Thermodynamics: Pressure Drop over a Valve

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SUMMARY

The discussion focuses on calculating the pressure drop over a valve in a thermodynamic system, specifically addressing the relationship between pressure, temperature, and entropy changes. Key equations mentioned include the Gibbs equation for entropy change (Δs = ∫CpdT/T - Rln(p2/p1)) and the first law of thermodynamics, which states that hin = hout for an adiabatic process. The participants emphasize the importance of understanding how the changing states of gas affect the calculations of enthalpy and entropy, particularly in relation to the ideal gas law and the behavior of gases under varying conditions.

PREREQUISITES
  • Understanding of thermodynamic principles, particularly the first law of thermodynamics.
  • Familiarity with the Gibbs equation for entropy change.
  • Knowledge of ideal gas behavior and the ideal gas law.
  • Ability to manipulate equations involving pressure, volume, and temperature in thermodynamic contexts.
NEXT STEPS
  • Study the derivation and application of the Gibbs equation in various thermodynamic processes.
  • Learn about the implications of adiabatic processes on enthalpy and entropy changes.
  • Explore the ideal gas law and its applications in real-world scenarios involving pressure and temperature changes.
  • Investigate the relationship between heat capacity and temperature in thermodynamic systems.
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Students and professionals in mechanical engineering, chemical engineering, and thermodynamics who are looking to deepen their understanding of pressure drop calculations and entropy changes in fluid systems.

  • #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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