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NPSHa Units
6

NPSHa Units

NPSHa Units

(OP)
When solving for NPSHa for a fluid other than pure water, do I need to adjust final answer to pure water?

Here's the problem.  I'm trying to calculate NPSHa for pump in saltwater.  The equation that I am using is:

NPSHa = 2.31(Patm-Pvp)/SG + h

Where,
Atmospheric Pressure, Patm = 14.7 psia
Vapor Pressure, Pvp = 0.95
Specific Gravity, SG = 1.02
Liquid Level above Pump Eye, h = 10 ft.

(The numbers Aren't that important, it's the principle that I'm stuck on)

I understand that when I input the SG into the first part of the equation I am accounting for the fact that I am analyzing saltwater.  Then I add the height of the saltwater above the pump eye.  So it seems to me the resulting NPSHa would be reflecting feet of saltwater head.  Now, if i intend on comparing this value to a NPSHr curve that was based on feet of pure-water head, wouldn't I need to multiply by the Specific Gravity?

-SB

      

RE: NPSHa Units

The use of the specific gravity term in your NPSHa equation put that calcuated value into "pure-water" terms so there isn't a need to do anything else relative to the NPSHr curve.

RE: NPSHa Units

(OP)
That's what I initially thought.  So I decided to solve the problem by just simply looking at it from a pressure point of view.  So find the pressure acting on the suction eye and then convert that pressure to head.  The answer comes out different.

Total pressure acting on suction = Patm + Pvp + P(due to static head)

Patm = 14.7 psia
Pvp = 0.95 psia
P(due to static head) = 10*1.02/2.31 = 4.42 psia

Total pressure acting on suction = 14.7 + 4.42 - 0.95 = 18.7 psia

Convert to foot-head: 18.7 x 2.31 = 41.97 ft

PREVIOUS METHOD:

NPSHa = 2.31 (14.7-0.95)/1.02 + 10 = 41.12 ft


RE: NPSHa Units

hmmm.... 18.7 x 2.31 = 43.197

Charlie
www.facsco.com

RE: NPSHa Units

3
I am afraid that there are two errors (not counting typo's). Firstly dtn6770 is wrong and the units in your equation are feet of pumped liquid and not of water, but he and FACS are correct in saying your numbers can be applied directly to the pump curve (see below). And secondly you have made an arithmetical error.

Your error is where you are converting the 18.17 psia to ft of liquid. (Note: you typed this as 18.7 psia.) If you multiply by 2.31 your answer would be in ft of water. To get the answer in ft of actual liquid you must include the SG. So the actual head is 18.17 x 2.31 / 1.02 = 41.15 ft. The difference to your other calc of 41.12 is just rounding error. The units of both calcs are ft of actual liquid.

Pump curves for centrifugal pumps are always expressed as head (i.e. feet) rather than pressure (i.e. psi) because that makes them applicable to a wider range of fluids. If the pump curve says you will get (say) 60 ft of head it means that with a fluid of SG=0.8 the differential pressure will be 60x0.8/2.31 = 20.8 psi, while with a fluid of SG=1.1 the diff press will be 60x1.1/2.31 = 28.6 psi. In other words, a given pump at a given speed will generate a fixed head (feet), irrespective of the fluid SG (within practical limits). Similarly, the NPSHr is expressed in ft of actual fluid, rather than as psi or ft of water.

Katmar Software
Engineering & Risk Analysis Software
http://katmarsoftware.com

RE: NPSHa Units

Looking back at your NPSHa equation...specific gravity doesn't influence atmospheric pressor or vapor pressor but it does impact the liquid column.

NPSHa = 2.31(Patm) - 2.31(Pvp) + h(SG)

also, in your post where you calculate the total pressure acting on the suction...don't subtract the 0.95 psia vapor pressure.  Vapor pressure doesn't act on anything.

There's a difference between available static head and NPSHa.  The availabel static head (Hst) is a component of NPSHa as is absolute pressure acting on the liquid surface (Ha), vapor pressure (Hvp), and suction line losses (Hfs).  For positive (flooded) suction...

NPSHa = Ha - Hvp + Hst - Hfs

Ha = 14.7 psia or 33.96 ft (not influenced by SG)
Hvp = 0.95 psia or 2.20 ft (not influenced by SG)
Hst = 10 ft => 10.2 ft (influenced by SG)
Hfs is ignored for this exercise

NPSHa = 33.96 - 2.2 + 10.2 = 41.96 ft

The static pressure at the suction will be the 14.7 psia atmospheric contribution plus the 10 ft adjusted for SG which would be 4.41 psi summing to 19.1 psia.  Vapor pressure doesn't come in to play.

RE: NPSHa Units

cbboards,
Some really good information above.  I think one point of confusion is your initial equation is a mixture of NPSH-r and NPSH-a terms.  

The only place I've ever seen vapor pressure included is NPSH-r.  When a pump vendor is developing an NPSH-r curve, they assume a fluid being pumped to determine if that particular fluid will approach its vapor pressure anywhere within the pump, and set the NPSH-r to include a safety factor to avoid that.  If you purchase a pump for sea water and then run hydrocarbon condensate through it, you will have a very different NPSH-r curve than the vendor sent you originally.

In calculating the NPSH-a, that is simply the accumulation of the various pressure sources upstream of the pump (i.e., atmospheric pressure, any extra applied pressure, and fluid column).  All of these numbers should be converted to ft of pure water to be able to match the NPSH-r curves apples to apples.  

Head is a pressure.  The units are "ft of pure water" or "psia".  It is unfortunate that the "pressure" units are expressed in "distance" units, but that is just one of those things you learn to live with and don't confuse in your mind.  Just like stating atmospheric pressure in inches of water or mm of Hg, either one is a pressure, but don't leave off the reference quantity.  You can't say that the head of 50 ft of pure water is the same as 50 ft of SG=1.02 seawater, it just isn't so.

David Simpson, PE
MuleShoe Engineering
www.muleshoe-eng.com
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

The harder I work, the luckier I seem

RE: NPSHa Units

Head is not pressure (apples). ie. the same head with different fluid sg yields different pressures.  Head is total energy and actually includes the v2/2g term, although it is usually neglected for NPSHa calculations.  NPSHA curves are stated in feet because the pump needs feet of head, not X psi.  A pump vendor does not care what fluid you pump (provided the chemistry is compatible with the pump materials).  No pump is bought only for seawater.  50 feet of head of sea water is the same as 50 feet of head of gasoline, although the NPSHa would be different due to the usually higher vapor pressure of gasoline vs seawater.

BigInchworm-born in the trenches.
http://virtualpipeline.spaces.msn.com

RE: NPSHa Units

BigInch,
I'm sorry, but head is pressure.  It is ft of pure water at a reference temperature.  If you have 20 ft of sea water then you have 20.4 ft of head.  It is truly unfortunate that the industry stupidly started using a distance term for a pressure term, but they did and now we live with it.  There is nothing magical about head, it is just a pressure (which is the total energy and includes a v^2/2g term).

Every pump I've ever purchased has started with a user data sheet and one of the first questions has always been expected fluid name and SG.  That experience is from buying 2-3 pumps a year for 15 years from a bunch of different vendors.

David

RE: NPSHa Units

Then please explain what fluid and SG you put on your data sheet when the PO is for a multiproduct jet, gasoline, diesel and crude pump and why, if I provide 22 feet of NPSHa on the suction side of each of those products I can meet the NPSHa criteria, but have widely different pressures.   The only pressures I use for NPSHA calcs are atmospheric or tank pressure head equivalents (if any) over the fluid in the tank added to suction head, and vapor pressure (head equivalent) subtracted from suction head.  That's from 30 years of buying pumps... most for multiple products.  All my pumps curves NPSHR and differential Discharge heads are given in feet of cold water, not sea water, gasoline or diesel.

BigInchworm-born in the trenches.
http://virtualpipeline.spaces.msn.com

RE: NPSHa Units

Actually I do believe that the sum of pressure and head can be expressed using mathematical equivalents of total energy, but due to the origins of each component of NPSHA for pumps (components of head, velocity and vapor pressure for instance), a given total from one system does not or may not necessarily represent or have the same physical meaning as the same total in another system.  Thus can you say the sum of X in one system is "equivalent" to same sum X in another system?

If both pressure and head had exactly the same physical significance, then suction pressures and differential pressures could be used equally as well to specify pump characteristics independently of the fluid being pumped.  Head is the only unit of measure that can do it.  Pressure units would equires more information to specify a pumping condition.

BigInchworm-born in the trenches.
http://virtualpipeline.spaces.msn.com

RE: NPSHa Units


A search of past threads with the key word NPSH would provide you with a plethora of information on the subject.

As for the SI units, if

p = pressure, given in N/m2
ρ= density, in kg/m3
g = acceleration of gravity, in m/s2

then,

p/ρ is mechanical energy per unit mass, in N.m/kg
p/ρg is pressure head, in m, similarly to v2/2g called velocity head.

since N = force, equates kg.m/s2

RE: NPSHa Units

Head vs Pressure
I have points of agreement and disagreement with both BigInch and zdas04. I have not thought of head as specifically including the V2/2g term. I have used the terms head and pressure virtually interchangeably - Examples (a) I talk of either friction head loss or friction pressure drop, (b) I use either static head or static pressure (c) but I must admit that I don't refer to velocity head as velocity pressure.

I would also take issue with the statement that head is ft of pure water. The term head comes from the relationship for the pressure at the base of a column of liquid i.e.
Pressure = density x gravity x column height

This is the same relationship which cbboards gave as
Pressure (psi) = HPSH (ft) x SG / 2.31
The value of 2.31 takes into account the acceleration of gravity plus the mixing of inches in psi with ft in head, plus it incorporates the horrific gc which corrects for the use of pounds force instead of poundals (which would be the correct unit of force in a consistent set of units). Does it sound like I am biased in favor of SI units???

So we call the column height the head, but strictly we should always give the density and gravity under which it was measured. We have all got a bit sloppy because 90% of the time we are talking of water (and 99.99% of the time we are on earth), but in my opinion it is equally correct to talk of head of gasoline or mercury. It's enough to get us all to develop a good head of steam! Historically, the best way to get an accurate measurement of pressure was to use a manometer, and so pressure has come to be measured in units of length. But it is important to understand how the practice arose.

Applicability of Head and Pressure to Centrifugal Pumps
In the above argument I have said that head and pressure are just two ways of measuring the same thing, but with pump curves we have to be a bit more specific in our usage.  I will leave it to the mechanical guys to explain why, but a centrifugal pump develops the same differential head, irrespective of the fluid pumped, when the head is measured in terms of the pumped fluid. So here it is much more useful to talk head than pressure, and a single curve can be used for a wide variety of fluids.

As zdas04 has pointed out, it is still essential to provide the SG (or density) on the pump spec sheet because although the pump will develop the same head for all fluids, it takes more power to do it for a high SG fluid than for a lower SG fluid. This is one weakness of the typical pump curves issued by manufacturers - the power requirements given are strictly applicable to water only.  But in a petrochem environment it is far more common to see an SG lower than water, so if you select the motor from the pump curve you have a bit of extra safety factor.

Does Fluid SG and Vapor Pressure affect NPSHr ?
Fortunately, the same convention of using head rather than pressure applies to the NPSHr as to the differential head, and curves for NPSHr are always given in units of length rather than pressure. The Hydraulic Institute quite categorically says "Specific gravity and vapor pressure do not change the HPSHr of a centrifugal pump." See http://www.pumps.org/content_detail.aspx?id=748    However, you will have to take these into account when you calculate the NPSHa.

If you really want to get picky in the calculation of NPSHr you could take the fluid's molecular weight into account. In volume terms, a high MW fluid generates less vapor from a given amount of liquid than water would, and so the bubbles have less effect on the pump capacity. But hey, we were talking sea water so let's not complicate the issue.


A wonderful resource for pump information is the McNally Institute.  The article on NPSH will tell you more than you ever wanted to know about the subject.  See http://www.mcnallyinstitute.com/11-html/11-12.html

Katmar Software
Engineering & Risk Analysis Software
http://katmarsoftware.com

RE: NPSHa Units

I missed the issues you have with BigInch. If I put 4 SGs on a spec sheet, I still get the pump and power curves back using SG = 1.  Was that it?  I agree bubbles can have a drastic effect on capacity, but temperature effects on vapor pressure is more significant to NPSHr-a than SG or MW.  SG I think is pretty much irrelevant when discussing NPSH, if you didn't have to convert applied (atmos or tank) pressure and vapor pressure to head.

Leaving the semantic issues aside, it is obvious that "head" is a far superior unit for characterizing pumps and system curves and additionally requires no inconvenient conversions when designing multiple product pumps and pipe systems such that a pump can get all products up to reach the given high point.  Very convenient.  Can you imagine the trouble and confusion it would cause, if pump curves were given in pressure units?

Power?  Multiproduct pumps usually have their power curve given in terms of a fluid with SG = 1.0 for which conversion for a power consumption by a paticular pumped product with a different SG is necessary and depends on what product is in the pump at any given time.  SG = 1 provides a convenient reference.  Since power = force x distance/time, a unit of head is more convenient for power too.

As viscosity also can affect pump curves, the curves are usually given for the viscosity of 1 cP for which conversions to a particular fluid's viscosity may be needed.

Some helpful quotes that I don't think have changed since the 32 years since published,

Quote:


"How to Design Piping for Pump-Discharge Conditions", "How to Size Piping for Pump-Discharge Conditions"
both by R. Kern, Hoffmann-La Roche Inc.
Chemical Engineering, May 1975.

Meaning of Net Psitive Suction Head

At the suction nozzle, velocity head is also a positive energy component.  However, this is not included in NPSH(r) diagrams for pumps made in the U.S. Consequently, in the calculations for NPSHa, velocity head does not have to be considered.

According to the Hydraulic Institude, "The NPSH requirements of centrifugal pumps are normally determined on the basis of handling water.  It is recognized that when pumping hydrocarbons, the NPSH to obtain satisfactory operation may be reduced for certain conditions."  However, it is customary to use manufacturers' data for the minimum required NPSH when handling hydrocarbons having specific gravities less than 1.  This means that (1) both available and required NPSH do not vary with the liquid's specific gravity, and (2) for a given centrifugal pump, one curve of rquired NPSH can be used with all normal specific gravities.

The effect of specific gravity can be explained as follows:  Looking at the analogy of NPSH as a force that pushes liquid into a pump, and assuming a pump that has and NPSH, of 8 ft at a given capacity, we can show that a head of 8 ft of water equals 3.45 psi, while a head of 8 ft of gasoline equals 2.60 psi.  It would appear that the force pushing liquid into the pump decreases with specific gravity.  However, the mass of liquid being accelerated has decreased by the same proportion.

In manufacturer's head capacity curves, the "Total Head" usually does not include velocity-head differences calculated between suction and discharge flanges.  Those diagrams that include velocity-head differences indicate "Total Dynamic Head" on the vertical scale.  In this case, velocity-head differences should be added to the total-head calculations.

"US pump manufacturers usually include velocity head differences between suction and discharge."

BigInchworm-born in the trenches.
http://virtualpipeline.spaces.msn.com

RE: NPSHa Units

BigInch, the only issue I had with your post was that you said that you felt that the V2/2g term was included in the concept of head.  On the other hand, I am happy to use terms like "friction head loss" and "static head" which clearly do not include V2/2g.

If you are talking specifically of the discharge head on a pump then V2/2g is part of the picture, even if it is zero at zero flow.  But I got the impression from your exchange with zdas04 that you were talking of head as a general concept, and in this case I would exclude V2/2g from the definition of head.

Kern wrote some great stuff - excellent balance of theory and practical application.  And 32 years ago was just the other day in my book.

regards
Harvey

Katmar Software
Engineering & Risk Analysis Software
http://katmarsoftware.com

RE: NPSHa Units

(OP)
Great discussion, and I appreciate all the attention you all have given this topic.  The points you bring up are the exact points I have been wrestling with and have sparked a bit of a controversy here in the office as well.  

The equation for NPSHa that I stated above (I apologize for the computational error which only serves to confuse things), which includes the specific gravity and vapor pressure terms, comes from the Hydraulic Institute American National Standard for Vertical Pump Tests.  And yes, I neglected the velocity head simply because the standard stated that in an open suction case the average velocity head is small enough to be neglected.

The 2.31 term simply comes from (144 in^2) / (62.4 lb/ft^3:(the density of fresh water)), that's why, when dealing with pure water the straight multiplication by 2.31 serves as an accurate conversion.  If the density of the liquid is different, say like seawater (63.2 lb/ft^3) you could either substitute (144 in^2) / (63.4 lb/ft^3)= 2.27 as your conversion factor or simply divide by the specific gravity of the liquid - either way you get the same result when converting from pressure to head.  

My way of thinking about the whole situation may be over-simplistic - that's the only reason I can come up with to explain why my way of thinking contradicts a published national standard created by people with far more intelligence than I.  I would just like to look at my pump sitting down there in the canal and say, ok, this pump needs to experience a certain minimum pressure at the suction eye, otherwise it will cavitate.  ACME pump company says that the pump needs X psi acting on the suction eye.  I know this because I have a required NPSH curve that came with the pump.  Assuming ACME tested my pump in freshwater conditions and created that curve, I can convert the curve to pressure.  If the NPSHr curve was based on a fluid other than fresh water, I guess they would have to tell me that.   Now, how high does the canal level need to be above my pump eye to create X psi?  Well, if the canal is a freshwater canal, the level will have to be higher than if the canal was saltwater.  

OK, this got me thinking...

I guess as the fluid enters the pump, the impeller naturally drops the pressure - this is why we worry about cavitation.  Now, if the fluid has a lower vapor pressure like saltwater, the pressure drop will have to be more for the fluid to vaporize and cause cavitation.  Ooh, so now I have I question...Is the pressure drop caused by the pump impeller constant regardless of what fluid it is pumping, or will different fluids experience different drops in pressure?

RE: NPSHa Units

I don't usually have this much difficulty making myself clear.  I think that my pedantic clinging to a definition that is different from the definition that the community uses is a waste of everyone's time.  

Just two parting questions--if a head gauge on a tank containing 0.8 SG oil reads 20 ft, what would a tape measure show?  Why does the inside of a head gauge have a Bourdon tube?

David

RE: NPSHa Units

The head loss as the impeller takes in fluid is due to the fluid being accelerated, ie. static head converted to v^2/2g velocity head.  Once the fluid is accelerated, static pressure is reduced, which when it approximately equals vapor pressure (or bubble point, if you will) begins cavitation.  Velocity head is not available to "resist cavitation", but note that it can be counted in the "Total" NPSHa, as any velocity entering the pump is a pre-acceleration that the pump would normally otherwise be required to provide, as in the case where entrance velocity is essentially absent such as as it would be when a pump is submerged in a static tank.

The head loss caused by acceleration into the volute would be due to the change in velocity experienced by the fluid being accelerated, and thus equal to differential_Velocity/2/g, (tangential velocity) as the fluid mounts the impeler, so since all fluids would be accelerated equally by the pump (at the same rpm), the head loss must essentially be constant for all fluids, however if that head loss were to be converted to an equivalent pressure loss, it would involve the SG term and gravity, so the pressure drop would vary for fluids of different SGs.

BigInchworm-born in the trenches.
http://virtualpipeline.spaces.msn.com

RE: NPSHa Units

My thoughts about the easiest and clearest way of thinking about the whole issue of head vs. pressure is to remember that a kinetic (centrifugal) pump produces head (and not pressure). Through the laws of nature, we may convert heads to pressures, but in the process of doing so, we must specify the fluid. The pump doesn't care.

Also, if you extend this line of thinking to positive displacement pumps, they produce capacity, and not pressure. A piston pump designed to pump 1 gpm will pass 1 gpm whether the pressure at its outlet flange is atmospheric or 1000 psig (assuming the pump can withstand this pressure).
Doug

RE: NPSHa Units

FYI
I believe on vedor data sheets, the pump vendor or architect enineer generally relates to NPSHA at the pump inlet flange.  
With regard to head.--  Terms used in fluid dynamics are Total Head, Total Dynamic Head etc.  Clearly when using the term Total it includes other than elevation head.

Regards

RE: NPSHa Units

"Centrifugal and Axial Flow Pumps", Stepanoff

Quote:

NPSHr is measured by a gage on suction, which is then translated to the standard reference point of the pump centerline, minus the head equivalent of the vapor pressure of the fluid at the NPSHr test temperature.

BigInchworm-born in the trenches.
http://virtualpipeline.spaces.msn.com

RE: NPSHa Units

BTW, total head including the term v^2/2/g wasn't my idea.  I robbed it from Bernoulli's, H = Z + ρ*g*h + v^2/2/g.

BigInchworm-born in the trenches.
http://virtualpipeline.spaces.msn.com

RE: NPSHa Units

(OP)
So bottom line this pump is going to put out a certain head.  And it will put out this same head regardless of the fluid.  If the fluid is saltwater, the pressure will just be greater than if the fluid is gasoline - but the head will be the same.

And on the flip side, if the pump requires 30 ft of head.  That means it needs 30 ft regardless of fluid, even though saltwater would produce higher pressures than gasoline.

RE: NPSHa Units

NPHSA is the potential energy of the fluid column at the entrance of the pump. There are two formulas: The first has a potential energy term, based on the conditions at the top of the fluid level.

The second formula has a kinetic energy term because the potential energy has been converted right at the last bit of enterance of the pump.

NPSHA stops at the pump enterance. Anything that goes on past this point is considered NPSHR, and is NOT the topic of the OP.

If in fact we are discussing NPSHR, then we are not ready to be discussing NPSHA, reasons obvious.

The only other formula for NPSHA includes acceleration head, and is reserved for Positive Displacement pumps, and again, is not the topic.

Head is directly converted to pressure.

Vapor pressure is always subtracted from NPSHA in the formula, because you can't use it as available head.

Charlie
www.facsco.com

RE: NPSHa Units

25362 (Chemical) 11 Jan 07 1:09
Just a clarification to others on where you were headed with SI units

Consider Bernoullis equation in the form
p/rho +  v^2/2 +  gz      units will be  m^2/s^2

N = force, equates kg.m/s2

units
p/rho= (N/m^2)/(kg/m^3)=m^2/s^2
V^2/2= m^2/s^2
gz= m^2/s^2    on earth  9.81m^2/s^2

Clearly dividing by g will give units in meters.

p/rho/g +  v^2/(2g) +z      units will be  meters


Regards

 

RE: NPSHa Units

BigInch
"Centrifugal and Axial Flow Pumps", Stepanoff


Quote:
NPSHr is measured by a gage on suction, which is then translated to the standard reference point of the pump centerline, minus the head equivalent of the vapor pressure of the fluid at the NPSHr test temperature.
BigInch-born in the trenches.
 
I hope Stepanoff has stated elsewhere in his text has indicated the following.-----
That gage on the pump suction flange should be a pitot-static tube so as to yield Total pressure  
(p/rho+V^2/2 or  p+rho*V^2/2  )
Clearly a regular pressure gage will not include the velocity head.

Regards


RE: NPSHa Units


Sailoday, you are right. I still think that Bernoulli's equation for frictionless flow of an incompressible fluid is meant to show us a balance of mechanical energy per unit mass of flowing fluid.

Thus I prefer to define P/rho as flow work per unit mass, N.m/kg, rather than m2/s2, which although dimensionally correct, it teaches us little.

In the same line of thought:

V2/2, kinetic energy per unit mass = mass×V2/(2×mass)

zg = potential energy per unit mass = mass×zg/mass

Any comment ?

RE: NPSHa Units

25362 (Chemical)
I agree that you can define KE and PE as you have.

Use of the terms should not be a problem as long as calculations and terms are well defined.

But with general notation, such as with TDH, etc p/rho would be in terms of ft or meters.

Regards

RE: NPSHa Units

sailoday,

Stepanoff has not stated any other qualifications and various standards exist.  Pump manufacturers have their choice as to what values they show in their curves, so they must also take care in labeling their curves accordingly.  It would normally be up to the application engineer to resolve any uncertainties.

BigInchworm-born in the trenches.
http://virtualpipeline.spaces.msn.com

RE: NPSHa Units

All:

For what it's worth, I'm with 25362 on the question of units.  If you think of head developed by a pump as work done by the pump per unit mass of fluid pumped, you will always be on theoretically sound ground on earth, on the moon, or even in space where's there's absolutely no gravity.  The same definition holds for compressors.

As others, including Katmar, have pointed out, the introduction of the dimensional constant gc mucks up the fundamentals pretty badly, especially in the English system of units.

I begin with two quotes from Wikipedia:
(1) "The value of g defined above is an arbitrary midrange value on the Earth, approximately equal to the sea level acceleration of free fall at a geodetic latitude of about 45.5°".
(2) "The units of acceleration due to gravity, meters per second squared, are interchangeable with newtons per kilogram. The quantity, 9.80665, stays the same. These alternate units may be more helpful when considering problems involving pressure due to gravity, or weight."

Once you grasp the idea that head is simply a way of defining pumping work per unit mass of fluid pumped, all the units confusion goes away.  Strictly speaking, from an engineering units point of view, when speaking of head:
(a) 1 m.kgf/kg is NOT the same as 1 m, and
(b) 1 ft.lbf/lb is NOT the same as 1 ft.

The numerical equivalence in (a) and (b) above turns out to only be approximately true at most places on earth since, on top of a high mountain or deep in a mineshaft, gc varies from the "standard" value.

For these reasons, in my professional work, I have quit even mentioning pump head in units of length.

All this may sound very pedantic, but I have seen serious engineering blunders as a result of such units confusion.

RE: NPSHa Units

UmeshMathur, you are correct that the value of the acceleration due to gravity will affect the conversion between head and pressure but I feel that you have not made the difference between g (the acceleration of gravity) and gc (the proportionality constant) sufficiently clear.

First of all we must note that gc does not exist in the British Engineering System or in the SI system of units (or indeed in any other consistent set of units).  I have heard people claim that it has a value of 1.00 in these systems of units, and obviously if you multiply anything by 1.00 you will still get the same answer but to put gc into these systems is to misunderstand what it is for.

It is only in the US Customary system of units that gc exists.  In the Foot-Pound-Second (FPS) system of units, which is a consistent set of units, the unit of mass is the pound and the unit of force is the poundal.  In the British Engineering System (BES) of units the unit of mass is the slug and the unit of force is the pound force.  In the US Customary units the unit of mass is the pound (taken from the FPS system), but instead of using poundals as the consistent unit for force they chose to use pounds force (from the BES units).  In order to overcome this inconsistency they had to include the gc proportionality constant, which is just the conversion factor from poundals to pounds force.

Unfortunately the conversion factor between poundals and pounds force is 32.2, which just happens to be the generally accepted value for the acceleration of the earth's gravity in ft/s2. And to make matters worse the symbols g and gc are very similar.  You could not design a more efficient recipe for confusion.

As you have said, g (the acceleration of gravity) varies from place to place, but I want to make it clear that gc never varies - it is a fixed value of 32.2.

This may sound even more pedantic than your post, but it is absolutely relevant here in the discussion of head vs pressure. As I noted in an earlier post, the pressure at the base of a column of liquid is calculated from
   Pressure = density x gravity x column height
In this equation "gravity" is the acceleration due to gravity and would normally have the value 32.2 ft/s2.  But because density includes the units "pounds" for the mass element while pressure includes "pounds force" for the force element the proportionality constant gc is required as well.

In cbboards post of 11 Jan 10:08 he showed how the constant 2.31 in his equation was calculated as 144/62.4.  It should be calculated as (144/62.4) x (32.2/32.2) which of course gives exactly the same answer, but explains why cbboards got the correct answer even though he ignored the gravity term.

Harvey

Katmar Software
Engineering & Risk Analysis Software
http://katmarsoftware.com

RE: NPSHa Units


One reasonable physical application for using length dimensions would be, for example, when draining open tanks to the atmosphere with
 
ΔP/ρg = 0

P = pressure, ρ = density, g = acceleration of gravity.

Still invoking Bernoulli (no friction) when equating velocity head to potential head h:
 
h = v2hole/2g ,

the operation of draining a homogeneous liquid from a hole near the base of the tank would show that the outflow speed is

vhole = √2gh.

This is the same result one would obtain by dropping an object through the distance h -and for the same reason: conservation of energy.

Just as the speed of a falling object is independent of its mass, so is the speed of the liquid independent of its density.

RE: NPSHa Units

Katmar:

Agreed on all points.  Unfortunately, we as a profession seem not to have bothered to clear up confusion in engineering units in our working documents.  You are quite right to mention the difference between g and gc.  By the way, I have never seen mass in units of slugs, except in the aerospace industry.

25362 is also on target concerning use of length units, especially when considering NPSHa - which was the original question posed by cbboards.

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