# Pressure drop saturated vapor (VLE)

I'm trying to find a formula for the pressure drop of a vapor flowing through an orifice which opens at a certain value of the temperature (or pressure). Let's assume that we have a system like this: If the system is in VLE, one can calculate the pressure of the system by using Antoine's equation. So if one heats up the system, the pressure will increase. But what happens with the pressure when the orifice above opens? The equilibrium conditions are not there anymore, so the Antoine equation is not valid anymore (or is it?). So I guess that the ideal gas law will apply, but how do I connect the two pressures, if I want to determine the mass flow due to the pressure difference?

I was thinking of something like this: $P = P_{sat} - \frac{m_vRT\theta}{V}$

where $m_v$ is the mass of the vapor, $R$ is the gas constant, $T$ is the temperature, $\theta$ is the vapor fraction and $V$ is the volume occupied by the vapor.

Is this formula correct, considering that $P_{sat}$ is dependent on the temperature (which does not go to 0, therefore vanishing the $P_{sat}$ term). Can anyone help me with this?

• if you have something more specific in mind than what you posted, such as your liquid exists in a flash drum with a rupture disk as a safety device, please say so. Without knowing how much "disengagement space" is available, I can't be certain that my answer matches your problem. However, don't go so far as to publish proprietary company information, as that's a big NO NO. – David White Nov 22 '16 at 16:55

## 1 Answer

I doubt that you will find a formula for the physical situation described in the posting. Before the orifice opens, the liquid and vapor are in equilibrium at the given temperature, and they exist above ambient pressure. Once the orifice is opened, the pressure in the container immediately drops, and the bulk liquid will be superheated by an amount that can be found from the Antoine equation vs. the bulk liquid temperature, at the pressure that exists in the container. This means that the bulk liquid will immediately boil, resulting in a mixture of bubbles and liquid in the space formerly occupied by liquid alone. Given this, there is a very high probability that you will get two-phase flow in the orifice. This will continue until enough liquid has boiled to drop the temperature back down to the point where the bulk liquid is at its boiling point at the new pressure in the container. In my experience, I have never seen an orifice calculation for two-phase flow, and in fact, I seriously doubt that any flow research has been done in this area because there are several different two-phase flow regimes, some of which are unstable (i.e., the ratio of liquid to vapor at any given instant is not constant).

• Thank you David. Now that you mentioned it, it does sound more complicated than I imagined :) I thought that there is someone who has some experience with this and has an "easier" solution. – Physther Nov 22 '16 at 18:32
• @Paul, I worked in industry for 21 years as a chemical engineer, so I do have experience with aspects of your problem. For pressure drop across an orifice, you need to know the flow rate. THAT is where the "rub" is ... there are probably no correlations that deal with two-phase flow across orifice plates. – David White Nov 22 '16 at 22:53
• Thank you very much for this information, David. It's important to know the limitations. – Physther Nov 22 '16 at 22:56