Physics 34.1 Bernoulli's Equation & Flow in Pipes (21 of 38) Flow with Pump***

In this video I will derive and explain the power-needed-from-a-pump=Pp=? To water from a lower reservoir to a higher reservoir.

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[yonasnegusse]

If there was a NOBEL PRICE for teaching...And if I were one of the committee that awards it...I would have given you a Nobel price for your ability, clarity, conciseness in your way of teaching. The day you decided to created this channel was a blessed day, as you continue to bless us everyday. I wish you a good health and strength to continue this. I also thank your family for their support to let you have all the time you need to give us this videos. I promise to donate as much as i can when i got the chance and reach there. God Bless you.

[fatmonk]

Absolutely fantastic series, thanks very much for your generous contribution.

[ahmedal-ebrashy]

finished the list, ofocurse I wish it is completed and I know how busy you guys are. Regardless thank you and thanks Mrs Biezen for your wonderful work

[ericgerard]

I had a question come up on my P.E. review that is very similar to this problem and I am hoping I can get a better explanation than what was provided by the reviewer. the problem statement is the following:

"A pump is set up at elevation 110 ft to transfer water to a reservoir at elevation 135 ft. With the pump running, the fluid velocity through the 12 in diameter discharge pipe is 5 ft/sec. Losses through the piping have been determined to account for 7 ft of head. What do you expect a pressure gauge at the pump discharge to read in psi during operation?"

I approached the problem by setting up Bernoulli's equation like what is shown in this video. The "gut check" part is when I am setting variables = to 0. I understand that I can set the Pressure at P1 = 0 because it is open to atmosphere. I am solving for P2 at the reservoir. I know/feel that the velocity at 2 can be = to 0 but its just not sitting right with me. Is there a better explanation than: " at the reservoir, after it is pumped, the velocity is 0". That is just not sitting right with me.

I appreciate any response, I can not find another example problem that will walk me through the "whys" when simplifying the equations.

Thanks in advance!

[IaMaSiNnEr]

Thank you for your lectures. I wish you continue the lecture series

[brendangolledge]

I don't see how this works. The pump won't increase the pressure in the reservoir if it is open to the air. And once the pump adds pressure to the water, isn't it already moving? So if you're at a point where the pump has already acted on the water, the velocity is not equal to 0. I don't see any place in this diagram where the pump has added pressure and the water is not moving.

Edit: I watched a previous lecture (this lecture being the first that I watched), and I think I understand how this can work. If the pump were off, then clearly, the energy immediately before the pump would be equal to whatever is after the pump. But the pump adds energy to the system, so the energy immediately after the pump (ignoring losses) must be equal to whatever the energy was before (in the reservoir), plus whatever the pump adds. So, the water being acted upon by the pump is not actually immobile (as suggested by the equation), but due to conservation of energy, you know that it has the same energy as the water in the reservoir (which is not moving), plus the energy of the pump.

[jonathanconstancio]

What if the lower tank is closed. Let’s say a bulk head. And you have two pumps which support a system in the manner of primary and secondary. One pump pushes fluid until completely empty than the other takes its place. Pump 1 would then fill while pump 2 is supporting the necessary system requirements. How would one solve for that?

[crispinweston]

Hi Michel - thanks for all your lectures on fluid dynamics, which I have found really interesting. Do you have any plans to finish this series? The particular problem I want some help on is where you have natural flow down a hillside from a reservoir, with branching pipes. The slope of each branch is then applying a different gravitational force to the liquid, sucking the water down the hillside, as well as presenting its own frictional resistance. If all the pipe has the same diameter, each new branch will also tend to slow the velocity of flow by increasing total cross-sectional area. I am having trouble seeing how all these different variables are going to relate to one another. Best wishes, and thanks again for what you have already posted. Crispin.

[hevoshullu]

Hello Michel! Your mechanical energy balance is incorrect. You should also consider the water in the suction tank. The correct height difference is between the surface of the lower tank and the pipe outlet.
When you add water into the suction tank it will increase the static suction head; eventually when you add enough, you will not need a pump (just for illustrative purposes). Really think you should fix this and not spread wrong information. Thanks!

[ritchiethomas]

How can we apply Bernoulli's equation to pumps? The equation can only be applied to fluid which move in streamline motion, right? And the impellers of the pump tend to make the flow turbulent.

[vanoski]

Sir, what if there is an elevation in the pump?

[meh]

Hi Michel, thank you so much for your clear and comprehensive explanations. You're the best. I can't find Bernoulli's equation from 22 of 38 up to 38 of 38. Can you please let me know if you have uploaded them on your channel? Thanks again. Wish you all the best. 🌼 🌸 🌻 🌹

[mattpfeiffer]

Hey Michel, similar to your other video, is the reason you are not including the height of the water in the lower tank as part of your h1 because you are simply ignoring it for this problem? Thanks

[andyoohhh]

At point 1 you took the velocity and preeusre at thr tank but not the height of thr tank level? Instead used pump height? Why?

[rol]

I think h1 should be at the height of where v1=0?

[rashel]

Thank you, if sometime you can do one in a closed system it will be nice (like in hydronic heating systems or solar systems )

[achrefabdelli]

Hey Mr Michel, I have some dificultes to solve a fluid problem, I'v been searching on net for a solution.. If I can borrow 10 minutes of your time to look for the exercice that I can send it to you via mail o whatever. thankyouuu

[pcgamingmaster]

Hi Prof. Michel, I'm Fahad from Indonesia and often watch your lectures, where is the rest of this videos? and do you still run the "Ask Michel Anything?", and if so, I would like to ask you to draw your life like the other Youtubers did, so we could know your background, and also thank you so much for these short lectures, I found it so useful :)

[Hansel.c]

Hi Michel,
Why is h1 = 0 when point 1 is taken at the top surface of the lower tank and you've set h=0 at the bottom of the lower tank.

[heinmt]

Can we ignore rho in frition head loss ( fh x rho x g )?
Because in my work, we find frition loss from respective losses chart.
If I use your equation, I need to multiply rho and losses value from chat.
Please help me explain Sir.

Thank you.