How far will the bottom gate of a reservoir open?

[jackkk_gatz]

Homework Statement

A reservoir contains a liquid of density ρ1 at a height H. At the bottom is a gate with width W, thickness L and density ρ2, attached to a hinge. Determine the angle of the gate with respect to the vertical

Relevant Equations

P=Patm + pgh
Bernoulli equation

<Moderator's note: Upload images to PF. Do not use an external image server.>

I got an expression for the angle taking that initially the gate is closed, so the gate has h, W and L as their dimensions. So there is a surface integral to find the resultant force applied on the gate, using the hidrostatic pressure. Then I supossed air plays no role on keeping the gate down, and the force applied on the gate remains constant until it is opened to a final theta angle, which is the angle im trying to find. Then I used the equivlance of Work done = gravitational potential energy, to find the expression the problem is asking.

The problem with this, I don't know exactly what role plays that the water starts flowing. I know the bernoulli equation, so the pressure doesn't remain constant throughout all the process of opening the gate, the thing is I can't find an expression for the pressure variaton. I would need to know the velocity at all times, that means I have two variables, pressure and velocity, all of this with respect to time.

Can I neglect the fact water starts flowing to find the expression? Can even a correct expression for the angle be found?

Can I neglect the atmospheric pressure? If not, does that means my force field isn't conservative?

The reasoning I presented at the beginning is flawed? If yes, in what part exaclty?

 

[erobz]

You can ignore atmospheric pressure because it acts on the surface of the reservoir and the outside surface of the gate.

I assume you are supposed to take the reservoir to be very large here - implying $H$ is constant.

You probably also assume that the velocity of the water is acting on the gate is negligible, so this is a hydrostatics problem. I say probably, in case this is some advanced graduate level fluid mechanics course.

 

[jackkk_gatz]

You can ignore atmospheric pressure because it acts on the surface of the reservoir and the outside surface of the gate.

I assume you are supposed to take the reservoir to be very large here - implying $H$ is constant?

You probably also assume that the velocity of the water is acting on the gate is negligible, so this is a hydrostatics problem. I say probably, in case this is some advanced graduate level fluid mechanics course.

It is a undergraduate level fluid mechanics course, everything I know is pretty basic. Hydrostatic pressure on different surfaces and very very basic uses of bernoulli equation.

If this is a hydrostatics problem, the pressure on the gate when it is opened, can be described like hydrostatic pressure?

 

[erobz]

It is a undergraduate level fluid mechanics course, everything I know is pretty basic. Hydrostatic pressure on different surfaces and very very basic uses of bernoulli equation.

If this is a hydrostatics problem, the pressure on the gate when it is opened, can be described like hydrostatic pressure?

Yeah, basically you are assuming the velocity of the flow is not large until after the gate. You are going to locate the center of pressure $y_{cp}$ and apply $F = \bar{p}A$. Combine that with Newtons Laws (summing moments about the hinge) to determine the angle of opening.

 

[haruspex]

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.

 

[erobz]

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.

I don't want to say that that is not the intent, but I don't have a single example of a problem like that in my entire fluid mechanics text. The OP says its introductory fluid mechanics.

 

[haruspex]

I don't want to say that that is not the intent, but I don't have a single example of a problem like that in my entire fluid mechanics text. The OP says its introductory fluid mechanics.

I see a couple of problems with the hydrostatic approach.
The fluid flow analysis says that the gauge pressure at the lower edge of the gate will be zero. Your hydrostatic approach says it will be almost $\rho _1gH$, yes.
Solving on that basis I get an expression for the sine of the angle which could exceed 1.

Clearly @jackkk_gatz is familiar with the Bernoulli equation, and this looks like a fluid flow problem, so the setter would expect the student to treat it as such, tough as it may be.

 

[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.

That really sounds complicated. I can find the exit velocity v, but using the conservation of mass V1A1=V2A2, where I need the angle formed to know the cross-sectional area, don't I? Or am I misunderstanding something? I have solved a problem really similar, that we did in class, to this one before, but it only asked for the volumetric flow, I used V1A1=V2A2

 

[haruspex]

That really sounds complicated. I can find the exit velocity v, but using the conservation of mass V1A1=V2A2, where I need the angle formed to know the cross-sectional area, don't I? Or am I misunderstanding something? I have solved a problem really similar, that we did in class, to this one before, but it only asked for the volumetric flow, I used V1A1=V2A2

The angle is just an unknown $\theta$. At distance x from the axle, measured down the line of the gate, what is the distance to the bottom? What equation does that give you for the velocity at that point?

 

[erobz]

Just out of curiosity what is the chapter you are working on in your book @jackkk_gatz ?

Also, are there values to this problem for $H,h,\rho_1,\rho_2,L$? When I solve for $\theta$ treating this as a hydrostatic problem these are the values that form the expression, I would like to see what kind of result I get.

 

[erobz]

Solving on that basis I get an expression for the sine of the angle which could exceed 1.

I'm not sure I follow this. How can the expression for the sine of the angle ever exceed 1? You must mean you can get an expression for the angle that can exceed $\frac{ \pi }{2}$.

 

[erobz]

I see a couple of problems with the hydrostatic approach.

I see a few problems with trying to use Bernoulli's to get the pressure acting on the gate as a function of $\theta$.

Bernoulli's is only valid along a streamline. What is the velocity of the flow along the streamline that is adjacent the gate? How is it calculated? In viscous flow the velocity of the flow adjacent the gate is zero. Which implies the hydrostatic condition.

If we are thinking of getting the total volumetric flow rate $Q$ as a function of $\theta$,and applying that to the cross-sectional area at some horizontal position to find the velocity of that position at the gate (assuming uniform velocity vertically), then that doesn't play nice with Bernoulli's. The flow velocities are not uniform vertically.

 

[jackkk_gatz]

Just out of curiosity what is the chapter you are working on in your book @jackkk_gatz ?

Also, are there values to this problem for $H,h,\rho_1,\rho_2,L$? When I solve for $\theta$ treating this as a hydrostatic problem these are the values that form the expression, I would like to see what kind of result I get.

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.

 

[jackkk_gatz]

I see a few problems with trying to use Bernoulli's to get the pressure acting on the gate as a function of $\theta$.

Bernoulli's is only valid along a streamline. What is the velocity of the flow along the streamline that is adjacent the gate? How is it calculated? In viscous flow the velocity of the flow adjacent the gate is zero. Which implies the hydrostatic condition.

If we are thinking of getting the total volumetric flow rate $Q$ as a function of $\theta$,and applying that to the cross-sectional area at some horizontal position to find the velocity of that position at the gate (assuming uniform velocity vertically), then that doesn't play nice with Bernoulli's. The flow velocities are not uniform vertically.

Actually my first try was to do something with forces, I took the assumption that on the whole surface of the gate there would be stagnation points, or something like that. That approach didn't worked, now revising my notes I notice I took the vector weight on the wrong direction, maybe there is where I got stuck and didn't get any result.
Just one question, the problem has to be solved with momentum? Or can I use forces, and would give out the same result? I know the summation of moments must be zero and also of forces, why to pick one or the other?

 

[haruspex]

the summation of moments must be zero and also of forces, why to pick one or the other?

Because there is an unknown force exerted on the gate by its axle. That force can be eliminated by considering torque about the axle.

 

[haruspex]

How can the expression for the sine of the angle ever exceed 1?

Easily. I got $\sin(\theta)=\frac {H\rho_1}{L\rho_2}$, which could well exceed 1. That is clearly wrong. Did you try your hydrostatic approach?

 

[haruspex]

Bernoulli's is only valid along a streamline. What is the velocity of the flow along the streamline that is adjacent the gate? How is it calculated? In viscous flow the velocity of the flow adjacent the gate is zero. Which implies the hydrostatic condition.

That objection would make Bernoulli always wrong when used in a pipe! So consider a streamline a little further from the gate. The pressure at points along that must be pretty much the same as right at the gate.
Bernoulli assumes non viscous, and for water that usually works out well enough.

The most obvious pattern to assume is straight line streamlines converging from the wide end to the narrow end. Since the pressure drop along the base line will be a bit greater than that along the line nearest the gate, it would flow a bit faster, but that doesn’t strike me as a difficulty.

Bear in mind that the actual flows are always a bit messier than Bernoulli pretends. In the "simple" case of a hole near the bottom of a tank, the flows do not really start from rest immediately before the hole and suddenly reach the exit speed. There would be flow lines from every point within the tank towards the hole, accelerating as they converge. See e.g. Borda mouthpiece.

Btw, where it says "width W" I assume it means depth W.

 

[jackkk_gatz]

Btw, where it says "width W" I assume it means depth W.

By that I meant, like the dimensions of the gate like hxWxL, length (h), width (W) and tickness or height L, my bad

 

[jackkk_gatz]

Easily. I got $\sin(\theta)=\frac {H\rho_1}{L\rho_2}$, which could well exceed 1. That is clearly wrong. Did you try your hydrostatic approach?

I tried the hydrostatic approach, don't know if I did it correclty. I had an expression that I had to square then use some trig. there were several things squared in the denominator and in the numerator only one multiplication, all that taking out the square root of the fraction. That expression can't exceed one, I think

 

[haruspex]

I tried the hydrostatic approach, don't know if I did it correclty. I had an expression that I had to square then use some trig. there were several things squared in the denominator and in the numerator only one multiplication, all that taking out the square root of the fraction. That expression can't exceed one, I think

Please post your solution.

 

[erobz]

I tried the hydrostatic approach, don't know if I did it correclty. I had an expression that I had to square then use some trig. there were several things squared in the denominator and in the numerator only one multiplication, all that taking out the square root of the fraction. That expression can't exceed one, I think

For the hydrostatic "solution" I have:

$$ \theta = \arcsin(\beta)$$

Where:

$$\beta = \frac{-b+ \sqrt{b^2 - 4 a c} }{2a}$$

$$ a = \left( L \rho_{gate} \right)^2 + \left( \frac{2}{3} \rho h \right)^2$$

$$ b = -\frac{4}{3} \rho \left( H- h\right) L \rho_{gate} $$

$$ c = \left( \frac{2}{3}\rho(H-h) \right)^2 - \left( \frac{2}{3} \rho h \right)^2$$

I did it earlier and haven't had time to give it a once over.

 

[haruspex]

By that I meant, like the dimensions of the gate like HxWxL, length (H), width (W) and tickness or height L, my bad

Ok, I see what has happened. h was not mentioned in the problem statement, whereas W is although it is irrelevant. So wherever I may have written W in an equation, read it as h.

 

[haruspex]

For the hydrostatic "solution" I have:

$$ \theta = \arcsin(\beta)$$

Where:

$$\beta = \frac{-b+ \sqrt{b^2 - 4 a c} }{2a}$$

$$ a = \left( L \rho_{gate} \right)^2 + \left( \frac{2}{3} \rho h \right)^2$$

$$ b = -\frac{4}{3} \rho \left( H- h\right) L \rho_{gate} $$

$$ c = \left( \frac{2}{3}\rho(H-h) \right)^2 - \left( \frac{2}{3} \rho h \right)^2$$

I did it earlier and haven't had time to give it a once over.

Reverse engineering, I get that you had the equation $3L\rho_g\sin(\theta)=2\rho_w(H-h(1-\cos(\theta)))$.
Since H>>h, that simplifies to $3L\rho_g\sin(\theta)=2\rho_wH$. Apart from a factor 2/3, that is what I wrote in post #16. Surely it is possible that $2\rho_wH>3L\rho_g$, which would make $\sin(\theta)>1$?

 

[erobz]

Reverse engineering, I get that you had the equation $3L\rho_g\sin(\theta)=2\rho_w(H-h(1-\cos(\theta)))$.
Since H>>h, that simplifies to $3L\rho_g\sin(\theta)=2\rho_wH$. Apart from a factor 2/3, that is what I wrote in post #16. Surely it is possible that $2\rho_wH>3L\rho_g$, which would make $\sin(\theta)>1$?

I definitely get some parameters sets that lead to negative results under the root. The thing is the function should not produce solutions for angles greater than $90$ in either case (static or dynamic). If the gate is thin, and the damn is high we should expect it to be pretty much horizontal.

 

[jackkk_gatz]

Please post your solution.

I made a mistake, but this is where I got, don't know what to do from there. Maybe I could use the approach of solving a quadratic equation, which I didn't knew of before reading erobz answer

At the end it is 1/2 not 2B, my bad

 

[erobz]

Reverse engineering, I get that you had the equation $3L\rho_g\sin(\theta)=2\rho_w(H-h(1-\cos(\theta)))$.
Since H>>h, that simplifies to $3L\rho_g\sin(\theta)=2\rho_wH$. Apart from a factor 2/3, that is what I wrote in post #16. Surely it is possible that $2\rho_wH>3L\rho_g$, which would make $\sin(\theta)>1$?

The way I see the accounting, at the very least the hydrostatic solution provides an upper bound for the dynamic solution you propose.

 

[haruspex]

I made a mistake, but this is where I got, don't know what to do from there. Maybe I could use the approach of solving a quadratic equation, which I didn't knew of before reading erobz answer

At the end it is 1/2 not 2B, my bad

As I explained, you have to use the torque balance. not a force balance.
Your integrals only produce forces, not torques. E.g. in line 1, the torque is $\rho_wWL\frac 12 h^2g\sin(\theta)$

Also, in the double integral, why $-y\cos(\theta)$ rather than "+"?

 

[jackkk_gatz]

As I explained, you have to use the torque balance. not a force balance.
Your integrals only produce forces, not torques. E.g. in line 1, the torque is $\rho_wWL\frac 12 h^2g\sin(\theta)$

Also, in the double integral, why $-y\cos(\theta)$ rather than "+"?

uuuh I don't know why I didn't put +, maybe I just did it by inertia, all the problems I had solved I applied -y, but this time isn't correct.
I forgot the torque is done from h/2, not h. I did use torque and eliminated h.

The torque done by the liquid, it is done also from h/2?

 

[erobz]

The torque done by the liquid, it is done also from h/2?

No, if you are using an integral and the pressure distribution

$$ \tau = \int y \, p(y) \, dA $$

With the hydrostatic distribution you can dispense with the integral for already computed results that relate the resultant force $F$ applied to location defined as the center of pressure $y_{cp}$:

$$y_{cp} F = \int y \, p(y) \, dA $$

It can be shown that for the hydrostatic distribution that:

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

Where:

$\bar y$ is the centroid of the area
$\bar I$ is the second moment of area

 

[haruspex]

I did use torque and eliminated h.

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

 

[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.