Find Max Height for a Siphon - Bernoulli and Continuity Equation Example Problem

By mini-lecture, experiment, and example problem - you’ll learn how to avoid sucking gasoline to start a siphon, what the max flowrate you can obtain will be, and how tall you’ll need to make it.
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The continuity equation example problem in this video will show that by varying the length of the siphon, you can control the flowrate. You can find the flowrate of the siphon using the Bernoulli Equation and Continuity Equation. As a Bernoulli equation example problem you’ll be comparing the surface of your tank to the exit of your drain pipe. When the siphon is used to draw water (or in this problem Ethanol), in order to siphon without cavitation occurring, you’ll need to find the saturation pressure of the siphon which will vary depending on the fluid.
Depending on which textbook you are using, Bernoulli Equation and Continuity Equation may be in Fluid Mechanics Chapter 4 or Fluid Mechanics Chapter 5. For this siphon experiment Bernoulli Equation doesn’t fully account for head loss at many points in the system, but is still a good starting point for main concepts. You’ll learn how to account for losses later in the course when studying continuity equation for flow in a conduit, Moody Diagram, Reynold’s Number, Minor Losses, etc.
TIMECODES
0:00 Introduction
0:48 How a siphon works
4:07 Easy Siphon Experiments
5:23 Bernoulli Equation and Continuity Equation
8:24 Siphon Example Problem

[@ObeyRoastMan]

This is hands down the best explanation on KZbin. You are a terrific communicator and your real world examples are great.

[@BrianBernardEngineering]

It gets a little silly when the cats get involved, so I'm glad the lesson still stuck.

[@WestExplainsBest]

Supremely underrated video and channel! Great explanation!

[@BrianBernardEngineering]

Thank you so much, that's really kind of you to say, and I know your TA Indiana appreciates it too.

[@IceKingProduc]

most underated video in the world . as a cat lover it makes it even more fun. xD

[@BrianBernardEngineering]

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.

[@bodeahmed1122]

The best explanation ever

[@BrianBernardEngineering]

thanks, I'm glad I could help

[@user-mi8sb2wc8u]

GREAT VIDEO MAN

[@BrianBernardEngineering]

Thank you so much. I couldn't have done it without your TA Indiana's help though.

[@sentinel8910]

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

[@BrianBernardEngineering]

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.

[@therocketman4494]

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.

[@BrianBernardEngineering]

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.

[@camdjkim]

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?

[@BrianBernardEngineering]

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.

[@freshmuff276]

how would you go about finding the velocity at the submerged end of the pipe and therefore the static pressure there? great vid

[@BrianBernardEngineering]

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.

[@mariamgirgis7623]

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.

[@BrianBernardEngineering]

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.

[@solank7620]

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.

[@BrianBernardEngineering]

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!

[@solank7620]

@@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).

[@BrianBernardEngineering]

@@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.

[@jaymzgaetz2006]

WHOA !!! I REMEMBER THIS DUDE...HE HAS A SON NAMED CHARLIE BROWN ! No mistaking that voice.

[@BrianBernardEngineering]

excellent memory, i never even showed my face in the cartoons. thanks for being such a long time fan.

[@captainamericawhyso5917]

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?

[@BrianBernardEngineering]

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.

[@captainamericawhyso5917]

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.

[@BrianBernardEngineering]

@@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.

[@glykerkele6651]

Thank you very much once again🙇‍♂️ I'll need to think this a little more to understand whats going on

[@huskers3120]

so fast man, slow it down

[@BrianBernardEngineering]

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.