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Ar (Gas) goes into spinner (Adiabatic Decompression): Power?

  1. Aug 9, 2017 #1
    1. The problem statement, all variables and given/known data

    Argo, in its gas form, is insterted into a spinner, with a rythm of 80.0 kg/min = 1.33 kg/s, temperature 800 C (= 1073.15 K), and pressure 1.50 MPa. It's decompressed adiabatically, and "escapes" the spinner with pressure 300 kPa.

    a) Temperature when it "escapes" ?
    b) Maximum Power of the spinner?
    c) Hypothesize that the spinner works as a machine that completes a "closedcircle" when a gas is inserted. What's the maximum efficiency of this machine?

    2. Relevant equations

    p = m/V
    PVγ = constant
    TVγ-1 = constant
    PV = nRT
    ec = 1 - Tc/Th

    γ = 1.67 (from the book)
    p = 1784 kg/m3

    3. The attempt at a solution

    a) That one I've solved:

    >Gas is inserted with a rythm of 1.33 kg/s. So, mi = 1.33 kg.

    >p = m/V = 1784 kg/m3

    >Therefore, Vi = 7.45 * 10-4 m3

    >PiViγ = PfVfγ <=> ... <=> Vf = 1.95 * 10-3 m3

    >TiViγ-1 = TfVfγ-1 <=> ... <=> Tf = 563 K

    The book's answer is 564 K, but to it flactuates between using Tk = Tc + 273.15 and Tk = Tc + 273,and it plays fast and loose with the Significant Digits. Not to mention that with Tf = 564 K, (c)'sresultis still not in line with the book's.

    b) That's where I'm stuck. I know that Power is given through P = W/Δt, but I've got no clue one how to find the Work. I figured I'd do this:

    W = ∫ViVfPdV, with P = PiViγ/Vγ, but I got a wholly different result.

    c) The machine will have a maximum efficiency if it works like a Carnotone, so:

    ec = 1 - Tc/Th = 1 - 563 K/1073.15 K = 0.475 = 47.5 %

    Any help is appreciated!
  2. jcsd
  3. Aug 9, 2017 #2
    Even though you got the correct answer for the exit temperature (563 K), this was not analyzed correctly. You can't just take the mass flow rate in a continuous flow turbine and treat the turbine as a closed system containing the amount of mass that flows through in one second. And the density of the gas is, of course, not 1784 kg/m^3; this is nearly twice the density of liquid water. The equation you should have used was $$\frac{T_f}{T_i}=\left(\frac{P_f}{P_i}\right)^{\frac{\gamma-1}{\gamma}}\tag{1}$$This can be derived from a combination of the ideal gas law with the equation $$P_fV_f^{\gamma}=P_iV_i^{\gamma}$$See if you can derive Eqn. 1.
    To get the power delivered by the turbine (rate of doing shaft work) in this continuous flow process (operating at steady state), you need to use the open system (control volume) version of the first law of thermodynamics. Are you familiar with this equation?
  4. Aug 9, 2017 #3
    Darn it.

    Yeah, it seemed weird to me as well, but I got the right result, so I figured I had to be doing something right. Generally I just thought "hey, as soon as I open the turbine this mass enters" and treated that as "the initial mass". Pretty faulty logic though, yeah.

    I looked it up on Wikipedia, but I messed up on the conversion part. Yeah, you're right.

    I tried, and I didn't make it, but I foundthe proof at Wikipedia, here: https://en.wikipedia.org/wiki/Adiabatic_process
    I don't get this part though: ΔU = aRnT2 - aRnT1 = aRnΔT. What is the "a"? Why does he use internal energy? My book doesn't mention anything about that (the whole thermodynamics section deals with ΔE & W). Are Q & ΔU connected in some way? I get the rest of the thought process.

    No, not really. I searched around a bit, but I didn't find anything that I recognized.
  5. Aug 9, 2017 #4
    I don't want to go off in a tangent on the derivation of the thermodynamics relationships for adiabatic reversible expansion of an ideal gas in a closed system. I want to keep the focus on the open system problem at hand. If you have questions about the derivation in Wikipedia, please start a new thread under General Physics.

    For the relationship I was referring to in my previous post starting with $$P_fV_f^{\gamma}=P_iV_i^{\gamma}$$
    we have (using the ideal gas law) $$P_f\left(\frac{RT_f}{P_f}\right)^{\gamma}=P_i\left(\frac{RT_i}{P_i}\right)^{\gamma}$$
    Simplifying this leads to $$P_f^{(1-\gamma)}T_f^\gamma=P_i^{(1-\gamma)}T_i^\gamma$$
    This leads to $$\frac{T_f}{T_i}=\left(\frac{P_f}{P_i}\right)^{\frac{\gamma-1}{\gamma}}$$

    This problem should never have been assigned to you if you have not yet learned about the open system version of the first law of thermodynamics. You really need it to solve this problem. And Physics Forums is not an appropriate location for the derivation; it is presented in virtually every elementary thermodynamics book. For good references, see Fundamentals of Engineering Thermodynamics by Moran et al and Introduction to Chemical Engineering Thermodynamics by Smith and Van Ness.
  6. Aug 11, 2017 #5
    Oh yeah, that makes sense. I just figured that since my book used integrations to prove TVγ-1 = constant, I figured I'd need to do the same here.

    Yeah, I searched again, but I really don't know it, and it's not in my book. Thanks for the sources though, I'll try and find them sometime.

    Anyway, thanks for all the help!
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