Vapour pressure and critical temperature 1

[22082002]

Hi All,
I am not clear about following information:
I am working on control valve process datasheet, and find kerosene having vapor pressure of 34.3 bara and critical pressure of 27 bara.
I was expecting it otherway around i.e. vapor pressure < critical pressure.

I would appreciate if you can clarify these and their interrelation(in general)/significance with regards to design of control valve.

Thanks in advance.
SAA

 

[rbcoulter]

No way that kerosene has a vapor pressure of 34 bar (about 34 atms) at standard temperature. It would be a gas!

 

[22082002]

Well, all the above values are as per simulation...

Regards,
SAA

 

[25362]

to 22082002, without considering the truth of your data, allow me to point that a petroleum fraction is not a pure hydrocarbon but a mixture of many HCs.

The critical point of the mixture is influenced by the components and their proportion. Let me give an example taken from the literature: a mixture of ethane and n-heptane:

A binary mixture having 26.5% mole of ethane (Pc=688 psia, Tc=90^oF) with n-C₇ (Pc=390 psia, Tc=512^oF) has a critical point at 475^oF and Pc= 640 psia.

Now, at 450^oF the VP of the mix is 715 psia, i.e., 75 psi above the critical pressure!

 

[TD2K]

The estimation of your critical pressure is reasonable. The vapor pressure is way off. You can get a rough estimate of the vapor pressure from the flash point. For kerosene, I would expect it to be between 125F to maybe 150F. For heavier hydrocarbons, the LEL is usually about 1 to 1.5 mol% vapor. Given that atmospheric pressure is 14.7 psia, that's a vapor pressure of about 0.15 to 0.23 psia.

Now, you have to adjust the vapor pressure for your operating temperature but, not withstanding 25362's point, I doubt it will be 34 bar(a).

What are the operating conditions for this valve?

 

[22082002]

I will try to confirm once again the vapor pressure.(before writing my first mail, I did get the same values as per Basic engineering document.)

Operating conditions on CV-Process datasheet:
Flow, max./Normal=138/110 m3/hr
Pre. drop = 25 / 26 barg
Upstream pressure = 34.0 / 34.3 barg
Temp.= 226 / 226 °C
Design P & T = 46.5 / 315°C
Operating density = 695 / 695 kg/m3
Viscosity = 0.177 / 0.177 cP

This valve is on the hot separator liquid(kero) line going to stripper. Corrosive to H2S..

Please let me know if you need additional information. I spoke to some of my collegues and got info that this valve is most complicated among the whole plants' Control valves and can have these conditions.

Look forward to your comments..

Thank you !

Regards,
SAA

 

[25362]

to 22082002, since you are speaking just of theoretical simulations, let me enlarge a bit what I have intended to say before. In principle, on a PT graph vapour pressures of individual hydrocarbons would be represented by a line stopping at the critical point (CP), as shown in the well-known Cox chart.

You are right, the CP for a pure species is the highest temperature and highest pressure at which vapour and liquid phases can coexist.

However, for a mixture it is in general neither.

The CPs of mixtures vary and are located on a curve (the locus) that changes with composition, under most conditions the CPs aren't located at the maximum values of P and T. This curve is the locus of the bubble points (to the left of the CP) and the dew points (to the right of the CP).

For a given composition the equilibrium curve around the CP bulges to the right, with the CP somewhere on the bulge, but -in general- neither at the highest temperature nor at the highest pressure. To the left of the CP, one single (vertical) temperature cuts the equilibrium lines at two points: the higher one representing the liquids' bubble point (i.e., VP); the lowest intersection being the vapour dew point. By being on a temperature between the bulge's maximum, and the CP, it is possible to cross the dew point line and to partly condense the vapour by reducing the pressure.

This is called retrograde condensation.

This phenomenon is of importance when operating some deep natural-gas wells where the prevailing conditions are to the right of the CP temperature of the mixture but still below the maximum saturated vapor dew point equilibrium temperature. When isothermally removing gas, the total pressure in the well drops and part of the nat-gas (butanes and heavier) liquefies, a fact that may lead to a reduction in the production of the well when the liquid is trapped in the sands. Repressurizing the well by lean gas, that has been removed from the condensing heavier species above-ground, is commonly used to return the underground reservoir to an elevated pressure and a more favourable equilibrium composition.

In summary, yes, vapour pressures (bubble points) can, in principle, be higher than the critical point pressures of a hydrocarbon mixture.

 

[rbcoulter]

The upstream pressure of your control valve is not equivalent to the vapor pressure of the fluid. Kerosene is normally a liquid, and since you didn't state otherwise this would be assumed. Even if this were a gas it is not customary to refer to a process pressure as &quot;vapor pressure&quot; since this is confused with the VLE parameters which are generally &quot;static&quot; attributes of fluids.
Usually &quot;vapor pressure&quot; refers to a the pressure imparted by a liquid in a tank or vessel and is a function of temperature.

 

[25362]

I wonder whether a wide-cut JP-4 (not a true kero) could have a VP of 34 bar at 226^oC. When plottting the highest RVP allowed a JP-4, 21 kPa, on a Cox chart it falls between n-hexane and n-heptane. Following their VP lines up to 226^oC one gets a VP lower than 34 bar, as 22082002 mentioned. Therefore, it appears that rbcoulter is right in saying the fluid is a liquid upstream the CV. However, it appears that it could partly vaporize downstream.

 

[22082002]

I am sorry for not stating the condition of the liquid.

You are right, it's Liquid. Further to this it is coming from a separator, in which H2, H2S, C1 to C4 get separated from this mexture, but still this liquid will have H2,H2S, C1-C4 as per equilibrium, at conditions mentioned above.
This liquid will go to a stripper.

Approximately 950 Sm3/hr gases will be stripped out from this stream.

I thank you for sparing time and look forward to hear more from you.

Warm regards,
SAA

 

[hacksaw]

with the fluid upstream of the valve coming, coming from a stripper will be saturated and contain dissolved gases. this is not quite the same as the situation you have with a single component fluid, but it is customary to allow for flashing effects in the valve sizing.