This is actually great. I failed fluid mechanics back in '93 and now got an itching to learn it again and do it right! Started with siphons because so many people seem to argue about how they work. I like this gentleman's style and explanation, although I nearly 'broke the tape' going back and replaying bits of it!! Thank you for a much appreciated video.
bro, that's a crazy, nice speech, also I was shocked by this stuff, I can't imagine this was a video 16 years ago, although the video quality is bad, the teacher quality is good!
Yes of course friction is negligible in this case. When you consider friction, you need to analyze the energy loss using a certain 'Energy Equation', an extension of the Bernoulli's equation I believe. All this is in another section called pipe analysis, which unfortunately I have no videos on it.
How much total flow rate we can have at maximum if we used running water of river ??we use the maximum limits of siphoning so we are able to make electricity by it ??how much maximum works done by river flow through the siphoning ??please tell?
Thank you Donny, I t was very helpful... I wish you explained more on the continuity equation and how we found the pressure difference. In addition, if you told us the assumptions like steady flow it would have been of great benefit.
Hey Dony, At point C, why is it that we choose the pressure to be 1 atm and not PB, the pressure inside of the pipe? That's the only thing that puzzles me. What's a good way to think about it? - Reuben
Never thought about it during fluids class, but if Za = Zc , then there is no velocity and the siphon doesn't work. The higher Za is until it reaches Zb, the quicker Vc is.
hi , that was really help full. but i have a q we assumed that Vc is much greater than Va , but you said the flow on Vc compared to Va is very slow. how is Vc greater Than Va then ?
There is a major error that is being ignored. You should label this as "Siphoning non-viscus fluids" , not "Siphoning Water ". I had some calculations to make for a working system, and IT WOULD NOT WORK.. In fact root (gh) gave results within 10% accuracy of tests.. For water, root (2gh) was off by more than 40%. I relied on these false formula. You may not realize it, but water is very viscus. After searching, I found formulas that work well for water in pipes less than 8cm diameter.
At 5:00 the equation is not correct. It should be [root of (.5GH)] or [root of (G H / 2)] IT SHOULD NOT BE [root of (2G H)] . GET A TUBE AND PAIL OF WATER AND DO THE TEST. the physics is wrong. I used several tubes of several sizes. The correct answer is 3.4, not 6.8 - - 6.8 is 100% higher than 3.4 - - therefore the answer is 100% wrong.- - Get the water out and do the test.
This is actually great. I failed fluid mechanics back in '93 and now got an itching to learn it again and do it right! Started with siphons because so many people seem to argue about how they work. I like this gentleman's style and explanation, although I nearly 'broke the tape' going back and replaying bits of it!! Thank you for a much appreciated video.
Damn I remember watching this during my first year first semester of college and now i'm finishing up my final semester in my 4th year... crazy
I took fluid mechanics 2 years ago and needed a refresher for a project I am working on that uses siphons. Thanks
bro, that's a crazy, nice speech, also I was shocked by this stuff, I can't imagine this was a video 16 years ago, although the video quality is bad, the teacher quality is good!
How inverted siphoning is differentiated from siphoning❓❓❓🙏🙏🙏
Yes of course friction is negligible in this case. When you consider friction, you need to analyze the energy loss using a certain 'Energy Equation', an extension of the Bernoulli's equation I believe.
All this is in another section called pipe analysis, which unfortunately I have no videos on it.
How much total flow rate we can have at maximum if we used running water of river ??we use the maximum limits of siphoning so we are able to make electricity by it ??how much maximum works done by river flow through the siphoning ??please tell?
What is discharge velocity in bell siphoning
In joule terms ?
What is mathematics for Bell siphoning ?
Will it give more pressure what if we use 15 meter long riser for better negative pressure ????
Thank you Donny,
I t was very helpful...
I wish you explained more on the continuity equation and how we found the pressure difference. In addition, if you told us the assumptions like steady flow it would have been of great benefit.
Do you also assume the friction between the moving water and the syphoning pipe is negligible ?
in pipe is inserted inward then from where the major head loss should be taken
Great! Love your enthusiasm as you explain
Hey Dony,
At point C, why is it that we choose the pressure to be 1 atm and not PB, the pressure inside of the pipe? That's the only thing that puzzles me. What's a good way to think about it?
- Reuben
Never thought about it during fluids class, but if Za = Zc , then there is no velocity and the siphon doesn't work. The higher Za is until it reaches Zb, the quicker Vc is.
hi , that was really help full. but i have a q
we assumed that Vc is much greater than Va , but you said the flow on Vc compared to Va is very slow. how is Vc greater Than Va then ?
What's the difference between A' and A??
what if the pipe is horizontal?
very very helpful! thanks a lot donylee! cheers from the philippines!
This guys talks at 100 miles an hour and I still understood it, how is that possible
Talvez algún día puedas explicar la nueva mecánica de fluidos de mi invención
What if point A to B has head and B to C again has a different head.
you are a very good teacher! Thanks!
you are a great teacher. Thanks
Thank you for the video, helped me a lot.
thanks brother...
Amazing.
Thanks.
Thanks
you are really helpfull but you speak too fast why is that rush :D thanks anyway u are a saver :)
you are too fast but this is really helpful
from chandanii chauk to china
root (2gh) is the correct answer
There is a major error that is being ignored. You should label this as "Siphoning non-viscus fluids" , not "Siphoning Water ".
I had some calculations to make for a working system, and IT WOULD NOT WORK.. In fact root (gh) gave results within 10% accuracy of tests.. For water, root (2gh) was off by more than 40%.
I relied on these false formula. You may not realize it, but water is very viscus.
After searching, I found formulas that work well for water in pipes less than 8cm diameter.
dude you are right, but he says some assumption like density is constant. dose non viscous liquid has constant density?
I am sorry but your 1.2 m is obscured...And i thought it was 2 m all the while..
Do you ever enjoy simply watching water? Or do you always have to understand, break down, analyze it?
At 5:00 the equation is not correct. It should be [root of (.5GH)] or [root of (G H / 2)]
IT SHOULD NOT BE [root of (2G H)] . GET A TUBE AND PAIL OF WATER AND DO THE TEST. the physics is wrong. I used several tubes of several sizes. The correct answer is 3.4, not 6.8 - - 6.8 is 100% higher than 3.4 - - therefore the answer is 100% wrong.- - Get the water out and do the test.
you talk like the actor for bruce lee in Once Upon a Time in Hollywood
not in a racist way. like you're a great guy sorry If I offended you
you say too fast :((
dude you speak so fast
great video but its equally hard to understand your accent!!
That is your problem, not his. Awesome video, thanks Donny
I agree, simple to understand. Great video!
Speaking English with very strong asian accent...very hard to understand!!
you are really helpfull but you speak too fast why is that rush :D thanks anyway u are a saver :)