg = 8.222 kJ Kg-1 K-1
    -------------------------------------------------------------------------------------------------------------------------
    Attempt at working out the thermal efficiency of the cycle:

    Sg = (1 - xg)Sf2 + xgSg2

    7.089 = (1 - xg) * 0.354 + xg 8.576

    xg = (7.089 - 0.345) / (8.576 - 0.345)

    xg = 0.819

    Now

    h2 = (1 - xg)hf2 + xghg2

    h2 = (1 - 0.819) * 101 + 0.819 * 2545

    h2 = 2103 kJ kg-1

    Specific work (W) = h1 - h2

    W = 3445 - 2103

    W = 1342 kJ kg-1

    And

    Q = h1 - hf2

    Q = 3445 - 101

    Q = 3344 kJ kg-1

    Eff (η) = W / Q

    η = 1342 / 3344

    η = 0.4013 or 40.13% efficient


    Am I looking at the right area to solve this?
     
  2. jcsd
  3. Jul 31, 2017 #2
    Hi,

    *edit* I've got efficiency=0.40 or 40% so what u did is fine

    but I've used those tables:
    https://www.physicsforums.com/attachments/thermodynamic_tables_si-pdf.88587/
     
    Last edited: Jul 31, 2017
  4. Jul 31, 2017 #3
    Looking at the values on the tables you've used they are very similar. Probably shouldn't give a difference in efficiency of 9%

    Is there anything in my method you can comment on? Have I used the correct method? Correct equations?
     
  5. Jul 31, 2017 #4
    Yeah I did it again and got correct 0.400 or 40% efficiency, My mistake was on 3rd point on superheated I looked by mistake upwards a bit on other bit of table.Everything seems fine
     
  6. Jul 31, 2017 #5
    Phew, I've struggled with this so glad to hear I've sussed it out.

    Thanks.
     
  7. Jul 31, 2017 #6
    Let 1= point after turbines when enters condenser

    Data for1: Pressure P1=3 KPa(KiloPascal)
    h1(entalphy1)=100.98 Kj/Kg
    v1(specific volume)=0.001003 m3/kg

    So we can use that Wpump(work that goes IN the pump)=v1(P2-P1)=h2-h1 (I)
    v1(P2-P1)=0.001003*(4000 KPA-3KPA) =4.008 m^3/kg
    from (I) 4.008=h2-h1 =>h2=h1(from above)+4.008 =>h2=104.988991 kj/kg

    For state 3

    We can say it is superheated because the saturation temp at 4 Mpa is 250.35 degrees celsius and we have 500 Celsius

    so h3=3446.0 kj/kg(table)
    s3=7.0922 kj/kg*K

    s3=s4(isentropic) => 7.0922=0.3543 +8.2222*x4 (use the fact that is 2 phase region)

    x4=0.8193=>h4=100.98+0.8193*(2443.9) =>h4=2103.26 kj/kg

    efficiency eta= wnet,in/qin=1-qout/qin=((h3-h4)-(h2-h1))/(h3-h2)
    =>efficiency = ((3446-2103)-(104.98-100.98))/(3446-104.98)= 0.40

    =>efficiency=40%
     
    Last edited: Jul 31, 2017
  8. Jul 31, 2017 #7
    Makes perfect sense that. Seem to get similar numbers at the various stages as well. Can both methods be used in this example?

    Or is one preferred?
     
  9. Jul 31, 2017 #8
    Hi,

    Both work well, choose and use the one you are more comfortable with, and don't forget to pay attention in state 3 (you might have superheated or saturated)

    Best,
    William
     
  10. Jul 31, 2017 #9
    Thanks.

    I will add this method to my notes as well.

    Cheers.
     
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