Thermodynamics: Pressure Drop over a Valve

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Discussion Overview

The discussion revolves around the thermodynamic analysis of pressure drop over a valve, focusing on the relationship between pressure, temperature, and entropy in a control volume involving an ideal gas. Participants explore various equations and concepts related to entropy change, enthalpy, and the behavior of gases under different conditions.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested
  • Homework-related

Main Points Raised

  • One participant has calculated T2 and m2 but is uncertain about relating the pressure outside the valve to the thermodynamic states inside the tank and the entropy generated.
  • Several participants inquire about the meaning of α4 and its relevance to the problem.
  • Participants discuss the equation for change in entropy for an ideal gas, with one proposing a manipulation of the Gibbs equation.
  • There is a suggestion that the process can be assumed adiabatic, leading to the conclusion that change in enthalpy is zero.
  • One participant questions whether temperature remains constant through the valve, considering the influence of pressure and volume.
  • Another participant asserts that for an ideal gas, the assumption of constant temperature through the valve is exact, not an approximation.
  • Entropy generation is discussed in terms of the number of moles passing through the valve and the relationship between pressure and temperature in the tank.
  • Participants explore how to express the number of moles in terms of pressure and temperature, with references to the ideal gas law.
  • There is a proposal to first solve the case with constant heat capacity before addressing temperature dependence.
  • One participant expresses difficulty in writing temperature in terms of pressure without using the ideal gas equation.
  • Another participant suggests using tabulated standard entropy values for improved accuracy in calculations involving temperature dependence.

Areas of Agreement / Disagreement

Participants express various viewpoints on the assumptions regarding temperature and enthalpy changes, leading to some disagreement about the implications of these assumptions. The discussion remains unresolved regarding the best approach to relate the changing inlet pressure and temperature to the overall analysis.

Contextual Notes

There are limitations regarding the assumptions made about constant heat capacity and the ideal gas behavior, as well as the dependence on specific definitions and relationships between variables. Some mathematical steps remain unresolved, particularly in expressing temperature in terms of pressure.

  • #31
TimoD said:
Thank you!

Could you possibly just clarify for me one last time; in post #11 where you derive an expression for pin, I see it comes from the equation ##Cp(T)\frac{dT}{T}=R\frac{dp}{p}##, which comes from ##ds = Cp(T)\frac{dT}{T}-R\frac{dp}{p}## where change in entropy is zero. Why is this so? Since we use pin and pout, surely the change in entropy is not zero between these points?
This would be the change in molar entropy for the gas in the tank. It establishes the relationship between the pressure and the temperature in the tank (and of the parcels of air that enter the valve).
Finally could you just confirm that you get an answer around 277kPa if you have worked it out?
This appears to be high. The exit pressure from the valve shouldn't be higher than the final pressure in the tank (200 kPa). Otherwise gas would be flowing back in the opposite direction. What would you get for the entropy change if you assumed that the pressure at the exit of the valve were at 200 kPa? (Maybe they gave you too small a value for the entropy change in the problem statement, not realizing that the exit pressure had to be less than the final pressure in the tank).

Chet
 
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  • #32
Chestermiller said:
This would be the change in molar entropy for the gas in the tank. It establishes the relationship between the pressure and the temperature in the tank (and of the parcels of air that enter the valve).

This appears to be high. The exit pressure from the valve shouldn't be higher than the final pressure in the tank (200 kPa). Otherwise gas would be flowing back in the opposite direction. What would you get for the entropy change if you assumed that the pressure at the exit of the valve were at 200 kPa? (Maybe they gave you too small a value for the entropy change in the problem statement, not realizing that the exit pressure had to be less than the final pressure in the tank).

Chet
Oh right I see, I misread p0 as pout.

Oh I thought that the average may be above even though the instantaneous is always below pin. With a 200kPa exit pressure though, I get an entropy change of 797.7J/K. Maybe my integral value is off? For that I get quite a high 674.4J/K.
 
  • #33
TimoD said:
Oh right I see, I misread p0 as pout.

Oh I thought that the average may be above even though the instantaneous is always below pin. With a 200kPa exit pressure though, I get an entropy change of 797.7J/K. Maybe my integral value is off? For that I get quite a high 674.4J/K.
I don't think so. I'm guessing that they meant 1000 J/K in the problem statement, rather than 100 J/K. But, just in case, please check all your units. I think the 277 kPa is close to the value you would get if you took the entropy change in the valve to be close to zero. What value would you get for pout it you used a value of 1000 J/K for the entropy generated?

Chet
 
  • #34
With your value of the integral and an assumed value of 1000 J/K for the entropy change, I get a value of about 185 kPa for pout, which sounds about right.

Chet
 
  • #35
Chestermiller said:
With your value of the integral and an assumed value of 1000 J/K for the entropy change, I get a value of about 185 kPa for pout, which sounds about right.

Chet
Yup I get that as well. Ok well since our method is sound I will note this in my assignment and see what happens. Thanks once again for all your help
 

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