Thermodynamics: Pressure Drop over a Valve

[Chestermiller]

Thank you!

Could you possibly just clarify for me one last time; in post #11 where you derive an expression for p_in, 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 p_in and p_out, 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

 

[TimoD]

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 p_out.

Oh I thought that the average may be above even though the instantaneous is always below p_in. 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.

 

[Chestermiller]

Oh right I see, I misread p₀ as p_out.

Oh I thought that the average may be above even though the instantaneous is always below p_in. 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 p_out it you used a value of 1000 J/K for the entropy generated?

Chet

 

[Chestermiller]

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 p_out, which sounds about right.

Chet

 

[TimoD]

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 p_out, 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