Theoretical calculation does not match the practical gauge reading on pump suction side 1

[SalvationTsai]

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!

 

[Artisi]

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.)

 

[BigInch]

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.

 

[SalvationTsai]

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.

 

[BigInch]

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 [&gamma;] * [&Delta;]Elev 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.

 

[LittleInch]

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

 

[SalvationTsai]

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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[LittleInch]

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

 

[SalvationTsai]

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.

 

[25362]

An interesting discussion on NPSH appears in thread798-129328.

 

[SalvationTsai]

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

 

[LittleInch]

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

 

[QualityTime]

"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?....."

[ul]
[li]1 meter of head = 0.099 974 399 331 kilogram-force/square centimeter[/li]
[li]If you are at sea level the atmospheric pressure is 10.3m[/li]
[li]Suction side water level = 1.5m above the centreline of the pump[/li]
[li]Vapor pressure at 80 deg C = 4.97m[/li]
[li]Specific gravity of water at 80 deg C = 0.972[/li]
[li]Assume the pipe loss is 2m (is this correct????)[/li]
[/ul]
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:
[ul]
[li]Pressure gauge is not calibrated[/li]
[/ul][ul]
[li]Pressure guage reading is not corrected to pump centreline[/li]
[/ul][ul]
[li]The assumed 2m pipe loss is incorrectly calculated[/li]
[/ul]

 

[QualityTime]

I will add one more thing to my last post:

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

 

[LittleInch]

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

 

[25362]

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 = (P_s - P_v)/[&rho;] + v²/2g

Where:

P_s = absolute suction pressure measured at the pump inlet
P_v = vapor pressure of the liquid at the pumped temperature
[&rho;] = 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.

 

[QualityTime]

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

 

[SalvationTsai]

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?).

 

[LittleInch]

Salvation, your posts always make me smile . 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

 

[LittleInch]

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

 

[SalvationTsai]

Dear good LittleInch:

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

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.