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Theoretical calculation does not match the practical gauge reading on pump suction side

Theoretical calculation does not match the practical gauge reading on pump suction side

Theoretical calculation does not match the practical gauge reading on pump suction side

(OP)
Greetings every senior engineers!
I have a basic question on how to exactly calculate the pressure on the pump suction side, and then matches the practical pressure gauge reading on pump suction side?

First I am using the NPSHa equation.
1atm+H(static)-H(vapor)-H(suction pipe loss)=NPSHa <absolute value>

1atm: I convert it to approximately 10m
H(static): the water source surface level is above the pump centerline for 1.5m
H(vapor): the water is 80 Celsius, so the vapor pressure is 4.83m
H(suction pipe loss): the length is about 2m, carbon steel pipe, and the inner diameter is 36mm, in this part I would like to assume the pipe loss is 2m.

So sum up the above values, 10m+1.5m-4.83m-2m=4.67m < absolute value>
Then 4.67m-10m=-5.33m=-0.533kg/cm^2 <gauge value>
But in the practical gauge reading on the pump suction side is +0.2kg/cm^2.
Thus I don't know why my theoretical value does not matches the practical gauge reading? and what makes the difference between it?

Later I found another question, I found that there is no any contribution by the pump itself, in my assumption there ought to be a force created by the pump which helps the pump to suck in the water.
Thus I would like to ask is it the factor I missed which cause my theoretical calculation can't matches the practical gauge reading on the pump suction side?

And if it does, how could I calculate the sucking in pressure generated by pump?
My centrifugal pump's rated Q is 4.8kg/hr, rated H is 120, 3600 rpm.

Thank you for reading my lengthy description of my question, truly thank you!

RE: Theoretical calculation does not match the practical gauge reading on pump suction side

What is the elevation in your area?

RE: Theoretical calculation does not match the practical gauge reading on pump suction side

(OP)
Dear good 1gibson, the elevation of my location is at the sea surface, thank you for your fast reply!

RE: Theoretical calculation does not match the practical gauge reading on pump suction side

Do I get this correctly, you are calculating NPSHa, and then trying to compare with the reading on a pressure gauge at the suction of the pump?

You cannot read NPSHa on a gauge.

RE: Theoretical calculation does not match the practical gauge reading on pump suction side

(OP)
Dear good Tenpenny, I am sorry for using wrong way to calculate the pressure of the pump suction side, and could you give me some hint to calculate it correctly?
And I am not not understand why NPSHa is a wrong way to calculate it.
Thank you for your teaching!

RE: Theoretical calculation does not match the practical gauge reading on pump suction side

Salvation. What you calcualted (NPSHA) is the amount of head in absolute terms that the fluid can supply at the pump suction flange. The pump needs to have a certain amount of head in absolute terms (even though it is negative in atmospheric terms) - NPSHR. NPSHR must be > NPSHA in order for the pump to be able to "suck" the liquid out of the pipe without it vapourising. If you're reading 0.2 barg, then it looks to me like either the static pressure flowing pressure reading or maybe your 2m of friction losses are exagerated as the guage will only measure pressure of the liquid and will not measure the vapour proessure as a negative quantity. The fact that it may be vapourising does not affect it's actual pressure.

If you think about a pan of water which is cold, then heated up until it boils. The pan of water once it boils doesn't weigh any less than the cold pan (within a few percent and before it all boils off of coourse). Therefore the pressure at the bottom of your pan of water stays the same. However if you tried to suck up the water through a straw (don't try this..) you would get mostly steam. Not sure if that helps, but in essence you're not measuring the same thing as noted above.

My motto: Learn something new every day

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

RE: Theoretical calculation does not match the practical gauge reading on pump suction side

SalvationTsai

Hydraulic Lesson 1. Pumps do not suck.

Lesson 2. A fully primed impeller is capable of lowering the pressure in the impeller eye.

Lesson 3. Flow will only happen if the pressure at the pump inlet (NPSHa)is higher than NPSHr.

In your calculations;
2 metres of pipe will not result in 2 metres of head loss - your flowrate has been given as 4.8kg/hr - at this flowrate the friction loss would be near enough to zero.
A rough estimate for your NPSHa is around 6 metres - what does the pump require at the flowrate.

The gauge reading you are seeing is the 1.5 metre static head less a few losses between the tank and your gauge.

It is a capital mistake to theorise before one has data. Insensibly one begins to twist facts to suit theories, instead of theories to suit facts. (Sherlock Holmes - A Scandal in Bohemia.)

RE: Theoretical calculation does not match the practical gauge reading on pump suction side

One difference between your measurement and NPHSa is the vapor pressure of water. The point of the NPSHa calculation is prevent going vapor in the eye of the impeller. After taking into account static head and friction, is the pressure at the impeller eye above vapor pressure? To keep it simple, the NPSHa calculation subtracts vapor pressure - to be sure the result is above zero (in theory). However, for real pumps the frictional losses near the impeller are difficult, if not impossible, to calculate. Therefore, manufacturers publish the NPSHr at the pump inlet, and the NPSHr value captures the friction in the pump inlet. So the net result is that when NPSHa is greater than NPSHr, the pressure at the impeller eye is greater than the vapor pressure of the fluid. One could just as easily not subtract vapor pressure in the NPSHa calculation, and say that NPHSa(without VP subtracted) > (NPSHr + vapor pressure). Or we could say NPSHa(without VP subtracted) - NPSHr > Vapor Pressure. All say the same thing. But subtracting the vapor pressure is the convention in the NPHSa calculation, and we all stick to it. The vapor pressure is just the minimum pressure below which we can't drop, and have the pump operate properly.

Which is just a long way to say the vapor pressure shouldn't be subtracted when calculating the actual inlet pressure.

And I suspect the assumption of 2m of friction losses in the 2m line are also too high.

RE: Theoretical calculation does not match the practical gauge reading on pump suction side

NPSHR values are NOT published for the pump inlet, they are "published" at the impeller centerline; if you are on the edge, you must take this elevation distance between the gauge and impeller centerline into account.

How accurate is your gauge anyway?

repeating Artisi....one of the cardinal rules, written in stone, in pump physice.....PUMPS DO NOT SUCK. If you start considering they do, you will surely walk down wrong paths.

I'm not sure where to even start with the "don't consider vapor pressure"....hopefully I just read that wrong.

RE: Theoretical calculation does not match the practical gauge reading on pump suction side

(OP)
Dear good LittleInch:
Your description on “NPSHR must be > NPSHA”, in my humble opinion, isn’t it should be NPSHA > NPSHR in order to avoid the cavitation?
And I really thank you for your vivid explanation (cold and hot water in pan) on why the vapor pressure should not be added in when calculating the gauge pressure reading, I do figure it out!

Dear good Artisi:
Thank you for your lesson 2 and 3, I will keep it in mind, but I don’t know why the lesson 1 said that the pump do not suck? In my humble opinion I thought a pump sucks up the water(theoretically 10m at best), and deliver the fluid to a higher level.
Could you please correct my concept?

Dear good 77JQX:
Thank you for your clear description on the relationship of whether the vapor pressure is within or without on the equation of NPSHa and NPSHr.
But it is indeed a clear explanation on why the vapor pressure shouldn’t be subtracted when calculate the actual inlet pressure.

I started to sum up the above three good seniors teaching. If my water source surface level is 2m, and H(suction pipe loss)=0m, and then my theoretical pressure 0.2kg/cm^2 will matches the practical gauge reading on the suction side!

But what will happen if my water source surface level is under the pump centerline about 3m?
In my hypothetical question, friction loss is zero, fluid type is a hot water 80 Celcius.
And then my NPSHa“1atm-H(static)-H(vapor)-H(suction pipe loss)” would be=
10m-3m-4.83m-0m=2.17m
And the practical gauge reading on the pump suction side should be -0.3kg/cm^2.
Thus if my pump’s NPSHr is 1m(less then the 2.17), then my pump could suck up the water and do the normal operating;
And if my pump’s NPSHr is 5m(higher then the 2.17), then my pump could not suck up the water and won’t be able to do the normal operating.
Is my above concept right?

Thank you for your patient for reading my question, thank you again!

RE: Theoretical calculation does not match the practical gauge reading on pump suction side

(OP)
Dear good DubMac:
I really don't know the concept on the "PUMPS DO NOT SUCK", could you please hint me on this?
Many of my book I found in the library do describe on the pump, said the pump have suction side and deliver side, please I really wish to figure it out, and son't want to stray on the pumping engineering.

RE: Theoretical calculation does not match the practical gauge reading on pump suction side

Pump suction is a WRONG term it has and does and will always cause a lot of confusion, the correct term is pump inlet. whenever coming across this term you should put your pen through the word and write pump inlet over it.

It is a capital mistake to theorise before one has data. Insensibly one begins to twist facts to suit theories, instead of theories to suit facts. (Sherlock Holmes - A Scandal in Bohemia.)

RE: Theoretical calculation does not match the practical gauge reading on pump suction side

You're quite right - I got that the wrong way round - sorry.

There are issues with having water below the suction inlet of the pump, as a centrifugal pump won't work at all unless you have flooded suction. You can get special centrifugal pumps that will self prime, but lets not go there at the moment.

A better example for the one you note above is that the pump is much further away and the losses in the pipework when you flow equal 3m. Then yes, you are essentially correct. With an NPSHR of 1m your pump should pump, but with 5m, will cavitate wildly and not work properly.

Beware that the NPSHR measurement is a function of performence and cavitation, which will eat your impellor in a matter of days, can occur at a few meters ABOVE NPSHR. Therefore you should always have a margin of 2-3 m min above NPSHR to avoid cavitation. Ask the pump supplier for his onset of cavitation curve if you are operating close the NPSHR margin.

Oh and a bit less of the "senior" bit... bigsmile

My motto: Learn something new every day

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

RE: Theoretical calculation does not match the practical gauge reading on pump suction side

A centrifugal CAN WORK even if the "inlet" fluid level is below the pump itself; and it doesn't necessarily have to be a designed self-priming pump. This concept is somewhat central to our discussion here.

The fluid is not "sucked" into the inlet, it travels there because it is moving from from a higher pressure zone to a lower pressure zone. The "mystery force" allowing this to happen is atmospheric pressure.

The fluid is not "sucked" in, it is pushed in. Take that and chew on it for awhile.......


RE: Theoretical calculation does not match the practical gauge reading on pump suction side

or suck on it for awhile....

It is a capital mistake to theorise before one has data. Insensibly one begins to twist facts to suit theories, instead of theories to suit facts. (Sherlock Holmes - A Scandal in Bohemia.)

RE: Theoretical calculation does not match the practical gauge reading on pump suction side

"The fluid is not "sucked" into the inlet, it travels there because it is moving from a higher pressure zone to a lower pressure zone. The "mystery force" allowing this to happen is atmospheric pressure."

Out of curiosity, how does this differ from any other 'sucking' motion, whether a vacuum cleaner or a woman with a straw?

RE: Theoretical calculation does not match the practical gauge reading on pump suction side

It doesn't differ - same principal applies, pressure is lowered by the "sucking" and flow takes place because of the "mystery force" - atmospheric pressure.

It is a capital mistake to theorise before one has data. Insensibly one begins to twist facts to suit theories, instead of theories to suit facts. (Sherlock Holmes - A Scandal in Bohemia.)

RE: Theoretical calculation does not match the practical gauge reading on pump suction side

Yes, there are many times the words sucking and blowing are used rather interchangeably, but I guess what really matters is who or what is responsible for initiating the sucking or blowing action. Thats about as far as I want to go with this discussion.

RE: Theoretical calculation does not match the practical gauge reading on pump suction side

I fully agree a centrifugal can work with liquid level below inlet level, just that unless you do it right, you can have problems starting the pump if the inlet line has drained down too far. Also it does tend to make vaporisation / cavitation issues come to the fore.

I did use the word suck in inverted commas simply to illustrate to the OP what I meant as this is the most easliy understood way of expressing this concept, even though not strictly correct. I will try the "inlet" instead of suction line/header/flange where I can.

My motto: Learn something new every day

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

RE: Theoretical calculation does not match the practical gauge reading on pump suction side

So we have just determined that pumps do, indeed, 'suck', insofar as anything 'sucks', because they create a lower pressure which allows atmospheric pressure to push the liquid into the pump.

RE: Theoretical calculation does not match the practical gauge reading on pump suction side

Guess the meaning of suck needs to be understood before applying it to describe the action of lowering the pressure internal to a pump so that atmospheric pressure can force liquid into the impeller. "if you suck something, you hold it in your mouth and pull at it with the muscles in your cheeks and tongue."

It is a capital mistake to theorise before one has data. Insensibly one begins to twist facts to suit theories, instead of theories to suit facts. (Sherlock Holmes - A Scandal in Bohemia.)

RE: Theoretical calculation does not match the practical gauge reading on pump suction side

Besides the other stuff...

I believe you get the different reading from your pressure gauge, because it does not measure absolute pressure. Does your gauge show a negative or almost no reading? If so, it is graduated from 0 to X kPa G.

Independent events are seldomly independent.

RE: Theoretical calculation does not match the practical gauge reading on pump suction side

(OP)
Dear good Artisi:
Thank you for your kind correcting me that the “suction” is a wrong concept, and using the “inlet” would be a more appropriate word for it, I will keep it in mind, and try to forget the wrong “suck” idea.

Dear good LittleInch:
Thank you for your verifying my two above hypothetical questions are essentially correct, and further teaching me the NPSHa should be better 2~3m above the NPSHr in order to have better performance and avoiding the cavitation.
You are indeed a respectful person! Really thank you for your valuable teaching!

Dear good DubMac:
After carefully reading your explanation, I could now have the concept that the water is pushed from a higher pressured zone to a lower pressured, instead of using the easy confused term “suck”.
But I am curious on one thing, in the propeller’s eye, the lower the pressure, the better the performance?
Or the pressure could only drop to 0KG/cm^2(gauge reading) in the propeller’s eye at best?

Dear good TenPenny:
Thank you for your precious opinion and comment on my post, it do greatly helps me more clear on the idea of the “mysterious force”.

Dear good BigInch:
My gauge is a compound one, -0.1~0 is measured in cm-Hg, and above zero is 0~4, measured in KG/cm^2.
And later I think my above hypothetical gauge reading “ -0.3kg/cm^2” on the very beginning of the pump inlet side is wrong.
In the situation of which the water source surface level is under the pump centerline, the gauge reading on the very beginning of the pump inlet side should be “0 kg/cm^2”, am I right on this hypothetical assumption?

Sum up the above information, I have few questions confused me once more.
First is in the situation when water source surface level is 3m above the pump centerline, make the H(pipe line loss)=0m, and H(inlet water velocity)=0, then the gauge reading on the very beginning of the pump inlet side would be +0.3kg/cm^2, and the gauge reading on the very center of the impeller eye would be -1kg/cm^2(vacuum state), and then the water would be pushed by the “mysterious force”, from a higher pressure zone, +0.3kg/cm^2, moving to a lower pressure zone -1kg/cm^2(vacuum state), am I right on this concept?

And the second is in the situation when water source surface level is 3m under the pump centerline, make the H(pipe line loss)=0m, and H(inlet water velocity)=0, then the gauge reading on the very beginning of the pump inlet side would be +0.0kg/cm^2, and the gauge reading on the very center of the impeller eye would be -1kg/cm^2(vacuum state), and then the water would be pushed by the “mysterious force”, from a higher pressure zone, +0.0kg/cm^2, moving to a lower pressure zone -1kg/cm^2(vacuum state), am I right on this concept?

Still thank you for every good, hospital, generous seniors who teaches my so kindly, despite my question is so basic and dumb, really thank you.

Best Regards.

RE: Theoretical calculation does not match the practical gauge reading on pump suction side

Glad to know you have a vacuum capable gauge. Some people get puzzled by that.

As velocity slows, static pressure increases. Are you entering a larger suction pipe diameter and not accounting for velocity head effects? It is customary to ignore velocity heads, but they can make small differences.

Can you make and post a diagram of the suction piping please? Click on ".. or upload your file to ENGINEERING.com" line down there at the bottom of this page.

It's not necessarily atmospheric pressure doing the driving. Flow is driven by whatever is causing any pressure differential you may have between any two points in the pipe that tries to drive the flow of the fluid inbetween. If connected to a closed tank, then it's tank pressure on one side and γ * ΔElev on the other - flow friction on the other side. The water column will move as it tries to balance the two pressures with its own weight plus the frictional forces if it does wind up actually flowing. At least if the system is in a gravitational field. If the system is in outer space, then it's flowrate will change until the pressure differential is balanced by the friction produced by the flowrate alone.

Independent events are seldomly independent.

RE: Theoretical calculation does not match the practical gauge reading on pump suction side

Water level level with inlet flange, at rest inlet pressure should be 0 barg if the tank is open to the atmosphere.

Water level at +3m, at rest inlet pressure will be 0.3 barg if tank open to atmosphere.

Water level at -3m, starts to get a bit tricky. If the inlet pipe is empty and water level in the pipe is the same as the tank, then at rest, inlet pressure will be 0 barg. If inlet pipe is full and discharge valve is closed, or you have a good non return valve on the outlet of the pump, then pressure would be -0.3 barg, as the column of water in the inlet pipe would exert a force due to gravity.

If inlet pipe is empty and you turn on the pump, will the pump, which is now trying to pump air instead of water, be able to lower the air pressure in the inlet line enough to allow atmospheric pressure to push the water into the pump. I don't know as centrifugal pumps should have a flooded suction before they start. Sometimes people instal a manual, or powered air pump to lower the pressure, but in reality, sometimes the pump can move enough air to do it, sometimes not. There are special self priming centrifugal pumps, but even they can't self prime more than about 5 m.

Don't get too hung up on what is happening in the centre of the pump, but no you can't get lower than 0 bara, -1 barg. In reality you need pressures higher than that to avoid cavitation.

Glad you learnt something and your questions aren't dumb.

My motto: Learn something new every day

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

RE: Theoretical calculation does not match the practical gauge reading on pump suction side

(OP)
Dear good BigInch and LittleInch, I am sorry for my late reply, because I am trying to understand to concept which you kindly lectured on me. I had searched the library and internet for theory which could help me clear my confusion, and then I find Bernoulli’s principle is quite fit for my question, but I am still seeking any other principle which could better explain or lead me to better calculate the pressure.

To dear good BigInch:

I could now more understand the relationship between velocity and static pressure by applying the Bernoulli’s principle, but now I am still confused on how the velocity head would affect on the total head, many of the reference often ignores it, but I will try to better understand it.

I will now trying to draw my diagram of the suction piping, because I still need more data of the practical piping.

I had made a hypothetical situation, in order to realize your concept of “tries to balance the two pressures”, please have a look, and kindly correct me for any wrong thinking I have, really thank you!






To dear good LittleInch:
True-hearted thank you for verifying my hypothetical assumption, when water source surface level is above or under the pump centerline for 3m.

But there is still difficult for me to understand the pressure, when the water source surface level is under the pump centerline for 3m and pipe is filled with water, I am puzzled on how it exert a force that could cause -0.3barg on the pressure gauge, thus I had also draw a hypothetical situation which confused me, please have a look of it, and please hint me if possible, I am really eager to know it, indeed thank you!
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RE: Theoretical calculation does not match the practical gauge reading on pump suction side

Salvation,

I think you're getting there. Your tank sketch and working out look good to me, but I'll let BigInch discuss that bit.

For the pump, the pressure at the NRV would be 0.3 barg (positive above atmospheric), but you can ignore the Xm value as this will be balanced by the Xm water pressure from the tank. This is on the basis that the pressure at the pump inlet is 0 barg.

If the NRV is located where you have it, in reality the pressure when you shut the discharge valve may be > 0 barg, in which case any residual pressure would be added to your NRV pressure.

Normally there is an NRV on the pump discharge and not one on the bottom of the inlet line in which case your inlet guage would read -0.3 barg. Having said that, I have seen NRVs on the submerged inlet line also, especially for a water level before pump inlet level.

When you are running the inlet pressure will also be -0.3 barg plus any friction losses so that the atmospheric pressure can push the water into the pump. Hence for your high water temp / vapour pressure, you could end up below NPSHR or even draw a vacuum / turns into steam

My motto: Learn something new every day

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

RE: Theoretical calculation does not match the practical gauge reading on pump suction side

(OP)
Dear good LittleInch:

Very thank you for your prompt reply, and thank you for teaching me the relative knowledge of non-return valve, but I am wondering that does a typical non-return valve is usually installed between the pump outlet and discharge valve?

And would there be any good or bad effect if the non-return valve is installed at the starting of the entrance piping (like my drawing)?
My thinking is while the pump stopped, the entrance piping will be filled with water because the submerged non-return valve blocked it return to the water, and then when restarting the pump, will it be easier for the pump to make the water flow(because the water is much closer to the impeller’s eye)? Or harder to make it flow because of the existing water in the entrance piping would be the extra burden (like the weight)?

I start to have the assumption, on how the pump makes the flow. (and I am using the absolute value)

First is when the water source surface is under the pump centerline,
For below 1m, the pump impeller eye’s pressure would be 0.9kg/cm^2, in order to make the flow, from 1kg/cm^2 (the pressure of the water source) to 0.9kg/cm^2.

For below 3m, the pump impeller eye’s pressure would be 0.7kg/cm^2, in order to make the flow, from 1kg/cm^2 (the pressure of the water source) to 0.7kg/cm^2.

For below 5m, the pump impeller eye’s pressure would be 0.5kg/cm^2, in order to make the flow, from 1kg/cm^2 (the pressure of the water source) to 0.5kg/cm^2.
But if my liquid is hot water in 80 Celsius (vapor pressure is 0.473kg/cm^2), so in theory my pump will have the cavitation problem, am I right on this?

I am looking forward for your valuable teaching, and really thank you for every word you teach on me!

Best and warmest regard.

RE: Theoretical calculation does not match the practical gauge reading on pump suction side

(OP)
Dear good 25362:
Very thank you for the reference you provided, I will study on it.

RE: Theoretical calculation does not match the practical gauge reading on pump suction side

Typically yes, an NRV appears on the discharge side of the pump. Locating it at the start as you show means that the pump is exposed to the static presusre and / or additional pressure in the discharge line which can result in seal leakage and seal damage. Many simple pump seals only work well when the shaft is rotating so they tend not to like high static pressure and normally only see the inlet pressure. So if your discharge line has a high static presusre (more than 3-4 barg) then your location of an NRV on the inlet side may not be good for the pump seals when the pump is stopped, even if you isolate it later, this presusre will be locked in.

If the pump inlet is flooded then it will make the starting easier, so in this instance it would help, but an NRV on the discharge line should do the same job. The only issue is that the seals may not stop air coming into the pump and slowly draining the inlet line of water.

Forget the impellor eye and just concentrate on the inlet flange pressure, but yes, your assumptions are then correct.

go back to your first calcualtion where you had the NPSHA calcualted as
1atm+H(static)-H(vapor)-H(suction pipe loss)=NPSHa <absolute value>

Hence in m you have, with x as the depth below your pump inlet and ssuming only a very short suction pipe so losses are estimated at 0.1m
10 - x - 4.83 - 0.1 = 5.07 - x = NPSHA

Your pump NPSHR is (let us assume) 3m. Therefore your max suction lift before you start having serious problems is therefore 2.07m. If NPSHR is 5m then you can't have any suction lift. In reality, as I said above, your pump might start to cavitate above the NPSHR value and therefore even if the NPSHR is 3m the limit is really 4m and hence your "lift" is limited to about 1m.

My motto: Learn something new every day

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

RE: Theoretical calculation does not match the practical gauge reading on pump suction side

"1atm: I convert it to approximately 10m
H(static): the water source surface level is above the pump centerline for 1.5m
H(vapor): the water is 80 Celsius, so the vapor pressure is 4.83m
H(suction pipe loss): the length is about 2m, carbon steel pipe, and the inner diameter is 36mm, in this part I would like to assume the pipe loss is 2m.

So sum up the above values, 10m+1.5m-4.83m-2m=4.67m < absolute value>
Then 4.67m-10m=-5.33m=-0.533kg/cm^2 <gauge value>
But in the practical gauge reading on the pump suction side is +0.2kg/cm^2.
Thus I don't know why my theoretical value does not matches the practical gauge reading? and what makes the difference between it?....."

  • 1 meter of head = 0.099 974 399 331 kilogram-force/square centimeter
  • If you are at sea level the atmospheric pressure is 10.3m
  • Suction side water level = 1.5m above the centreline of the pump
  • Vapor pressure at 80 deg C = 4.97m
  • Specific gravity of water at 80 deg C = 0.972
  • Assume the pipe loss is 2m (is this correct????)
Suction Pressure = (10.3/0.972)+ 1.5 - 4.97 - 2 = 5.13m = 0.5 kg/cm2
Your pressure gauge read 0.2 kg/cm2

This must be a very small pump if the suction pipe diameter is 36 mm. The difference could be due to the following:
  • Pressure gauge is not calibrated
  • Pressure guage reading is not corrected to pump centreline
  • The assumed 2m pipe loss is incorrectly calculated

RE: Theoretical calculation does not match the practical gauge reading on pump suction side

I will add one more thing to my last post:

The pressure gauge is not tapped right at the pump suction flange

RE: Theoretical calculation does not match the practical gauge reading on pump suction side

Qualtytime,

Please read the posts above. The main reason for the discrepancy is that you can't measure the vapour pressure with a guage. It just forms part of a calculation to discover if the liquid will vapourise inside the pump / inlet pipe or not. We have already established that his 2m losses in the pipe are far too big.

My motto: Learn something new every day

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

RE: Theoretical calculation does not match the practical gauge reading on pump suction side

Gage reading measures only a mechanical stress (a pressure) on the liquid as said by 77JQX.

NPSH is a totally different concept, as explained by all thread participants, and is not read on the gage. It is calculated as:

NPSH = (Ps - Pv)/ρ + v2/2g

Where:

Ps = absolute suction pressure measured at the pump inlet
Pv = vapor pressure of the liquid at the pumped temperature
ρ = density of the liquid
v = velocity of the liquid at the section where measurement is taken
g = acceleration owing to gravity

As you can see, this equation gives NPSH in unis of liquid height.

RE: Theoretical calculation does not match the practical gauge reading on pump suction side

LittleInch you are right....not sure what I was thinking about last night when I did the calcs...done this calculation many times before....got mixed up between doing a NPSHa calculation and a pressure gauge reading calculation

•1 meter of head = 0.099 974 399 331 kilogram-force/square centimeter
•If you are at sea level the atmospheric pressure is 10.3m
•Suction side water level = 1.5m above the centreline of the pump
•Vapor pressure at 80 deg C = 4.97m
•Specific gravity of water at 80 deg C = 0.972
•Assume the pipe loss is 2m (is this correct????)

Suction Pressure = 1.5 - 2 = -0.5m = -0.05 kg/cm2
Your pressure gauge read 0.2 kg/cm2

RE: Theoretical calculation does not match the practical gauge reading on pump suction side

(OP)
Dear good LittleInch:

I am now further understand that the reason why we typically installed the NRV at the pump’s discharge side.
It is because that when pump stopped, and my discharged water surface level is higher than the pump centerline for more than 30~40m (as you kindly mentioned previously “more than 3-4 barg”), then the water will tend to flow back to the pump, and thus cause the seals (on both inlet side and discharged side?) leakage and damage, then installed the NRV at the discharged side would greatly solved this problem, am I correctly understanding your lecture?
But I didn’t well grasp the meaning on the sentence of your “even if you isolate[?] it later, this pressure will be locked[?] in.” and “The only issue is that the seals may not stop air[?] coming into the pump and slowly draining the inlet line of water.”, could you please hint me again on this?

And I really thank you for your calculation of NPSHa and pipe loss assumption; I do believe it should be a very small value of pipe loss, and I will do the calculation on NPSHa (using the equation you taught) again on my actual pump (it is a boiler feed water pump and the hot water source surface level is higher than the pump centerline for 1.5m~2m), and I am trying to find the NPSHr on the pump (or could I calculate it?).


Dear good QualtyTime and 25362:

I sincerely thank you for your calculation and good opinion on my post, and I will try to check the suggestion you kindly provided, but I now think the gauge should be fine, the problem ought to be the bad assumption of pipe loss I assumed, like the good LittleInch and QualtyTime said, the pipe loss should much small, the total length is no more 2m, and the flow rate 4.8kg/hr and the diameter is 36mm (but have flanges, elbows, will it make big pipe loss?).

RE: Theoretical calculation does not match the practical gauge reading on pump suction side

Salvation, your posts always make me smile afro2. Yes you understand correctly.

If you have a seal on the inlet, stop the pump, then isolate the pump on the discharge side, the pressure from the discharge will be locked in the pump.

Some seals don't work as well with negative pressure so the pump and inlet line might slowly empty due to air passing the seal in the reverse direction to normal sealing pressure.

Npshr is a vendor supplied figure from the pump curve as it changes with each pump and flowrate.

Good luck and glad we've been of help. LI.

My motto: Learn something new every day

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

RE: Theoretical calculation does not match the practical gauge reading on pump suction side

Second paragraph, "seal" should read "non return valve".

My motto: Learn something new every day

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

RE: Theoretical calculation does not match the practical gauge reading on pump suction side

(OP)
Dear good LittleInch:

It is a big relief to know that my questions makes you smile but not headache. :D

And I have draw a concept of your teaching, and still I am wondering my understanding is correct or not understand it properly.

Please have a look on it, and still I am very very thank you for your kind and generous of teaching on me.

Best and warmest regards.

RE: Theoretical calculation does not match the practical gauge reading on pump suction side

Salvation,

You're getting very close. Pump systems usually incorporate an nrv in the system to prevent back flow through the pump, which is a bad thing.

First diagram yes,.

Second one, also correct, but if the nrv doesn't leak (some seal very well), then the pump would sit at high pressure with possible damage to and leakage from the seals.

If the nrv was on the discharge side and no inlet nrv, then the water might drain out as air leaks into the pump via seals which are designed to keep liquid from getting out as opposed to air getting in. This might be very slow or not happen, but you always need to think about "what if" situations including drain down or loss of liquid level.

Having reread what I wrote I can see why you were a bit confused, hope the above helps.

LI

My motto: Learn something new every day

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

RE: Theoretical calculation does not match the practical gauge reading on pump suction side

(OP)
Dear good LittleInch:

Thank you for verifying my 2 diagrams, and I had re-drawn the second diagram(with the NRV at discharge side), please kindly see it once more, thank you very much!
But I don't quite know where & how the air comes from? I thought the pipe is full of water, or is it leaked (from the outside atmosphere) into the seal(NRV)?

I do sometimes confused, but it is due to my poor English and novice at pumping engineering, but I really feel fortunate that a lot of people are willing to teach me, especially dear good you. :D
Best and warmest regards.

RE: Theoretical calculation does not match the practical gauge reading on pump suction side

Salvation,

You're getting very very close now.

It's also me assuming you know things. Pumps normally have seals around the shaft to allow it to rotate while full of fluid. The seals are various types - mechnical seals, chevron seals, stuffing boxes etc - which are designed to stop liquid getting out, not air getting in.

IF you have a situation like your last drawing with an inlet pipe liquid level lower than the pump and no NRV on the inlet line, then the pump at rest will be at a lower pressure than atmospheric. Normally then someone isolates the pump on the discharge side. In that instance, a small amount of air can leak into the pump through the pump shaft seal and slowly the pump and inlet line will empty. I told you this just so that if it happened to you you could see why the pump might not re-start after some time because all the fluid had leaked out due to a small air leak in the pump, despite being isolated from the discharge line and no leaks in normal operation.

My motto: Learn something new every day

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

RE: Theoretical calculation does not match the practical gauge reading on pump suction side

(OP)
Dear good LittleInch:

Thank you for your clear explanation on the seals of the shaft, I used to assume the seal is on the NRV!
I am really happy and glad I could now finally clear on this "idea of a possibility of air leaks into the pump shaft's seal", then in the very end, water in the inlet pipe line will be fully emptied, and the pressure in the pump should be equal to the atmosphere!

Really thank you, and because of your kind teaching, I do learn something new today and the last few days, sincerely thank you!

Best and Warmth regards.

RE: Theoretical calculation does not match the practical gauge reading on pump suction side

A pleasure. It's sometimes not easy to explain things by posts and many times people don't respond, so your feedback and questions have been very good.

LI

My motto: Learn something new every day

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

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