Water flowing out of a tank through a hole

[brotherbobby]

Homework Statement

Water stands up to a height $h$ in an open tank which has a hole in the bottom when the experiment begins, as shown in the figure below.

Explain whether one can apply the continuity equation for the top surface of water (marked $\mathbf 1$) and for the hole at the bottom (marked $\mathbf 2$) where the water flows out, assuming water to be incompressible.

Relevant Equations

Continuity equation : $A_1 v_1 = A_2 v_2$, where the $A's$ and the $v's$ are the areas and velocities at positions 1 and 2 respectively.

I assume the water to start flowing from rest at position 1, hence $v_1 = 0$. Applying the continuity equation, $A_1 v_1 = A_2 v_2$, we find the (wrong) result that the velocity at position 2 is $v_2 = 0$ also! (We assume that $A_2$ is small but finite)

Hence, to answer the question, the continuity equation cannot be applied for points 1 and 2. A possible reason for this is there is no continuity for point 1 - no water is flowing into it. "Continuity" demands no loss or gain of liquid at a given point, which is violated for point 1.

Is my reasoning correct?

 

[jbriggs444]

I assume the water to start flowing from rest at position 1, hence $v_1 = 0$. Applying the continuity equation, $A_1 v_1 = A_2 v_2$, we find the (wrong) result that the velocity at position 2 is $v_2 = 0$ also! (We assume that $A_2$ is small but finite)

The argument here is correct. However, the conclusion you draw is not.

Why do you think that the result of zero [initial] velocity is wrong?

 

[brotherbobby]

The argument here is correct. However, the conclusion you draw is not.

Why do you think that the result of zero [initial] velocity is wrong?

Am assuming the liquid to be at rest to begin with. Torricelli's principle states that a column of liquid open at both ends to air can be treated to be like a stone falling from rest, both acquiring the same speed ($\sqrt{2gh}$) opon falling a height $h$. Of course the crucial difference is that the water would have that speed in all directions, while for the stone it is vertically down.

Hence the water starts flowing from zero speed. Is the reasoning correct?

 

[brotherbobby]

The argument here is correct. However, the conclusion you draw is not.

Why do you think that the result of zero [initial] velocity is wrong?

Oops, did you mean zero [final] velocity?

Well, surely the result is wrong as we know liquids to flow out with a finite speed $v_2 \neq 0$.

 

[jbriggs444]

Oops, did you mean zero [final] velocity?

I said what I meant and I meant what I said.

You have assumed an initial state with zero motion at the top surface and correctly concluded that the principle of continuity then requires that the initial state also have zero fluid motion at the hole.

Well, surely the result is wrong as we know liquids to flow out with a finite speed $v_2 \neq 0$.

We know that they attain a finite speed, yes.

 

[jbriggs444]

Am assuming the liquid to be at rest to begin with. Torricelli's principle states that a column of liquid open at both ends to air can be treated to be like a stone falling from rest, both acquiring the same speed ($\sqrt{2gh}$) opon falling a height $h$. Of course the crucial difference is that the water would have that speed in all directions, while for the stone it is vertically down.

Hence the water starts flowing from zero speed. Is the reasoning correct?

No. This reasoning is not correct.

You started with the premise: "assuming the liquid to be at rest to begin with"
You ended with the conclusion: "hence the water starts flowing from zero speed".

You've simply restated your assumption. Torricelli's principle is irrelevant. That argument has no force.

 

[brotherbobby]

I cannot see a way out. One of the two has to give.

1. Either we abandon continuity (because point 1 has no water flowing "into" it) and find the velocity of outflow $v_2$ using Bernoulli's equation. This was my feeling to start with since continuity was giving the result $v_2 = 0$. But I am assuming that the liquid has zero speed at point 1, an intuitive point but might well be untrue.

2. Or we accept continuity and assume a vanishingly small but finite speed at position 1 ($v_1 \neq 0$). But that would fly in the face of intuition - the water has a small speed to begin with at position 1? How come?

 

[jbriggs444]

I cannot see a way out. One of the two has to give.

1. Either we abandon continuity (because point 1 has no water flowing "into" it) and find the velocity of outflow $v_2$ using Bernoulli's equation. This was my feeling to start with since continuity was giving the result $v_2 = 0$. But I am assuming that the liquid has zero speed at point 1, an intuitive point but might well be untrue.

Please restate what you think the continuity principle says. It does not state that the flow velocity across a particular interface stays constant over time.

Also, who says that point 1 has no fluid flowing into it? It most certainly does. Air is a fluid.

Are you, perhaps, reasoning that if we have a bucket with a closed top and a hole in the bottom then no fluid will flow out of the hole? If so, that reasoning is perfectly correct. The result can be demonstrated experimentally to be true. I've done that experiment many times (most easily done with an inverted drinking glass and a playing card). Water indeed does not flow out.

That is one reason that they put vent holes in gasoline cans.

 

[brotherbobby]

Please restate what you think the continuity principle says. It does not state that the flow velocity across a particular interface stays constant over time.

Also, who says that point 1 has no fluid flowing into it? It most certainly does. Air is a fluid.

Are you, perhaps, reasoning that if we have a bucket with a closed top and a hole in the bottom then no fluid will flow out of the hole? If so, that reasoning is perfectly correct. The result can be demonstrated experimentally to be true. I've done that experiment many times (most easily done with an inverted drinking glass and a playing card). Water indeed does not flow out.

That is one reason that they put vent holes in gasoline cans.

According to continuity, the net (volume) of liquid flowing through a a section is constant over time, assuming liquid is not compressibe. That is $\dot V = \text{constant}$, and since $\dot V = Av$, we have $A_1 v_1 = A_2 v_2$.

"Are you, perhaps, reasoning that if we have a bucket with a closed top and a hole in the bottom then no fluid will flow out of the hole? If so, that reasoning is perfectly correct. The result can be demonstrated experimentally to be true. I've done that experiment many times (most easily done with an inverted drinking glass and a playing card). Water indeed does not flow out. "

I am unaware of this example, fascinating though it looks. Please explain it to me when you can. But for now, am stuck with this problem without seeing a way out.

 

[brotherbobby]

Also, who says that point 1 has no fluid flowing into it? It most certainly does. Air is a fluid.

Yes. But air has a density different from water. Does continuity apply for an interface involving two different fluids?

I am aware that in the event of compressibility, continuity comes in with the density ($\dot m = \text{constant}$) : $\rho_1 A_1 v_1 = \rho_2 A_2 v_2$. However, I assume this is for the "same" liquid, compressed or expanded to have a different densities. I am not sure this is valid for the complicated case where one fluid moves into a section (or interface) and a different fluid flows out.

 

[jbriggs444]

Yes. But air has a density different from water. Does continuity apply for an interface involving two different fluids?

Fair enough.

Instead of directly considering a plane where the top surface of the water lies, consider a plane that is fixed in place a small distance below where that surface starts.

Would you agree that the continuity equation applies between this surface and the egress hole?

 

[brotherbobby]

Fair enough.

Instead of directly considering a plane where the top surface of the water lies, consider a plane that is fixed in place a small distance below where that surface starts.

Would you agree that the continuity equation applies between this surface and the egress hole?

Let me make a small drawing of the plane you mention.

Yes, continuity would apply to the plane immediately below the one from where the water begins to flow.

 

[haruspex]

I cannot see a way out. One of the two has to give.

1. Either we abandon continuity (because point 1 has no water flowing "into" it) and find the velocity of outflow $v_2$ using Bernoulli's equation. This was my feeling to start with since continuity was giving the result $v_2 = 0$. But I am assuming that the liquid has zero speed at point 1, an intuitive point but might well be untrue.

2. Or we accept continuity and assume a vanishingly small but finite speed at position 1 ($v_1 \neq 0$). But that would fly in the face of intuition - the water has a small speed to begin with at position 1? How come?

As far as I can make out, your argument says that you cannot make an iron rod move by pushing on one end because the velocity starts at zero at the other end. (Maybe with the ends reversed from you bucket of water model, but the principle is the same.)
The resolution is that nothing is completely rigid or incompressible. It will take some minute fraction of a second for the interatomic forces to ripple along.

 

[kuruman]

The problem is asking, "Explain whether one can apply the continuity equation for the top surface of water (marked 1) and for the hole at the bottom (marked 2) where the water flows out, assuming water to be incompressible. " My interpretation is that the author omitted to say "given that water flow has already been established". I base this interpretation on the figure that shows a puddle of water in front of the tank and a drop in the process of emerging from the nozzle. I suspect that the point of the question is "Is the continuity equation applicable when there is a difference in height between A₁ and A₂?"

 

[brotherbobby]

The problem is asking, "Explain whether one can apply the continuity equation for the top surface of water (marked 1) and for the hole at the bottom (marked 2) where the water flows out, assuming water to be incompressible. " My interpretation is that the author omitted to say "given that water flow has already been established". I base this interpretation on the figure that shows a puddle of water in front of the tank and a drop in the process of emerging from the nozzle. I suspect that the point of the question is "Is the continuity equation applicable when there is a difference in height between A₁ and A₂?"

There is a way out of this, if I am right in what I write below.

Let us calculate the efflux velocity (initially at $t = 0$) $v_2$ using the more reliable method of Bernoulli's theorem. The well known answer is $v_2 = \sqrt{2gh}$. This gives the velocity at the open (top) surface $A_1$ as $v_1 = \frac{A_2}{A_1} \sqrt{2gh}$. Hence, assuming $A_2 \lll A_1$, $v_1 \approx 0$. The diagram alongside shows my calculations.

Note, the crucial approximation in the end. In the event the top and bottom areas ($A_1\; \text{and}\; A_2$) are comparable, as the case will be if we made the same hole at the bottom of a glass tube of the same height as that of the tank, the starting velocity at the top $v_1$ would be comparable (though smaller) to $v_2$.

I make a diagram of a test tube containing water. Note the open area $A_1$ of the test tube is much less than that of the tank but the hole area at the bottom $A_2$ is the same. The two velocities are comparable.

 

[kuruman]

There is a way out of this, if I am right in what I write below.

The way out of what? I don't understand what you have shown by invoking the Bernoulli equation. The Bernoulli equation is a statement of conservation of mechanical energy per unit volume in the case of an ideal incompressible fluid. The continuity equation is a statement of mass conservation, namely that if μ mass per unit time (kg/s) of fluid flows through area A₁, the same μ kg/s of fluid must flow through area A₂ under the assumption that the fluid is completely rigid and incompressible. I think it is safe to answer the question under this assumption.

 

[brotherbobby]

The way out of what? I don't understand what you have shown by invoking the Bernoulli equation. The Bernoulli equation is a statement of conservation of mechanical energy per unit volume in the case of an ideal incompressible fluid. The continuity equation is a statement of mass conservation, namely that if μ mass per unit time (kg/s) of fluid flows through area A₁, the same μ kg/s of fluid must flow through area A₂ under the assumption that the fluid is completely rigid and incompressible. I think it is safe to answer the question under this assumption.

I started this thread with a doubt. Is the continuity equation applicable for an open tank filled with water and a hole in the bottom? How can it be, for it will mean the velocity of outflow $v_2 = 0$, as, supposedly, $v_1 = 0$, the velocity at the top surface of the liquid. This assumption seemed reasonable because the top surface of liquid is at rest initially. But clearly something is wrong here because we know $v_2 \neq 0$.

I have since progressed to understand that yes the continuity equation is valid for this situation, but that $v_1 > 0$, contrary to what I felt earlier. My last post is an attempt at clarifying the matter.

 

[jbriggs444]

I started this thread with a doubt. Is the continuity equation applicable for an open tank filled with water and a hole in the bottom? How can it be, for it will mean the velocity of outflow $v_2 = 0$, as, supposedly, $v_1 = 0$, the velocity at the top surface of the liquid. This assumption seemed reasonable because the top surface of liquid is at rest initially. But clearly something is wrong here because we know $v_2 \neq 0$.

I have since progressed to understand that yes the continuity equation is valid for this situation, but that $v_1 > 0$, contrary to what I felt earlier. My last post is an attempt at clarifying the matter.

It is not clear that you have come to grips with the idea that $v_1$ can start out zero and increase and that $v_2$ can also start out zero and increase without ever violating continuity.

Your initial doubt seems to have stemmed from comparing $v_2$ after a while with $v_1$ right away and deciding that the mismatch meant something.

 

[kuruman]

I will try to explain the continuity equation one more time. Fill a conical funnel with water and plug the outlet hole with your finger (it is safe to try this at home). Let's say that the water area at the surface is A₁ and the area of the outlet hole A₂. Note that no water is flowing, i.e. v₁=v₂=0. The continuity equation becomes A₁x0=A₂x0. That certainly is true.

Now unplug the outlet hole. The continuity equation says A₁v₁=A₂v₂. Note that all quantities in the equation except A₂ are functions of time. Considering that the continuity equation is applicable immediately after you remove your finger at t = 0 (before the water starts flowing), do you think that there is a later time at which the continuity equation stops being applicable?

 

[brotherbobby]

I will try to explain the continuity equation one more time. Fill a conical funnel with water and plug the outlet hole with your finger (it is safe to try this at home). Let's say that the water area at the surface is A₁ and the area of the outlet hole A₂. Note that no water is flowing, i.e. v₁=v₂=0. The continuity equation becomes A₁x0=A₂x0. That certainly is true.

Now unplug the outlet hole. The continuity equation says A₁v₁=A₂v₂. Note that all quantities in the equation except A₂ are functions of time. Considering that the continuity equation is applicable immediately after you remove your finger at t = 0 (before the water starts flowing), do you think that there is a later time at which the continuity equation stops being applicable?

No, the continuity equation is valid for all values of time $t >0$.

My confusion was about the time $t = 0$. Assuming $v_1 = 0$, continuity gives $v_2 = 0$ also. Clearly that is not true since the water has a starting speed at the 2nd end but not at the first. I cannot see how continuity can be used to solve this.

But as we know, it can be solved using Bernoulli's equation.

 

[haruspex]

the water has a starting speed at the 2nd end

I don't see why you are so sure of that.
If you consider a thin disc of water at the hole, it has a mass and a force acts on it. That creates a nonzero acceleration, but it can still be an acceleration from rest.

 

[brotherbobby]

I don't see why you are so sure of that.
If you consider a thin disc of water at the hole, it has a mass and a force acts on it. That creates a nonzero acceleration, but it can still be an acceleration from rest.

Yes, I agree to that. You mean that at $t = 0$, both $v_1\; \text {and} \; v_2 = 0$?

 

[jbriggs444]

the water has a starting speed at the 2nd end

The water has a starting speed of zero at both ends. [Or non-zero at both, depending on your starting setup].

 

[haruspex]

Yes, I agree to that. You mean that at $t = 0$, both $v_1\; \text {and} \; v_2 = 0$?

Yes.

 

[kuruman]

Here is yet another way to look at the continuity equation. It says $Av=\text{constant}$. At $t=0$, before the water starts flowing, the value of the constant is zero. After the water flow has started, the product $Av$ is still constant but non-zero at different points along the path at every instant. You can figure out its value by multiplying the speed of the fluid by the area through which it flows at any point along the fluid flow.