At 7:30, why is does the atmospheric pressure at D not equal 0 and disappear. The final equation should be P1/pg + h1 = 0? Also it was proved that siphons can work in a vacuum, where p atm is negligible.
I was using absolute pressures instead of gage pressures, is why I used atm as 101kpa instead of 0. This problem could be completed in terms of gage pressure if you wanted, you would reach the same answer. But be careful when looking up vaporization pressures for your liquid, since those will probably be published as absolute pressure. I've seen the vacuum experiments (papers about them). There are some really cool siphon experiments that reject easy explanations. My favorite is the Flying Drop experiments. Very cool.
It's probably not harmful to think of it that way, even if technically not fully correct. Similar to in electrical circuits, we think of current flowing from positive to negative, even though we now know its electrons that are moving, and they move the other way. There's a bunch of these sorts of things throughout engineering curriculums, that while not entirely correct, it's still kind of ok to think that way, because it may still be useful in understanding a bigger picture, even if the details are a bit off.
@@BrianBernardEngineering Thanks for the reply. I agree. We should keep in mind as many possibilities as possible for some conclusion in order to be able to see what works to what extent. It should not be exclusive. Doubt about correctness leads to the right answer. I don't speak English very well, sorry if there are mistakes.
question regarding the example problem. we know that Vd=sqrt(2gh2) and yet we found h3 instead and used that. is that allowed per say? got kinda confused there
When finding maximum discharge rate Q (which is directly related to maximum velocity Vd, we ARE using h2 as you said. I think I skipped a part ... to achieve maximum velocity, h1 would be zero. This means that h3 the peak height, would be equal to h2, the height of the tank siphoning from. If h2 were 12.2m, and h3 a little higher, 12.5m, you would get cavitation and the siphon could break. If h3 were 12.2m and h2 were a little lower, only 11m, it would barely not cavitate, but velocity would be a little slower. You get maximum exit velocity when initial height h2 is at the max height h3, which is why I used the same height for both.
Is the diameter of the pipe influences the speed of the water flow? If so...whats the formula to count if i want to make the instalation fo water flow in 1000 Liter/ hour?
diameter does not affect "speed" as in meters/second, but it does affect "volumetric flow rate" as in liters/hour. Use Bernoulli eqn to find speed by comparing the top of the tank to the bottom exit from the tube. Then multiply speed by area of the pipe, pi r^2. You'll then have m/s*m^2 which gives you m^3/s, volumetric flow rate. Then unit conversion to liters per hour. Warning - this flowrate only applies at this moment in time. As the tank empties out, the flowrate will slow down, since the height gets lower as it empties, so the exit velocity will decrease, so flowrate decreases. I have another video called "Find time taken to empty a tank" which addresses this exact warning and how you can overcome it, if you are really interested in finding time, not just flowrate.
@@BrianBernardEngineering thank youuuu ❤️ its fine...because i will use power head pump with power 1000liter/hhour to return the water from bottom tank move to the over tank...to make the water will flows continuosly...so the water in over tank will not deacrease or going to empty ❤️❤️❤️ Now i find the missing key that diameters doesnt effect the speed of the flows as in meter/second ...big thanks for you ❤️❤️❤️
Hmmm, let's check. From Bernoulli Eqn, the exit velocity would not change based on exit diameter, it's only a function of the drop in height. Same velocity, but double the diameter, would mean 4x the volumetric flow rate (area*velocity, and area is proportional to diameter squared, and 2^2 = 4).
I'm trying to extend the siphon duration. I cant change the pipe size which is 3/4 in. if I add a 1/4 inch coupler to the outlet will the siphon still work?
Let's work it out. If both ends are at atmospheric pressure, the siphon really just converts PE on inlet into KE on outlet. And good news, kinetic energy is only based on velocity, not area. Same PE at inlet, same KE at outlet, so same velocity leaving, but with smaller area, means smaller volumetric flowrate. Volumetric flowrate Q is VA, and area is based on diameter squared. Your coupler reduces diameter by 3x, so it reduces area by 9x. If area is 9x smaller, then at same velocity, volumetric flowrate will be 9x lower. This should make time 9x longer to empty. Yes, I think your plan should work, and by making the outlet 3x smaller diameter, you should increase time to empty by 9x. This is an approximation since we aren't including head loss, but is hopefully good enough for your purpose.
This is an amazing explanation Dr. Brian. I have been scratching my head for a while now when it comes to siphons but your explanation was very clear! I have a burning question: does the siphon work better if the outlet end is submerged? Or does it not matter? (My thinking is- if it’s not submerged, air can get in more easily and break the siphon especially at low rates… but then again, if it’s not submerged, the pressure difference remains the same as time progresses. For example, if you let the outlet end of the siphon spill onto the floor, it will spill out until it’s empty, right? If submerged, the pressure difference decreases and the velocity decreases as time progresses, and a state of equilibrium is reached as you explained.)
Main difference with submerged exit is increased pressure. However far under the surface, that's extra pressure. Bernoulli Eqn from inlet to outlet, you would then have a lower change in pressure between inlet and outlet, same change in height between inlet and outlet, therefore velocity difference would have to be smaller. Flow rate would be slower.
Does the pressure of the water depth have any effect on siphon height? For instance say the siphon was located deep in the ocean. And assume the other end was still below relative. Somehow it’s magically emptying into the abyss or something. Could you increase the maximum height of the siphon then, beyond 10 meters? Not that there would be any particular use for this. I’m just wondering about the mechanics of the maximum height.
Yes, Pressure at the exit DOES make a difference. This video focused on examples where both ends are at atmospheric pressure, since that's the normal scenario we see most often, like when draining water out of a large aquarium. But if the exit is above atmospheric pressure, then the siphon height CAN be higher than the max I stated in this video. This is a really good sign, when you are able to learn something, and you then can find edge cases like this where it's a little bit different - shows you are understanding the bigger picture and not just memorizing details. Good work!
@@BrianBernardEngineering Great thanks a lot for the reply and the educational content :) Does the water pressure at the entrance make any difference? Or in other words, the water depth of the end of the siphon (the higher end, the one where water will enter from).
@@solank7620 Yes, you're right, both inlet and exit pressures matter, it's really the difference in pressure between inlet and outlet that is important. Regular siphon, same atm pressure both ends, the difference in height inlet to outlet is what determines the velocity in the siphon. When inlet and exit pressures are different, that difference also contributes to velocity in the siphon, just like difference in height. If inlet lower pressure than outlet, velocity will be lower than it would be based on height alone. Once you know velocity, then you can get max height. And you should get same height whether you measure above inlet or above outlet.
@@BrianBernardEngineeringthat is so interesting because it is a bit counterintuitive - that the siphon height increases with higher pressure at the outlet of the siphon. But looking at the equation, that is true. But still I can’t get my head around it!
Working with my TAs is the best part of the job. My most recent upload (Inclined Manometer) has footage of Indy running on his treadmill, so that was definitely fun to shoot.
Why couldn't I compare points A and C to get the maximum height of the siphon? knowing the pressures at both points and assuming velocity at point C is zero at the saturation pressure, I tried to solve it that way but it gave me a bigger value of height and I don't get why it's wrong.
Velocity wouldn't be zero at the top. From the continuity equation, velocity needs to be equal at every point in the tube, otherwise you are accumulating mass somewhere instead of mass in = mass out for each control volume. Restated - the conservation of mass requires velocity to be the same at the top as at the exit (since cross sectional area is the same at both points, really its mass flow rate that has to be equal at both, but in this problem that makes velocity the same at both spots). From conservation of energy, since you didn't include kinetic energy at the top, the height you found was too large since the gravitational potential energy included energy that should have gone towards v.
Great video, there's something that confuses me though What's the hydrostatic pressure at C and D ? Is there even hydrostatic pressure at those points? If there is, point D is lower than point B so how come water flows from lower pressure to higher?
C (top of the siphon) to D (exit of the siphon): Point D is exposed to the atmosphere, so if hydrostatic pressure is rho*g*h, point D is at h=0. It's not submerged at all, it's at the surface. I'd call this 0 gage pressure (or atmospheric absolute pressure). Point C ... this point has a negative depth from D. It's hydrostatic pressure is rho*g*-h, since it's not at a depth at all, it's actually above point D. Electricity flows from higher voltage to lower voltage. Heat flows from higher temperature to lower temperature. These are Potentials. Flow is generally from higher potential to lower potential. Fluid flow is often from high pressure to low pressure - but it doesn't have to be. There are other sources of potential that are relevant besides pressure, ie gravity. Water prefers to flow downhill, not uphill. In this siphon, pressure and gravity fight each other, which is why the result is partially counterintuitive.
Thank you so much sir, for taking the time to reply🙏. So..when you mention point D is at zero height, which height do you mean, from where do you measure that h ? If we were to calculate the hydrostatic pressure in the first container then height zero whould be at point A. And sure point C whould have negative height, but point D wouldn't. As the 2 containers are filled with water they are basically connected through the siphon, so can we calculate the hydrostatic pressure as if we had one large container? If we do so and point A is zero height then point D has a very large hydrostatic pressure, larger than any point in the first container.
@@captainamericawhyso5917 We're playing a little loosey goosey with the term "hydrostatic" in this discussion. Static = not moving. The difference between A and D is velocity, A is modeled as stationary (or moving very slow approximately zero), but fluid at D is moving. You must account for the dynamics, not just the statics. That's why D isn't at higher pressure than A. The gravitational potential energy at A didn't become pressure at D, it became velocity. For C and D, Even though fluid is moving between C and D, it's moving at same velocity, which allows a hydrostatic analysis between them, since the velocity terms cancel out. The hydrostatic pressure equation is just the Bernoulli Eqn, with velocities either zero, or equal to each other.
Thank you so much. For some reason I mistook the sigh p as pressure instead of densety and couldn't for the life of figure out how to get the Max height.
very easy mistake to make. Same with all the different 'v's, volume, velocity, specific volume...I always try to really exaggerate the slant in the density character to visually distinguish it from a p. That helps me out. Of course the best way is to just always write out all your units, and you can catch these mistakes since the units won't match up ... but yea, I'll probably never actually follow through with that.
I try to thread a fine balance - I want to be as short as possible to respect your time, but you're right, it's possible to go too fast that it's hard to understand if I move on before the last part has sunk in. I'll keep working on it.
This is hands down the best explanation on TH-cam. You are a terrific communicator and your real world examples are great.
It gets a little silly when the cats get involved, so I'm glad the lesson still stuck.
At 7:30, why is does the atmospheric pressure at D not equal 0 and disappear. The final equation should be P1/pg + h1 = 0? Also it was proved that siphons can work in a vacuum, where p atm is negligible.
I was using absolute pressures instead of gage pressures, is why I used atm as 101kpa instead of 0. This problem could be completed in terms of gage pressure if you wanted, you would reach the same answer. But be careful when looking up vaporization pressures for your liquid, since those will probably be published as absolute pressure. I've seen the vacuum experiments (papers about them). There are some really cool siphon experiments that reject easy explanations. My favorite is the Flying Drop experiments. Very cool.
Nice and logical explanation. 1:15 In some schools the operation of the siphon is still explained as the result of atmospheric pressure.
It's probably not harmful to think of it that way, even if technically not fully correct. Similar to in electrical circuits, we think of current flowing from positive to negative, even though we now know its electrons that are moving, and they move the other way. There's a bunch of these sorts of things throughout engineering curriculums, that while not entirely correct, it's still kind of ok to think that way, because it may still be useful in understanding a bigger picture, even if the details are a bit off.
@@BrianBernardEngineering Thanks for the reply. I agree. We should keep in mind as many possibilities as possible for some conclusion in order to be able to see what works to what extent. It should not be exclusive. Doubt about correctness leads to the right answer. I don't speak English very well, sorry if there are mistakes.
Supremely underrated video and channel! Great explanation!
Thank you so much, that's really kind of you to say, and I know your TA Indiana appreciates it too.
question regarding the example problem. we know that Vd=sqrt(2gh2) and yet we found h3 instead and used that. is that allowed per say? got kinda confused there
When finding maximum discharge rate Q (which is directly related to maximum velocity Vd, we ARE using h2 as you said. I think I skipped a part ... to achieve maximum velocity, h1 would be zero. This means that h3 the peak height, would be equal to h2, the height of the tank siphoning from. If h2 were 12.2m, and h3 a little higher, 12.5m, you would get cavitation and the siphon could break. If h3 were 12.2m and h2 were a little lower, only 11m, it would barely not cavitate, but velocity would be a little slower. You get maximum exit velocity when initial height h2 is at the max height h3, which is why I used the same height for both.
Is the diameter of the pipe influences the speed of the water flow? If so...whats the formula to count if i want to make the instalation fo water flow in 1000 Liter/ hour?
diameter does not affect "speed" as in meters/second, but it does affect "volumetric flow rate" as in liters/hour. Use Bernoulli eqn to find speed by comparing the top of the tank to the bottom exit from the tube. Then multiply speed by area of the pipe, pi r^2. You'll then have m/s*m^2 which gives you m^3/s, volumetric flow rate. Then unit conversion to liters per hour. Warning - this flowrate only applies at this moment in time. As the tank empties out, the flowrate will slow down, since the height gets lower as it empties, so the exit velocity will decrease, so flowrate decreases. I have another video called "Find time taken to empty a tank" which addresses this exact warning and how you can overcome it, if you are really interested in finding time, not just flowrate.
@@BrianBernardEngineering thank youuuu ❤️ its fine...because i will use power head pump with power 1000liter/hhour to return the water from bottom tank move to the over tank...to make the water will flows continuosly...so the water in over tank will not deacrease or going to empty ❤️❤️❤️
Now i find the missing key that diameters doesnt effect the speed of the flows as in meter/second ...big thanks for you ❤️❤️❤️
what if the output tube diameters is diuble from the input. and the down loop is a a downward arch! It should increase the transport speed.
Hmmm, let's check. From Bernoulli Eqn, the exit velocity would not change based on exit diameter, it's only a function of the drop in height. Same velocity, but double the diameter, would mean 4x the volumetric flow rate (area*velocity, and area is proportional to diameter squared, and 2^2 = 4).
I'm trying to extend the siphon duration. I cant change the pipe size which is 3/4 in. if I add a 1/4 inch coupler to the outlet will the siphon still work?
Let's work it out. If both ends are at atmospheric pressure, the siphon really just converts PE on inlet into KE on outlet. And good news, kinetic energy is only based on velocity, not area. Same PE at inlet, same KE at outlet, so same velocity leaving, but with smaller area, means smaller volumetric flowrate. Volumetric flowrate Q is VA, and area is based on diameter squared. Your coupler reduces diameter by 3x, so it reduces area by 9x. If area is 9x smaller, then at same velocity, volumetric flowrate will be 9x lower. This should make time 9x longer to empty. Yes, I think your plan should work, and by making the outlet 3x smaller diameter, you should increase time to empty by 9x. This is an approximation since we aren't including head loss, but is hopefully good enough for your purpose.
This is an amazing explanation Dr. Brian. I have been scratching my head for a while now when it comes to siphons but your explanation was very clear! I have a burning question:
does the siphon work better if the outlet end is submerged? Or does it not matter?
(My thinking is- if it’s not submerged, air can get in more easily and break the siphon especially at low rates… but then again, if it’s not submerged, the pressure difference remains the same as time progresses. For example, if you let the outlet end of the siphon spill onto the floor, it will spill out until it’s empty, right? If submerged, the pressure difference decreases and the velocity decreases as time progresses, and a state of equilibrium is reached as you explained.)
Main difference with submerged exit is increased pressure. However far under the surface, that's extra pressure. Bernoulli Eqn from inlet to outlet, you would then have a lower change in pressure between inlet and outlet, same change in height between inlet and outlet, therefore velocity difference would have to be smaller. Flow rate would be slower.
Does the pressure of the water depth have any effect on siphon height?
For instance say the siphon was located deep in the ocean.
And assume the other end was still below relative. Somehow it’s magically emptying into the abyss or something.
Could you increase the maximum height of the siphon then, beyond 10 meters?
Not that there would be any particular use for this. I’m just wondering about the mechanics of the maximum height.
Yes, Pressure at the exit DOES make a difference. This video focused on examples where both ends are at atmospheric pressure, since that's the normal scenario we see most often, like when draining water out of a large aquarium. But if the exit is above atmospheric pressure, then the siphon height CAN be higher than the max I stated in this video. This is a really good sign, when you are able to learn something, and you then can find edge cases like this where it's a little bit different - shows you are understanding the bigger picture and not just memorizing details. Good work!
@@BrianBernardEngineering Great thanks a lot for the reply and the educational content :)
Does the water pressure at the entrance make any difference? Or in other words, the water depth of the end of the siphon (the higher end, the one where water will enter from).
@@solank7620 Yes, you're right, both inlet and exit pressures matter, it's really the difference in pressure between inlet and outlet that is important. Regular siphon, same atm pressure both ends, the difference in height inlet to outlet is what determines the velocity in the siphon. When inlet and exit pressures are different, that difference also contributes to velocity in the siphon, just like difference in height. If inlet lower pressure than outlet, velocity will be lower than it would be based on height alone. Once you know velocity, then you can get max height. And you should get same height whether you measure above inlet or above outlet.
@@BrianBernardEngineeringthat is so interesting because it is a bit counterintuitive - that the siphon height increases with higher pressure at the outlet of the siphon. But looking at the equation, that is true. But still I can’t get my head around it!
most underated video in the world . as a cat lover it makes it even more fun. xD
Working with my TAs is the best part of the job. My most recent upload (Inclined Manometer) has footage of Indy running on his treadmill, so that was definitely fun to shoot.
Thanks, helped a lot!
You're welcome!
Why couldn't I compare points A and C to get the maximum height of the siphon? knowing the pressures at both points and assuming velocity at point C is zero at the saturation pressure, I tried to solve it that way but it gave me a bigger value of height and I don't get why it's wrong.
Velocity wouldn't be zero at the top. From the continuity equation, velocity needs to be equal at every point in the tube, otherwise you are accumulating mass somewhere instead of mass in = mass out for each control volume. Restated - the conservation of mass requires velocity to be the same at the top as at the exit (since cross sectional area is the same at both points, really its mass flow rate that has to be equal at both, but in this problem that makes velocity the same at both spots). From conservation of energy, since you didn't include kinetic energy at the top, the height you found was too large since the gravitational potential energy included energy that should have gone towards v.
Thnq sir for this video its very usefull to my studies 🙏🇱🇰
You're very welcome. I hope you do great!
Great video, there's something that confuses me though
What's the hydrostatic pressure at C and D ? Is there even hydrostatic pressure at those points? If there is, point D is lower than point B so how come water flows from lower pressure to higher?
C (top of the siphon) to D (exit of the siphon): Point D is exposed to the atmosphere, so if hydrostatic pressure is rho*g*h, point D is at h=0. It's not submerged at all, it's at the surface. I'd call this 0 gage pressure (or atmospheric absolute pressure). Point C ... this point has a negative depth from D. It's hydrostatic pressure is rho*g*-h, since it's not at a depth at all, it's actually above point D. Electricity flows from higher voltage to lower voltage. Heat flows from higher temperature to lower temperature. These are Potentials. Flow is generally from higher potential to lower potential. Fluid flow is often from high pressure to low pressure - but it doesn't have to be. There are other sources of potential that are relevant besides pressure, ie gravity. Water prefers to flow downhill, not uphill. In this siphon, pressure and gravity fight each other, which is why the result is partially counterintuitive.
Thank you so much sir, for taking the time to reply🙏.
So..when you mention point D is at zero height, which height do you mean, from where do you measure that h ?
If we were to calculate the hydrostatic pressure in the first container then height zero whould be at point A. And sure point C whould have negative height, but point D wouldn't.
As the 2 containers are filled with water they are basically connected through the siphon, so can we calculate the hydrostatic pressure as if we had one large container? If we do so and point A is zero height then point D has a very large hydrostatic pressure, larger than any point in the first container.
@@captainamericawhyso5917 We're playing a little loosey goosey with the term "hydrostatic" in this discussion. Static = not moving. The difference between A and D is velocity, A is modeled as stationary (or moving very slow approximately zero), but fluid at D is moving. You must account for the dynamics, not just the statics. That's why D isn't at higher pressure than A. The gravitational potential energy at A didn't become pressure at D, it became velocity. For C and D, Even though fluid is moving between C and D, it's moving at same velocity, which allows a hydrostatic analysis between them, since the velocity terms cancel out. The hydrostatic pressure equation is just the Bernoulli Eqn, with velocities either zero, or equal to each other.
Thank you very much once again🙇♂️
I'll need to think this a little more to understand whats going on
how would you go about finding the velocity at the submerged end of the pipe and therefore the static pressure there? great vid
based on continuity, velocity will be the same at every point of the pipe, so the submerged inlet will have same velocity as the exit of the pipe.
GREAT VIDEO MAN
Thank you so much. I couldn't have done it without your TA Indiana's help though.
The best explanation ever
thanks, I'm glad I could help
Thank you so much. For some reason I mistook the sigh p as pressure instead of densety and couldn't for the life of figure out how to get the Max height.
very easy mistake to make. Same with all the different 'v's, volume, velocity, specific volume...I always try to really exaggerate the slant in the density character to visually distinguish it from a p. That helps me out. Of course the best way is to just always write out all your units, and you can catch these mistakes since the units won't match up ... but yea, I'll probably never actually follow through with that.
WHOA !!! I REMEMBER THIS DUDE...HE HAS A SON NAMED CHARLIE BROWN !
No mistaking that voice.
excellent memory, i never even showed my face in the cartoons. thanks for being such a long time fan.
so fast man, slow it down
I try to thread a fine balance - I want to be as short as possible to respect your time, but you're right, it's possible to go too fast that it's hard to understand if I move on before the last part has sunk in. I'll keep working on it.