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Control volume, steady- state and steady-flow devices, enthalpy

  1. Aug 13, 2012 #1
    1. The problem statement, all variables and given/known data

    Air is to be compressed from 120 kPa and 310 K to 700 kPa and 430 K. A heat loss of 20 kj/kg occurs during the compression process. For air: cv=0.7165 kj/kg.K and R=0.287 kj/kg.K .

    A) Neglecting kinetic energy changes determine the power input required for a mass flow rate of 90 kg/min.


    3. The attempt at a solution

    m = 90 kg/min = 1.5kg/s

    Apply 1st law of thermodynamics - energy conservation

    Ein = Eout

    Win + Mh1 = Qout + mh2

    Therefore:

    Win + mCpT1 = Qout + mCpT2

    Cp = 1.005 Kj/Kg.K

    Win + 1.5kg X 1.005 KJ/Kg.K X 310K = 20KJ/Kg/K + 1.5Kg/s X 1.005 KJ/Kg.K X 430K=

    Win = 201Kw.


    Is this correct?
     
  2. jcsd
  3. Aug 13, 2012 #2

    rude man

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    What is your justification for this equation? It's not the 1st law. Enthalpy equals heat only at constant pressure, which is not the case here. Or, the other way, enthalpy is not conserved since this is not a throttling process.

    How about going back to the real 1st law? U = Q - W

    Hint: if the air can be considerd an ideal gas, this problem is trivial. If not, I'm not sure myself how to tackle it at the moment.
     
  4. Aug 13, 2012 #3
    If air is to be considered and ideal gas here, how would i go about the problem with that in mind?
     
  5. Aug 13, 2012 #4

    rude man

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    Like I said, start with the first law. What are the changes in U and Q?
     
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