NPSH
NPSH
(OP)
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
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





RE: NPSH
Is this schoolwork?
rmw
RE: NPSH
So the Head loss in the suction line will be the frictional head loss which is the 1.2m value given in the question..
I am actually on a work placement and it is part of a project!
Hope all is well,
Mojo
RE: NPSH
Haven't checked all your calculations. Although you didn't mention it, it is assumed all data refers to the same (pumping) temperature, otherwise results change.
Since liquid level in the tank could vary and may go down to zero, its elevation over the pump center line is usually neglected.
The velocity head, however small, has been miscalculated, you used (1.1 kg/s)2 instead of using volume:
4000/(846×3600) = 1.313×10-3 m3/s. If the cross-section area is 1.963×10-3m2 then you"ll get a different result for u2/2g.
RE: NPSH
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
A much better figure!
Thanks