by the way.. it's my understanding that if you hook up a hose right at the base of the Water Tank... then let the water flow though the hose... the water goes a distance "d" ... and if you then hook up a Spray Nozzle to the end of the hose.. and set the nozzle for a Jet spray... the water only shoots out a distance of "d" .. the same.. SAME Distance!!.. however, at home... you will find that a Nozzle (or pinching the hose) WILL Increase the distance!!.. apparently it has to do with Water Friction and the SPEED of water as it courses through the Pipes and hose. The Pressure at your end of the Line is Lower if the water is flowing Fast. I cite the professor Lewis Carroll Epstein from City College of San Francisco in his Book, "Thinking Physics, second edition" .. what a Surprise.. I never knew that...
How are you professor , hope you are well , it's really a cherishing moment and fun to watch your lectures ! Really appreciate your way of never letting the students feel bad or low when they sometimes are unable to answer correctly or precisely !
so is it only the height that affects the velocity or is it the volume. So if you have a 50 gallon tower versus a 500 gallon tower and the height of both towers are the same the velocity of the water coming out would be the same?
I think that we can use the A1 V1 = A2 V2 to solve for V1 then substitute into Bernoulli's Equation. We can assume that v1 = 0 but we don't necessarily have too. As long as you know A1 = Pi r1^2
Hi.if i may ask ,if the water tank is on the rooftop tower at level 40 and goes down to level 27 up to level 32 hotel rooms..do we still need a booster pump? it supply at least 500 hotel rooms..thank you
Thanks. The video is great. There is one thing that's not clear to me: Why do we assume P2 is the atmospheric pressure ? What happened to the hydrostatic pressure of the water column ?
p2 is at the very end of the spigot. The hydrostatic pressure inside the water tower drops to the atmospheric pressure as the water moves through the spigot.
I have a doubt Initially u said we cannot apply continuity equation as we don't know v1 but while using the B-equation u neglected it assuming the tank to be very wide Please clarify
in my opinion, when u using continuty eq, you don't know exactly the V1, and if we neglected it, it become A1. 0 = A2 . V2, so we can't find the V2 because A1 . 0 = 0 *CMIIW
How to find drain size which is require for particular tank... Because as level of tank decrease the velocity is also decrease, Then how i calculate drain size for constant outlet flow Q = A V
Dixit, Per your question as h decreases: If you want Q to be constant, then as v decreases by sqrt (h), then A must increase by the same factor (1/sqrt(h)). Thanks for the comment, and keep up with the physics! You might also like my new website: www.universityphysics.education Cheers, Dr. A
How much temperature increases this velocity So we got biggest model ever possible It's less then 100 degrees which is vapor pressure against atmospheric pressure So if we do this experiment under a pressurized chamber then we got very high speed at discharge Try it
@@yoprofmatt combined ideal gas law with heat formula with Bernoulli equation Q= msdelta t Bernoulli equation Ideal gas laws Coulombs law Gravity laws Quantum physics laws Einstein equation All together For finding out the maximum power possible by this siphoning in herons fountain ?
P0 at the top of the tank is (nearly) identical to P0 at the bottom of the tank. At least the difference is very small when compared to the other terms in the equation. Cheers, Dr. A
@@yoprofmatt i have a question, why is P2 also atmospheric pressure? i get why P1 would be that, but for P2, wouldnt there be additional pressure to the atmospheric pressure, from the water being "pushed" out/pressurised due to the weight of the water above it?
I don't get it, it looks like trying to reinvent the wheel. Why don't you just apply Torricelli's law? Basically you demonstrate Torricelli's law using Bernoulli's equation
So, no one is going to appreciate his super human capability of writing like that ?
Not super. Possibly human. Secrets here: www.learning.glass
Cheers,
Dr. A
The video is probably reversed (right to left) in editing. IRL it would look backwards.
dude and I was just wondering why nobody was mentioning the way he was writing lmao
@@MrNeutross no He was teaching his students facing towards them so I doubt it was edited that way
Thanks Wayne, more videos coming every week.
Cheers,
Dr. A
Excellent work, to the point and clear as hell.
Please keep the good work going. We need more lectures.
that practice falling from the top of the tank was perfect
Dr. A,
You took my understanding of physics on a whole new level. Thanks for the videos.
Pun intended? Level. As in water level? Okay, maybe not. Nonetheless, thanks for the comment.
Cheers,
Dr. A
@@yoprofmatt lmao😂 love it when teachers try to make jokes and they are actually not cringy.
another great Matt Anderson Video!! Thank you.......
by the way.. it's my understanding that if you hook up a hose right at the base of the Water Tank... then let the water flow though the hose... the water goes a distance "d" ... and if you then hook up a Spray Nozzle to the end of the hose.. and set the nozzle for a Jet spray... the water only shoots out a distance of "d" .. the same.. SAME Distance!!.. however, at home... you will find that a Nozzle (or pinching the hose) WILL Increase the distance!!.. apparently it has to do with Water Friction and the SPEED of water as it courses through the Pipes and hose. The Pressure at your end of the Line is Lower if the water is flowing Fast. I cite the professor Lewis Carroll Epstein from City College of San Francisco in his Book, "Thinking Physics, second edition" .. what a Surprise.. I never knew that...
How are you professor , hope you are well , it's really a cherishing moment and fun to watch your lectures !
Really appreciate your way of never letting the students feel bad or low when they sometimes are unable to answer correctly or precisely !
You just clarified something I struggled in lab for the past hour haha ty
Excellent good vibes physics lectures!! Making it easy to comprehend.
so helpful! thanks a lot Dr.A. I appreciate that.
+علي مصطفی الموسی
Thanks, علي مصطفی الموسی , glad to be of help.
Dr. A
Fantastic!
Respect! Besides the content....I am amazed at his ability to write backwards...wow!
amazing teacher.. keep it up sir! it helps me
fantastic
Thank you so much for this video ...I tried to find out similar videos but I couldn't get
you are great, sir.now i got a new teacher ,that's u. plz upload new, new videos. it's help full to me
Amazing. Thank you Sir
thanks a lot sir your videos are a great help !
You are most welcome! Glad you're finding them useful.
Cheers,
Dr. A
Thanks!
wish i had teachers like him.
Great work professor.
Thanks. Keep up with the physics!
Cheers,
Dr. A
Hi sir, thanks for the video. Can you please explain how much time it is required to empty a tank.
That's incredibly interesting
Great video, very informative. Can we do volumetric flow rate at exit dia of the tank?
does the size of the outlet pipe not matter?
what about as the water level drops?
Great
Thanks.
Dr. A
Would 'h' be measured from to the top, bottom, or centerline of the spigot?
Superb
Thanks professor
prof I just like your tutorial
good presentation
am an engineering student Uganda east Africa.
Great to hear. I love finding out where my Fandersons are from. Keep up with the physics!
Cheers,
Dr. A
so is it only the height that affects the velocity or is it the volume. So if you have a 50 gallon tower versus a 500 gallon tower and the height of both towers are the same the velocity of the water coming out would be the same?
In this problem why we not consider the centre of mass of water ?
I think that we can use the A1 V1 = A2 V2 to solve for V1 then substitute into Bernoulli's Equation. We can assume that v1 = 0 but we don't necessarily have too. As long as you know A1 = Pi r1^2
You would also need A2 in order to have a 2x2 ecuation that can be solved
What if the smaller pipe runs through the ground? Is it still opened to the atmosphere?
Doesn't the velocity also depend on the diameter of the pipe?
Dr. A,
Could you kindly make videos on quantum mechanics please? Your videos are so much helpful!
Thanks
I'm working on it. Might be a bit before I get to quantum, however.
Cheers,
Dr. A
Hi.if i may ask ,if the water tank is on the rooftop tower at level 40 and goes down to level 27 up to level 32 hotel rooms..do we still need a booster pump? it supply at least 500 hotel rooms..thank you
😑😑😑😑😑
very helpful
Thanks. The video is great. There is one thing that's not clear to me: Why do we assume P2 is the atmospheric pressure ? What happened to the hydrostatic pressure of the water column ?
p2 is at the very end of the spigot. The hydrostatic pressure inside the water tower drops to the atmospheric pressure as the water moves through the spigot.
@@rasmushochreuter2102 Thank you.
matt thnks for effort....but how do u write like that
I have a doubt
Initially u said we cannot apply continuity equation as we don't know v1 but while using the B-equation u neglected it assuming the tank to be very wide
Please clarify
in my opinion, when u using continuty eq, you don't know exactly the V1, and if we neglected it, it become A1. 0 = A2 . V2, so we can't find the V2 because A1 . 0 = 0 *CMIIW
he looks oddly like Ryan Reynolds in a professor kind of way
Hi, can you make videos on waves
Here you go: ua-cam.com/video/bwreHReBH2A/v-deo.html
Cheers,
Dr. A
good videos but what if the tank is closed from above?
@Mitchell Garrity actually,it wouldnt be the same.You would have to use a different variable
It will be on gauge pressure
How to find drain size which is require for particular tank...
Because as level of tank decrease the velocity is also decrease,
Then how i calculate drain size for constant outlet flow Q = A V
Dixit,
Per your question as h decreases: If you want Q to be constant, then as v decreases by sqrt (h), then A must increase by the same factor (1/sqrt(h)). Thanks for the comment, and keep up with the physics!
You might also like my new website: www.universityphysics.education
Cheers,
Dr. A
Matt Anderson
Thanks for the reply..
Surely i will like and subcribe your channel.
How much temperature increases this velocity
So we got biggest model ever possible
It's less then 100 degrees which is vapor pressure against atmospheric pressure
So if we do this experiment under a pressurized chamber then we got very high speed at discharge
Try it
Yep, haven't discussed temperature at all yet. Which of course comes into it with the ideal gas law.
Cheers,
Dr. A
@@yoprofmatt combined ideal gas law with heat formula with Bernoulli equation
Q= msdelta t
Bernoulli equation
Ideal gas laws
Coulombs law
Gravity laws
Quantum physics laws
Einstein equation
All together
For finding out the maximum power possible by this siphoning in herons fountain ?
why omit P0 tong when barometric pressure is equal to 1 and above altitude difference
P0 at the top of the tank is (nearly) identical to P0 at the bottom of the tank. At least the difference is very small when compared to the other terms in the equation.
Cheers,
Dr. A
@@yoprofmatt i have a question, why is P2 also atmospheric pressure? i get why P1 would be that, but for P2, wouldnt there be additional pressure to the atmospheric pressure, from the water being "pushed" out/pressurised due to the weight of the water above it?
hello in reality can water in a pipe reach 1000ms???
you look like Dr. Strange :)
You look like Yvonne.
Cheers,
Dr. strAnge
Sir is this also called Torricelli's Law?
Yes.
Cheers,
Dr. A
water psi in a tower is 1/2 psi per foot of height
Sounds about right, since about 30 feet of water = 1 ATM = 14.7 psi.
Cheers,
Dr. A
I don't get it, it looks like trying to reinvent the wheel. Why don't you just apply Torricelli's law? Basically you demonstrate Torricelli's law using Bernoulli's equation
Excellent catch. Yes, Torricelli's law is a particular case of Bernoulli's equation.
Cheers,
Dr. A
I feeel like METeorolgists use BERNOULIS eq all the time
Try to grow a beard sir and your name will be Professor/Doctor Strange. jk, anyway I learned a lot from watching your lectures thanks
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