NPSH for Separator Water Stream

[gregq8]

I am working on a project to replace the pumps that are used to pump a water stream leaving a three-phase separator. The three phase separator operates at 60 deg C and 111kPaa. I have assumed that the water-rich stream being pumped is at equilibrium with the oil and vapor streams leaving the separator and as such the water stream should be saturated with small amounts of hydrocarbons.

For determining the NPSH available, I would like to know what value should I use for the water-rich stream vapor pressure. Is it reasoable to assume pure water (i.e. Pvp ~ 20kPaa) or is it better to take the vapor pressure equal to the separator operating pressure (i.e. 111kPaa)?

Many thanks

Gregq8

 

[roker]

hello,

what is the source of the 3 phase stream?
did the existing pumps have operating problems?
can you recalculate the available NPSH of the existing pumps? at what level above the pumps suction the separator is located?

regards,
roker

 

[gregq8]

Hello Roker,

The feed to the separator is from several oil wells.

The existing pumps are positive displacement pumps that are fed using charge pumps. We are looking to replace these pumps with higher capacity centrifugal pumps.

The static head available is approx 4m of water whereas the NPSHR is 7.3m. If I assume that the vapor pressure of the water stream to be the surface pressure in the separator then there is insufficient NPSH available for the new pumps.

Since the amount of hydrocarbons dissolved in the water stream is small, it it reasonable to assume the vapor pressure to be that of pure water at the separator temperature?

Gregq8

 

[TrevorP]

You can estimate the vapour pressure of the stream assuming it is an ideal mixtures (Raoult's and Dalton's law). It won't be an ideal mixture, but it'll probably be close enough. Easier still, (and more accurate) would be to use a software package like Aspen, HYSIS etc., if you have this. I would guess it's going to be fairly close to pure water, but it depends on what you have in there.

 

[SeanB]

The conservative approach would be to assume that the vapor pressure is equal to the operating pressure in the drum. Since you are buying new pumps this may be a good way to go. If you are able to find a suitable pump selection using the more conservative method, you know that in reality the vapor pressure cannot not be any higher so you should never have an NPSH problem.

 

[roker]

hello again,

I agree to use the vapor pressure of water at pumping temperature, by the way a NPSHR of 7.3 m is quite high, did you try to work at lower rpm?

regards,
roker

 

[SeanB]

Your NPSHr is somewhat high - what is the rated flow for the pumps in question and like Roker suggested, have you considered a lower RPM pump (1800 vs 3600)?

 

[TrevorP]

The conservative approach would be to assume that the vapor pressure is equal to the operating pressure in the drum.

I agree that this is the conservative approach, but overly so. With a 4m head, gregg8 will require a NPSHR of 4m less frictional loss, i.e. the pumps will be inadequate. The NPSHA is (111 - ~20)/(sg * g)+ 4 - frictional loss, i.e. the pumps will undoubtedly be adequate.
The NPSHR does seem rather high, but it would depend on the pump head and flowrate (and hence pump speed) that is required.

 

[SeanB]

I agree in this case that by assuming that vp is equal to the operating pressure is probably going to cause difficulty in finding a pump. However, I still think that you want to use a conservative estimate of the vp in cases like this where you are not sure of the value (if you under estimate your pump is going to cavitate). I have in cases like this assumed that the vp was somewhere in the middle of the operating pressure and the vp of pure water. If you choose something like 55 kpaa, I allowed for 2 psi (sorry for mixing units but I am used to imperial units) and a 3 ft safety margin, you would still have ~26 ft available for your pump.

 

[MortenA]

I think that people here are beeing way too conservative. I assume that separator works OK (the water stream is much more water that "water rich") and that there is a low level switch that will switch off the pump - should the water level become too low.

The low low switch would most likely cut the flow alread when the water become more "emulsion" - so that pumping the emulsion wont happen.

I would assume the vapour pressure of water at 60 deg C!

Best regards

Morten

 

[DickRussell]

Since the water was in "equilibrium" with some vapor stream, it is theoretically at its bubble point, and its vapor pressure is that of the vapor above it. However, depending on what that vapor is, if it is just hydrocarbons and things like nitrogen, with no appeciable solubility in the water, then dropping the pressure of the water after its removal from the vessel can't result in real vaporization - it won't really "boil," so to speak, because there isn't enough volatile stuff to come out. You'd need a microscope to see any tiny bubbles. So it has a "effective" bubble point much higher than the 60 C. Now, if there is much dissolved stuff that can evolve, that would be different, but I wouldn't expect so.

 

[gregq8]

Thank you all for your feedback.

I found the following papers that describes how to address this issue:

1) Tsai, M.J. "Accounting for Dissolved Gases in Pump Design", Chemical Engineering, July 26, 1982, pp65-69

2) Chen C.C. et al, "Cope with Dissolved Gases in Pump Calculations", Chemical Engineering, October 01, 1993

3) Wood D.W. et al, "Pumping Liquids Loaded with Dissolved Gases", Chemical Engineering, July 01, 1998

The approach that is recommended is to use an 'effective vapor pressure' that lies between the process pressure and the the liquid vapor pressure. The 'effective vapor pressure' is defined as the pump-eye pressure at which the volume of flashed gas is 2.5% of the total volume.

At 60 deg C, the 'effective vapor pressure' calculated using the above approach is approx 42kPaa whereas the vapor pressure of water is ~20kPaa.

Best regards

Gregq8