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Homework Help: Computing the work of a turbine.

  1. Oct 11, 2013 #1
    I am preparing for a re-exam, this is a problem from the exam I took, but I can't see what I did wrong and why.

    1. The problem statement, all variables and given/known data

    Case: A turbine that is producing work dW is powered by compressed air (treated as a diatomic ideal gas).

    Known quantities:
    [itex]P_0, P_f, T_0[/itex]

    Wanted quantity:
    [itex]dW[/itex] per mole air.

    It is also known that the process is adiabatic, so [itex]dQ = 0[/itex], and that the flow is stationary.

    2. Relevant equations

    3. The attempt at a solution

    By the first law of thermodynamics, and the adiabatic property:
    [itex]\Delta U = dQ + dW = dW[/itex].

    The energy content of a diatomic ideal gas is given by:

    [itex]U = \frac{f}{2} nRT[/itex], where f = 5 (the amount of quadratic degrees of freedom)

    Thus, ΔU should equal the change in energy content of the air before and after the turbine:
    [itex]\Delta U = U_0 - U_f =\frac{f}{2} nR (T_0 - T_f) = dW [/itex].

    To compute the unknown [itex]T_f[/itex] we again use the fact that the process is adiabatic, so the following should hold
    [itex]P_0^{-\frac{2}{f+2}} T_0 = P_f^{-\frac{2}{f+2}} T_f \iff T_f = T_0 (\frac{P_f}{P_0})^{\frac{2}{f+2}}[/itex].

    Inserting this result into the prior equation gives
    [itex]\Delta U = \frac{f}{2} R T_0 ( 1 - (\frac{P_f}{P_0})^{\frac{2}{f+2}}) = dW[/itex] per mole (with n=1).

    However, the solution sheet states that the term f/2=5/2 should be 7/2 (rest unchanged). I can't see why. They use a different technique as well, which I don't understand.

    The correct solution:

    Stationary flow implies that [itex] H_0 = W + H_f[/itex] by the first law of thermodynamics. The enthalpy [itex]H=C_P T[/itex]. So, [itex]W=C_P (T_0 - T_f)[/itex]. Furthermore,

    [itex]T_f = T_0 (\frac{P_f}{P_0})^{1-1/\gamma}[/itex] and

    [itex]C_P = 7/2 nR[/itex] by the properties of diatomic ideal gases.


    [itex]W/n = \frac{7}{2} R T_0 ( 1 - (\frac{P_f}{P_0})^{1-1/\gamma})[/itex]

    which is not what I get:

    [itex]\Delta U = \frac{f}{2} R T_0 ( 1 - (\frac{P_f}{P_0})^{\frac{2}{f+2}}) = dW[/itex] per mole (with n=1).

    Gamma is defined as (f+2)/f. Also observe that their notation is W instead of dW ( which here is not meant to be read as a change in work, but a quantity of work ).
    Last edited: Oct 11, 2013
  2. jcsd
  3. Oct 11, 2013 #2
    Have you learned about the form of the first law applicable a continuous flow open system? For steady state operation, it says that the change in enthalpy per unit mass passing through the system is equal to the "shaft work" per unit mass passing through the system. Please go back and restudy the section in your textbook on the first law for continuous flow open systems. This will tell you why the "correct solution" is correct.
  4. Oct 12, 2013 #3
    Yes I will, however, I prefer using as few formulae and extra definitions as possible, including concepts like enthalpy, even if it may make things harder computationally, unless explicitly required. I want to understand what makes my solution wrong. What am I computing and why do I get less energy than what I'm supposed to?
  5. Oct 12, 2013 #4
    [itex]\Delta U = dW = - P\Delta V + dW_{Other}[/itex]

    Is the [itex]-P\Delta V[/itex] the problem here?

    Is turbine work limited to [itex]dW_{Other}[/itex]? Would my solution hold if we had a machine where that wasn't the case?

    The gas has to push away atmosphere on exit, due to increased volume? But how does that not reflect on the amount of energy you can get out of the turbine? The only way I can get less energy in dU than I get out of the turbine seems to be when the pressure of the gas is higher than that of the atmosphere. But wouldn't that be dependent on the atmospheric pressure, which could be any value?
  6. Oct 12, 2013 #5
    You're close to having it. First of all, the equation should be [itex]\Delta U = dW = - \Delta (PV) + dW_{Other}[/itex]. If you want to do it as a closed system, take as your closed system the contents of the turbine at any time plus a small parcel of gas about to enter the turbine. In the next instant, the small parcel of gas has entered, and and another small parcel of equal mass has exited at the low pressure end. Since the system is at steady state, the internal energy of the gas within the turbine has not changed between the initial and final states. Only the internal energy of the parcel that leaves is different from the internal energy of the parcel that entered. The work done by the gas behind the inlet parcel in forcing it into the turbine is the upstream pressure times the volume of the parcel. The work done by the downstream parcel on the gas ahead in leaving the turbine is the pressure downstream times the volume of that parcel. The rest of the work is the "shaft work", which you call dWother. So the total change in internal energy per unit mass entering the system is [itex]\Delta U = - \Delta (PV) + dW_{Other}[/itex], where all the quantities in this equation are per unit mass entering (and leaving).
  7. Oct 13, 2013 #6
    Ok, so basically the gas loses more energy than is put into turbine work, due to the net expansion of the parcels of gas and them having to push away existing gas, which consumes energy?

    I would have to add [itex]-\Delta(PV)[/itex] to what I've got in other words. Can I do this by computing [itex]-(P_0 V_0 - P_f V_f)[/itex]?

    With the enthalpy solution that part seems to disappear by the use of [itex]H = C_P T[/itex]. Is constant pressure assumed because of the steady state?
  8. Oct 15, 2013 #7
    Not exactly. The gas actually does more work on the turbine than the expansion work. The work done by the gas behind the entering gas in pushing it into the turbine is greater than the work by the exiting gas in pushing away the gas ahead of it. The extra work done is [itex](P_0 V_0 - P_f V_f)[/itex].

    One way to think about this is to replace the gas by an incompressible fluid (which doesn't expand). If the inlet pressure is higher than the outlet pressure (and, if viscous drag is negligible), the fluid causes the turbine to rotate and do work (just like blowing on a pinwheel). So, even without a gas expanding, work is done.

    No. For an ideal gas, dH is always equal to C_pdT, irrespective of whether the pressure is constant.
  9. Oct 21, 2013 #8
    Ok, I try to visualize the turbine as two containers/parcels of gas at different pressures and with a connection between them that contains a turbine. The flow pushes the turbine, whereupon the system loses more energy or pressure than it would if there was just a connection and no turbine to push.

    As soon as the pressure is (almost immediately) equalized I open both containers to let them equalize with the atmosphere, I close them and open the container that is connected with the high pressure air. During equalization with the atmosphere and during equalization with the high pressure reservoir work is done. In the first case by the gas and in the second case on the gas. The difference between them is the work I won't get as W_other? Is that correct?

    I get that dH = const * dT by using PV = NkT and U = f/2 NkT as long as N is fixed. But I have trouble showing that const = dQ/dT (= C_p) without assuming that the pressure is fixed.
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