Can someone clarify how to factor in Height vs Depth in Bernouli's equation?

[manohman]

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So by bernoulis' principle,

P1+1/2pv^2+ pgH = constant

But part of pressure is the depth of a fluid. How do you distinguish the depth in a fluid vs the height of the fluid so you dont factor in depth or height twice?

 

[Cawolf]

The height should be the distance from the Earth's surface. It is the term for gravitational potential energy.

There is no "depth" term in that equation.

 

[manohman]

The height should be the distance from the Earth's surface. It is the term for gravitational potential energy.

There is no "depth" term in that equation.

But what about for something like this?

The pressure is different at different depth here. Do you just ignore the depth component because its always taken into account by the gravity component? Becaue the formula for height is pgH whereas for pressure at a depth its Pressure = pg*depth. But i guess when you compare the two sides of the equation the difference for either way of thinking is the same, but thats relaly confusing to think of both methods.

 

[Cawolf]

This would use Toricelli's theorem, which is really just a derivation of the fluid continuity expressions. It is obtained by approximating the velocity of the much larger top opening to be zero.

Here you would use the difference in height between the top and the hole to obtain that v = sqrt(2gh) with h being the difference in height from the top of the tank to the spout.

The pressure we are dealing with is the same at both holes, atmospheric! 🙂

 

[manohman]

This would use Toricelli's theorem, which is really just a derivation of the fluid continuity expressions. It is obtained by approximating the velocity of the much larger top opening to be zero.

Here you would use the difference in height between the top and the hole to obtain that v = sqrt(2gh) with h being the difference in height from the top of the tank to the spout.

The pressure we are dealing with is the same at both holes, atmospheric! 🙂

I see, so here still it's just height that takes into account the pressure difference at the different depths.

What about if youre comparing from inside the enclosure at the bottom of the tank to just after the water leaves the tank? The surface area is so large so velocity is zero, but the two are at the same height, but their pressures are different. The only way to take the difference in pressure into account is to cosnider depth right?

Is this a limiting case? Or should we consider pressure at depths if there is fluid above the point in question we are looking at? which makes sense becaue we do the same for calculating the pressure just as the water leaves the tank to be 1 atm (air as the liquid above, but with atmospheric pressure we say its basically the same at different depths).

 

[Cawolf]

I see, so here still it's just height that takes into account the pressure difference at the different depths.

Well in the picture you posted, the comparison of pressure at the top and spout is null. Both spots are open to the outside of the tank and have the same pressure, atmospheric.

What about if youre comparing from inside the enclosure at the bottom of the tank to just after the water leaves the tank? The surface area is so large so velocity is zero, but the two are at the same height, but their pressures are different. The only way to take the difference in pressure into account is to cosnider depth right?

So you are saying the same height, but before it leaves and after? That is the conversion of gravitational potential energy to kinetic energy. You would use depth to compare pressures when the tank is not open to the atmosphere, such as in a pipe or if there was no hole at the bottom of the pictured tank.

Is this a limiting case? Or should we consider pressure at depths if there is fluid above the point in question we are looking at? which makes sense becaue we do the same for calculating the pressure just as the water leaves the tank to be 1 atm (air as the liquid above, but with atmospheric pressure we say its basically the same at different depths).

I am not really sure what you are asking. There is a pressure gradient as you go down the tank and a greater height of water exists above some point.

 

[manohman]

Well in the picture you posted, the comparison of pressure at the top and spout is null. Both spots are open to the outside of the tank and have the same pressure, atmospheric.



So you are saying the same height, but before it leaves and after? That is the conversion of gravitational potential energy to kinetic energy. You would use depth to compare pressures when the tank is not open to the atmosphere, such as in a pipe or if there was no hole at the bottom of the pictured tank.



I am not really sure what you are asking. There is a pressure gradient as you go down the tank and a greater height of water exists above some point.

Yeah so if its at the same height as the opening and youre trying to find the velocity of the water as it leaves. If you used a height from the surface of the earth, both points of inteerest have the same height. If you look at the pressure, then its not just atmospheric pressure but also the depth of water,that go into the pressure value of bernoulis equation. BUT you could also take the pressure from the water into account if you make the bottom of the tank height = 0 meters and do the mpH from that point.

 

[Cawolf]

You lost me with that paragraph, sorry.

I think you are right. . . like I said, Toricelli's theorem is an approximation that only applies to the exit velocity of the spout.

There is pressure exerted by the water at every other point, except when it leaves.