Modeling Pressure Change in a Pressurized Chamber

[mattie]

I'm trying to model the rate of change of the pressure in a pressurized rigid chamber with normal air (assumed to be an ideal gas).
It has an air outflow V̇ (m^3/s) with ρ1.
What's the change in p between state 1 and state 2?
My assumption is that it can be modeled like an adiabatic expansion.
this would lead to

, leading to:

or

Though I'm not sure of this assumption.
I would like to either back this up, or change the assumption.
Any suggestions?

 

[Chestermiller]

If the heat transfer between the vessel and the gas is negligible, then your equation is OK, provided the V's are specific volumes (volume of vessel, a constant, divided by mass of gas in vessel at any time).

But your final equation is incorrect.

 

[mattie]

If the heat transfer between the vessel and the gas is negligible, then your equation is OK, provided the V's are specific volumes (volume of vessel, a constant, divided by mass of gas in vessel at any time).

But your final equation is incorrect.

Thanks for your reply.

The heat transfer is indeed assumed to be negligible.

As the density of the air flowing out, as well as the density of the air in the vessel are the same, I've assumed that ΔV̇/V = Δv̇ /v
Is this assumption also correct?

Also: On what grounds can I justify using an adiabatic process here?

 

[Chestermiller]

Thanks for your reply.

The heat transfer is indeed assumed to be negligible.

As the density of the air flowing out, as well as the density of the air in the vessel are the same, I've assumed that ΔV̇/V = Δv̇ /v
Is this assumption also correct?

Are you saying that the volume rate of flow out of the tank based on the initial pressure and temperature are constant? On what basis are you making such a radical assumption. Other possibilities are (1) the volume rate of flow based on the outside pressure and current discharge temperature, or just (2) the mass rate of flow out of the tank. So why have you made such an unusual choice, and how could you control it?

Also: On what grounds can I justify using an adiabatic process here?

If the discharge is relatively fast so that the heat transfer between the vessel mass and the gas doesn't have time to respond.

 

[mattie]

Are you saying that the volume rate of flow out of the tank based on the initial pressure and temperature are constant? On what basis are you making such a radical assumption. Other possibilities are (1) the volume rate of flow based on the outside pressure and current discharge temperature, or just (2) the mass rate of flow out of the tank. So why have you made such an unusual choice, and how could you control it?

I'm making this assumption as this is a part of a more complex system, which is not of importance to this part in particular. The air flow is modeled as a function of pressure and air density. \dot{V}=A\cdot C \sqrt{\frac{2(p_{in}-p_{out})}{\rho}}
The system is solved using a Matlab ODE solver, so the input will be known for each small interval
For this to work, I need to know the change in pressure as a function of outflow of air, and total volume (which is assumed to be constant at a small time-frame).
The volume of the chamber itself changes in a prescribed manner, which in itself adds another (adiabatic as well) component to the rate of change in pressure.

If the discharge is relatively fast so that the heat transfer between the vessel mass and the gas doesn't have time to respond.

What I meant to ask was: Why is this not similar to a free expansion, for which PV=constant?

 

[Chestermiller]

I'm making this assumption as this is a part of a more complex system, which is not of importance to this part in particular. The air flow is modeled as a function of pressure and air density. \dot{V}=A\cdot C \sqrt{\frac{2(p_{in}-p_{out})}{\rho}}
The system is solved using a Matlab ODE solver, so the input will be known for each small interval
For this to work, I need to know the change in pressure as a function of outflow of air, and total volume (which is assumed to be constant at a small time-frame).

Does your equation calculate the volume flow rate at the inside pressure and density or the outside pressure and density? What you really need is the mass flow rate.

The volume of the chamber itself changes in a prescribed manner, which in itself adds another (adiabatic as well) component to the rate of change in pressure.

The volume of the chamber doesn't change.

What I meant to ask was: Why is this not similar to a free expansion, for which PV=constant?

This differs from free expansion in that the gas within the tank is doing work to force the gas leaving ahead of it out of the tank.

 

[mattie]

Does your equation calculate the volume flow rate at the inside pressure and density or the outside pressure and density? What you really need is the mass flow rate.

It calculates the volume flow rate using the inside pressure, outside pressure, and density inside of the chamber.
The air density in the chamber is assumed to be known.

The mass flow rate would be density multiplied by the volumetric flow, and the specific volume would be 1 divided by the density.

The volume of the chamber doesn't change.

Not as a part of the interaction posted here, but it does in a separate one. This interaction also adds a component to the rate of change in pressure.

This differs from free expansion in that the gas within the tank is doing work to force the gas leaving ahead of it out of the tank.

Ah, I think this is the part I was looking for!

 

[Chestermiller]

It calculates the volume flow rate using the inside pressure, outside pressure, and density inside of the chamber.
The air density in the chamber is assumed to be known.

The mass flow rate would be density multiplied by the volumetric flow, and the specific volume would be 1 divided by the density.

So if you wanted to calculate the mass flow rate using your volume flow rate equation, would you multiply by the density inside the chamber or the density outside the chamber? You are aware that these two results would be different, right?

Not as a part of the interaction posted here, but it does in a separate one. This interaction also adds a component to the rate of change in pressure.

Huh? Your diagram shows that the rigid container has constant volume. So what are you talking about?

 

[mattie]

So if you wanted to calculate the mass flow rate using your volume flow rate equation, would you multiply by the density inside the chamber or the density outside the chamber? You are aware that these two results would be different, right?

Yes, I use the one inside of the chamber, because it's this air that moving outwards.

Huh? Your diagram shows that the rigid container has constant volume. So what are you talking about?

I split the interaction into 2 processes. 1 process reduces the size of the chamber following a pre-determined path. And then there's the one I posted here, which consists of air flowing outwards.
Both have a certain influence on the air pressure difference, which I calculate by adding them together.

 

[Chestermiller]

I split the interaction into 2 processes. 1 process reduces the size of the chamber following a pre-determined path. And then there's the one I posted here, which consists of air flowing outwards.
Both have a certain influence on the air pressure difference, which I calculate by adding them together.

That doesn't make sense to me. If m represents the mass of gas within the (constant volume) vessel at any time, then the pressure within the vessel at any time is given by: $$p=p_0\left(\frac{m}{m_0}\right)^{1.4}$$ where the subscript 0 refers to the pressure initially. So the rate of change of pressure is related to the rate of change of mass by: $$\dot{p}=1.4\frac{p_0}{m_0}\left(\frac{m}{m_0}\right)^{0.4}\dot{m}=1.4\frac{p}{m}\dot{m}$$

 

[mattie]

That doesn't make sense to me. If m represents the mass of gas within the (constant volume) vessel at any time, then the pressure within the vessel at any time is given by: $$p=p_0\left(\frac{m}{m_0}\right)^{1.4}$$ where the subscript 0 refers to the pressure initially. So the rate of change of pressure is related to the rate of change of mass by: $$\dot{p}=1.4\frac{p_0}{m_0}\left(\frac{m}{m_0}\right)^{0.4}\dot{m}=1.4\frac{p}{m}\dot{m}$$

Ah, that makes it clearer for me.
the reason I split up the interactions is because both processes work independently from each other. one of them reduces the volume, and the other reduces the mass of the gass.
And since each small time step is calculated individually, the volume and pressure are considered to be constant while calculating their change.
Could you explain why the last formula I wrote in my initial post is wrong?

 

[Chestermiller]

Yes, I use the one inside of the chamber, because it's this air that moving outwards.

I split the interaction into 2 processes. 1 process reduces the size of the chamber following a pre-determined path. And then there's the one I posted here, which consists of air flowing outwards.
Both have a certain influence on the air pressure difference, which I calculate by adding them together.

Let me simplify it for you further. From your equation \dot{V}=A\cdot C \sqrt{\frac{2(p_{in}-p_{out})}{\rho}} it follows that the rate of change of mass within the tank is at any time is $$\frac{dm}{dt}=-AC\sqrt{2\rho (p-p_{out})}=-AC\sqrt{\frac{2m (p-p_{out})}{V}}\tag{1}$$where p and $\rho=\frac{m}{V}$ refer to the pressure and density of the air within the tank at any time.

From my equation $$p=p_0\left(\frac{m}{m_0}\right)^{1.4}$$ it follows that$$\frac{m}{m_0}=\left(\frac{p}{p_0}\right)^{1/1.4}\tag{2}$$where the subscript 0 refers to the values of the mass and pressure at time zero. If I take the time derivative this equation, I obtain: $$\frac{dm}{dt}=\frac{m_0}{1.4p_0}\left(\frac{p}{p_0}\right)^{\frac{1}{1.4}-1}\frac{dp}{dt}=\frac{m}{p}\frac{dp}{dt}\tag{3}$$From Eqns. 1-3, you can eliminate both m and dm/dt to obtain an equation for dp/dt exclusively in terms of p. Let's see what you get.