Calculation of gas pressure increase

[DJB82]

There is a liquid in a reactor and the headspace is occupied by nitrogen. The gauge pressure in the headspace is 0barg. If the liquid then emitted a known quantity of e.g. CO2 into the headspace how would I work out the pressure increase in the headspace?

 

[jari001]

If you know the amount of gas emitted and the state variables of the system, you can calculate the partial pressure of the gas using an equation of state.

 

[DJB82]

So dividing the volume of gas emitted in litres by the free headspace volume in the vessel would be an incorrect way to work out the pressure increase in barg?

 

[LittleInch]

That's not correct.

To use the gas laws you need to work in absolute pressure and absolute temperature for starters.

Assuming your temp is the same and the compressibility of your gases are the same then

P2= P1 x ( V1/V2).

V2 is the new volume of the original gas now it has been pressurised by the additional gas entering your fixed gas space in the reactor.

Hence V2 = (V1 / (V1 + Vg) where Vg is the volume of your CO2.

So lets say your Vg is twice the original volume, then V2 = 0.333, so pressure P2 = 3 bara / 2 barg

It is was 4 times then P2 =5 bara / 4 barg

Note that this also assume ( very big assume) that the amount of gas released by the reaction is not pressure dependent. In most reactions the pressure will have an impact on gas produced so it becomes an iterative function to balance pressure with emitted volume of gas.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[DJB82]

Thanks for the excellent reply.

Can you explain why V2 = (V1 / (V1 + Vg))), rather than V2 = V1 - Vg, I'm a little confused on this. Thanks

 

[LittleInch]

I forget to say that the volume figure you use must be at the same pressure, hence use of standard volumes.

The equation I'm using is essentially saying that the original volume V1, is now being squeezed (pressurized) into a smaller volume by the introduction of more gas. If you think of this in terms of mass might help and assume the density of the gases is the same.

Then the original mass is fixed. The additional mass is then added to the original mass. Thus the ratio of original mass/volume to the new volume is original divided by original plus new.

Your equation doesn't work because you're thinking about the volume of Vg at the final pressure, i.e. the actual volume. Like I said at the beginning here, your volume need to be at the same reference pressure otherwise you're not working on the same basis.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[adammal44]

It's worthwhile to note here that nitrogen gas is slightly soluble in water. As your hydrogen gas evolves into the gas space and pressure increases, the increase in pressure will force some nitrogen back into solution. This is a slow process. This is useful to note, because the opposite can happen as well --- as you try to vent out, the fluid will also have effervescence attributable to nitrogen bubbles. You can estimate the solubility using Henry's law.

As to your real question, you can Van der Waal's equation of state to determine the final pressure of your system.

P = [ R * T / (V_molar - b)] - [a / V_molar²]

P is the system pressure (atmospheres)
R is the Ideal Gas Law Constant
T is the system temperature (Kelvin)
a is a Van der Waals constant (specific to the gas)
b is a Van der Waals constant (specific to the gas)
V_molar is the gaseous molar volume of your specific gas at a certain pressure/temperature condition

You are searching for your system pressure. You have two linked equations (the equation above for the hydrogen part of your system and the equation above for the nitrogen part of your system). You change V_molar to a real fraction: (Volume / moles) and you have two equations now. Assume the volume of your headspace is a function in both equations as well (if your tank is filled with a relatively incompressible liquid and your pressure change isn't too huge, though you can iterate on that later], with a special feature that we’ll get to in a second.

P_system = [0.08205 L*atm/mol*k]*[T]/[Volume_headspace / moles_nitrogen - b_nitrogen] - [a_nitrogen / (Volume_headspace / moles_nitrogen)]

P_system = [0.08205 L*atm/mol*k]*[T]/[Volume_headspace / moles_hydrogen - b_hydrogen] - [a_hydrogen / (Volume_headspace / moles_hydrogen)]

The trick here is that Volume_headspace in both equations is the following function:
Volume_headspace = molar volume_nitrogen (P_system) * moles of nitrogen + molar volume_ (P_system) * moles of hydrogen.

In this case, find a curve of molar volume of these gases and fit an equation to them. Insert that equation.

Solve the linked equations for a solution to P_system.

 

[adammal44]

Or, just use the ideal gas law.

PV = nRT

Make sure you're using absolute pressures and appropriate units. If your temperatures and pressures aren't outrageous, it's a good starting point while you work your way up to more exotic equations of state.

If you're OK with some inaccuracy, you can use the Van der Walls equation of state above without the linked equations by just treating your system as a single-gas system with a gas that looks like part-hydrogen and part-nitrogen. Just create new a/b values, by using the molar rations of your gas to weight each gases' individual constants to make a constant for your "new single gas". Then solve for P - assume that number of moles is the sum of moles of both gases.

If your pressures are high, there is a slight modification of the Ideal Gas Law that includes a compressibility term...

 

[zdas04]

LittleInch's formula is a tiny bit misleading. It really is (at constant head-space physical-volume, and constant temperature with no change in compressibility):

P₂=P₁(m₁/(m₁+m_gas))

The reason that the formula given can (and does) work is that it requires rolling the mass back to a volume at standard conditions which is an excellent surrogate for mass. It is easier to get your head around it if you assume that the volume of the head space is maintained at a constant level (by adding mass) while the gas evolves into the head space.

Doing it this way, you don't have to guess a final pressure to determine the density at actual conditions to allow you to convert to standard conditions. For an experiment you can rearrange this equation to determine how much gas evolved from the process for a measured pressure change.

m_gas=(P₁*m₁/P₂)-m₁

[bold]David Simpson, PE[/bold]
MuleShoe Engineering

In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual. Galileo Galilei, Italian Physicist

 

[DJB82]

Hi David,

Correct me if I'm not right but should that formula be P1=P2(m1/(m1+mgas)) rather than P2=P1(m1/(m1+mgas))? If not then that would mean that P2<P1

Thanks

 

[zdas04]

DJB82,
I think you are right.

[bold]David Simpson, PE[/bold]
MuleShoe Engineering

In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual. Galileo Galilei, Italian Physicist