# How to calculate the velocity of water streaming into the boat?

• Lotto
Lotto
Homework Statement
I have a problem with understanding the solution of the problem below. At the beginning, there is a formula for the speed of water streaming inside the boat. But I don't know how to calculate it.
Relevant Equations
##v=\sqrt{2g(h_2-h_1)}##
The problem:

Its solution:

I am sure that the velocity ##v## can be calculated by using Bernoulli's equation, but I fail to calculate it.

I don't know what pressure is at the hole - is it ##h_2 \rho g## or ##(h_2 - h_1) \rho g##? The pressure of the water "under" the hole is ##h_2 \rho g## and the pressure just "above" the hole is ##h_1 \rho g##. So what is the pressure at the hole? Is it its difference?

And I am also a little bit surprised that I can calculate the buoyancy when there is a hole in the boat. I thought it could be calculated only when the submerged body is without holes - when a liquid is only around the body, not inside of it. Why can I ignore the holes and calculate the buoyancy as if there were none?

There is a pressure a little way outside the hole and a lower one a little way inside. That pressure difference accelerates the water through the hole, according to the Bernoulli equation.
Since the velocity of the incoming water is constant, the rate of descent of the boat is constant. No acceleration means the net force on the boat is zero, and its weight has not changed.

You can apply Bernoulli's (steady flow) between the pond surface and the water inside the boat (at its surface) for a relatively slow leak. That is how they get ## v^2 = 2g\Delta h##. The mass of the boat and its contents( people, etc...) is constant, but the buoyant force of the boat is slightly increasing as it sinks (when the boat is fully submerged its at its largest.

erobz said:
the buoyant force of the boat is slightly increasing as it sinks
The rate of flow of water into the boat is determined by ##\Delta h##, that in turn determines the rate of sinking, and ##\Delta h## determines the buoyant force.
Since the hole is small, the boat sinks slowly, i.e. at a terminal velocity.
Thus, ##\Delta h## is constant and the buoyant force remains pretty much equal to the weight of the boat+contents.

haruspex said:
There is a pressure a little way outside the hole and a lower one a little way inside. That pressure difference accelerates the water through the hole, according to the Bernoulli equation.
Since the velocity of the incoming water is constant, the rate of descent of the boat is constant. No acceleration means the net force on the boat is zero, and its weight has not changed.
Ok so how to calculate the velocity from Bernoulli's equation?

The pressure above the water surface is ##p_a## and pressure at the hole is, I would say, ##(h_2-h_1)\rho g +p_a##. Let's suppose that velocity of the water surface is zero and that the "potential energy" is measured from the water surface of the pond. So then I would write the equation like this:

##p_a - (h_2-h_1)\rho g = p_a + (h_2-h_1) \rho g -h_2 \rho g +\frac 12 \rho v^2##.

Apparently, this isn't correct. So where is the mistake?

Lotto said:
The pressure above the water surface is ##p_a## and pressure at the hole is, I would say, ##(h_2-h_1)\rho g +p_a##.
The pressure below the hole is ##(h_2)\rho g +p_a## and the pressure just above it is ##(h_1)\rho g +p_a##.

Lotto said:
Ok so how to calculate the velocity from Bernoulli's equation?

The pressure above the water surface is ##p_a## and pressure at the hole is, I would say, ##(h_2-h_1)\rho g +p_a##. Let's suppose that velocity of the water surface is zero and that the "potential energy" is measured from the water surface of the pond. So then I would write the equation like this:

##p_a - (h_2-h_1)\rho g = p_a + (h_2-h_1) \rho g -h_2 \rho g +\frac 12 \rho v^2##.

Apparently, this isn't correct. So where is the mistake?
I would forget the hole ( I think you would need to find out how thick the boat is on its bottom - perhaps it is to be negligible thickness?) Apply Bernoulli's between the surface of the pond and the surface of the water in the boat. I don't think hole geometry has relevance when applied this way?

Maybe I'm missing something.

erobz said:
I would forget the hole ( I think you would need to find out how thick the boat is on its bottom - perhaps it is to be negligible thickness?) Apply Bernoulli's between the surface of the pond and the surface of the water in the boat. I don't think hole geometry has relevance when applied this way?

Maybe I'm missing something.
That would mean the hole might as well be the full width of the boat, in which case it would sink immediately.

haruspex said:
That would mean the hole might as well be the full width of the boat, in which case it would sink immediately.
But ##Q## is ##Q##? The velocity of the surface of the water in the boat water is ##v = \dot h##. The pond height change assumed negligible, where ##\dot h ## is the rate of change of water height in the boat. As far as I can see the hole is given dimension just so we can assume the steady flow Bernoulli holds.

At the pond surface everything is zero implying.

$$0 = \Delta h + \frac{\Delta \dot h ^2}{2g}$$

?

Last edited:
erobz said:
But ##Q## is ##Q##? The velocity of the surface of the water in the boat water is ##v = \dot h##.
Relative to the sinking boat, but it is stationary relative to the surrounding water.
erobz said:
As far as I can see the hole is given dimension just so we can assume the steady flow Bernoulli holds.
Yes, but only until the water has entered the boat. Bernoulli is basically a conservation of work equation. The water is accelerated through the hole, but then what?
Initially , in principle, it has the right KE to reach the surface level of the surrounding water. But even if so, it would just fall back in the boat. As the water builds up in the boat, the new water entering just swashes around with that, so mechanical energy is lost.
Hence you can only apply Bernoulli for the transition through the hole.

haruspex said:
Relative to the sinking boat, but it is stationary relative to the surrounding water.

Yes, but only until the water has entered the boat. Bernoulli is basically a conservation of work equation. The water is accelerated through the hole, but then what?
Initially , in principle, it has the right KE to reach the surface level of the surrounding water. But even if so, it would just fall back in the boat. As the water builds up in the boat, the new water entering just swashes around with that, so mechanical energy is lost.
Hence you can only apply Bernoulli for the transition through the hole.
So if I understand it well, I can't apply Bernoulli's equation on the whole water in the boat - so I can't work with the water surface for example.

So I tried to calculate the velocity from the equation on the photo:

Is it correct?

Lotto said:
View attachment 346327

Is it correct?
The first equation looks right, but you then dropped a factor of 2.

haruspex said:
The first equation looks right, but you then dropped a factor of 2.
Yes, sorry, my fault.

But I am still a little bit confused that the pressure under the hole is ##p_a+h_2\rho g##. Because what about the Pascal's law? Why don't we add ##h_1\rho g## to it, the pressure generated above the hole?

Lotto said:
Yes, sorry, my fault.

But I am still a little bit confused that the pressure under the hole is ##p_a+h_2\rho g##. Because what about the Pascal's law? Why don't we add ##h_1\rho g## to it, the pressure generated above the hole?
For the same reason we do not add ##h_1## to ##h_2## to find the depth just below the hole.

haruspex said:
For the same reason we do not add ##h_1## to ##h_2## to find the depth just below the hole.
Ok.

But what I still don't understand is the magnitude of the buoyancy force. Why is it ##Sh_2\rho g##? Because the buoyancy force is the net hydrostatic force acting on the body, and the net hydroststic force should be ##(S-A)h_2\rho g##, because the surface the force is acting on is ##S-A##. Or not? Why?

Lotto said:
Ok.

But what I still don't understand is the magnitude of the buoyancy force. Why is it ##Sh_2\rho g##? Because the buoyancy force is the net hydrostatic force acting on the body, and the net hydroststic force should be ##(S-A)h_2\rho g##, because the surface the force is acting on is ##S-A##. Or not? Why?
When the hole first appears, yes it is reduced to S-A. As a result, the boat accelerates downwards. But as the water level rises in the boat, the water accelerated through the hole exerts pressure on the boat+water system. As this restores the buoyancy to somewhere close to its original value, the acceleration downwards of the boat declines.

haruspex said:
When the hole first appears, yes it is reduced to S-A. As a result, the boat accelerates downwards. But as the water level rises in the boat, the water accelerated through the hole exerts pressure on the boat+water system. As this restores the buoyancy to somewhere close to its original value, the acceleration downwards of the boat declines.
Well, I am not sure that I understand it. The water streaming through the hole exerts pressure, ok, but how do you know the pressure is ##Ah_2\rho g##?

Lotto said:
Ok.

But what I still don't understand is the magnitude of the buoyancy force. Why is it ##Sh_2\rho g##?
The area of the crack is only 0.06% of the area of the bottom of the boat.
##A/S=10/(200X80)##

Lnewqban said:
The area of the crack is only 0.06% of the area of the bottom of the boat.
##A/S=10/(200X80)##
So you say it is just an approximation - OK, I agree we can neglect the hole. But if I understand haruspex well, we don't have to neglect it because the pressure exerted by the flowing water restors the buoyancy...

Lotto said:
Well, I am not sure that I understand it. The water streaming through the hole exerts pressure, ok, but how do you know the pressure is ##Ah_2\rho g##?
##(h_2-h_1)\rho g## is an upper bound for the pressure difference through the hole. That puts an upper limit on the velocity of water through the hole, and hence an upper limit on the rate of sinking. Hence the net force must tend to zero.

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