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Difference btwn NPSH and NPSHA 3

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tbarkerjr

Mechanical
Oct 31, 2008
20
I came across these two definitions for Net Positive Suction Head and Net Positive Suction Head Available:

NPSH = p(s)/y + v(s)^2/2g - p(v)/y

and

NPSHa = p(atm)/y - h(e) - h(l) - p(v)/y

where:

p(s) = static pressure in fluid
y = specific weight of fluid
v(s) = velocity of fluid
g = acceleration of gravity
p(v) = vapor pressure
h(e) = elevation from surface to pump
h(l) = head loss

Correct me if I am wrong, but these two formulas should be representative of the same quantity, theoretically. This is, of course, the vessel from which the fluid is being pumped is at atmospheric pressure. Could someone confirm or explain this?

p(s)/y + v(s)^2/2g = p(atm)/y - h(e) - h(l)

I appreciate all the help.



 
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Yeah, these were the formulas they outlined. However, there is no distinction between the two. Given that the tank is exposed to atmospheric pressure and fluid is being pumped therefrom, the two formulas should be equivalent. The static head plus the velocity head should be the same as the leftover head from the atmospheric minus the losses (since it is in suction); they represent the same energies. Correct me if I'm wrong. Engineering toolbox does not explicitly describe this. I just want to know that my reasoning is sound.
 
Forget NPSH; that's just a mathematical definition.

The only two that are important are NPSHa and NPSHr. You don't calculate NPSHr, you get it from the mfgr's pump data, so now there's only one you have to calculate, NPSHa, which must be >= NPSHr.

Use absolute pressures. (If tank is open or closed, it makes no difference to the equation, as long as you calculate the tank's absolute pressure correctly). Vapor pressure is always absolute.

NPSHa = p(on_fluid_surface)/y - p(v)/y + v(s)^2/2g + h(e) - h(l)

I've substituted absolute pressure on_fluid's_surface for p(atm).

The sign convention for h(e) is positive when the fluid surface is above the pump and negative when below the pump CL.

Dropping the velocity term is common as it's almost always relatively small, not worth calculating, and yields a conservative result.




**********************
"Pumping accounts for 20% of the world’s energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies)
 
Thanks for the help Biginch, that's what I figured. I understand the concept I just got thrown off by the apparent differentiation NPSH and NPSHa.
 
tbarkerjr, if done correctly the two equations you mention in your first post will give identical results.

BigInch, my understanding is that the velocity term is by definition not part of the equation you mention.

Cilliers
 
CJKruger, thanks for confirming that. I appreciate it.
 
By definition it is part of NPSHa, but it is often neglected (for good reason, especially since cavitation can often occur at values well above NPSHr).

**********************
"Pumping accounts for 20% of the world’s energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies)
 
Actually velocity head is included in NPSHr by virtue of the test method used to determine the pump's NPSHr curve, so it can obviously be included in NPSHa calculation too, if the engineer wants to consider it. During the test, the velocity head is a part of the total head measurement.

**********************
"Pumping accounts for 20% of the world’s energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies)
 
tbakerjr,
Although it might not be applicable for your application you should keep that in mind: If you have a reciprocating pump installed the acceleration head must be considered, too. Although acceleration is part of the "pump's problem", the acceleration head is part of the NPSHA calculation.
 
BigInch, just to make sure I am not confused: Do you agree that strictly speaking the velocity term should not be part of the equation you posted 12 Jan 10 13:39 ?
 
No I do NOT agree that it should be removed from my previously posted equation. That equation is correct to my understanding, which I can further reference by,

Velocity head is included in the definition of Total Suction Head, Total Discharge Head, Total Head, and by reference to Total Suction Head in the definition of NPSH, is also included in NPSH as well. This is found in the Pump Handbook - Second Edition by I.J. Karassik, W.C. Krutzsch, W.H. Fraser, J.P. Messina, Published by McGraw-Hill 1986, Section 13.3 in which they also say conforms to the same definitions and units of the Hydraulic Institute.

Velocity head is included in the definition of NPSH in "Centrifugal and Axial Flow Pumps", Second Edition, Section 2.6 by A.J. Stepanoff published by John Wiley & Sons 1957.

**********************
"Pumping accounts for 20% of the world’s energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies)
 
BigInch, I agree with the references you quote, but think what they say is different from the equation.

1) What references say:
NPSH = Total suction pres - Vap pres
= P_suction/y + v_suction^2/2g - P_sat/y

So you add the velocity per the references.

2) Usual method of doing calc:
P_suction/y + v_suction^2/2g + H_loss = P_vessel/y + H_Elev

Thus:
NPSH = P_vessel/y + H_Elev - H_loss - P_sat/y

Your equation can be correct if H(l) includes the recoverable velocity to pressure conversions (for example if entrance K=1.5 instead of 0.5), but this is not usually done.

Cilliers
 
Here's what I think they say. In other words the same as the equation given in ione's hyperlink to the toolbox.

For a point after all losses have been accounted for.

hs = ps/? + vs2/(2g)
where
hs = suction head close to the impeller
ps = static pressure in the fluid close to the impeller
? = specific weight of the fluid
vs = velocity of fluid
g = acceleration of gravity



**********************
"Pumping accounts for 20% of the world’s energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies)
 
BigInch,

It is not just your understanting, it's the way things are.

CJKruger,
Usually BigInch's understanding is the "right" understanding.
 
I always ignore velocity head for suction design purposes, but it is there in the definitions and the equation. For troubleshooting suction problems, it may be prudent to include it just to help to understand what is going on there. It can often explain an small reductions in pressure experienced when the suction line diameter or some other restriction, straightener, valve etc has changed the velocity nearby.

**********************
"Pumping accounts for 20% of the world’s energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies)
 
BigInch, I think the equation in your post dated 12 Jan 10 13:39 is incorrect. But lets agree to disagree.

ione, did you actually compare the equation posted by BigInch on 12 Jan 10 13:39 to the ones in the link posted by you ?

Cilliers
 
CJK,

You probably know by now that I never agree to disagree. If there is some error in my understandings, I prefer to get it correct. Please be clear in your comments, because at this point I have little idea of what you mean by all the questions and what your point eventually might be. Additionally if you could point out my supposed error using either my or ione's reference equations, rather than typing out new ones (esp. without supplying definitions of all variables), it would be of help.

Thanks

**********************
"Pumping accounts for 20% of the world’s energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies)
 
BigInch,

The equation posted by you is:
NPSHa = p(on_fluid_surface)/y - p(v)/y + v(s)^2/2g + h(e) - h(l)

I believe the correct equation is:
NPSHa = p(on_fluid_surface)/y - p(v)/y + h(e) - h(l)

Cilliers
 
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