Bernoulli's Equation --- Velocity out of a Spout

[Dr Gerrard]

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All right, here is another question. The first one is mostly just a semantics issue, which I still am curious about, but this is actually a conceptual question.

When you have fluid flowing out of a container at some point, through a spout, why does V = square root of 2gh?

This equation includes the assumption that the pressure of the fluid coming out of the spout is the same as the pressure at the surface of the fluid, which would be atmospheric pressure. However, I do not see how the fluid at the spout is the same as atmospheric pressure.

1) There is a height difference, and P = Patm + rho*g*h
2) The area of the spout is smaller, thus increasing the velocity. No matter where you read, it says that the greater the velocity, the smaller the pressure.

 

[kobe200LATE]

This is just an application of Bernoulli's law. Lets say point A is at the top of the water tank and point B is at the bottom where the spout is. The height difference between points A and B is h.

So for point A, we have Patm + (1/2)rho*Va^2 + rho*g*h

For point B, we have Patm + (1/2)rho*Vb^2 + rho*g*h

Setting these two values equal, we have Patm + (1/2)rho*Va^2 + rho*g*h = Patm + (1/2)rho*Vb^2 + rho*g*h, with both Patm's cancelling out

If we make point B our zero point of reference for h, then the rho*g*h term = 0 for point B.

Also, since the area at point A is so much greater than the area at point B, we can assume that the velocity Va = 0

This leaves us with the equation: rho*g*h = (1/2)rho*Vb^2

Solving for Vb gives us the square root of 2gh

Also, you are confusing the hydrostatic pressure with the dynamic pressure of the fluid. The hydrostatic pressure would be greater at point B because it is at a higher depth, but the dynamic pressure (the one involved with Bernoulli's equation) is completely different.

 

[Dr Gerrard]

I think the last two sentences is what I was looking for.

I mainly did not understand why pressure at point A and point B is the same.

You mentioned dynamic pressure, can you explain how that is different from hydrostatic pressure?

Is Hydrostatic pressure = Patm + rho*g*h? Or is this dynamic pressure?

Is dynamic pressure constant everywhere in a fluid? If so, what is the point of it even being in the equation if it is always constant?

 

[kobe200LATE]

IIRC, the pressure at points A and point B are both equal because they are both exposed to the atmosphere. Therefore, their pressures are both equal to the pressure of the atmosphere.

Dynamic pressure is different from hydrostatic pressure because dynamic pressure is the pressure that you use when a fluid is moving. Hydrostatic pressure is the pressure of a fluid when it is not moving (ie when its static). This means that the Patm + rho*g*h is due to hydrostatic pressure.

I also don't think dynamic pressure is constant everywhere in a fluid. It is in this case, but not all cases. For example, when two fluids are at the same height, the fluid that has a higher velocity will have a lower dynamic pressure while the fluid with a lower velocity will have a higher dynamic pressure. The dynamic pressure basically makes up for the difference between the velocity and position of the fluid in Bernoulli's equation.

Overall, I think you just need to stop thinking about hydrostatic pressure for this problem. Hydrostatic pressure equations only apply when the fluid is not moving, but in this case it is clearly moving. Therefore, you must use Bernoulli's equation instead and not let hydrostatic pressure interfere with your thinking.

Hope this helps