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Thermodynamics. Turbine, hwo to use enthalpy to find mass flow?

  1. Nov 22, 2011 #1
    1. The problem statement, all variables and given/known data

    The problem (tried my best to translate it):
    A small high speed turbine is operating on compressed air. It deliveres dW/dT=100 W. At the inlet, the pressure is 400 kPa and the temperature 50*C.

    At the exit, the pressure is 150 kPa and the temperature -30*C.

    Neglect the velocity and assume an adiabatic process. Find the necessary mass flow of air through the turbine.


    2. Relevant equations
    First law for control volume
    Definition ofg enthalpy

    3. The attempt at a solution
    I have derived the first law for a control volume:
    dE/dt=(dQ/dt)-(dW/dt)+∑m_i (h_i+0.5v_i^2+gz_i)-∑m_e (h_e+0.5v_e^2+gz_e)

    Where t is time, and m_i og m_e is rate of change of mass flow at the inlet and exit, respecitively.
    Assumed steady state: dE/dt=0.
    Adibatic dQ/dt=0.
    Also m_e´=m_i´=m´

    By neglecting kinetic and potential energy associated with gravity, i end up with:

    dW/dt=m(h_i-h_e) <=> m=(dW/dt) / (h_i-h_e)

    So far so good, but now I need to find the change of enthalphy. We were supposed to solve this task without the use of steam tables. I have tried to use the definition of constant volume heat capacity, but no luck so far. Any input?

    Thanks in advance. :)
     
  2. jcsd
  3. Nov 22, 2011 #2

    Andrew Mason

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    I think all you have to do is apply the first law to an adiabatic process (ΔQ=0) so:

    ΔU = -W => dU/dt = -dW/dt

    So start by determining the internal energy change of 1 mole of air going from 323K to 243K. [This is not a reversible process so the change in pressure does not help you determine the work done]. Assume air is an ideal diatomic gas with a Cv of 5R/2.

    AM
     
  4. Nov 23, 2011 #3
    Let me ask, are you using temperature/pressure tables for air?

    If so then with the known temp/pressure at the inlet (T1/P1) and the known temp/pressure at the outlet (T2/P2) then you can refer to the tables to find the inlet and outlet enthalpies, h1 and h2 respectively.

    Thus Q = m(dot)*(h1 - h2)

    Solve for m(dot)

    Not sure if this is what your looking or but that is my two cents
     
  5. Nov 23, 2011 #4

    Andrew Mason

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    My suggestion to use change in internal energy is not correct because the work done on the turbine by the gas is not the total work that is done. Since the volume of the air doubles there is also the expansion work done against the atmosphere. This is taken into account in the change in enthalpy but not with the change in internal energy.

    Since atmospheric pressure is constant you can determine the work done on the atmosphere: Watm = PatmΔV

    By the first law,

    ΔQ = 0 = ΔU + W where W is the total work done by the air = Wturbine + Watm

    ΔU = -Wturbine + -Watm → Wturbine = - (ΔU + Watm) = -(ΔU + PatmΔV)

    AM
     
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