NPSH calculation consideration

[alpha0125]

If we have 50 psi seperator with gas, water and natural gas condensate(oil), the gas seperates and goes out from the seperator from top. The water and liquid comes out from the bottom. If i need to transfer liquid from the seperator to 200 miles far to the tank, then i need pump to transfer the liquid because of pressure drop on the pipe.
if there is a 15 feet tall seperator, and the top level of water is at 12 feet and and top level of oil is at 15 ft (which is total of 3 ft oil). I want to pump all 3 feet of oil and i want to pump 5 ft of water everytime pump kicks on. The specific gravity of water is 1.14 and specific gravity of oil is .742.
I know how to calcualte the NPSH.
how much head do i have to use to calcualte NPSH? I think i need to calcuate NPSH for oil and NPSH for water and add it. For NPSH for oil , what is my suction head becuase i only have 3ft of oil and i want to pump all that. if i stop my flow when water level gets to 10 ft that means i do not have any oil left. I thought my suction head is whatever the differnce in the liquid level ( for water case, 1O ft becuase i stopped my pump at when water is at 10 ft).
1) What is the for the oil case?
2) My question is since i have water and oil in the vessel, the vapor pressure of of both od them is 50 psi. if oil is saturated at that pressure, what happens to the vapor pressure?


Sorry if it is dump question, i am recently graduated engineer.

 

[TD2K]

There are two NPSH, NPSHA and NPSHR. NPSHR is what you see on the pump curve for a given flow. NPSHA is what the system provides.

For this system, I would assume both the oil and water is a saturated fluid or in other words, they are in equilibrium with the gas in the separator. Since it sounds like you first pump out the water (that has settled) and then the oil, it's a moot point because the oil will always be saturated in your setup.

If you re-arrange the equation for NPSHA in this case, you will find that system pressure and vapor pressure are the same and cancel out. NPSHA simplifies to static head of liquid minus line losses. NPSHA is simply the pressure at the suction of the pump minus the vapor pressure of the liquid in ft of liquid.

Assuming you have a horizontal centrifugal, you need an estimate of the height to the centerline of the pump and then the height from grade to the surface of the liquid inside the vessel. Let's say the bottom of the vessel (horizontal) is 10' above grade and minimum liquid height in the vessel is 3'. Let's say the distance to the centerline of the pump is also 3'. Maximum NPSHA = 10' + 3' - 3' = 10 ft. From this, you have to calculate and subtract the suction line losses. Let's say that is 1.2 ft of liquid. NPSHA is 8.8 ft. Compare this to the NPSHR, remember to leave a suitable margin.

 

[EmmanuelTop]

I would just add that vapor pressure of water at process temperature will likely be marginal, as well as the amount of dissolved gas.

If you intend to use the same pump for pumping both oil/condensate and water, make sure you design for the worst case NPSH (oil in equilibrium with gas) and size the motor for the heaviest fluid (water).

Dejan IVANOVIC
Process Engineer, MSChE

 

[alpha0125]

Hello TDK2,

So my vapor pressure and the system pressure for oil and water case is same. or only on the oil case and why?

 

[SNORGY]

The vapour pressure of the water will be Psat at the pumping temperature. Presumably, the vapour pressure of the condensate will be somewhat close to or perhaps a bit lower than the absolute pressure in the vessel, at least in the worst case. It depends on the nature of the condensate and whether or not the gas is entrained with the mixture or, in part, flashed from the condensate. Maybe take the HC vapour pressure as similar to that for an unstabilized condensate and assume it is somewhat above RVP = 13 PSIA?

Without an oil analysis, the condensate vapour pressure will be a guess.

 

[LittleInch]

For both situations - water and oil, I would assume without other evidence, that the effective vapour pressure, or dissolved gas pressure is equal to the separator pressure. hence the only way to achieve NPSH is to have the pump inlet X feet below the top liquid level. With a small allowance for friction this often means either a can pump or a deep pit next to the separator. Basically exactly what TD2K says.

I can't quite envisage what you're doing, but my guess is that your outlet nozzle is at a point 8 ft below top liquid level??? You then pump out the 5 ft of water followed by the 3 ft of condensate?? A sketch or drawing would be SOOOO useful here.

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

 

[alpha0125]

here are my attachment

 

[LittleInch]

The numbers on your sketch are different to the OP, but on the basis that you have one nozzle and are pumping out to the 5ft level, this should then be all in oil. Your NPSH is then 4 ft oil above the nozzle, plus 3 ft from nozzle to centre line of pump minus friction. The pressure on the top equals vapour pressure, so minimum NPSHa of around 6.5ft.

Why is oil saturated?, because that's what happens with methane plus there could be volatile compounds still in the"oil"

NPSH calculation is a normal, but pressure on liquid cancels out vapour pressure.

At atmospheric pressure the gas will evolve from the oil and volatile compounds boil off, so you normally end up with less liquid than you started with. Vapour pressure will stabilise.

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

 

[alpha0125]

Thank you all