Bernoulli's Equation Applied to Two Free Surfaces

[tjm2444]

and p_2=\rho g y_2, the hydrostatic pressure.
\rho g y_1 = \rho g y_2 + \dfrac{1}{2}\rho v_2^2
So solving for v_2
v_2 = \sqrt{2 g h}
This is the velocity of the water moving through the pipe.

But aren't you saying here that the pressure in the pipe is simply the hydrostatic pressure? Is the water moving at point 2? If so, then the pressure at 2 is lower than the hydrostatic pressure.

 

[256bits]

One last time, yes it does.

See image.

You are using 2 separate datums, which is why this is not the way to apply Bernoulli.

P1 + pgh1 + 1/2 p v1^2 = P2 + pgh2 + 1/2 p v2^2

Using a datum level as the surface of liquid 2:

v1 = -v2
P2 = 0
h2 =0

the equation reduces to
P1 + pgh1 + 1/2 p v1^2 = P2 + pgh2 + 1/2 p v2^2
P1 + pgh1 + 1/2 p v1^2 = 0 + 0 + 1/2 p v2^2

And 1/2 p v1^2 = 1/2 p v2^2

So P1 = pgh1

 

[boneh3ad]

First, thank you for understanding my question, bone3ead!

Most texts say that Bernoulli's Equation applies to any two points along a streamline. Is there no streamline that can be traced from the free surface on one side to that on the other? And if there is (which I believe is the case) then why does Bernoulli's Equation fail in this case?

The motion is occurring everywhere. The left surface is moving down and the right up.

I am sure you could find a streamline somewhere that would connect two points such as those you are using, but in reality, if the tank is much larger than the pipe connecting the two, most of the motion will be in the region of the pipe. At any rate, the entire problem is unsteady, so Bernoulli's equation doesn't really apply. There is an unsteady analog to Bernoulli's equation, but then you get significantly more complicated. There are much easier ways to analyze the flow than in that case.

The solution is to say that the surfaces aren't changing much (i.e. D >> d from my previous post) and approximate the situation as steady. Again, it is only an approximation though.

But aren't you saying here that the pressure in the pipe is simply the hydrostatic pressure? Is the water moving at point 2? If so, then the pressure at 2 is lower than the hydrostatic pressure.

The static pressure at one opening is the hydrostatic pressure from that tank and the static pressure at the other opening is the hydrostatic pressure from the height of the water in that tank.

The pressure at point 2 is zero (gauge). There is no pressure other than atmospheric at that point. Same with point 1. The potential energy then that leads to the motion between tanks is gravitational in nature and the flow rate between the two (and thus velocity of the surfaces) is tied to the height difference. However, it is between the inlet and outlet of the pipe that you really need to apply Bernoulli's equation to get your estimate. Picking two points at each surface loses any information about what occurs between them because both points are essentially in equilibrium.

 

[Studiot]

So P1 = pgh1

That's an interesting assertion for a free surface of a liquid.

Would you like to explain it?

 

[256bits]

That's an interesting assertion for a free surface of a liquid.

Would you like to explain it?

P1 is the pressure of the liquid in container 1 at the same level as the surface of container 2.
P1 is NOT the pressure of the free surface of container 1 which is causing the confusion here.

The energy level at the two free surfaces 1 and 2 are of course not going to be equal.
By how much - by the value P1 = pgh1.