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!
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!
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????
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.
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!
This is excellent! Your instructions are perfect and easy to follow. Thank you!
Glad it was helpful!
Best video solves all points with clarity 👌🏼👌🏼
It can't be better than this.thanks for your work
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.
Amazing explanation! Thank you!!
Great video, thanks a lot!
Sooo informative! Thank you!
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!
Thanks! Awesome video.
Thank you!
Excellent!
best explanation
smart! thank u
Better than my fluids class! Thank you!
Thank you friend
Master you are too good❤
15:29 15:29
If you have a hose going up a hill as showed in your picture would that produce minor losses?
is it necessary to draw the parobolic line each time?
Does the arch above hb matter when calculating hb? e,g, what if the arch was far higher?
can we predict VSD or DOL type of pump by this cure?
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!
can we find system curve for closed system ???
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.
What is the unit of pump head?
Sir... could you explain, why hpump equation sudently become (1+f(L/D))... why vB^2/2g was replaced by 1?
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
Why not taking point B on surface of up tank?
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
I would recommend to use z for elevation not another h
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????
I remember putting the energy in to friction. Then I remembered fuck it.