1. Pay attension v2 is just the point velocity at point 2, not mean sectional velocity. So we should use velocity distribution to obtain the mean velocity V and then calculate Q by AV. 2. If point 1 and 2 are on a same horizontal line, which means they are not on the same sreamline due to different tube size, we can not apply Bernoulli Eq for them.
Nice example! One thing to be sure for the flow rate, the area of the end of the tube is pi*r^2, but the D on the diagram is mean Diameter. So the Q Should be pi * (D/2)^2*V2 right？
Why is the fluid in the left tube in static equilibrium but not the fluid in the right tube? And if both tubes are in static equilibrium shouldn't V1 and V2 both be zero?
Dude you make that equation look so elegant and yet it still frys my brain :-) The worst part is the online calculator agrees with my logic a bit but I still don't trust the answer :-p thanks for the great share, its helped :-)
@@SimmySigma alot thanks :-) I'm still wrapping my head around the relationship between flow speed and volume and over flow and laminar flow :-) I must admit I don't think I have covered that much ground so far to the point I will rewatch this video and try another round :-p I'm just shocked water can be so complicated and bernoulli so clever :-D
Very good video. I was confused at first when you wrote (-h2 - 0) but then I realized where you drew your datum (it is different from the datum point you began the video with). Really enjoyed this video. Thanks.
You are right. But when I square the (D/2), it becomes (D^2)/4, because I have to square the denominator as well. I factor out the (1/4) and thats where i get pi/4. :)
Same question actually. If the other hand was exposed to the atmosphere, it'd make sense because the pressures would cancel each other. But that's not the case
The idea comes from the fact that the atmosphere act everywhere in all direction so the atmospheric pressure cancels out anyways if you add it on the equation Patm+P1= Patm+pgh notice how Patm cancels out atmospheric pressure is always present sort of like gravity unless ur in a vacuum and it’s about 15psi believe
Awesome Vid Simmy, How would your answer differ if you used the original Datum (the one used for the z's in the bernoulli equation) for all the temporary datum you used earlier in the video. Please, Please reply.
Ogbonna Chris Thank you! If we used the original datum, the answer would still be the same. Take, for an example, P1. If we used our original datum, the pressure at point 1 would be -gamma*(0 - h2) = gamma*h2. Datums are just reference points relative to the points you are interested in. Try it out on paper and see if you get the same result! Hope this helps!
Sajeda, look at where he establishes his datum (y = 0). Since they share a common datum, z1 and z2 in the energy equation cancel out. His datum varies throughout his solution, but when he comes to the energy equation his datum is the point furthest to the right, from where h1 and h2 are referenced.
Samuel, he establishes the elevations of pt. 1 and pt. 2 at the datum to the far right; therefore, z1 and z2 share a common datum and there is no elevation difference between the two points. His reference frame is correct.
![Basics of Linear Momentum: Part 1 [Fluid Mechanics #27]](http://i.ytimg.com/vi/lz7hqpqm4aQ/mqdefault.jpg)
![Basics of Linear Momentum: Part 1 [Fluid Mechanics #27]](/img/tr.png)
Best channel for learning fluid mechanics
Dani Barker the last i time i used this channel two years ago.
I've got A on fluid mechanics.
1. Pay attension v2 is just the point velocity at point 2, not mean sectional velocity. So we should use velocity distribution to obtain the mean velocity V and then calculate Q by AV.
2. If point 1 and 2 are on a same horizontal line, which means they are not on the same sreamline due to different tube size, we can not apply Bernoulli Eq for them.
Nice example! One thing to be sure for the flow rate, the area of the end of the tube is pi*r^2, but the D on the diagram is mean Diameter. So the Q Should be pi * (D/2)^2*V2 right？
Why is the fluid in the left tube in static equilibrium but not the fluid in the right tube? And if both tubes are in static equilibrium shouldn't V1 and V2 both be zero?
Dude you make that equation look so elegant and yet it still frys my brain :-) The worst part is the online calculator agrees with my logic a bit but I still don't trust the answer :-p thanks for the great share, its helped :-)
Aw, thank you for the comment!! I’m glad it helped. 😊
@@SimmySigma alot thanks :-) I'm still wrapping my head around the relationship between flow speed and volume and over flow and laminar flow :-) I must admit I don't think I have covered that much ground so far to the point I will rewatch this video and try another round :-p I'm just shocked water can be so complicated and bernoulli so clever :-D
Do you have some PDFs ....best channel...am loving it!
Very good video. I was confused at first when you wrote (-h2 - 0) but then I realized where you drew your datum (it is different from the datum point you began the video with). Really enjoyed this video. Thanks.
You are right. But when I square the (D/2), it becomes (D^2)/4, because I have to square the denominator as well. I factor out the (1/4) and thats where i get pi/4. :)
why v1=0? cant get it
Because of stagnation. Theres essentially no fluid movement at that area
Burnouli equations holding how much maximum energy flow under NTP STP 🙏🙏🙏❓❓❓❓❓
Great work with units I can get behind! :) Thank you!
What is differential or integrated formula for this theorem??
Thanks for the upload!
Why velocity at 2 is not zero since velocity at 1 is zero ?
Horizontal velocity is not affected, but vertical velocity would be 0.
how is P4=0? ist it the atmospheric pressure?
Same question actually. If the other hand was exposed to the atmosphere, it'd make sense because the pressures would cancel each other. But that's not the case
The idea comes from the fact that the atmosphere act everywhere in all direction so the atmospheric pressure cancels out anyways if you add it on the equation Patm+P1= Patm+pgh notice how Patm cancels out atmospheric pressure is always present sort of like gravity unless ur in a vacuum and it’s about 15psi believe
If your pressure are gauge pressures then atmospheric pressure is zero
Awesome Vid Simmy, How would your answer differ if you used the original Datum (the one used for the z's in the bernoulli equation) for all the temporary datum you used earlier in the video. Please, Please reply.
Ogbonna Chris Thank you! If we used the original datum, the answer would still be the same. Take, for an example, P1. If we used our original datum, the pressure at point 1 would be -gamma*(0 - h2) = gamma*h2. Datums are just reference points relative to the points you are interested in. Try it out on paper and see if you get the same result! Hope this helps!
Can you please explain why z1 and z2 cancel out? How is h2 = h1 ?
Sajeda, look at where he establishes his datum (y = 0). Since they share a common datum, z1 and z2 in the energy equation cancel out. His datum varies throughout his solution, but when he comes to the energy equation his datum is the point furthest to the right, from where h1 and h2 are referenced.
Are we supposed to use the big diameter (0.1m) to find Q?
It would have been a nice video if you explained what gamma represented.
pg
density times gravitational acceleration
it is a good example but you should have to write giving space to see again legibly
Why can u say the velocity at 1 is equal to 0 whereas velocity at 2 isn’t ?
cause the point 1 a stagnant point
Thanks ❤️❤️❤️
Nice video. Out of curiosity (perhaps it is american notation) why do you use Gamma for density as opposed to Rho?
Craig Schofield Gamma denotes specific weight rather than density.
Craig, gamma = rho (density) x g (9.81 m/sec^2).
why dont you consider the pressure from the atmosphere?
here the gauge pressure is equal to 0
l didnt really understand why Z1 and Z2 became equal...isnt Z1 equal to H2 and Z2 equal to H1??
Samuel, he establishes the elevations of pt. 1 and pt. 2 at the datum to the far right; therefore, z1 and z2 share a common datum and there is no elevation difference between the two points. His reference frame is correct.
Apologies, I've just realised you're using Gamma for weight density.
Classic example...
incorrect