How far will the bottom gate of a reservoir open?

[haruspex]

Okay I see
$v_1(h-x\cos(\theta)) = (h-h\cos(\theta))\sqrt{2gH}$
Is this what you meant?

If this is correct

$v_1 =\frac {(h-h\cos(\theta))\sqrt{2gH}} {(h-x\cos(\theta))}$

Then I use Bernoulli, P1 being at the top of the reservoir

$P_atm =P(x) + v(x)p/2+pgh(x)$

A couple problems with that last line.
On the right, you forgot to square the velocity.
On the left, you need the corresponding terms somewhere on the same streamline. That can be the surface of the flow at exit from the gate, where you know the gauge pressure is zero, or just before entry to the gate, at height h, where the velocity is zero.

 

[jackkk_gatz]

A couple problems with that last line.
On the right, you forgot to square the velocity.
On the left, you need the corresponding terms somewhere on the same streamline. That can be the surface of the flow at exit from the gate, where you know the gauge pressure is zero, or just before entry to the gate, at height h, where the velocity is zero.

$P+pg(H-h) =P(x) + v^2(x)p/2+pgh(x)$
why the velocity at h is zero? I would think at the top the velocity is zero since it is not moving, not at H-h

 

[haruspex]

$P+pg(H-h) =P(x) + v^2(x)p/2+pgh(x)$
why the velocity at h is zero? I would think at the top the velocity is zero since it is not moving, not at H-h

You don't need that P (which I assume is atmospheric pressure). We can take all pressures as gauge, i.e. relative to atmospheric.
The usual model for fluid leaving a tank is that just inside the tank the pressure is as though there were no exit and the velocity is zero. Going into the exit combines a sudden drop in pressure with a sudden jump in speed.
Of course, in reality, it isn’t quite like that. There is a steady drop in pressure (compared with what would apply statically at that depth) and a steady gain in speed as streamlines within the tank converge on the exit. But unless you care about those details the simple model works fine.

 

[erobz]

Take your first point at the surface of the reservoir, under the assumption that area is large in comparison to the discharge cross-section. Continuity implies that the velocity at the surface will be approximately…?

PS the variable for density is “\rho” I feel like $P$ and $p$ are going to be confusing choices.

 

[haruspex]

Take your first point at the surface of the reservoir, under the assumption that area is large in comparison to the discharge cross-section. Continuity implies that the velocity at the surface will be approximately…?

PS the variable for density is “\rho” I feel like $P$ and $p$ are going to be confusing choices.

Unfortunately we have given conflicting advice on where to take the first point , but either will serve. At the surface, p = 0, z=H, v=0; at the hinge, just inside the reservoir, p=g(H-h)$\rho$, z=h, v=0.
Either way, $\frac 12v^2+gz+\frac p{\rho}=g(H-h)$.
Or maybe it's not unfortunate - seeing the equivalence helps to illustrate how Bernoulli works.

And yes, since lowercase p is usually used for pressure (at least in statements of the Bernoulli equation) it is important to use $\rho$ for density.

 

[erobz]

Unfortunately we have given conflicting advice on where to take the first point , but either will serve. At the surface, p = 0, z=H, v=0; at the hinge, just inside the reservoir, p=g(H-h)$\rho$, z=h, v=0.
And yes, since lowercase p is usually used for pressure (at least in statements of the Bernoulli equation) it is important to use $\rho$ for density.

I just thought it helps skirt the whole " I believe its approximately zero at the surface, but not down at the top of the gate" issue.

 

[haruspex]

@jackkk_gatz , how are you going with this? Do you need more help?