How do we use Bernoulli's Principle in this situation?

In summary: So that's where the sudden drop of (1/2)ρv2 occurs. But that's also where the pressure drops from ρgh to atmospheric. In summary, the conversation discusses a large water tank with a small hole in the bottom and the speed of the water leaking from the hole when the water level is 30 meters above the bottom of the tank. It is determined that the pressure at the top and bottom of the tank is approximately the same due to the negligible density of air, and the sudden drop in pressure and increase in velocity occurs at the aperture of the bottom hole.
  • #1
eprparadox
138
2

Homework Statement


A large water tank, open at the top, has a small hole in the bottom. When the water level is
## 30## ##m## above the bottom of the tank, the speed of the water leaking from the hole:
A. is ##2.5## ##m/s##
B. is ##24## ##m/s##
C. is ##44## ##m/s##
D. cannot be calculated unless the area of the hole is given
E. cannot be calculated unless the areas of the hole and tank are given

Homework Equations



## p + \rho g y + \frac{1}{2}\rho v^2 = constant ##

The Attempt at a Solution



## p_1 + \rho g y_1 + \frac{1}{2}\rho v_1^2 = p_2 + \rho g y_2 + \frac{1}{2}\rho v_2^2 ##

We can set our reference from ## y_1 = 0 ## and because the large opening at the top is so much larger than the small opening, ## v_2 = 0 ##

This gives us

## p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \rho g y_2 ##

Here's where I get confused. What exactly are ## p_1 ## and ## p_2 ##?

I know in solutions to this problem, they're assume to be the same since both openings are open to the atmosphere. And because of this, ## p_1 ## and ## p_2 ## cancel out and we can solve for ## v_1 ##.

But that doesn't make sense to me.

These two holes are 30 meters apart so the pressures should be different.

I'm not sure how to reconcile what ## p_1 ## and ## p_2 ## are.

Any thoughts?
 
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  • #2
The pressure IS higher just above the bottom but at that point v is still small. When the water flows out of the bottom orifice the pressure suddenly drops from ρg(30m) to atmospheric. The atmospheric at the surface and at the outflow cancel, giving you your simple equation which just says change in p.e. = change in k.e.
 
  • #3
Thanks so much for the response.

Just to set the notation: I consider point 2 to be at the top opening and 1 to be at the bottom small hole opening.

Let's say this tank is sitting on a table in some room. Then I agree that at the opening of the tank at the bottom is atmospheric pressure. But if the top is 30 meters higher, then the pressure there has to be lower than that bottom point by ## \rho g h ##, no?

That is, ## P_2 = P_1 - \rho g h ##.

And if you plug this into Bernoulli's equation, then you get this extra ## -\rho g h ## term.

I'm not sure if what I'm saying makes sense, but thanks any thoughts.
 
  • #4
eprparadox said:
But if the top is 30 meters higher, then the pressure there has to be lower than that bottom point by ## \rho g h ##, no?
Yes, but with the density of the air, which is negligible.
 
  • #5
@Orodruin ohh, so what you're saying is that because the density of air is so low, ## P_1 \approx P_2 ##
 
  • #6
eprparadox said:
@Orodruin ohh, so what you're saying is that because the density of air is so low, ## P_1 \approx P_2 ##
Right.
 
  • #7
eprparadox said:
Let's say this tank is sitting on a table in some room. Then I agree that at the opening of the tank at the bottom is atmospheric pressure. But if the top is 30 meters higher, then the pressure there has to be lower than that bottom point by ## \rho g h ##, no?.
See, there are two "bottoms".. there is the bottom of the tank where p is indeed ρgh higher than at the top. But just beneath the bottom is the aperture, and right at the bottom of the aperture v suddenly changes from near-0 to where 1/2 ρ v2 = ρgh as it spurts out of the bottom.
 

1. How does Bernoulli's Principle apply to the flight of an airplane?

Bernoulli's Principle states that as the speed of a fluid (such as air) increases, its pressure decreases. In the case of an airplane, the shape of the wing causes air to move faster over the top of the wing than underneath it, creating a difference in air pressure. This difference in pressure creates lift, allowing the airplane to stay in the air.

2. Can Bernoulli's Principle explain how a curveball is thrown in baseball?

Yes, Bernoulli's Principle can be used to explain how a curveball is thrown. As the ball spins, it creates different air pressures on either side of the ball. This difference in pressure causes the ball to curve in the direction of the lower pressure, allowing pitchers to throw a curveball.

3. How is Bernoulli's Principle used in the design of race cars?

In race car design, Bernoulli's Principle is used to create downforce, which helps keep the car on the ground and improve its handling. The shape of the car, particularly the wings and spoilers, is designed to create different air pressures that produce downforce.

4. Can Bernoulli's Principle explain how a shower curtain moves inward when the shower is turned on?

Yes, Bernoulli's Principle can explain this phenomenon. As the water from the shower hits the air, it causes the air to move faster, creating a lower pressure area. This lower pressure on the inside of the shower curtain causes it to move inward.

5. How does Bernoulli's Principle apply to the flow of blood through arteries?

In the human body, Bernoulli's Principle helps explain how blood flow through arteries. As the heart pumps blood through narrow arteries, the speed of the blood increases, causing a decrease in pressure. This difference in pressure helps push the blood through the arteries and maintain proper circulation.

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