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How to find maximum work done in an adiabatic process?

  1. Aug 2, 2011 #1
    My Question - Steam at 200 psi and 600 F flows through a turbine operating adiabatically and exits at atmospheric
    pressure. For every kilogram of steam flowing through the turbine there are 150 Btu of shaft work
    delivered.
    (a) What is the final condition of the exit steam?
    (b) What is the maximum amount of work that could be obtained per kg of steam for adiabatic operation between these two pressures?

    Relevant Equations
    1 psi = 6894.76 Pa
    T2^k / P2^(k-1) = T1^k / P1^(k-1)
    (T2/T1)^k = (P2/P1)^(k-1)
    W = n R (T1 - T2) + m L

    My Approach
    (a) By final condition, do we mean phase (solid/liquid/gas)?
    P1 = 200 psi = 200 x 6894.76 Pa = 1378952 Pa
    P2 = atmospheric pressure = 101325 Pa
    T1 = 600 F = (5/9) * (600 - 32) C = 315.56 C or 588.56 K

    T2^k / P2^(k-1) = T1^k / P1^(k-1)
    (T2/T1)^k = (P2/P1)^(k-1)
    For steam, k = 1.3
    (T2/588.56)^1.3 = (101325/1378952)^0.3
    (T2/588.56)^1.3 = 0.45693
    T2/588.56 = 0.54745
    T2 = 322.2 K = 49.2 C

    This shows that the the steam will convert into liquid. But the problem says that steam exits. This means whole of steam has not condensed into liquid.
    But I am not sure how to find how much of steam will condense.

    (b) Mass of steam = 1 kg = 1000 g
    Number of moles of steam n = 1000/18
    R = 8.314 in SI units
    Latent heat of vaporization of water L = 540 cal/g = 540 * 4.186 J/g
    Final pressure is 1 atm. Steam condenses at 100 C (=373 K) at 1 atm
    Therefore T2 = 373 K

    W = n R (T1 - T2) + m L
    W = (1000/18) mol * 8.314 J.mol^-1.K^-1 * (588.56 - 373) + (1000 g) * (540 cal/g) * (4.186 J/cal)
    W = 2.36 x 10^6 J or 2.36 MJ

    But I am not sure if I have solved it correctly. If yes, then is it possible to make the solution shorter?

    Note: This is not for my homework. I finished student life some years back. But sometimes I read some things out of my own hobby and collect problems from different sources. I was reading this topic and got stuck in this problem.
    So I need help. Thank you.
     
  2. jcsd
  3. Aug 3, 2011 #2

    rock.freak667

    User Avatar
    Homework Helper

    Using your steam tables you can get the enthalpy h1 @ 200 psi, 600F and you are given the shaft work. Applying the steady state equation you will see that

    w=h1-h2 so you can solve for h2.

    and you know that h2=hf+x2hfg (these are at atmospheric pressure as stated).

    The maximum work will require what of all the steam? (i.e. what should x be)
     
  4. Aug 3, 2011 #3
    Use steam tables as mentioned or better yet a Mollier Diagram (visual diagram of the steam tables) to locate the initial conditions which are clearly super heated steam.You should be able to determine h1 and s1 knowing T1 and p1. From first law -W(turbine)= Mass(Dot)(h2-h1). Calculate new h2. Assuming adiabatic, reversible s1=s2 We don't know if the final condition is saturated of superheated. Knowing s1=s2 and h2 on your Mollier diagram you can see that you are still in the super heated region and you can determine the final T2 and p2. The maximum work output of the turbine will occur when you can further condense the steam to a saturated gas at p2. Find h2 at p2 when steam is a saturated gas and calculate turbine output -max.
     
  5. Aug 3, 2011 #4
    After reconsidering my answer to part B it appears that the maximum work produced by the turbine occurs at the greatest enthalpy difference possible between these pressures. This would occur at the highest temperature the turbine could withstand at 200 psia and the comparable temperature at p2, assuming s1=s2.
     
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