Does hose length affect NPSHA calculation for centrifugal pump?

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
4 replies · 3K views
pray4mojo
Messages
4
Reaction score
0
At an oil refinery, a mobile centrifugal pump is used to pump warm crude oil from a storage tank. Liquid level in tank at elevation of 3 metres above pump. Suction side of pump is fed with a flexible hose that is 10 metres long.
Crude oil density: 846 kg/m3
Crude flow 4000 kg/h
Internal diameter of flexible hose: 50mm
Vapor Pressure of crude oil: 0.0654 bar (a)
Friction head in suction line: 1.2 m

NPSA Available = [P(atm) - Pv] / [density * grav] + hs - hf

Velocity Head: H(v) = u^2/2g
[where u = Q/pi*r^2]

I have plugged in my values into the two above equations to find out NPSHa and Velocity head.
However, the 10m value that is mentioned for the flexible hose i did not plug in.Does this 10m value need to be used in the solution or is it just for the purpose of a sketch?
My solution is attached, any help will be kindly appreciated.
Thanks,
Mojo
 

Attachments

  • npsh.jpg
    npsh.jpg
    12.1 KB · Views: 469
Physics news on Phys.org


Think about a term for the friction in the inlet piping...You already show the term for it.
 


So the Head loss in the suction line will be the frictional head loss which is the 1.2m value given in the question..

Also The formula is H(v) = u^2/2g
[where u = Q/pi*r^2]

Therefore when calculating Q, i must divide by 3600 and also multiply by radius squared and Pi and After converting Q to m3/s,

My final answer for the velocity head is 9 x 10-3
 


Your NPSHa looks good. I get .022 for the velocity head though.
 


Thanks Fred,

I reworked my solution to the velocity head and got an answer of 0.02281m also..

Converting my u value to m3/hr involved 4000/846 then multiplying by 3600 and the area