Rate of change of pressure - gases

1. Feb 4, 2005

shrikeerthi

Hai,

I have the following situation: I have a closed container with a certain gas at a certain temperature(Tg) and pressure(pg). Now I open the container. The gas will escape through the opening to the atmosphere in order to create a pressure balance. There will be no significant increse in the pressure in the atmosphere but there will be noticible pressrue drop in the container. Does anyone know the formula to calculate the change in pressure over time in the container? Thanks in advance for any kind of help.

With Regards,
sk

2. Feb 4, 2005

Clausius2

Let's employ the conservation equations:(NOTE: variables without any subindex mean conditions inside the container.)

1) Continuity:

$$\dot m=-V\frac{d \rho}{dt}$$ is the mass flow.

2) Energy:

$$\frac{d(\rho e V)}{dt}+ \dot m (e + \frac{U^2}{2} + \frac{P}{\rho})=0$$

which means that the time change of internal energy per unit mass (e) is due to the outflow of stagnation enthalpy, because:

$$h_o=e + \frac{U^2}{2} + \frac{P}{\rho}=c_p T$$ is the stagnation enthalpy of the reservoir.

$$\frac{d(\rho c_v T V)}{dt}-V\frac{d\rho}{dt}c_p T=0$$

$$\frac{1}{\gamma-1}\frac{d(PV)}{dt}-V\frac{d\rho}{dt}c_p T=0$$

$$\frac{1}{\gamma-1}\frac{dP}{dt}-\frac{\gamma}{\gamma-1}\frac{P}{\rho}\frac{d\rho}{dt}=0$$

and we come to a surprise checking that the last equation is:

$$\frac{dP}{dt}=\frac{\gamma P}{\rho}\frac{d\rho}{dt}$$

which is the equation of an isentropic line. Remind I have written the energy equation neglecting terms of heat flux (which at $$Re_D Pr >>1$$ are in fact negligible. Integrating the last equation we have:

$$\frac{P}{\rho^\gamma}= \frac{P_o}{\rho_o^\gamma}$$ which is a constant value and depends on initial values. This equation doesn't provide a value of P(t) yet.

The evolution of P(t) has a strong dependence on the orifice shape. For simplicity, we will assume it is a convergent nozzle, with subsonic flow everywhere. So that, the mass flow is not constant, and will depend on the inner conditions. The mass flow through a convergent nozzle can be written as:

$$\dot m(t)=-V\frac{d \rho}{dt}=\rho_s U_s A_s$$ where "s" are the conditions just at the outlet section of the nozzle. Pay attention to this change of variables:

$$\dot m(t)=\rho a A_s \frac{\rho_s}{\rho}\frac{U_s}{a_s}\frac{a_s}{a}$$

The variable $$a$$ is the Sound Speed. The Sound Speed is different inside the container ($$a$$) and just at the outlet ($$a_s$$) because the gas has been accelerated and thus its static temperature has changed along the streamline that goes from the inner part (at rest) to the outlet (at movement with velocity Us).

By means of the definition of total enthalpy:

$$h_o=h+\frac{U^2}{2}$$

one can calculate those fractions as a function of the Mach Number at the outlet section: $$M=\frac{U_s}{a_s}$$. In particular, the pressure evolution from inwards to the outlet can be written as:

$$\Big(\frac{P}{P_a}\Big)^{\frac{\gamma-1}{\gamma}}=1+\frac{\gamma-1}{2} M^2$$ where P_a is the ambient pressure.

Substituting this in the mass flow last equation:

$$\dot m(t)=-V\frac{d \rho}{dt}=\rho a A_s \Big(\frac{P}{P_a}\Big)^{-\frac{\gamma+1}{2\gamma}}\Big[\Big(\frac{P}{P_a}\Big)^{\frac{\gamma-1}{\gamma}}-1\Big]^{1/2} \Big(\frac{2}{\gamma-1}\Big)^{1/2}$$

So that, this last equation with the isentropic line will give you the evolution P(t), because there are two equations for two unknowns. Remind that $$\rho a=\sqrt{\gamma P \rho}$$.

You have checked that calculating the evolution of the pressure inside a container is not a trivial task. Energy and Continuity of mass considerations must be employed. Also, the shape of the nozzle is very important, due to the elipticity of the subsonic problem the information of boundaries must be propagated inside the container. Without the equation of the nozzle mass flow, you'll never reach a P(t) solution.

Last edited: Feb 4, 2005
3. Feb 4, 2005

FredGarvin

Are you assuming a constant volumetric flow rate?

4. Feb 5, 2005

Clausius2

Fred, I don't know if your question is directed towards me, but anyway I have quoted myself to clear it up. Also, it is not convenient to talk about volumetric rate. The compressibility of the flow let mass flow to be more convenient. The mass flow is not constant if the conditions along the entire nozzle are subsonic. And that's what I assumed. Otherwise, the problem will be simplified a bit more.

5. Feb 5, 2005

FredGarvin

Sorry. I was asking you Clausius. The reason I ask is from your initial equation for mass flow. I guess I'm not used to your notation. What is V? Because I am not seeing how the units for mass flow work out in that equation unless V is volumetric flow. That's all. I agree with your assumption of a non-constant mass flow simply because the dP and density will be changing rapidly.

I prefer to talk in mass flow terms, but when we do have to talk in volumetric terms we always reference back to standard conditions.

6. Feb 5, 2005

Clausius2

V is the volume of the container (constant).

$$\dot m=-\frac{dm}{dt}=-V\frac{d\rho}{dt}$$

If the inner pressure would be greater than that which enhances a sonic flow through the nozzle, the mass flow will remain constant due to the hyperbolic behavior of the flow.

7. Feb 7, 2005

shrikeerthi

Thank you Clausius. I think the conditions just outside the exit is different from that of the atmosphere. If that is so, then is there some way to calculate those variables representing the condition at the exit from the initial data that I have in hand.

With Regards,
sk

8. Feb 7, 2005

Clausius2

I have solved it as a function of any external pressure $$P_a$$

9. Feb 8, 2005

shrikeerthi

Thank you Clausius. Could you tell me how to calculate the evolution of pressure in the container if the container is also injected with another gas at a constant mass flow rate during the time when the opening is kept open.

10. Feb 9, 2005

Clausius2

The procedure is straightforward, take a look and try to understand what I posted and you'll be able to formulate it again with another conditions.