Vapor pressure in NPSH calcuation

[farahnaz77]

sometimes I make doubt about basics, one of them is as follows:
Assume that we have a pressurized vessel containing e.g. LPG and N2 is used for blanketing and its operating pressure is 8 bara at 20deg.C, so vapor pressure is 8 bara?
it means that in NPSHa calculation Pa=Pv. Is it true? Don't we use its vapor pressure at 20 deg c? what about a storage tank with working pressure between -20 to 50 mmH2O?g
I'm confused,help me please.

thanks in advance

 

[itdepends]

NPSHA (NSPH available) = 8bar - vapour pressure of LPG at 20degC

(I am assuming your operating pressure is 8bar absolute. If it is 8bar gauge then NPSHA = 9bar - vapour pressure of LPG at 20degC)

Note that further reductions in NPSHA result from frictional losses in the pump/compressor suction piping.

i.e. NPSHA = tank pressure (absolute) - vapour pressure of fluid - frictional losses in suction piping.

Cheers

Itdepends

 

[25362]

A small correction to NPSHa as given by itdepends, add velocity head (usually small and negligible) and the geodetic suction head.

 

[TrevorP]

I'm a little confused about your original post. You appear to be saying that you believe that the vapour pressure is the same as the upstream pressure in a vessel. It is not. The vapour pressure is a physical property of any given fluid, and is constant at any given temperature. It does not change with the pressure of the system. If the vapour pressure is higher than the system pressure it will vaporise.
Is this what you were asking, or have I misunderstood?

 

[25362]

To farahnaz77, TrevorP is right. The VP of LPG at 20 deg C is determined by its composition, whether it is in the presence of a permanent (inert) gas or not.

For example, VP of butane, 0.3 bar; of propane, about 8 bar. For a mixture of both, somewhere in the middle. Commercial LPG (liquefied petroleum gas) is known to include other hydrocarbons (lighter and heavier, also of olefinic character) and traces of odorants.

In the absence of extraneous components and gases, LPG may be in a state of equilibrium within the vessel, as a boiling liquid would. In this (common) case its VP equals the pressure in the vessel and the NPSHa is generated mainly by the geodetic head.

 

[25362]

To farahnaz77, you may find thread407-59805 is very instructive.

 

[nakagawa]

To farahnaz77, This is a very interesting broblem I have been struggling with for years. question is "Assume that we have a pressurized vessel containing e.g. LPG and N2 is used for blanketing and its operating pressure is 8 bara at 20deg.C, so vapor pressure is 8 bara?" "it means that in NPSHa calculation Pa=Pv. Is it true? Don't we use its vapor pressure at 20 deg c?"
My answer is: We can assume Pa (operating press)=Pv only if liquid in the vessel is equilbrium with the vapor phase blanketed by N2. or more precisely the liquid at the pump suction is in equilibrium with vapor phase.
however, if the both phases are not in equilibrium, the Pv is lower than Pa. I believe no one can tells whether N2 blanketted LPG is in equilbrium (or saturated with N2)or not. Therefore I will be forced to assume that equilibrium conditon exist and assume Pa=Pv just to be on the safe side when calculating NPSHa.
I think this assumption(Pa=Pv) is qvery much questionable because;
1. It will will take a long time to establish equilibrium condition (in a LPG vessel by padding with N2). If the vessel is a surge tank receiving LPG and Pumping out LPG,it is hard to believe equilbrium can be reached except when pumping out rate is high enough to put the tank pressure on decreasing trend to reliese vapor(bubbling) at the level assumed as base elevation for NPSHa calculation.
2. Even if NPSHa<NPSHr I do not believe harmful cabitation takes place. Vapor bubbles may evolve somewhere inside pump, however the vapor is very rich in N2 and it cannot be in equilibrium with LPG surrounding the vapor bubbles. the bubbles can not be collapsed. the evolved bubbles decrease the average density and can lower the dischage pressure somewhat but this may not be a problem thou I didn't check the possible amount vapor volume within reasonable assumption.
The question exist whenever we deal with liquid or liquid mixture in equilibrium with very light components, say boiling point difference of more than 200deg C.

 

[CHD01]

The vapor pressure of the liquid is dependent upon the liquid temperature only, its that simple.

The more you learn, the less you are certain of.

 

[sailoday28]

25362 (Chemical) Nov 1, 2004 states
A small correction to NPSHa as given by itdepends, add velocity head (usually small and negligible) and the geodetic suction head.



I agree with all of the above, IF, the fluid is incompressible (generally reasonable assumption) AND isothermal flow.

 

[brainstorming]

Vapor pressure

If a quantity of liquid is placed in a closed container, the molecules of the liquid that are escaping from the surface are not removed from the space above the liquid. Thus, the system will arrive at a steady state where the rate of evaporation is equal the rate of condensation. A state of equilibrium is than said to be established. After this equilibrium state is established, the vapor molecules remain in constant concentration in the space above the liquid, and they will exert a constant definite pressure, at the system temperature. This pressure is called the vapor pressure of the liquid.


Then if equilibrium exists in your system, u can use the 8 bara as vapour pressure in your calc.

Regards