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Thermodynamics: Pressure Drop over a Valve

  1. Oct 22, 2015 #1
    Hey Guys! Anyone able to help out here?

    upload_2015-10-22_10-12-24.png

    I have already happily solved for T2 = 369.91K and m2 = 11.492kg

    However, for question 2.3, I'm terribly stuck. I'm not even sure what to make my control surface.
    How can I relate pressure outside the valve with the two thermodynamic states I have inside the tank, as well as with the entropy generated?

    Many thanks in advance for any advice!

    PS α4 = 6
     
    Last edited: Oct 22, 2015
  2. jcsd
  3. Oct 22, 2015 #2
    What is ##\alpha_4##?
     
  4. Oct 22, 2015 #3
    My apologies, in my case it is 6, so V = 6.1m3. Just a randomiser to prevent students copying directly.
     
  5. Oct 22, 2015 #4
    What is the equation for the change in entropy per mole of an ideal gas in terms of the initial and final temperatures and pressures?
    If you choose the inlet plane and the outlet plane to the valve as the boundaries of your control surface, for any parcel of air passing through this control volume, what is the change in enthalpy? What is the change in temperature of the parcel for an ideal gas? Do all parcels of gas enter the valve at the same temperature and pressure?

    Chet
     
  6. Oct 22, 2015 #5
    Ok:
    I know how to find change in specific entropy using a Gibbs equation manipulation ( Δs = ∫cvdT/T + Rln(v2/v1) ).

    Then, using the first law for a single flow system, I get that hin = hout. I think it is correct to assume an adiabatic process here? So I guess change in enthalpy is zero.

    As to the change in temperature, Tout - Tin = (Poutvout - Pinvin)/R... I think??

    Finally no, since the state inside the tank in constantly changing.

    Am I doing anything right?
     
  7. Oct 22, 2015 #6
    The equation I had in mind was ##\Delta s=\int{C_p}\frac{dT}{T}-R\ln \frac{p_2}{p_1}##
    Correct.
    Not quite. If the change in enthalpy for an ideal gas is zero, and the enthalpy is a unique function of temperature, what is the change in temperature through the valve?

    Given the answer to this question, what is the change in entropy per mole in passing through the valve (using pin to represent the pressure in the tank upstream of the valve and pout to represent the pressure coming out of the valve (assumed constant according to the problem statement)?
    Correct. The pressure and temperature of the gas entering the valve change with time. We will take this into account.
    Yes!!!
     
  8. Oct 22, 2015 #7
    Well if enthalpy is strictly a function of temperature then it seems temperature is constant through the valve also. Is assuming this a bit of an approximation, since h = u + Pv, clearly pressure and volume have an influence too?

    In that case though, ##\Delta s=-R\ln \frac{p_{out}}{p_{in}}##

    Where to from here though? How do I involve the changing inlet pressure and temperature?
     
  9. Oct 22, 2015 #8

    No. For an ideal gas, there is no approximation involved. This is exact. So Δ(Pv)=0.
    Suppose that the number of moles of gas remaining in the tank at any instant of time is m. Then the number of moles of gas passing through the valve over any short time interval -dm. So, during the time that -dm passes through the valve, the amount of entropy generated is:
    $$dS=-dmΔs=R\ln\left(\frac{p_{out}}{p_{in}}\right)dm$$
    Now, from the ideal gas law for the tank, how is m related to pin, Tin, and V?

    Chet
     
  10. Oct 22, 2015 #9
    ##m=\frac{p_{in}V}{RT_{in}}##
     
  11. Oct 22, 2015 #10
    Good. Now before proceeding with the case where the heat capacity is a function of temperature, I recommend that we continue by first solving for the case of constant heat capacity (to see how that plays out and to make life simpler for ourselves). You need to eliminate Tin and express m exclusively in terms of pin. You can do this from the adiabatic expansion relationship for the gas in the tank. So, what is Tin as a function of pin, and then, what is m as a function of pin? Then, what is dm?

    Chet
     
  12. Oct 22, 2015 #11
    I can also help you get started for the case in which the temperature dependence of the heat capacity is included. In the part 2.1, you must have used the equation:
    $$C_p(T)\frac{dT}{T}=R\frac{dp}{p}$$
    Integrating this equation from the initial condition, you get:
    $$\int_{T_0}^{T_{in}}{\frac{C_p(T')}{R}\frac{dT'}{T'}}=\ln\frac{p_{in}}{p_0}$$
    where T' is a dummy variable of integration.
    So, solving for pin gives:
    $$p_{in}=p_0\exp\left[{\int_{T_0}^{T_{in}}{\frac{C_p(T')}{R}\frac{dT'}{T'}}}\right]$$
    See what happens when you substitute this into the equations for dS and Δs.

    Chet
     
  13. Oct 23, 2015 #12
    Ok I think I see where this is going, but I can't seem to find a way to write Tin in terms of pin without using the ideal gas equation?

    As for the integral of cp with temperature dependence, we have an appendix of tabulated standard entropy values which we are encouraged to use due to their improved accuracy. I think they would apply here, so could I write ##p_{in} = p_{0}exp[s_{T_{in}}-s_{T_{0}}]##?
     
  14. Oct 23, 2015 #13
    OK. I think you mean here "without assuming constant heat capacity," right? Let's skip the simplifying approximation implied in post # 11, and move on to post #12.
    Oh. I wasn't aware that this was the case. So you are supposed to work with the table. So in Part 2.1, you just used s(pfinal,Tfinal)=s(p0,T0) to obtain the final temperature?
    Let's come back to this after you've answered my question above.
    Chet
     
  15. Oct 23, 2015 #14
    In part 2.1 I used ##\Delta s=\int{C_p}\frac{dT}{T}-R\ln \frac{p_2}{p_1}##, and then substituted standard entropy values for the cp integral, following which I solved for sT2 (because everything else is known, and ##\Delta s= 0##) and then used tabulated data to interpolate and solve for T2.
     
  16. Oct 23, 2015 #15
    OK. I want to make sure we are on the same page. When you use the term "standard entropy," you mean the entropy at temperature T and pressure 1 atm., correct? This entropy is usually signified by using a superscript 0 to indicate that it is at 1 atm. So,
    $$s^0(T)=s(1 atm, T)$$
    Please confirm.

    Chet
     
  17. Oct 23, 2015 #16
    Yes exactly that. Sorry still getting used to the syntax for equations
     
  18. Oct 23, 2015 #17
    OK. Thanks.

    Let's go back to your post #12. The equation you wrote was ##p_{in} = p_{0}exp[s_{T_{in}}-s_{T_{0}}]##. But you're missing an R in the denominator of the exponent. So,
    $$p_{in} = p_{0}exp \left[\frac{(s_{T_{in}}-s_{T_{0}})}{R}\right]$$
    Do we agree on this?

    Chet
     
  19. Oct 23, 2015 #18
    Ah yes of course! My mistake.

    Ok so I have an expression for pin, and I also know that

    ##\Delta s = \int\frac{Q}{T} + S_{gen} = 100J/K##

    since the process is adiabatic. I'm still struggling though to find an expression for dm that I can use in our equation $$dS=-dmΔs=R\ln\left(\frac{p_{out}}{p_{in}}\right)dm$$
     
  20. Oct 23, 2015 #19
    This is not the correct equation to use. But, don't worry, I'll get you there.
    This is the correct equation to work with. As I said, don't worry, I'll get you there. We're almost done. We are going to integrate this equation to get ΔS which it the total amount of entropy generated in the valve, namely, 100 J/K. From this, we are going to determine pout.

    The next step is to substitute the equation for pin from the previous post into$$dS=-dmΔs=R\ln\left(\frac{p_{out}}{p_{in}}\right)dm$$
    Please show me what you get.

    Chet
     
  21. Oct 23, 2015 #20
    After simplifying I get $$dS=-dmΔs=R(\ln(p_{out}) - \ln(p_{0}) - \frac{s_{T_{in}}-s_{T_{0}}}{R})dm$$
     
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