5 replies to this topic

[mojo4king]

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

Attached Files

•  npsh.jpg   105.58KB   36 downloads

[Art Montemayor]

Mojo:

Your main problem is that you are depending on an "equation", instead of concentrating on the problem using your engineering sense. The equation is useless unless you know what it is that it represents and why it is composed of various parameters.

You need to study the basic material first, before attempting to attack it with equations. Net Positive Suction Head Available (NPSHA) has a basic definition and that definition is represented by the correct equation employed to obtain the numerical value. The absolute liquid pressure available to a pump at its “eye” has to be positive. That means that the total head on the suction side of the pump must overcome all the resistances that it confronts before it gets to the “eye” – and that, by common sense, means it must overcome the resistance of any hose that is used as its suction line. It is that simple.

For a good explanation and derivation of the expressions used for Net Positive Suction Head, go to:

http://www.engineeri...head-d_634.html

[demank]

Yup2

Beware in using the equation.
Let me explain the basic equation:

NPSHa = hp – hvpa + hst – hf –hvh – ha

Where:
hp = absolute pressure head on the surface of liquid, m (of fluid)
hvpa = absolute vapour pressure of liquid at suction temperature, m (of fluid)
hst = static head due to liquid level, m (of fluid)
hf = friction head due to flowing friction, m (of fluid)
hvh = velocity head, m (of fluid)
ha = acceleration head, m (for centrifugal pump, ha = zero)

If the tank is atmospheric, hp - hvpa = [P(atm) - Pv] / [fluid density * grav].

Then you ask about the 10 m value, where should this valus be put in?
The answer is in the hf part (friction head in the suction line),

[mojo4king]

Hello denmank,

The Head loss in the suction line will be the frictional head loss which is the 1.2m value given in the question? The length of the line is surely irrevelant. The head loss through the suction hose (whatever length it might be) is what is needed?

Thanks,
Mojo

[demank]

Hello denmank,

The Head loss in the suction line will be the frictional head loss which is the 1.2m value given in the question? The length of the line is surely irrevelant. The head loss through the suction hose (whatever length it might be) is what is needed?

Thanks,
Mojo

Hello,,

Yes, of course.
If frictional head already stated, then 10 m long pipe no need to use in the equation.

[mojo4king]

Thanks Denmank,

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

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