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Isentropic Efficiency and Entropy Production Rate of Turbine

  1. Aug 2, 2017 #1
    1. The problem statement, all variables and given/known data
    Water vapor at 6 MPa, 600C enters a turbine operating at steady state and expands to 10 kPa. The mass flow rate is 2 kg/s, and the power developed is 2626 kW. Stray heat transfer and kinetic and potential energy effects are negligible. Determine (a) the isentropic turbine efficiency and (b) the rate of entropy production within the turbine, in kW/K.

    2. Relevant equations
    dmcv / dt = ∑ mi - ∑ me
    dEcv / dt = Qcv - Wcv + ∑ mi (hi + Vi2/2 + gzi) - ∑ me (he + Ve2/2 + gze)
    dScv / dt = ∑ Qj / Tj + ∑ misi - ∑ mese + σcv

    3. The attempt at a solution
    From the problem statement, I know that entry mass flow rate and exit mass flow rate are the same, because it is steady-state. I also know that, because of the steady-state operation, the energy balance equation simplifies to:
    Wcv = m(h1 - h2)

    Using the steam tables, I found the enthalpy and entropy values at state 1 to be 3658.4 kJ/kg and 7.1677 kJ/kg⋅K, respectively. I can use this enthalpy value, along with the given mass flow rate and work output values to find h2, but what I need to find is h2s. Once I find that, I feel like I can crack this problem wide open, but I'm having issues on how to do just that. Since the working fluid isn't an ideal gas, I can't use the relative pressure and volume relations, so I haven't the slightest clue on how to find h2s. If anyone can steer me in the right direction, that would be really appreciated.
     
  2. jcsd
  3. Aug 3, 2017 #2
    ##h_2s##?? That's the exit enthalpy value assuming constant entropy?
     
  4. Aug 3, 2017 #3
    Yes, the way our professor taught us, all the exit values under constant entropy are marked with subscript s.
     
  5. Aug 3, 2017 #4
    Then what you do is look up in your steam tables the state where the pressure is 10 kPa and the entropy is 7.1677. You might need to interpolate. You need to find the enthalpy for this state.
     
  6. Aug 3, 2017 #5
    So I think I found the enthalpy value, assuming constant entropy, which worked out to be around 2217.17 kJ/kg. I had to find quality under constant entropy first in order to do so. Consequently, I also found the isentropic turbine efficiency to be around 91.1%. Does this sound like I'm on track?
     
  7. Aug 3, 2017 #6
    Yes
     
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