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Converting head to presssure (psia or psig?)

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JBrou

Mechanical
Dec 16, 2014
7
First off this may be a stupid question but I cant seem to figure this out from any online sources. I'm pretty new (currently interning) but I feel like I have a pretty good grasp of NPSHa/NPSHr and how to calculate them until my mentor ran across this diaphragm pump that he says was incorrectly speced out and has an NSPHa problem.

The pump shows:
NPSHa = 10.3 ft
NPSHr = 9 psia

So my question is: When you convert from head to pressure how do you know if it is absolute or gauge pressure?

Obviously, if you look at a pump curve it would just be psi since it's a differential pressure, but what if you just want to know the pressure at the bottom of a tank with a water level say 10 ft high? I would initially think it would be 10'/2.31 = 4.33 psig but now I'm thinking twice about that assumption.

Any advice would be greatly appreciated!
 
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Unfortunately it is a matter of convention. Most NPSHr values are given in feet of liquid absolute. You can then choose whether to convert to gauge or absolute pressure.

Both suction head and discharge head SHOULD be expressed in feet liquid absolute.

The differential head is just that.

No pump curve should ever be submitted in pressure units. It is bad practice, will cause numerous errors, and I would reject any such vendor submittal.

Any pump head calculation has to take into account the pressure at the liquid surface at suction and discharge conditions.

Your assumption about the head calculation is correct. As long as your pressure in the tank is atmospheric.

If you keep the units in feet of liquid, there will be less cause for error.

Take a look at some other threads for NPSH calcs. Here is one source:

 
When doing NPSH calculations, it is far more logical and easier to understand if you use ABSOLUTE pressures. NPSHr is given in absolute units (of cool water), so it eases the understanding.

If the tank is not pressurized, then at the surface of the tank you have 15 psia, if you are at sea level. If the tank is closed and has a pressure of20 psig, then add 20 to 15 to get 35 psiA, the pressure at the surface.

We'll use the 15 psiA pressure at the surface.
At the nozzle 10 feet below, the pressure is 15 psia + 10ft x SG x 62.4pcf/144in2/ft2 = 19.3 psiA. If the tank has water, the specific gravity, SG = 1.0
If it's gasoline then SG = 0.54, etc.

NPSH at the inlet to tank discharge nozzle = 19.3 x 144 / 62.4 = 44.62 ft.

Remember to subtract nozzle discharge coefficients, fitting losses and pipe friction losses too for everything between there and the pump suction.



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BTW the 19.3 was using SG = 1, water. And assuming water's vapor pressure is 0. (it's not)

If it was gasoline then
you would have 17.3 PSIA at the nozzle.
Say gasoline's vapor pressure is 7.3 psiA

sSo subtract the vapor pressure = 17.3 - 7.3 and pressure at nozzle is = 10 psiA
NPSH = 10 * 144/62.4 = 23.07 ft

now subtract all the fitting and pipe losses to pump suction from that.





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I suppose NPSHr = 9 psia is what you got from the diaphragm pump supplier? It seems to be a reasonable value for a diaphragm pump although I'm not aware of any supplier who expresses NPSH in psi instead of m or ft. When somebody gives me a NPSH value in psi I always think this person does not know what NPSH means (and usually I'm right).
 
micalbrch,

Yes that value was received by the supplier. And I'm assuming an NPSHr value of 9 psia (~ -5.7 psig) means this pump can pump the specified fluid even under partial vacuum at the suction without cavitating? It would make sense to me because it's a positive displacement pump.
 
GHartmann,

So you're saying most NPSHr values on a pump curve are given in absolute feet of head. How can you tell? Is it just accepted to be absolute head? The reason I'm asking is that I don't think I've ever seen that specifically called out on a pump curve (it just says ft).

Also, that link you attached was the one source I could find some examples on also. But look at the example where they have a pump pumping from a tank under vacuum. Below are the givens, a link to a picture of the system, and their calculations.

Given:
•Gage pressure = - 20 inches of vacuum
•Atmospheic pressure = 14.7 psi
•Liquid level above pump centerline = 5 feet
•Piping = a total of 10 feet of 2 inch pipe plus one 90° long radius screwed elbow.
•Pumping = 100 gpm. 68°F fresh water with a specific gravity of one (1).
•Vapor pressure of 68°F water = 0.27 psia from the vapor chart.
•NPSHR (net positive suction head required) = 9 feet

System Layout:

Now for the calculations:
NPSHA = Atmospheric pressure(converted to head) + static head + surface pressure head - vapor pressure of your product - loss in the piping, valves and fittings
•Atmospheric pressure = 14.7 psi x 2.31/sg. =34 feet
•Static head = 5 feet
•Gage pessure pressure = 20 inches of vacuum converted to head ◦inches of mercury x 1.133 / specific gravity = feet of liquid
◦-20 x 1.133 /1 = -22.7 feet of pressure head absolute
•Vapor pressure of 68°F water = pressure x 2.31/sg. = 0.27 x 2.31/1 = 0.62 feet
•Looking at the friction charts: ◦100 gpm flowing through 2.5 inch pipe shows a loss of 17.4 feet or each 100 feet of pipe or 17.4/10 = 1.74 feet loss in the piping
◦The K factor for one 2 inch elbow is 0.4 x 1.42 = 0.6 feet
•Adding these two numbers together: (1.74 + 0.6) = a total of 2.34 feet friction loss in the pipe and fitting.
NPSHA (net positive suction head available) = 34 + 5 - 22.7 - 0.62 - 2.34 = 13.34 feet. This is enough to stop cavitation also.

So here are some questions I have for this...
[ul][li]Do they add atmospheric pressure converted to head in the NPSHA equation they list above to convert the static head to absolute head? The static head alone would be gauge head?[/li][/ul]
[ul]
[li]As for the pressure inside the tank. Can you even have a negative value for absolute head?[/li]
[/ul]
[ul]
[li]They throw in some piping loss in there, but it this absolute head? Taking the friction loss through the 2" elbow: 0.6 ft (assuming absolute head) = 0.6 ft*1/2.31 = 0.26 psia (correct?) = -14.4 psig of friction loss across the pipe? So it would gain pressure though the elbow...that doesn't even make sense.[/li]
[/ul]
 
bimr,

Thanks for the link but it says "The page cannot be found" when I open it.
 
BigInch,

That all makes absolute sense.

So what is the advantage of using absolute pressure? You say it eases the understanding, but it always seems easier to me to think in terms of gauge pressure. Maybe because like you and GHartmann stated, most NPSHr values are given in terms of absolute head. Then, it's just easier to go from absolute pressure to absolute head? Like I asked GHartmann how do you know for sure it is in abolsute head when you look at the pump curve, because I can't recall seeing absolute head called out on a pump curve. Or is it always given in absolute units?

I know I have little experience and a lot of questions. Thank you all for taking the time to give your advice!
 
"And I'm assuming an NPSHr value of 9 psia (~ -5.7 psig) means this pump can pump the specified fluid even under partial vacuum at the suction without cavitating?" Be careful! Do not mix up NPSH with pressure. That is a common mistake. It possibly means that that this pump can pump the specified fluid even under partial vacuum without cavitating but it does not have to mean that. It depends upon the fluid.
 
micalbrch,

So, here's my thought process on why I'm thinking that. Please correct me if I'm wrong anywhere.
[ul]
[li]NPSH is the head required to prevent cavitation right?[/li]
[li]With the fluid's specific gravity, that head can be converted to a pressure. This pressure is the pressure required to keep that liquid from cavitating at the pumps suction since NPSHr takes the fluid's vapor pressure in consideration. When you say don't mix up NPSH with pressure your talking about how two different fluids will give you a different pressure readings at the bottom of the same column of fluid correct? (just one example that comes to the top of my head from Flowserve's Cameron Hydraulic Data)[/li]
[li]If that pressure is < 0 psig, wouldn't that mean the pump's suction can ideally be under partial vacuum without cavitating?[/li]
[/ul]
If the NPSHr is converted to 9 psia (given on the data sheet), doesn't that mean you need to have over 9 psia (-5.7 psig) at the suction inlet in order to not cavitating? So it can operate at say 12 psia (-2.7 psig - partial vacuum) for example without cavitating? I'm sure you would want more than a 3 psi margin though, is there a standard for how much excess you would like to have?

Thanks for your comments
 
Head is used when discussing centrifugal pumps because it is the most suitable units based on how they work. Diaphragm pumps work differently and pressure units are actually more appropriate. Using head units implies that you are working with a centrifugal pump.
 
Head is neither absolute or gauge, basically because it is not a pressure at all and head is not referenced to some differential between that and atmospheric, gauge, nor to absolute pressure, zero. Head is a unit of energy. Head can be given an equivalent numerical value to pressure, but only because you can assume some fluid density and then convert to psi. Seriously, any rocket scientists wouldn't waste his time using gauge pressure designing a pump for rocket fuel, would they now. Hummm gauge pressure at launch does not equal gauge pressure at 244 miles high. What's the gauge pressure of a rocket fuel tank?

Pumps add energy to a fluid. How much? Equal to the differential head at the time the fluid is passing through it and it's added to each unit of mass passing along. What's the energy upon discharge? Well you take the energy at the suction flange at the time and add the pump's differential energy added ([&Delta;]H) Pump head is energy per unit mass. That's why the differential head that a pump can add to a fluid decreases with increasing flow. A fixed amount of available maximum power divided by ever increasing mass flow rate = less and less energy added to the fluid per unit mass. Curve is highest at shutoff flow and lowest at maximum flow rate.

Total energy = mass x height above a datum + 1/2 mv^2
If m = 1 unit mass, and you know and remember that, then for convenience we can express all this energy in terms of a linear measurement in feet, or meters alone. But you must remember that a pump adds energy to each and every unit of mass that passes through it. We normally neglect the kinetic energy part, the 1/2 mv^2 bit, because the velocity is so small we can usually get away with such a simplification.

Using the previous example with water, and realizing that the pump doesn't know, or care what's doing the pushing into it, it just needs to know how much push its getting, that's atmospheric "head" at the surface of the liquid + any additional energy it receives from gravity acting on the fluid's mass, m x g at the suction flange, which is 10 ft static head, so it's 19.3 x 144/62.4 = 44.54 ft x whatever the mass/unit volume of the fluid has in the column, and remember we will ignore the kinetic part caused by velocity in the suction pipe. (you can consider it if you like). That's the energy at the suction flange. To get the energy at the discharge, take the energy at the suction flange and add the pump differential head, say 200 ft, to get the total energy per unit volume of fluid exiting the pump, 244.54 ft. Some little bit of that will be kinetic, which we can calculate if you like from knowing the volumetric flow rate and the pipe diameter, we can figure the velocity and the kinetic component. Subtract that from the total energy and you have remaining the pump's discharge static component of energy. If you want to figure out the pump's discharge pressure, subtract the head equivalent of atmospheric pressure at the discharge, convert that head to an equivalent pressure to find out the discharge gauge pressure. If you want to know the discharge absolute pressure, don't, just use the 244.54 ft - the kinetic part.

Now, just for fun, pretend that it's a long pipe that pumps fuel from the top of Mt Everest to the International Space Station (ISS), don't use any atmospheric pressure at the pipe outlet, and figure out what discharge pressure is there. Be sure to consider the actual value of gravitational accelerations and atmospheric pressures at the top of Mt Everest and at the ISS. You can locate pumps anywhere you need one, or more, along the pipeline. How much pressure do you have at pump exit when gravitational acceleration is zero and there is no atmosphere. What is the exit velocity? Gauge pressure is equal to absolute pressure. When you can work that out, you will have your head fully wrapped around pumping problems.

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Wow...that Mcinally example is certainly a source of confusion

Subtracting the gauge vacuum pressure from atmospheric is just a way to arrive at the absolute pressure at the liquid surface. Just stay in absolute units.

Expressing NPSHr in terms of absolute head is just convention.

Pressure or head differential is neither gauge nor absolute, thus the term psi

The Hydraulic Institute gives recommendations on head margins for NPSHr





I will say it again: use your pressures and friction losses and convert them to feet of liquid absolute (only the suction and discharge surface pressures are in absolute units, the others are just pressure or head). Using absolute head has kept me out of trouble for more than 30 years and I am thankful for being taught that lesson.

Your Cameron pump book will be a good resource. If you can find an old Durco book, it is equally good.
 
Im sorry if my post from yesterday unnecessarily extended this thread. What I wanted to point out towards the OP is that a NPSH value < 10 m (30 ft) does not automatically mean that there is a partial vacuum in front of the pump. It is possible, but not a law. It all depends upon the fluid. And Compositepro is right: For PD pumps like diaphragm pump one uses pressure, not head but apart from that NPSH has a unit and it is meter or foott. psi is not a unit for NPSH and using psi for NPSH can lead to misunderstaning like NPSH = 9 psia = partial vacuum. I would always question when a vendor or client gives me a NPSH in psi.
 
So am I correct here...

If you have a liquid (SG = 1, VP = 0 psia, and no friction loss in the pipe for simplicity) column that is 2.31 feet high....

NPSHa at the bottom of that column is: NPSHa = 2.31 ft (liquid height) + 33.957 ft (atm pressure in ft) - 0 (line loss) - 0 (vapor pressure) = 36.267 ft??

But if you look at it pressure-wise, 2.31 ft will give you an additional 1 psi (differential pressure) at the bottom of the column compared to the pressure reading without the liquid?

I think I was so confused about this because I couldn't stop thinking of the two could basically be interchangeable equation wise.
 
Yes that is correct.

You have 14.696 + 1 = 15.696 psia

15.696 psia = 36.27 ft (absolute)

We are telling you, based on our experience, stop thinking of gauge pressures except to convert to absolute.

Utilize feet of liquid in calculating your pump hydraulic requirements

TRUST US [bigsmile]
 
G Hartman,

I adhere with most things you say, but as BI points out, the is no such thing as head absolute or guage, there is only head which is why NPSH is normally quoted in feet or metres.
It's often best to work in psia throughout which is what the flow assurance endings tend to do so you don't get confused.

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way
 
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