I'm studying for my Water Resources PE and, having been out of uni for 4 years and never having to touch a pump in my career, this is exactly the refresher I needed. Thanks!
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.
i am slightly confused, if engineering company design a centrifugal pump for viscosity other than water (i.e. other fluid) how to link this pump with a vendor pump tested by water only!
1. I can't understand why the head of pump will be larger the head of B after increasing the Q? 2. Besides, why the connection point of the system curve and the pump curve is the operating point? Why don't we adjust the operating point along the pump curve and choose the point nearest to the 0.82????
5:32 souldn't the equation have another term equal to (v^2)/2g which corresponds to the hydraulic losses at the output of the pipe? Having something like Hb=Zb+ hL being hL the primary hadraulic losses.
That’s the “1” that’s added to f L/D. We can either look at the surface of the water, in which case those are hydraulic losses, or at the exit of the pipe, in which case it’s kinetic energy.
Dear sir in some other videos , they consider velosity head as helping the pump so they subtract its value from the total head the pump should give to the fluid
I have two questions 1- why is the velocity in the friction head is the same as the velocity in point b? 2- Is the velocity in point B the same in the impeller? If so, why?
1 - The friction head is due to the velocity in the pipe. At point B, the assumption is that the diameter of the pipe is the same as everywhere else. Since the diameter is the same, and flow rate has to be the same, we know that velocity is the same. 2 - The velocity immediately before and after the pump is the same as point B. However, inside the pump, things are a lot more complicated, as the impeller is causing a lot of rotational movement while the area is also changing. At this level, we don't try to look too deeply into what's happening inside the pump. That's what the curve is for!
The best video I could find on this topic. Subscribed.
literally couldnt agree more
best video om the topic. been studying this for momths amd the dude just gave it all to me in less than 20 minutes ✌😅. great video.
I've saved this video in a playlist and I watch it everytime before my fluid exams. Must say this is an awesome piece of work here.
Better than my fluids class! Thank you!
Amazing explanation! Thank you!!
I'm studying for my Water Resources PE and, having been out of uni for 4 years and never having to touch a pump in my career, this is exactly the refresher I needed. Thanks!
Sir... could you explain, why hpump equation sudently become (1+f(L/D))... why vB^2/2g was replaced by 1?
Bro that what I do when selecting equivalent pump from the existing which isn't in good condition anymore on site
Does the arch above hb matter when calculating hb? e,g, what if the arch was far higher?
If you have a hose going up a hill as showed in your picture would that produce minor losses?
This video is gold, thank you!
how did you guess Hp and correspondence value of Q? : Is there mathematics formula?
I think Q&H should be determined by the application; then find the pump gives the good efficiency in the region. Not the other way around.
I tried many ways to better understand TDH concept. Finally This video helped me.Thank you very much sir. Subscribed👍
Thanks for the kind words!
I would recommend to use z for elevation not another h
Thank you friend
smart! thank u
best explanation
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.
i am slightly confused, if engineering company design a centrifugal pump for viscosity other than water (i.e. other fluid) how to link this pump with a vendor pump tested by water only!
1. I can't understand why the head of pump will be larger the head of B after increasing the Q?
2. Besides, why the connection point of the system curve and the pump curve is the operating point? Why don't we adjust the operating point along the pump curve and choose the point nearest to the 0.82????
This is excellent! Your instructions are perfect and easy to follow. Thank you!
Glad it was helpful!
What is the unit of pump head?
Great video, thanks a lot!
I remember putting the energy in to friction. Then I remembered fuck it.
can we predict VSD or DOL type of pump by this cure?
is it necessary to draw the parobolic line each time?
Thank you!
can we find system curve for closed system ???
Why not taking point B on surface of up tank?
Thanks! Awesome video.
Master you are too good❤
15:29 15:29
Best video solves all points with clarity 👌🏼👌🏼
It can't be better than this.thanks for your work
Sooo informative! Thank you!
5:32 souldn't the equation have another term equal to (v^2)/2g which corresponds to the hydraulic losses at the output of the pipe? Having something like Hb=Zb+ hL being hL the primary hadraulic losses.
That’s the “1” that’s added to f L/D. We can either look at the surface of the water, in which case those are hydraulic losses, or at the exit of the pipe, in which case it’s kinetic energy.
Dear sir in some other videos , they consider velosity head as helping the pump so they subtract its value from the total head the pump should give to the fluid
Excellent!
I have two questions
1- why is the velocity in the friction head is the same as the velocity in point b?
2- Is the velocity in point B the same in the impeller? If so, why?
1 - The friction head is due to the velocity in the pipe. At point B, the assumption is that the diameter of the pipe is the same as everywhere else. Since the diameter is the same, and flow rate has to be the same, we know that velocity is the same.
2 - The velocity immediately before and after the pump is the same as point B. However, inside the pump, things are a lot more complicated, as the impeller is causing a lot of rotational movement while the area is also changing. At this level, we don't try to look too deeply into what's happening inside the pump. That's what the curve is for!
@@PostcardProfessor so I guessed right. Thanks a lot