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

[eprparadox]

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?

 

[rude man]

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.

 

[eprparadox]

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.

 

[Orodruin]

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.

 

[eprparadox]

@Orodruin ohh, so what you're saying is that because the density of air is so low, $P_1 \approx P_2$

 

[Orodruin]

@Orodruin ohh, so what you're saying is that because the density of air is so low, $P_1 \approx P_2$

Right.

 

[rude man]

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.