Help with NPSHA Calc - my numbers need to be checked

[Rvanpelt]

I am attempting to calculate the NPSHA for a centrifugal pump in a system I am designing. I have checked over my calculations and come up with a number that looks to be technically correct but does not seem intuitively correct. I want to get a second opinion to see if I did something wrong.

Inputs:

Liquid is water at 80F and sea level

Flow is 4.2 l/min

Pump inlet is 1” ID

Tank level height above the pump inlet varies from 3 ½ feet to ½ foot

The run from the tank to the pump is in 2 sections:

First section is 15 feet of 1” ID Buna hose

Second section is 2 feet of 1” 316 SS pipe with (1) 90 degree elbow and (1) ball valve which is fully open when the system is running

The numbers I came up with are Average Re = 4100 and NPSHA = 36 feet.

The one area I am unsure of is the supply tank. The tank is a 30 gallon cylinder lying on its side. It is filled through an opening at the top which is closed with a screw-on lid. The lid is not airtight so I included atmospheric pressure in the above calculation. If I use 0 feet for atmospheric head then the NPSHA comes out 2.29 feet. So a more intuitive answer would be something less than 34 feet of atmospheric head (34 feet atmospheric = 36 feet NPSHA).

 

[katmar]

Atm pressure is usually taken as 32.2 ft rather than 34 ft (unless you are working at the bottom of a mine). The screw on lid worries me. If air cannot get in as fast as the water is pumped out you will not have full atm pressure on the water surface.

You should use your minimum level of 0.5 ft for the tank level.

At these flowrates in a 1" pipe your pressure drop will be negligible.

The vapor pressure of water at 80F is about 1.2 ft and needs to be subtracted from the NPSHa.

On this basis I get a worst case NPSHa of 31.5 ft (=32.2+0.5-1.2)

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[gr2vessels]

I would also substract some 0.5 ft for those line losses and outlet nozzle losses, perhaps consider a vortex breaker if your actual level could go down to 6".
cheers,
gr2vessels

 

[Rvanpelt]

Thank you for the replies. I double checked and the cap for the tank is vented so the air should be entering at the same rate the liquid is evacuating, especially at these low flow rates.

 

[TD2K]

NPSHA is pressure in the vessel minus vapor pressure plus head minus line losses.

If you are at sea level, you are taking a maximum head of 34 feet. Subtact the vapor pressure of water at 80F which is 0.5 psia or 1.16 feet. Depending on line losses (which I haven't calculated) and your liquid head (0.5 to 3.5 feet), 36 feet NPSHA is not impossible

 

[25362]

The only comment I would think of refers to dissolved air. If the liquid pressure at the pump's eye is lower than that at the source, dissolved air may desorb and create problems when the volume of flashed gas is greater than, say, 3%.

There was an interesting article in the Chemical Engineering issue of July 26, 1982, titled Accounting for dissolved gases in pump design worth reading.

 

[Rvanpelt]

I was given another recommendation of a simple test to determine if there will be sufficient liquid supplied to the pump inlet that seems to work if the NPSHA or NPSHR are not easily obtainable.

I set up the system exactly as designed except that the pump was not connected. I simply opened the valve and let the liquid flow out the pipe that would be connected to the pump inlet and measured the flow rate over the liquid height range. I measured flow rates from 8 gal/min at 16 inches above the pump inlet to 5 gal/min at 10 inches above the pump inlet. My application is requiring a maximum pump output of 1.1 gal/min so this seems to verify that the system is more than sufficient.

The pump curve indicates a NPSHR of 10 feet at the low end of the flow rate range (which is still a factor of 2 higher than my application) so the difference between my NPSHA and the NPSHR correlates the simple flow test so I think I am comfortable with the set-up.

Thanks for the responses!