Hydrodynamics:Bernoulli's equation

[woollyyak]

Homework Statement

Torricelli's thereom is v = sqroot(2g(y2-y1)) where the velocity at the top is negligable compared to the velocity at the bottom (v).
If the velocity at the top is considered (v2) then v1A1=v2A2 is substituted into Bernoulli's to give

v1 = sqroot(2g(h2-h1)/1-(A1/A2)^2)

Homework Equations

The above equation says that if A1=A2 then the velocity at V1 is infinite. Is that right? How should this be interpreted?
To increase the velocity of v1 then I can increase the area of A1(bottom) to be closer to the area of A2(top) or increase the height (h2-h1). I could do this by adding an extended pipe the size of A1 down from A1. Is that correct? I am a little surprised that the velocity increases with the area (A1) increasing.
Thanks.

The Attempt at a Solution

 

[alphysicist]

Hi woollyyak,

Homework Statement

Torricelli's thereom is v = sqroot(2g(y2-y1)) where the velocity at the top is negligable compared to the velocity at the bottom (v).
If the velocity at the top is considered (v2) then v1A1=v2A2 is substituted into Bernoulli's to give

v1 = sqroot(2g(h2-h1)/1-(A1/A2)^2)

Note here that you have canceled the pressures out of Bernoulli's equation, which means the pressures were equal to each other. In other words this would apply to a pipe that is open to the air at both ends, which is important in the questions you ask below.

Homework Equations

The above equation says that if A1=A2 then the velocity at V1 is infinite. Is that right? How should this be interpreted?

No, it's not infinite. Remember that the rules of algebra prohibit dividing by zero, so the above expression should read:

<br /> v_1 = \sqrt{\frac{2g(h_2-h_1)}{1-(A_1/A_2)^2}}\mbox{ unless }A_1=A_2<br />

To see what's going on, go back a step in your calculations:

<br /> \frac{1}{2} v_1^2 \left(1-\left(\frac{A_1}{A_2}\right)^2 \right) = g (h_2-h_1)<br />

Remember again that you said the pressures were equal at both ends of the pipe; what happens when the areas are equal? This equation is telling you that if A_1=A_2, then h_2=h_1. In other words, the only way the areas are the same is if the heights are the same, and the only way the heights can be the same is if the areas are the same. In other words, equal areas means a horizontal pipe.

To increase the velocity of v1 then I can increase the area of A1(bottom) to be closer to the area of A2(top) or increase the height (h2-h1).

You cannot increase the area A_1 at the bottom in this case. I think what is leading you in the wrong direction is simpy that you have a wrong idea about what the area refers to. The areas in these equations is not the area of the pipe, it's the area of the fluid in the pipe. Sometimes these are the same.

But in your case it's not. If you look at a steady water flow from a faucet, the stream narrows as the water goes down. The same thing happens in your pipe if it is vertical. As the water falls, the flow narrows, and it's not completely filling the pipe anymore. So the area at the bottom is not the area of the pipe.

I could do this by adding an extended pipe the size of A1 down from A1. Is that correct? I am a little surprised that the velocity increases with the area (A1) increasing.
Thanks.

Increasing the height that the water falls would definitely increase the speed at the bottom. Just like if you drop a ball into a hole, it's falling faster when it hits the ground (bottom of the hole) than if the hole wasn't there. If you add more pipe to the end, then you're redefining "bottom" in the same way, and the speed will be faster when it reaches the new bottom.

(Once again, everything is based on the fact that the pressures were the same at each end for your pipe.)

 

[woollyyak]

Thanks a lot alphysicist. Would a cone design have the same exiting velocity as a cubic design, both with the same top and bottom area and with equal height?

 

[alphysicist]

Thanks a lot alphysicist. Would a cone design have the same exiting velocity as a cubic design, both with the same top and bottom area and with equal height?

Sorry, I don't know what you're referring to by a cone design or a cubic design. If you're talking about effects of the pipe shape on the fluid flow, I think you would need a better treatment than just Bernoulli's equation/equation of continuity. There are a great many simplifications that go into deriving an introductory physics treatment, which means important properties of real fluids get ignored.

 

[woollyyak]

Just to make sure, I'm referring to a reservoir.
The example in the text uses a reservoir of cubic shape with a circular exit at the bottom, A1. If the shape was a funnel of cone shape would V1 be the same if A1, A2 and the height difference was the same as with a cubic reservoir?
Thanks.

 

[alphysicist]

I hope I'm visualizing what you say correctly. If so, then what you say sounds right to me.

 

[arildno]

WARNING:
You cannot apply Bernoulli naively for arbitrary choices of surfaces areas, and the immediate fallacy you are falling into when doing that is NOT the assumption that real fluids has viscosity, say, but instead the assumption behind Bernoulli that the velocity field can be regarded as STATIONARY.

When the ratio between the areas is huge, then the water level is practically constant at the site with the greatest surface area, which IS compatible with the condition of stationary flow.

When that ratio is close to 1, however, the water level is NOT constant at all, meaning that the actual fluid domain is continually changing, which requires a NON-stationary analysis to solve properly.

 

[alphysicist]

Hi arildno,

WARNING:
You cannot apply Bernoulli naively for arbitrary choices of surfaces areas, and the immediate fallacy you are falling into when doing that is NOT the assumption that real fluids has viscosity, say, but instead the assumption behind Bernoulli that the velocity field can be regarded as STATIONARY.

When the ratio between the areas is huge, then the water level is practically constant at the site with the greatest surface area, which IS compatible with the condition of stationary flow.

When that ratio is close to 1, however, the water level is NOT constant at all, meaning that the actual fluid domain is continually changing, which requires a NON-stationary analysis to solve properly.

Thank you for adding that to the discussion. When he said reservoir I was assuming that meant that the shape had an upper surface area that was open to the atmosphere and much larger than the exit hole area (and also since the velocity v2 was not mentioned I believe he was making the same assumptions and coming to the correct conclusion). However for woollyyak's benefit I should have mentioned that explicitly to make sure we are visualizing the same thing.

 

[woollyyak]

Thanks guys. I appreciate your input.