How far will the bottom gate of a reservoir open?

[jackkk_gatz]

Now I think I got it right

Then should I solve a quadratic equation? Is it the only way to solve the last equation?

 

[jackkk_gatz]

You did? Maybe too soon. In your double integral there should be another factor y, leading to a $\frac 13h^3$ term, not $\frac 12 h^2$.

Yes now I notice I did, I realized that the torque performed by the fluid is an integral as well, I remembered another expression is used as erobz pointed it out.

 

[haruspex]

Yes now I notice I did, I realized that the torque performed by the fluid is an integral as well, I remembered another expression is used as erobz pointed it out.

Taking the sixth equation and applying the simplifying assumption H>>h, gives $\sin(\theta)=\frac{\rho_wH}{\rho_gL}$. That's the same as I got for the hydrostatic solution in post #16. @erobz got something similar, just a factor 3/2 difference.
As I keep pointing out, this has the disastrous feature that for $\rho_wH>\rho_gL$ it makes $\sin(\theta)>1$. It is clearly not a valid treatment.

Please attempt the fluid flow solution I outlined in post #5. At least it gives sensible torque expressions.

 

[erobz]

That's the same as I got for the hydrostatic solution in post #16. @erobz got something similar, just a factor 3/2 difference.

That factor you are mentioning bothers me (they shouldn't be different):

$$\bar p = \frac{ \rho g \left( H - h \right) + \rho g \left( H - h \cos \theta \right) }{2} = \frac{\rho g }{2} \left( 2H - h\left( 1+ \cos \theta \right) \right)$$

Then determine the moment arm of the resultant force $F= \bar p A$:

$$y_{cp} = \bar y + \frac{ \bar I }{ \bar y A}$$

$\bar y = \frac h 2$
$A = hW$
$\bar I = \frac{1}{12} W h^3$

$$ \implies y_{cp} = \frac h 2 + \frac{ \frac{1}{12} W h^3 }{ \frac h 2 W h } = \frac h 2 + \frac h 6 = \frac 2 3 h $$

Then summing the moments:

$$ \rho_{gate} \cancel{hW}L \cancel{g}\sin \theta \frac {\cancel{h}} {\cancel{2}} = \frac 2 3 \cancel{h} \frac{\rho \cancel{g} }{ \cancel{2}} \left( 2H - h\left( 1+ \cos \theta \right) \right) \cancel{h W}$$

$$ \rho_{gate} L \sin \theta = \frac 2 3 \rho \left( 2H - h\left( 1+ \cos \theta \right) \right)$$

It must be missing from your result because of your assumption $h \ll H$, which effectively neglects the change in hydrostatic pressure across the gate.

 

[haruspex]

That factor you are mentioning bothers me (they shouldn't be different):

$$\bar p = \frac{ \rho g \left( H - h \right) + \rho g \left( H - h \cos \theta \right) }{2} = \frac{\rho g }{2} \left( 2H - h\left( 1+ \cos \theta \right) \right)$$

Then determine the moment arm of the resultant force $F= \bar p A$:

$$y_{cp} = \bar y + \frac{ \bar I }{ \bar y A}$$

$\bar y = \frac h 2$
$A = hW$
$\bar I = \frac{1}{12} W h^3$

$$ \implies y_{cp} = \frac h 2 + \frac{ \frac{1}{12} W h^3 }{ \frac h 2 W h } = \frac h 2 + \frac h 6 = \frac 2 3 h $$

Then summing the moments:

$$ \rho_{gate} \cancel{hW}L \cancel{g}\sin \theta \frac {\cancel{h}} {\cancel{2}} = \frac 2 3 \cancel{h} \frac{\rho \cancel{g} }{ \cancel{2}} \left( 2H - h\left( 1+ \cos \theta \right) \right) \cancel{h W}$$

$$ \rho_{gate} L \sin \theta = \frac 2 3 \rho \left( 2H - h\left( 1+ \cos \theta \right) \right)$$

It must be missing from your result because of your assumption $h \ll H$, which effectively neglects the change in hydrostatic pressure across the gate.

It is not valid to take the average pressure and pretend that is applied equally across the plate. You have to integrate the torques.

 

[erobz]

It is not valid to take the average pressure and pretend that is applied equally across the plate. You have to integrate the torques.

It is perfectly valid for hydrostatic pressure distribution (linear) to apply the resultant force $F = \bar p A$ at the center of pressure $y_{cp}$.

If you would like I can share the derivation in my text book?

 

[Lnewqban]

I'm working with chapter 2 and chapter 3-Elementary fluid Dynamics. The book I worked on class was Fluid Mechanics by Munson. My professor did not specify which topic is the subject of this problem, he only said that the topics that would be relevant to the problems, would be hydrostatics and bernoulli.
And there are no values, just variables.

As shown, it seems to me that this problem has problems.

If the hinged gate at the bottom is initially in a vertical position, as represented in your linked diagram, you can’t have “a reservoir containing a liquid”: that reservoir will be profusely leaking and unable to keep “a height H”.

Being vertically hanging from its top hinge, those “width W, thickness L and density ρ2” do not mean anything, regading its closing capability via torque due the gate’s weight.

This hinged gate could be an automatic relieve valve, which remains closed by gravity for open surface levels below H, but crack-opens to avoid overflow at the top of the tank, if the surface level goes above the H-limit.

If the above is true, the professor may be asking to “determine the angle of the gate with respect to the vertical” that will prevent any increase of the level in the reservoir (actually “containing a liquid”) above “a height H”.

 

[haruspex]

It is perfectly valid for hydrostatic pressure distribution (linear) to apply the resultant force $F = \bar p A$ at the center of pressure $y_{cp}$.

If you would like I can share the derivation in my text book?

Ah, that's what $y_{cp}$ means. But the centre of pressure is only going to be 2/3 of the way along if it is zero at one end. As H increases relative to $h\cos(\theta)$, the centre of pressure will move towards h/2.

 

[haruspex]

If the hinged gate at the bottom is initially in a vertical position, as represented in your linked diagram

I would not assume the diagram is that accurate. We are not given separate variables for the depth of the gate and the depth of the vertical aperture.

Being vertically hanging from its top hinge, those “width W, thickness L and density ρ2” do not mean anything, regading its closing capability via torque due the gate’s weight.

The depth, h, was missing from the initial post.

This hinged gate could be an automatic relieve valve, which remains closed by gravity for open surface levels below H, but crack-opens to avoid overflow at the top of the tank, if the surface level goes above the H-limit.

Unless there is something restraining it, it will always open a bit, until the reservoir is empty.

the professor may be asking to “determine the angle of the gate with respect to the vertical” that will prevent any increase of the level in the reservoir (actually “containing a liquid”) above “a height H”.

That would be impossible to answer without knowing the rate at which water is entering the reservoir.

 

[erobz]

Ah, that's what $y_{cp}$ means. But the centre of pressure is only going to be 2/3 of the way along if it is zero at one end. As H increases relative to $h\cos(\theta)$, the centre of pressure will move towards h/2.

Yeah, I agree with that. My $y_{cp}$ is wrong.

$$ \bar y = H - \frac h 2 \cos \theta $$

$$ y_{cp} = H - \frac h 2 \cos \theta + \frac{ \frac {1 }{12}W h^3}{ \left( H - \frac h 2 \cos \theta \right) Wh}$$

There is no way all that is 1 though. That part has to be a product of your assumption of $h \ll H$

 

[haruspex]

Yeah, I agree with that. My $y_{cp}$ is wrong.

$$ \bar y = H - \frac h 2 \cos \theta $$

$$ y_{cp} = H - \frac h 2 \cos \theta + \frac{ \frac {1 }{12}W h^3}{ \left( H - \frac h 2 \cos \theta \right) Wh}$$

There is no way all that is 1 though. That part has to be a product of your assumption of $h \ll H$

That expression cannot be right for the centre of pressure on the gate as a distance from the hinge. It must be between 0 and h.

 

[erobz]

That expression cannot be right for the centre of pressure on the gate as a distance from the hinge. It must be between 0 and h.

Yeah I saw something was off!

$$ \bar y = H - h( 1 - \frac 1 2 \cos \theta )$$

...I think

so

$$ y_{cp} = H - h( 1 - \frac 1 2 \cos \theta ) + \frac{ \frac {1 }{12}W h^3}{ \left( H - h( 1 - \frac 1 2 \cos \theta ) \right) Wh}$$

 

[haruspex]

Yeah I saw something was off!

$$ \bar y = H - h( 1 - \frac 1 2 \cos \theta )$$

...I think

so

$$ y_{cp} = H - h( 1 - \frac 1 2 \cos \theta ) + \frac{ \frac {1 }{12}W h^3}{ \left( H - h( 1 - \frac 1 2 \cos \theta ) \right) Wh}$$

From first principles I get that the centre of pressure, as a distance from the hinge, is $\frac h2\frac{3(H-h)+2h\cos(\theta)}{2(H-h)+h\cos(\theta)}$.
For H>>h, that reduces to h/2, as it should, and for H=h it gives 2h/3, as it should.

 

[jackkk_gatz]

I got a little lost, what happened? Is there a consensus on what approach is correct? Or is there still no definitive answer?

 

[erobz]

I got a little lost, what happened? Is there a consensus on what approach is correct? Or is there still no definitive answer?

I think the consensus is try the dynamic approach.

 

[jackkk_gatz]

You are not asked to consider the process of the gate opening from the closed position. You only need to treat it as equilibrium, i.e. find at what angle the torques balance.

You can start by considering the end-to-end process. Water that is essentially static just before the gate eventually exits the gate, back to atmospheric pressure. Solve to find the exit velocity v.
What is the linear velocity at an arbitrary distance through the gate (measured along the gate from axle) in terms of v?
Use Bernoulli again to find the pressure at an arbitrary distance through the gate and the torque an element there exerts about the gate axle.
Integrate to get the total torque (messy) and equate to the gravitational torque.

Let me see if I understood this correctly, I can find the velocity at the outlet of the gate as a free jet? Then use this velocity to find a function for the pressure at the gate? Then use this function to get the total torque, like I did previously when I used the static pressure?

 

[haruspex]

find the velocity at the outlet of the gate as a free jet?

Yes.

Then use this velocity to find a function for the pressure at the gate?

No, use the velocity at exit and the known way that the height h(x) varies down the length of the gate (x) to find the velocity as a function of x.
From that, use Bernoulli to find the pressure as p(x).
Then you are in a position to calculate the net torque on the gate from the water.

 

[jackkk_gatz]

No, use the velocity at exit and the known way that the height h(x) varies down the length of the gate (x) to find the velocity as a function of x.
From that, use Bernoulli to find the pressure as p(x).
Then you are in a position to calculate the net torque on the gate from the water.

I guess I have to find h(x) with V1A1=V2A2, then use the bernoulli equation to find p(x). So I have

$v_1x\cos(\theta) = (h-h\cos(\theta))\sqrt{2gH}$

Then I use this on the bernoulli equation to find p(x), the thing is, wouldn't I end up having two pressures? The first one being p(x) and the second one the pressure at the exit?

 

[haruspex]

So I have
$v_1x\cos(\theta) = (h-h\cos(\theta))\sqrt{2gH}$

Not quite.
What is the height of the gate at distance x down it from the hinge?
(You can check your answer by putting in x=0 to get h, x=h to get h-hcos(theta).)

 

[jackkk_gatz]

Not quite.
What is the height of the gate at distance x down it from the hinge?
(You can check your answer by putting in x=0 to get h, x=h to get h-hcos(theta).)

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

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

 

[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?