Hi Pat, Great video. Just wanted to ask about how increasing flow rate increases the suction pressure. I’d have thought that increasing the flow would reduce suction pressure because the pressure loss in the suction pipe work would increase thus reducing the suction pressure? Or is the fact that you increase the suction throttle valve opening which reduces the pressure drop across this valve which therefore increases the suction pressure? Or is it a combination of the two effects?
Thanks Sean! You're spot on on all front. By opening a valve on the suction I AM increasing the pressure drop in the suction line (which should drop the suction pressure); however, the pressure drop across the valve is lower because I've opened it, so the net effect is an increase in suction pressure. Had I not had a valve on the suction and only used the one on the discharge then opening that discharge valve would have caused the suction pressure to drop, as you correctly note.
@@ProcesswithPat thanks for your reply Pat - that makes total sense! Just as a follow up - can you control flow in a system using a valve at any point in the system? For example at the most upstream point in a system (say at the inlet to a pump) or at the most downstream point of a system (say if the pump led into a long length of pipe work which had a control valve near the outlet). Both valve locations could theoretically be used to control downstream/upstream flow respectively? (Hopefully that makes some sense haha!)
What I like about your videos is that you not only explain things theoretically using equations and charts, but you validate it by practically doing it, that's the way to go, that the way to teach, keep up the good job.
Good video, three observations: a) You correctly point out that your suction lines are uhmmm, sub-optimal. In general suction lines are designed quite large with low dP characteristics (minimal bends and fittings) so that changes in flow do not really affect the pressure at the pump suction. b) Increased suction pressure does not necessarily lead to higher flow. In your case this is true because your throttling the suction valve. However, when filling up a suction drum of say 1m to 2m head, and still controlling the flow on the outlet of the pump as is typical in most setups, the suction pressure goes up -> flow stays the same -> discharge head goes up correspondingly by 1m. Normally flows from centrifugal pumps are controlled on the discharge. c) Many, especially bigger pumps, have a long flat portion of the pump curve at low flowrates. I.e. in this range there is basically no difference in the differential head when changing the flow.
Thanks Daniel! Nice to see someone is paying attention! a) No excuses there, other than I used what I had. I have said my 50 Hail Perrys... b) True. The message was meant to be if all I did was increase suction (and not anything else) what would happen? The thing is the idea of "only changing one thing" can be a little misleading because other things do change, and you one would need to specify what else you'd have to do to keep other variables constant (e.g. throttle discharge to keep constant flow as you put it). Controllers add another layer. I was actually contemplating a pump that is driven by a steam turbine - at least a motor keeps a constant speed, a turbine-driven pump speed may speed up or slow down depending on which controllers you've put in auto/manual... Messy stuff to try and generalise! c) This is also true, but I wanted to stick to the general principle. You're right that this effect would not be as appreciable with flat pump curves, and I actually experienced something similar when I throttled the hose too much and created a very steep system curve. Then the differential pressure stayed relatively constant because flow could not change much.
@@ProcesswithPat Didn't mean to be critical, just some observations. I totally get the purpose of your videos - understanding engineering principles in a day to day setting. Young engineers lack this understanding. I really enjoy your videos - Perry, Liebermann and McCabe would be proud!
Thank you for your videos!! They are very helpful for me in understanding the fluid dynamics behind the design. I'm in the SCADA business and very familiar with the electronic side using PIT's, FIT's and VFD's. Enjoy the fundamental hydraulic analysis.
Okay, great. But what will happen to the shut-off head in case you increase the suction pressure? For example, if there is a pump with a suction pressure of 0.2 bar and a discharge pressure of 9.5 bar, and the shut-off head is 10 bar, what will happen to the pump shut-off head when you increase the suction pressure to 1.5 bar? Will it increase or remain constant?
It seems like it depends where you are along the curve. If you're at the top beginning of the curve it's going to be close to a 1:1 relationship. Toward the top end of the pump's flow rate it's going to result in a near zero increase.
Great video.. So here is a question. Using variable speed centrifugal pumps for swimming pools. When the pump is above the pool level (suction lift no gravity feed) sometimes pumps loose prime. Will installing smaller diameter pipes increase pressure to the pump head preventing loss of prime. So 50mm pipes to pump head 63mm discharge?
If I have LPG Tank as supply reservation and three multistage centrifugal pumps installed on common manifold together as a one or two working together to provide me with Q more if run one pump, put when I operate 3 pumps the flow decreased to a very low level to stopped completely. Can you interpret or explain it why? Thank you
Hello Pat! It is really an interesting video and got some great insights. I wonder if this question really makes some sense but please do let me know. As relief design engineers we often look if vessels are being overpressurized with liquid for our relief valve design. For a given impeller size we take a look at the total dynamic head aka differential head provided by the pump at zero flow that is dead head pressure. But if we find that the dead head pressure is less than MAWP of the vessel downstream*110%, we move on to add the level in the vessel upstream of pump converting into pressure. Is it the right thing to do? Can we compare the MAWP of the vessel downstream by adding up the dead head pressure and the level within the vessel upstream of pump?
Apologies for the late reply, but I'm going to approach this topic with extreme caution, because I do not want to be giving design advice with real-life project implications. So take all disclaimers as given and always consult a non-internet based expert! ;) I presume that if you are in the industry, then you are familiar with API 2000, 520 & 521 for venting & pressure relief. A relief valve on top of a vessel is typically relieving a vapour (either air, an inert gas blanket, or the vapour phase of a volatile substance). You need to differentiate between normal inbreathing and outbreathing, and emergency pressure/vacuum relief. When you are pumping a liquid into a vessel, the maximum LIQUID flow rate of the pump will tell you about the necessary relief VAPOUR flow rate, because the liquid is displacing the vapour at round about the same rate. But you are not actually relieving liquid but rather a vapour. This is NORMAL OUTBREATHING, and your safety device should not be trying to relieve this continuously. Rather, there should be another vent/pressure control that handles this. The PSV is there in case of failure of the regular breathing system. But that talks about relief flow rate. Relief pressure is determined by the design conditions of the vessel, which is also a degree of freedom, and this is the gist of your question. Based on your comment, it sounds to me as if you are trying to sum all possible pressure sources to consider in the relief pressure of the vessel, but it sounds a little backwards. The design pressure and MAWP should be based on this; you sound like you're starting with a given MAWP and working backwards. I understand how this can happen in the project, and it isn't necessarily wrong/bad. Also, if you are working on brown-fields, modifications, or revamps, you cannot always change the design pressure you've been given. The point is, you should not reach a point where you have zero vapour in your system because it it totally liquid-filled and you are pissing liquid out of your PSV into your vent system. To prevent such a scenario you need overfill protection in the form of adequately SIL-rated high level switches. What you are describing is only ONE relief case for which the PSV is sized. In reality, there are other cases which may be more severe. A possible example is the "fire case" - where you assume that there is a fire in the vicinty of the tank, and the radiant heat and boils the contents of the tank for which you are sizing your relief valve. It is highly dependant on application - pumping water, solvents, or petroleum products need to be assessed individually. It get's complicated and what I am learning is that there is no "correct" answer, because at the end of the day you are not getting away from doing a risk assessment. At the end of the day YOU need to determine what is acceptable risk, and you will not find a silver bullet in this regard. It's a tough thing to accept, but the onus is evetually on the designer and the user to determine what acceptable risk means.
@@ProcesswithPat Thank you Pat for detailing. I was specifically considering overpressure with liquid case in the question mentioned above assuming it to be the only governing relief case and a pressure vessel in specific and not a low pressure storage tank to be precise. We do take total dead head pressure which is the sum of shut off head and suction pressure as a criteria to determine overpressure with liquid scenario.
Try think about it in reverse… can you think of a situation where you increase the pressure at the inlet to something, and the flow rate through it decreases? That’d be counterintuitive, no? Pressure is a form of energy (think Bernoulli)
Hi Pat, if I am opening the suction valve.. this means that system curve's gradient is reduced meaning that it intersects the pump curve at a lower head. Therefore, the pressure that the pump discharges is reduced. Does my understanding make sense? I suppose I'm asking if the suction valve impacts the system curve. Thanks!
I know this is a dumb question, but I'm having an argument at work with someone who allegedly knows more than me…. If a pump is not running (VFD@0hz) it is impossible for discharge on a centrifugal pump to be higher than suction pressure, right? That's indicative of the gauges being gimp, right?
(Eliminating the possible factor of head pressure, the cooling tower supplying the suction is 14 floors up and is discharging to feed a chiller's condenser on the same level as the pump)
Why preasure of discharge is greater than suction side ??? because of the diameter ,the diameter of suction greater than discharge diameter i cant underestand that but i know when diameter increase ,preasure increase and velosity decrease 😢
@@ProcesswithPat ….Double Suction Horizontal Centrifugal Pump rated for 6500 m3/hr, 32” Suction line first 6 meters then 32” to 50” Eccentric Reducer to 50” long radius elbow inside a water basin, for a total suction pipe length of 8 meters before elbow. Centreline of Pump Suction is 3.3 m above bottom of water basin which 5.5 m tall filled with water. Note eye of impeller is 750 mm above centreline of Suction piping.
Thanks man, I'm honestly shocked you haven't lost patience with me, I feel like I'm wasting your time with my inability to understand basics. Anyway, so pressure (whether it be suction or discharge), comes from elevation right? If the pump inlet was supplied by a freely flowing river, you'd have 0 suction pressure? If your pump was located above the top of your inlet source, that would be 0 suction pressure too right? My understanding from other videos is that a pump doesn't actually add pressure, because the discharge pressure (and thus differential pressure) depends on the elevation that the water is being pumped against. Discharge pressure = Suction pressure + Pressure resistance due to elevation that is being pumped against. So what is the correct terminology for what the pump is really adding? Energy? In your experiment, I'm assuming the discharge was just shooting out the hose onto the grass or something, so why was the discharge pressure greater than the suction pressure if there was no resistance on the discharge side?
First of all, pumps transfer energy to a fluid. This energy is translated into a combination of pressure and kinetic energy. Second, I think it would help you to demystify the 0 suction pressure. This is simply zero GAUGE pressure - there is absolutely no reason that a pump can run with _negative_ suction pressure (gauge, obviously). The only concern with a very low suction pressure is that you risk cavitation if you are pumping a volatile fluid. Look at my video when I start registering my pressures - the discharge pressure goes up a bit before it appears as though the suction pressure moves. I only note zero suction when I see suction gauge pointer move. Before it moves it looks like it is zero but it negative, but the gauge is not able to show this. If I took that the same pump I used in the video, and placed in next to a swimming pool and had the hose run into my swimming pool, then the pump would be generating "suction lift" and the inlet pressure would be negative. The outlet pressure pressure would be determine by downstream resistance, and the differential would be determined by the flow (all these are inter-related as you are hopefully seeing, changing one variable changes them all). A thing to note with my swimming pool example is that centrifugal pump is not self-priming. This means it would not be able to suck up fluid from my swimming pull if the pump started off filled with air. But this is a separate issue.
Hi pat I have question How to increase outlet flow ? It means inlet flow was low Example Inlet flow was 0.5 bar &I want outlet flow above 2 bar Is it possible
Okay, great. But what will happen to the shut-off head in case you increase the suction pressure? For example, if there is a pump with a suction pressure of 0.2 bar and a discharge pressure of 9.5 bar, and the shut-off head is 10 bar, what will happen to the pump shut-off head when you increase the suction pressure to 1.5 bar? Will it increase or remain constant?
Hi Pat,
Great video.
Just wanted to ask about how increasing flow rate increases the suction pressure.
I’d have thought that increasing the flow would reduce suction pressure because the pressure loss in the suction pipe work would increase thus reducing the suction pressure? Or is the fact that you increase the suction throttle valve opening which reduces the pressure drop across this valve which therefore increases the suction pressure? Or is it a combination of the two effects?
Thanks Sean! You're spot on on all front. By opening a valve on the suction I AM increasing the pressure drop in the suction line (which should drop the suction pressure); however, the pressure drop across the valve is lower because I've opened it, so the net effect is an increase in suction pressure. Had I not had a valve on the suction and only used the one on the discharge then opening that discharge valve would have caused the suction pressure to drop, as you correctly note.
@@ProcesswithPat thanks for your reply Pat - that makes total sense!
Just as a follow up - can you control flow in a system using a valve at any point in the system? For example at the most upstream point in a system (say at the inlet to a pump) or at the most downstream point of a system (say if the pump led into a long length of pipe work which had a control valve near the outlet). Both valve locations could theoretically be used to control downstream/upstream flow respectively?
(Hopefully that makes some sense haha!)
Aaaand there you go, 2 hours of pump operations explained in an 8min video... Thanks Pat!
What I like about your videos is that you not only explain things theoretically using equations and charts, but you validate it by practically doing it, that's the way to go, that the way to teach, keep up the good job.
You have an excellent ability to take difficult topics and create a simple example to explain it.
The practical explanations will never not be interesting, fabulous video once again Pat!
Thanks a lot!
Good video, three observations:
a) You correctly point out that your suction lines are uhmmm, sub-optimal. In general suction lines are designed quite large with low dP characteristics (minimal bends and fittings) so that changes in flow do not really affect the pressure at the pump suction.
b) Increased suction pressure does not necessarily lead to higher flow. In your case this is true because your throttling the suction valve. However, when filling up a suction drum of say 1m to 2m head, and still controlling the flow on the outlet of the pump as is typical in most setups, the suction pressure goes up -> flow stays the same -> discharge head goes up correspondingly by 1m. Normally flows from centrifugal pumps are controlled on the discharge.
c) Many, especially bigger pumps, have a long flat portion of the pump curve at low flowrates. I.e. in this range there is basically no difference in the differential head when changing the flow.
Thanks Daniel! Nice to see someone is paying attention!
a) No excuses there, other than I used what I had. I have said my 50 Hail Perrys...
b) True. The message was meant to be if all I did was increase suction (and not anything else) what would happen? The thing is the idea of "only changing one thing" can be a little misleading because other things do change, and you one would need to specify what else you'd have to do to keep other variables constant (e.g. throttle discharge to keep constant flow as you put it). Controllers add another layer. I was actually contemplating a pump that is driven by a steam turbine - at least a motor keeps a constant speed, a turbine-driven pump speed may speed up or slow down depending on which controllers you've put in auto/manual... Messy stuff to try and generalise!
c) This is also true, but I wanted to stick to the general principle. You're right that this effect would not be as appreciable with flat pump curves, and I actually experienced something similar when I throttled the hose too much and created a very steep system curve. Then the differential pressure stayed relatively constant because flow could not change much.
@@ProcesswithPat Didn't mean to be critical, just some observations. I totally get the purpose of your videos - understanding engineering principles in a day to day setting. Young engineers lack this understanding.
I really enjoy your videos - Perry, Liebermann and McCabe would be proud!
Not at all! Please do be critical! Nobody is perfect and I’m certain I speak a lot of nonsense sometimes.
High praise indeed!
Thank you for your videos!! They are very helpful for me in understanding the fluid dynamics behind the design. I'm in the SCADA business and very familiar with the electronic side using PIT's, FIT's and VFD's. Enjoy the fundamental hydraulic analysis.
Excellent video , continue making such cool, scientific stuff.
Amazing channel, subscribed. Thanks for this, please don't stop posting quality content like this! Looking forward to your next videos.
I really appreciate that Claus!
Okay, great.
But what will happen to the shut-off head in case you increase the suction pressure?
For example, if there is a pump with a suction pressure of 0.2 bar and a discharge pressure of 9.5 bar, and the shut-off head is 10 bar, what will happen to the pump shut-off head when you increase the suction pressure to 1.5 bar? Will it increase or remain constant?
It seems like it depends where you are along the curve. If you're at the top beginning of the curve it's going to be close to a 1:1 relationship. Toward the top end of the pump's flow rate it's going to result in a near zero increase.
Great video..
So here is a question.
Using variable speed centrifugal pumps for swimming pools. When the pump is above the pool level (suction lift no gravity feed) sometimes pumps loose prime.
Will installing smaller diameter pipes increase pressure to the pump head preventing loss of prime. So 50mm pipes to pump head 63mm discharge?
Your videos are awesome👍
Keep posting.. Following all video.
Lots of respect from India.
A bit late to the party but is there a way to find the relationship between the decrease of differential head vs the flow?
Marvelous tutorial. Please could you explain what is stream
If I have LPG Tank as supply reservation and three multistage centrifugal pumps installed on common manifold together as a one or two working together to provide me with Q more if run one pump, put when I operate 3 pumps the flow decreased to a very low level to stopped completely. Can you interpret or explain it why? Thank you
great video Pat many thans
Thank you so much
You should have also mapped by how much flow rate increases when you increase the inlet pressure.
Great videos, very educational.
Hello Pat! It is really an interesting video and got some great insights. I wonder if this question really makes some sense but please do let me know. As relief design engineers we often look if vessels are being overpressurized with liquid for our relief valve design. For a given impeller size we take a look at the total dynamic head aka differential head provided by the pump at zero flow that is dead head pressure. But if we find that the dead head pressure is less than MAWP of the vessel downstream*110%, we move on to add the level in the vessel upstream of pump converting into pressure. Is it the right thing to do? Can we compare the MAWP of the vessel downstream by adding up the dead head pressure and the level within the vessel upstream of pump?
Apologies for the late reply, but I'm going to approach this topic with extreme caution, because I do not want to be giving design advice with real-life project implications. So take all disclaimers as given and always consult a non-internet based expert! ;)
I presume that if you are in the industry, then you are familiar with API 2000, 520 & 521 for venting & pressure relief.
A relief valve on top of a vessel is typically relieving a vapour (either air, an inert gas blanket, or the vapour phase of a volatile substance). You need to differentiate between normal inbreathing and outbreathing, and emergency pressure/vacuum relief.
When you are pumping a liquid into a vessel, the maximum LIQUID flow rate of the pump will tell you about the necessary relief VAPOUR flow rate, because the liquid is displacing the vapour at round about the same rate. But you are not actually relieving liquid but rather a vapour. This is NORMAL OUTBREATHING, and your safety device should not be trying to relieve this continuously. Rather, there should be another vent/pressure control that handles this. The PSV is there in case of failure of the regular breathing system.
But that talks about relief flow rate. Relief pressure is determined by the design conditions of the vessel, which is also a degree of freedom, and this is the gist of your question.
Based on your comment, it sounds to me as if you are trying to sum all possible pressure sources to consider in the relief pressure of the vessel, but it sounds a little backwards. The design pressure and MAWP should be based on this; you sound like you're starting with a given MAWP and working backwards. I understand how this can happen in the project, and it isn't necessarily wrong/bad. Also, if you are working on brown-fields, modifications, or revamps, you cannot always change the design pressure you've been given.
The point is, you should not reach a point where you have zero vapour in your system because it it totally liquid-filled and you are pissing liquid out of your PSV into your vent system. To prevent such a scenario you need overfill protection in the form of adequately SIL-rated high level switches.
What you are describing is only ONE relief case for which the PSV is sized. In reality, there are other cases which may be more severe. A possible example is the "fire case" - where you assume that there is a fire in the vicinty of the tank, and the radiant heat and boils the contents of the tank for which you are sizing your relief valve.
It is highly dependant on application - pumping water, solvents, or petroleum products need to be assessed individually.
It get's complicated and what I am learning is that there is no "correct" answer, because at the end of the day you are not getting away from doing a risk assessment. At the end of the day YOU need to determine what is acceptable risk, and you will not find a silver bullet in this regard. It's a tough thing to accept, but the onus is evetually on the designer and the user to determine what acceptable risk means.
@@ProcesswithPat Thank you Pat for detailing. I was specifically considering overpressure with liquid case in the question mentioned above assuming it to be the only governing relief case and a pressure vessel in specific and not a low pressure storage tank to be precise. We do take total dead head pressure which is the sum of shut off head and suction pressure as a criteria to determine overpressure with liquid scenario.
Thank you for the video's. You deserve more subscribers!
I appreciate that!
Great video as usual. Kudos! Btw, does this apply to a close loop system?
Hi Pat, thanks for the viedo. Why inreasing the suction pressure will also increase the flow rate ?
Try think about it in reverse… can you think of a situation where you increase the pressure at the inlet to something, and the flow rate through it decreases? That’d be counterintuitive, no? Pressure is a form of energy (think Bernoulli)
@@ProcesswithPat Ok I see the logic. Thank you.
Hi Pat, if I am opening the suction valve.. this means that system curve's gradient is reduced meaning that it intersects the pump curve at a lower head. Therefore, the pressure that the pump discharges is reduced. Does my understanding make sense? I suppose I'm asking if the suction valve impacts the system curve. Thanks!
Excellent video. Is TDH same as "differential head"?
Thanks.
I know this is a dumb question, but I'm having an argument at work with someone who allegedly knows more than me…. If a pump is not running (VFD@0hz) it is impossible for discharge on a centrifugal pump to be higher than suction pressure, right? That's indicative of the gauges being gimp, right?
(Eliminating the possible factor of head pressure, the cooling tower supplying the suction is 14 floors up and is discharging to feed a chiller's condenser on the same level as the pump)
Great explanation!
Why preasure of discharge is greater than suction side ??? because of the diameter ,the diameter of suction greater than discharge diameter i cant underestand that but i know when diameter increase ,preasure increase and velosity decrease 😢
thank you man
Hi Pat….could you please help me with a complicated NPSHa calculation for a Pump?
Sure thing. Send it to me and let me see
@@ProcesswithPat….do you have a place I can send you photo?
Describe it here.
@@ProcesswithPat ….Double Suction Horizontal Centrifugal Pump rated for 6500 m3/hr, 32” Suction line first 6 meters then 32” to 50” Eccentric Reducer to 50” long radius elbow inside a water basin, for a total suction pipe length of 8 meters before elbow. Centreline of Pump Suction is 3.3 m above bottom of water basin which 5.5 m tall filled with water. Note eye of impeller is 750 mm above centreline of Suction piping.
Water Temperature is 27C and atmospheric pressure is 10.26
Thanks man, I'm honestly shocked you haven't lost patience with me, I feel like I'm wasting your time with my inability to understand basics.
Anyway, so pressure (whether it be suction or discharge), comes from elevation right? If the pump inlet was supplied by a freely flowing river, you'd have 0 suction pressure? If your pump was located above the top of your inlet source, that would be 0 suction pressure too right?
My understanding from other videos is that a pump doesn't actually add pressure, because the discharge pressure (and thus differential pressure) depends on the elevation that the water is being pumped against. Discharge pressure = Suction pressure + Pressure resistance due to elevation that is being pumped against. So what is the correct terminology for what the pump is really adding? Energy? In your experiment, I'm assuming the discharge was just shooting out the hose onto the grass or something, so why was the discharge pressure greater than the suction pressure if there was no resistance on the discharge side?
First of all, pumps transfer energy to a fluid. This energy is translated into a combination of pressure and kinetic energy.
Second, I think it would help you to demystify the 0 suction pressure. This is simply zero GAUGE pressure - there is absolutely no reason that a pump can run with _negative_ suction pressure (gauge, obviously). The only concern with a very low suction pressure is that you risk cavitation if you are pumping a volatile fluid.
Look at my video when I start registering my pressures - the discharge pressure goes up a bit before it appears as though the suction pressure moves. I only note zero suction when I see suction gauge pointer move. Before it moves it looks like it is zero but it negative, but the gauge is not able to show this.
If I took that the same pump I used in the video, and placed in next to a swimming pool and had the hose run into my swimming pool, then the pump would be generating "suction lift" and the inlet pressure would be negative. The outlet pressure pressure would be determine by downstream resistance, and the differential would be determined by the flow (all these are inter-related as you are hopefully seeing, changing one variable changes them all).
A thing to note with my swimming pool example is that centrifugal pump is not self-priming. This means it would not be able to suck up fluid from my swimming pull if the pump started off filled with air. But this is a separate issue.
How to increase the suction power at the inlet of pump
Not sure what you mean by "suction power".
Hi pat
I have question
How to increase outlet flow ?
It means inlet flow was low
Example
Inlet flow was 0.5 bar &I want outlet flow above 2 bar
Is it possible
I don’t understand your question. You’re mixing up pressure and flow…
Fabulous
thats a regenerative turbine pump in the demo. not really a centrifugal pump. just sayin
Ye video Hindi me chahiye
Your pump isn't centrifugal pump
Does it matter for proof of concept and not scale to 1
No such thing as suction pressure, suction causes vacuum
Okay, great.
But what will happen to the shut-off head in case you increase the suction pressure?
For example, if there is a pump with a suction pressure of 0.2 bar and a discharge pressure of 9.5 bar, and the shut-off head is 10 bar, what will happen to the pump shut-off head when you increase the suction pressure to 1.5 bar? Will it increase or remain constant?