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Throttling a Gas Cylinder 
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#1
Oct3012, 10:03 PM

P: 4

How do I write an expression that describes how the pressure and temperature in an insulated tank changes with time after a valve is opened? This is an adiabatic, reversible process with fixed volume V, pressure P, and temperature T, I've been thinking and thinking but I don't know how to solve this problem if I'm not given final V,P, or T.



#2
Oct3112, 01:18 AM

P: 32

Your question needs more detail. It is not apparent which quantities are being held fixed here. If it is adiabatic expansion then the volume V is obviously changing. There is no thermodynamic process (except diffusion) where P,V, and T all remain constant.
I'll assume you have an insulated gas expanding due to some valve being opened: The relation you need to use is [itex]P_1{V_1}^\gamma=P_2{V_2}^\gamma[/itex], assuming the gas is ideal. The derivation of this relation can be found at http://en.wikipedia.org/wiki/Isentropic_process 


#3
Oct3112, 01:29 AM

P: 4

Okay so let's say we are given an initial P, T, and V of a gas inside of a tank. When the valve is opened the volume is expands because the contents inside are higher pressure/temperature than the outside, how do I derive P(t) and T(t), pressure and time with respect to time?
The wikipedia page has the right ideas, I just have no clue how to write P and T as a function of time. 


#4
Oct3112, 01:51 AM

P: 32

Throttling a Gas Cylinder
Ok, so it IS adiabatic expansion. The formula I have written in my last post relates P and V at different times. However, you should also remember that the equation of state PV=NRT is always valid for an ideal gas. So you can easily transform equation [itex]P_1{V_1}^\gamma=P_2{V_2}^\gamma[/itex] in terms of P and T as [itex]\frac{{P_1}^{\gamma1}}{{T_1}^\gamma}=\frac{{P_2}^{\gamma1}}{{T_2}^\gamma}[/itex]. And you are done.



#5
Oct3112, 01:43 PM

P: 4

I got this but how do you relate pressure and temperature with TIME, not with each other. I know how to do that, how would I change the equation so that the time variable is present and I can find out the pressure after a certain time or the temperature at a certain time.



#6
Oct3112, 02:01 PM

P: 32

[itex]\frac{P^{\gamma1}}{T^\gamma}[/itex] is a conserved quantity for the adiabatic process. No matter what instance of time you take during the process, the instantaneous pressure and temperature will yield a constant value of [itex]\frac{P(t)^{\gamma1}}{T(t)^\gamma} = \frac{{P_1}^{\gamma1}}{{T_1}^\gamma} [/itex].
Also we know [itex]P(t)V(t)=NRT(t)[/itex]. You know values of V(t), [itex]P_1[/itex] and [itex] T_1[/itex]. You have two equations in two unknowns. Solve for P(t) and T(t). 


#7
Oct3112, 05:50 PM

P: 4

OH, i got it now, it's pretty messy but I think it works, thanks a bunch for all of the help



#8
Nov112, 08:14 PM

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P: 5,063

The number of moles in the tank is changing with time. To do what you want to do, you need to know the pressure drop/flow rate relationship for the value when it is open a certain amount. ΔP = C (ρv^{2})/2, where ΔP is the pressure in the tank minus the atmospheric pressure outside, ρ is the gas density in the tank, v is the gas velocity coming out of the valve, and C is a constant which depends on how far open the valve is. Knowing the velocity v, the area of the valve exit, and the density determines the mass flow rate out of the tank. You then know the molar flow rate out of the tank.



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