Fluid Mechanics: Centrifugal Pump Characteristics (21 of 34)
ฝัง
- เผยแพร่เมื่อ 10 ส.ค. 2018
- Note: At 44:52, the equation should be Q = V*A, not Q = V/A.
0:00:15 - Introduction to centrifugal pumps, measuring pump head
0:04:09 - Pump performance curves
0:16:18 - Net positive suction head (NPSH)
0:21:46 - Example: NPSH
0:25:30 - Pump selection using pump specific speed (Ns)
0:33:50 - Example: Centrifugal pump in a pipe network
0:56:15 - Comments on homework
Want to see more mechanical engineering instructional videos? Visit the Cal Poly Pomona Mechanical Engineering Department's video library, ME Online: www.cpp.edu/meonline
This lecture series was recorded live at Cal Poly Pomona during Spring 2018. The textbook is White "Fluid Mechanics (8th edition)."
I am utterly amazed watching these videos...i am a 50 year old guy who was never good in math and science but have an overabundance of curiosity about how things work. I feel lucky to live in a time where i can look anything up on youtube and get some undestanding about how the physical world works around me. Also, i had no idea how involved pumps were...meaning the engineering behind the pump.
We feel lucky to live in a time where it is possible to leverage our efforts to help others well-beyond the classroom.
@@CPPMechEngTutorials thank you for your posts. I feel lost sometimes watching these things but rely on osmosis. Knowledge helps us all thanks
What a great teacher. Wish i could've attended to his lectures. Most of this stuff i had to figure out on my own and with the books
What a great teacher! I am electrical engineer but just enjoy listening and understanding...
Feel like this prof is more like a coach, you don't wanna let him down. I bet students work their tail off for him because he's engaged in the learning process.
Never let Prof. Biddle down... NEVER!
Muchas gracias. Excelente explicación. I am an electrical engineer and these concepts are very useful for me. Thank you very much.
Am over 2 years in pump industry as engineer. Splendid, crystal clear sir. Thank you.
All I can say is brilliant and thank you.
Thank you for sharing!!
Our pleasure.
Excellent lecture, Dr. Biddle. My professor doesn't even try to explain and solve these problems. Thank you very much.
Thanks!
It´s such an amazing class, thank you so much for this video , I got a brief summary about centrifugal pumps which I learned through the university.
Thank Prof for the great lectures!
I haven't enjoyed a lecture as i have enjoyed this one!👏🏾👏🏾
Excellent Lecture. Pretty Understandable
Thanks for including time stamps in these videos!
No problem. We are starting to add time stamps to the lecture series to make it easier for viewers to find exactly what they need. Hopefully the timestamps will save people a lot of time and frustration.
I left my class for this topic,but luckily i found this video,it's about 2 days bfr exam,,i watched it,,excellent lecture,,hope my final exam become good
just loved the lecture, gorgeous explanation, for anyone wondering how is H vs Q is plotted it is drawn by EULERS MOMENTUM EQUATION
Your welcome
Great Teacher ,nice channel, improve my listening and English on engineering.
This video is great, a lot of good information presented in a clear manner. Thanks!
Our pleasure!
Great lecture, Thanks Professor.
this is great info, I'm studying for my next wastewater certification and notice a few differences. Total dynamic head and Head in general is always referd to as the verticle distance the water has to be pumed (measured in feet). But it sounds like head is just any force acting to restrict flow?
Also Cavitaion is described only happening at the impeller near the inlet or outlet and not in the pipe itself. this is due to the pressure difference across the impeller. the bubbles are formed by rapid pressure loss and on implotion they take part of the impeller with them and they sound like BBs in the pump.
again, great video. i enjoyed getting away from the text book for a bit.
Thank you prof. what a simple straight forward outstanding explanation!..
Thanks!
Thank you sir for sharing this valuable informations !
Best lecturer. Thanks
Thank you very much . Very helpful lecture videos.
Beautiful Class Thanks Prof
Great professor. Thank you
This is very useful for engineering student thank you sir😍👍👍
Great lecture, remembering when I was in university. Now I work in small business servicing pump, and this lecture is realy helpfull. Thanks Prof
We're glad it was helpful.
Informative and pretty good.
Excellent explanation
Very much interesting and informative lecture. Thank you Professor.
Well explained...thanks a lot
Excellent lecture
Thanks a lot Dr Biddle for sharing this is a really helpfull !
Glad it helped!
Thank you very much sir... really appreciate it 🙏...
This is a beautiful culmination of my pumps course. the clarity of it brings tears to my eyes.
It is crazy how I can watch those lessons for free. Thank you very much.
The world is a crazy place. :)
If you try and do the math on minute 44:48, on the replacement of V, Q=AV, so you should have V=Q/A. Thus, V=4Q/(pi*D^2). V=4Q/pi with D=1.
Yup. We put that message in the comments.
should i square it like the v?
Thank you sir.. I like ur way
Great explains 👍
Thanks for your effort. One tip though, unlike the case in the pump sketch you showed, for centrifugal pumps and compressors, the suction side is always one standard size larger than the discharge size. This is to minimize losses and, consequently, reduce cavitation possibilities.
Thanks for the tip.
Thank you for sharing
thank you very much professor!!!
You're welcome!!!
Thank you very much
Appreciated the clear explanations and real-world practical focus.
Thanks!
The equation provided to calculate delta z based on NPSH is for the case that the water level at the supply reservoir is below the water level at the demand reservoir. So, for the case which is shown in the video, the water level at the downstream reservoir should be lower than that the water level at point 1 and for this case: delta z (minimum water level difference between two reservoirs)= (Pa-Pv)/gamma - NPSH (for available pump) -HL (all head losses) as shown in the video. It is not surprising that if the difference in water levels reduced to values less than delta Z, cavitation is expected to take place. For the other case if the water level at the demand or downstream reservoir is greater than the water level at supply reservoir (most cases), delta Z (Max different in water levels)= NPSH (of the available pump)- (Pa-Pv)/gamma +HL (all head losses) and in this case if the water level at supply reservoir or the well is dropped to lower depth, than the difference in the water levels increased to values more than computed delta Z and we expect cavitation will take place in this situation.
Thank you for your lacture.. I learnt a lot and but why the H vs Q curve by manufacturer is decreasing while system head curve is increasing..However, I know, at 0 flow, head will be maximum and if flow will increase, head will be go down because pressure will be reduced but please make me clear on what basis or data manufacturer plot H vs Q.
How to find his lectures about Turbine & compressor
LEGEND Dr. Biddle
Hey it would be great if the handout was posted for this video.
Thank you
Very good vid
great tutorial for a EE....At 25.00 regarding NPSH, visually the water tank is below the pump, the Z is the height between the two. How would the analysis change, if the pump was below and the water tank was above? I would think the ABS | Patm -Pv | may be a more general solution to avoid an answer with a -Z value.
Thanks good video
Great class, what's the professor's name?, he mentions 'his book' and I'd like to check it out....thanks CPPMechEngTutorials guys !
love this subject 😊
We love it too!
thank you
God bless you
Hi teacher
I would like to thank you for your wonderful explanation
and I want to know why in the pump efficiency chart, when the height increases, the discharge decreases,please
may i know which textbook you guys used? and where can i get the copy?
This professor is a legend, I wish I had him as teacher at University 😮
There can be only one Professor Biddle.
Thanks for some great videos! This is probably a shot in the dark, but could anyone help me reference to the book he is using. I’m in the middle of a thesis but I cant seem to find alot of good theoretical references on the subject.
Hello Sir,
Can you drive the specific speed formoula?
How we have got the range for every type?
I got an exam tomorrow. My prof sucks. You are great. Thank you
my supervisor recommend this video i was in search of topic for research.
Thank you for the great video. I have a question about the effect of temperature on the characteristics curve of the pump.
How does the characteris curve of a centrifugal pump change with the decrease in temperature provided that the pump power is constant.
A reference material to read would of great help.
Thank you
I can tell you that increasing tempreture will increase the risque of cavitation; unfortunantlly i don't know really it's effect on HQ curve.
Thank you sir for the explannation, it helped understand the fundamentals better. Also, I would like to add a note on vapor pressure. Vapor pressure of Water @ 15degC unit shall be read in Pa (i.e 1666 Pa)
Would cavitation also occur if boiling coolant (car's cooling system) is flowing through the pump? Pump would be at the lowest point in the system. I'm really curious about this
Great lecture indeed but I'm confused where did the professor get 175 and 140ft to generate the pump head curve?
Thank yu sir
How can we calculate the flow rate of a split type centrifugal pump if we know the refrigerant flow rate through the chiller's evaporator?
Dr. Biddle, Good day to you. Which textbook are you using?
please Sir If the inlet fluid is subcold ( at low temperature ) but the liquid is entering the pump at low pressure and the impeller will start swirling the liquid and create bubbles just as when you move your hand quickly on water surface you will create water bubbles though the inlet pressure is above the vapor pressure is that possible
Thank you so much for the excellent lecture prof. I have one question: why the efficiency form on previous graph is different from those of pump performance map?
Yeahh, the same question crossed to my mind !!! Even I'm confused for it
Dear Prof! speaking about elevation of the pump above reservoir surface you said:
Well, as you get the pump higher and higher in the flow rate, that pressure is going to start drop.
I think that could be:
Well, as you get the pump higher and higher in the elevation, that pressure is going to start drop.
What do you think about that? Thank you very much.
I just want to know that shouldn't the peak of efficiency be below the horsepower curve?
Got it sir
Is there a theoretical derivation for the shape of a pump curve?
I guessed maybe a centrifugal pump imparts constant energy per unit mass to whatever fluid passes through it. Then you'd get E = constant (for any given fluid density) = P + (1/2)dv^2 (d is density), or h = E/(dg) - Q^2/(2gA^2). This has the shape of an upside down parabola (right basic shape), but when I try to compare it to real pump curves, the real curves turn downwards something on the order of 100 times faster than my theoretical model. Do you know why this might be? Am I making some unit conversion error, or is the whole thought process just wrong?
If the real pump curves turned down something like 10-50% faster than the theoretical one, I'd be happy, because I guess the pump doesn't conserve energy perfectly and it doesn't have the same efficiency at all operating ranges. But getting a result 10,000% different than the theory is unsatisfying.
Actually V=Q/A (Cross sectional area of pipe)
ΦΙΛΙΕ ΣΕ ΕΥΧΑΡΙΣΤΩ, ΔΕΣ ΠΩΣ ΚΑΝΟΥΝΕ ΜΑΘΗΜΑ ΣΕ ΠΑΡΑΚΑΛΩ. Do you know If I could find the document that was given to the students? Thank you
Which textbook does he use?
dear DR.
AT 45 MIN there is a little error Q=VA so V=Q/A SO it well be V=Q 4/ π D^2
thanks lot ...
He assumed D = 1
You're right.
Good
Hello,
could anyone tell me how pump curve is determined at a certain speed however the input power = ( torque x angular velocity).
so, to get more discharge we need more input power by increasing velocity
and then by applying the efficiency formula (output power / input power) we can draw an efficiency curve.
so, how this curve could be at a certain velocity?
How do you arrive to the equation which include the term Hp?
Which Text book he is referring to?
what is the textbook title which you kindly are using please?
That information is in the video notes.
Sir what will happen if pump is run at very low head than rated head... There will be low efficiency i know. What other practical problems will occur.
47:42, when Q=0, Hs=150. But how at increased flow rate Q=500, Hs is also increasing to 150.5? Because according to the curve at increased flow rate head decreases. Isn't it?
Where i can find the explaination of chapter 10 ?
Is NPSH used for the single purpose of determining the maximum elevation which the liquid can be raised?
No, because NPSH changes with the change of Q. The higher is the flow rate, the higher is the pressure drop at the suction side. Therefore it's rather used to get to know the maximum flow rate available.
Also the temperature plays the key role, since it influences the vapor pressure.
What a great teacher, where do you teach, i want to join one of your classes
Sorry. Only Cal Poly Pomona students are supposed to be in the classroom.
Great class. One would intuitively think that the velocity of the fluid at the discharge of the pump is higher than the suction velocity. Apparently, this is not true, because the professor is neglecting that difference. I would like to understand better the reason behind it. Any advice?
Usually, suction pipe is one size higher than the discharge pipe size, hence the velocity in discharge pipe is higher than that of the suction pipe
When you run the pump at a constant speed, how do you increase the flowrate through the pump?
Open the flow control valve downstream from the pump.
Raise the fluid temperature to change viscosity, density? (Not practical but I'm an electronic guy working at a farm and my boss keeps trying to make me do things that are not correct. My solution, learn about fluid dynamics on my own, not to be right but to respect energy and knowledge). Thank you professor, much respect for your life's work, reminding me as always I am standing on the shoulders of giants
Change the impeller size as per manufacturers impeller data
@@CPPMechEngTutorials
Awesome
Love this defining of all abbreviations .... thats unfotunately not standart any more...
how can we increase Q at constant V ? Area? I mean with velocity diagrams we can calculate everything then why we need graphs for?
Since Q = VA, we cannot increase Q at constant V (assuming the fluid is incompressible).
For centrifugal pumps and compressors, the suction side is always one standard size larger than the discharge size. So yes, area will be a factor.
I have a question,
If a centrifugal pump flow limit is from 60 LPM to 535 LPM at 3500 rpm, does the low flow limit of 60 LPM changes with running the pump at lower rpm? Does this limit changes goes down when we run the pump at lower rpm?
Surely changing the flow rate from 60 LPM to 535 LPM would mean a change of impeller, assuming constant rpm.
If you have a 60 LPM impeller, on a 3500 rpm motor, and you reduce the speed of the motor - the flow rate will reduce.
Professor: why is the velocity difference between point 1 and 2 very small? Shouldn’t the velocity at point 2 be very high because impeller blade is pushing the liquid and velocity should increase significantly.
The pump definitely provides Kinetic energy to the water which intutively make us think the increase in velocity of flow at outlet, but because of volute casing of pump that kinetic energy of fluid gets converted into pressure energy. Thus the velocity remains same. Moreover velocity is governed by the pipe diameter. Since pipe diameter at inlet and outlet of pump is assumed to be same and flow rate also going to be same to maintain steady state ( because whatever amount of water is entering , the same amount of water will move out of pump). Hope it will help!!
on NPSH example problem, Patm defined by value of 101000 kPa. Should it be valued 101 kPa if Pv also in kPa? CMIIW
Yes you're right. He denoted it with the coma. However the value of Pv at 15C according to my steam table is equal to 1,7 kPa.
@@dawidpietrucha5333 Same, from Table A.5 in the class textbook, interpolating you'd get 1.782 kPa for vapor pressure.
do you have videos for closed loop system.
Some of the earlier videos in the lecture series involve the energy equation for pipes.
Check me out but at 25:10 the 101kpa makes it 101,0000 Pa which is NPSH is 7.4 METERS then that makes sense with the whole equation to make the units workout.
But he wrote that vapor pressure is 1666 kPa and then out it in the equation as that value so there's a mistake on that one way or the other. Either it was incorrectly labeled with the kPa units initially or he should've left the Patm in 101 kPa form and converted later.
Can someone confirm?
Yeah exactly. Patm is 101 kPa and the Pv at 15C according to the steam tables is equal to 1.7 kPa.
@@dawidpietrucha5333 lol too long ago I can't tell if you're agreeing or disagreeing
He's agreeing....
@ 23:13 the Pv should have been in Pa not kPa, looking at chart now and its interpolated between 1.227 and 2.337 kPa or 1227 Pa and 2337 Pa respectively. Therefore 1666 should be is Pascals units not kilo-pascals.
Also, should be 1.782 kPa for vapor pressure after interpolating.