Thermodynamics: Pressure Drop over a Valve

  • #26
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Great. Before we start integrating, let's check to make sure that the numbers in your table are correct. Is the final pressure in your table 200 kPa? Is the difference in mass between beginning and end consistent with what you calculated in part 2.2? Also, let's be sure that the units are consistent when we do the integration.

To do the integration numerically, the integral over each interval of mass is just calculated as ##\frac{I(m)+I(m+\Delta m)}{2}\Delta m##, where I(m) is the value of the integrand at mass m.

Chet
Yes I think the values are close enough.

Ok so if I use that parallelogram approximation to find the integral, then I can solve for pin? What about the iterative method that the question speaks of?
 
  • #27
21,488
4,864
Yes I think the values are close enough.

Ok so if I use that parallelogram approximation to find the integral, then I can solve for pin?
I think you mean the trapazoidal rule.
What about the iterative method that the question speaks of?
I don't know. It doesn't seem like an iteration is needed. Maybe they were using equal increments in pressure, rather than temperature, in which case they would have to iterate to get the temperature (and thus the mass) at each step. Or maybe they were using the word iterate to signify numerical integration. Or maybe they didn't realize that they could solve for the exit pressure explicitly. It's hard to say. All we can do is have confidence in what we did.

Chet
 
  • #28
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I think you mean the trapazoidal rule.

I don't know. It doesn't seem like an iteration is needed. Maybe they were using equal increments in pressure, rather than temperature, in which case they would have to iterate to get the temperature (and thus the mass) at each step. Or maybe they were using the word iterate to signify numerical integration. Or maybe they didn't realize that they could solve for the exit pressure explicitly. It's hard to say. All we can do is have confidence in what we did.

Chet
Yes that's the one sorry.

Chestermiller thank you so much for your step by step help through all of this as well as your patience through all my blunders. I really appreciate it! God bless you sir
 
  • #29
21,488
4,864
Yes that's the one sorry.

Chestermiller thank you so much for your step by step help through all of this as well as your patience through all my blunders. I really appreciate it! God bless you sir
You've done very well. This was not an easy problem. I think you have a real knack for understanding thermodynamics.

Chet
 
  • #30
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You've done very well. This was not an easy problem. I think you have a real knack for understanding thermodynamics.

Chet
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?

Finally could you just confirm that you get an answer around 277kPa if you have worked it out?
 
  • #31
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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
 
  • #32
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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
21,488
4,864
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
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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
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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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