What time is needed to move water from a pool to a container?

[songoku]

Homework Statement

A person wants to move all the water from pool (on the left side) to a container (on the right side). He uses pipe with cross sectional area 5 cm square. If h = 2 m and H = 20 m, what is the time needed? Take the area of the pool 50 m square

Relevant Equations

Bernoulli equation

Flow rate

I take position 1 as the surface of the pool and position 2 as the surface of the container so the value of $P_1 = P_2 = P_{atm}$ and $v_1=0$ and $h_2=0$

$P_1 + \rho gh_1 + \frac{1}{2} \rho {v_1}^2 = P_2 + \rho gh_2 + \frac{1}{2} \rho {v_2}^2$

$\rho gH = \frac{1}{2} \rho {v_2}^2$

${v_2}$ = 20 m/sThen, $\frac{\text {volume of water in pool}}{\text {time}} = \text {area of pipe} . {v_2}$

I get time = 10 000 seconds but the answer key is 600 seconds

Where is my mistake? Thanks

 

[haruspex]

Where is my mistake?

Two mistakes.
A mistake In conversion of units (I'm guessing) led to 10000 instead of 100.
The other mistake is that you assumed the height difference is constant.

But there is a flaw in the question. You are not given the cross sectional area of the container on the right. That would be ok if it were very wide, but it does not appear to be. Maybe assume it is very deep into the page.

 

[Delta2]

A mistake In conversion of units (I'm guessing) led to 10000 instead of 100.

I must be doing the same mistake then cause i am also getting 10000...

 

[haruspex]

I must be doing the same mistake then cause i am also getting 10000...

Whoops, you are right.

But there is a mistake I missed. What is atmospheric pressure, and why is that important here?
In view of this, I withdraw my charge that the question is flawed. The correct answer is about 64000 seconds.

 

[songoku]

The other mistake is that you assumed the height difference is constant.

How to account for the change of H every second? The value of H will change from 20 m to 18 m (18 m is when the pool is empty) so I calculate the value of $v_2$ for H = 20 m and H = 18 m and I take the average of the speed?

But there is a mistake I missed. What is atmospheric pressure, and why is that important here?

The pressure at the surface of the water in the pool and the surface of the water in the container is equal to atmospheric pressure but they will cancel out in Bernoulli equation. Is this not correct?

Thanks

 

[haruspex]

they will cancel out in Bernoulli equation.

Not here they won't. If it is atmospheric pressure at the open end, what will the pressure be 10m further up?

 

[songoku]

Not here they won't. If it is atmospheric pressure at the open end, what will the pressure be 10m further up?

You mean I have to use barometric formula such in this link: https://www.math24.net/barometric-formula/ ?

Thanks

 

[haruspex]

You mean I have to use barometric formula such in this link: https://www.math24.net/barometric-formula/ ?

Thanks

No, not 10m higher up outside the pipe, 10m higher up inside the pipe.

Here's a different hint: how high can you pump up water using a pump positioned at the top of the pipe? What is pushing the water up?

 

[songoku]

No, not 10m higher up outside the pipe, 10m higher up inside the pipe.

Here's a different hint: how high can you pump up water using a pump positioned at the top of the pipe?

I am not sure. Maybe using conservation of energy, so the maximum possible height is $\frac{v^2}{2g}$ where $v$ is the speed of water entering the pump?

What is pushing the water up?

Force from the pump?

Thanks

 

[haruspex]

Force from the pump?

No, the pump is above the water. In physics, there is no such thing as suction.

 

[songoku]

No, the pump is above the water. In physics, there is no such thing as suction.

Maybe the difference in pressure? Pressure on the end of pipe at pool is higher than pressure at the end of pipe at the container side so water is being pushed to the container?

Thanks

 

[haruspex]

I'll try something else..
Consider two points in the flow on the right hand side, one at the top of the pipe and one at the bottom (the open end).
How do the velocities compare? How do the heights compare? What does that tell you about the pressures at those two points?

What is the pressure at the open end? So what is the pressure at the top point?

 

[songoku]

I'll try something else..
Consider two points in the flow on the right hand side, one at the top of the pipe and one at the bottom (the open end).

By "on the right hand side", you mean the part of pipe closer to the container?

How do the velocities compare? How do the heights compare? What does that tell you about the pressures at those two points?

The velocity will be the same because the area of pipe is constant all along the pipe?

The height will decrease as the container filled with water?

For the pressure at those two points:
$P_{\text{top}} + \frac{1}{2} \rho {v_{\text{top}}}^2 + \rho gh_{\text{top}} = P_{\text{bottom}} + \frac{1}{2} \rho {v_{\text{bottom}}}^2 + \rho gh_{\text{bottom}}$ , where $v_{\text{top}}=v_{\text{bottom}}$ and $h_{\text{bottom}}=0$

$P_{\text{top}} + \rho gh_{\text{top}} = P_{\text{bottom}}$

Pressure at the bottom (open end of pipe at the container) will be bigger compared to pressure at top part of the pipe

What is the pressure at the open end? So what is the pressure at the top point?

Pressure at open end = $P_{\text{atm}}$ so pressure at the top point is $P_{\text{atm}}- \rho gh_{\text{top}}$

Thanks

 

[haruspex]

so pressure at the top point is $P_{\text{atm}}- \rho gh_{\text{top}}$

Right, and what will that be, given the value of H?

 

[songoku]

Right, and what will that be, given the value of H?

Taking $g=10$ m/s² for simplification, density of water = $1000 ~ kg/m^3, H = 20$ m , $P_{atm}=1 \times 10^5$ Pa

pressure at the top point = 1 x 10⁵ - 1000 x 10 x 20 = - 1 x 10⁵ Pa?

The pressure is negative?

Thanks

 

[jbriggs444]

The pressure is negative?

Bingo.

Can pressure be negative? Or, phrased another way: What happens to water if pressure goes negative or gets very small?

 

[songoku]

Can pressure be negative?

I don't think so

Or, phrased another way: What happens to water if pressure goes negative or gets very small?

Water will flow from high pressure position to lower pressure position but I don't think the water will flow up towards the top part of the pipe.

Or maybe you mean the water will turn to gas?

Thanks

 

[jbriggs444]

I don't think so

Right. The water pressure cannot go negative. As you say, the water will turn to gas. "Boil" is a term that we use to describe this.

Water will flow from high pressure position to lower pressure position but I don't think the water will flow up towards the top part of the pipe.

The pressure at the top of the pipe may not be negative. But it may be very small when the upper entrance to the pipe is at atmospheric pressure. What happens in that case?

 

[Lnewqban]

... The pressure is negative?

Did you create this sketch yourself?

Please, see:
https://en.m.wikipedia.org/wiki/Siphon

The text of the problem does not specify the way in which the water flows.

"A person wants to move all the water from pool (on the left side) to a container (on the right side). He uses pipe with cross sectional area 5 cm square."

 

[haruspex]

I don't think soWater will flow from high pressure position to lower pressure position but I don't think the water will flow up towards the top part of the pipe.

Or maybe you mean the water will turn to gas?

Thanks

So this sets a limit on the height of the column of water in the pipe on the right hand side. If the bottom 10m is filled with water then the space above that is a near vacuum.
Taking that into account, what do you get for the time to empty?

 

[songoku]

The pressure at the top of the pipe may not be negative. But it may be very small when the upper entrance to the pipe is at atmospheric pressure. What happens in that case?

The water will move upwards from the pool to the pipe because the water is being pushed by atmospheric pressure?

Did you create this sketch yourself?

No, I post the original question

Please, see:
https://en.m.wikipedia.org/wiki/Siphon

This is really helpful. I do not know this set up is called siphon

The text of the problem does not specify the way in which the water flows.

"A person wants to move all the water from pool (on the left side) to a container (on the right side). He uses pipe with cross sectional area 5 cm square."

The person wants to move the water from pool (on left side) to container (on right side) so the water will flow from left side to right side

So this sets a limit on the height of the column of water in the pipe on the right hand side. If the bottom 10m is filled with water then the space above that is a near vacuum.
Taking that into account, what do you get for the time to empty?

Sorry I am still not sure if I understand how this set up works.

1. Shouldn't I need to consider how deep the pipe in the pool on the left side? In this question, I just ignore it because the question did not give the information and I just assume the pipe is at the top of the water surface of the pool?

2. The pressure at the end of the pipe in the pool is higher than the top part of the pipe so the pressure will push the water to flow in the pipe?

3. After reaching the top part of the pipe, the water will flow down due to gravity.

4. Will the water turn to gas when it moves to the top of the pipe due to the pressure at the top of the pipe being zero?

Thanks

 

[haruspex]

Shouldn't I need to consider how deep the pipe in the pool on the left side? In this question, I just ignore it because the question did not give the information and I just assume the pipe is at the top of the water surface of the pool?

Yes, you do need to consider the depth of water in the pool. Since this varies, you need to get a differential equation relating the depth to the rate of flow.
The depth to which the hose goes in the pool doesn’t matter as long as it is always below the surface. Take it as always being at the bottom of the pool.

The pressure at the end of the pipe in the pool is higher than the top part of the pipe so the pressure will push the water to flow in the pipe?

Yes.

After reaching the top part of the pipe, the water will flow down due to gravity.

Yes.

Will the water turn to gas when it moves to the top of the pipe due to the pressure at the top of the pipe being zero?

The amount of water that turns to gas is very small. Even at 30C, the SVP of water is only 1/30 of an atmosphere, so there will only be enough water vapour to create that pressure in that volume, at that temperature.
This won't change until the pool empties; once that column of vapour has been created at the start of the process, as water flows over the apex it will run down without any further evaporation.
You can safely ignore it, treating the vapour region as a vacuum.

 

[songoku]

Yes, you do need to consider the depth of water in the pool. Since this varies, you need to get a differential equation relating the depth to the rate of flow.
The depth to which the hose goes in the pool doesn’t matter as long as it is always below the surface. Take it as always being at the bottom of the pool.

Yes.

Yes.

The amount of water that turns to gas is very small. Even at 30C, the SVP of water is only 1/30 of an atmosphere, so there will only be enough water vapour to create that pressure in that volume, at that temperature.
This won't change until the pool empties; once that column of vapour has been created at the start of the process, as water flows over the apex it will run down without any further evaporation.
You can safely ignore it, treating the vapour region as a vacuum.

So I need to set up bernoulli equation for point 1 (which is at the bottom of the pool) and point 2 (which is at the surface of the container).

The depth of water in the pool will decrease to it will affect the value of $P_1$ (because $P_1= P_{\text{atm}}+ \rho gh$ and the value of $h$ is not constant)and also for $P_2$ to be always equal to $P_{atm}$ I need to consider the rate of increase of water level in the container?

 

[haruspex]

point 2 (which is at the surface of the container).

No. As we have established, there is not a continuous stream of water between those two points, so you cannot apply Bernoulli between them.

You know the pressure at the apex of the tube.

 

[songoku]

No. As we have established, there is not a continuous stream of water between those two points, so you cannot apply Bernoulli between them.

You know the pressure at the apex of the tube.

So I take position 1 as the left end pipe in the pool and position 2 as the apex of the tube.

$P_1+ \frac{1}{2} \rho {{v_1}}^2 + \rho gh_1=P_2 + \frac{1}{2} \rho{{v_2}}^2 + \rho gh_2$

$P_{\text{atm}}+\rho gh+ \frac{1}{2} \rho {{v_1}}^2 + \rho g(H-h)=P_2 + \frac{1}{2} \rho{{v_2}}^2 + \rho gH$

$h$ is function of time , $P_2=0$

Can I take $v_1$ as 0?

Thanks

 

[jbriggs444]

$$\rho g(H-h)$$

What does this term represent in the physical picture?

Do we actually have a continuous stream of water of density $\rho$ in the downstream side of the pipe?

 

[songoku]

What does this term represent in the physical picture?

$H-h$ is the height of the left end of the pipe (the one in the bottom of the pool) measured with respect to the container

Do we actually have a continuous stream of water of density $\rho$ in the downstream side of the pipe?

I am not setting equation for the downstream side of the pipe. I am setting up bernoulli equation for the leftmost end of the pipe (in the bottom of the pool, I set this as position 1) and the apex of the pipe (as position 2)

Thanks

 

[jbriggs444]

I am not setting equation for the downstream side of the pipe. I am setting up bernoulli equation for the leftmost end of the pipe (in the bottom of the pool, I set this as position 1) and the apex of the pipe (as position 2)

I am not getting it. If you are considering the left hand portion of the pipe then the right hand side and, in particular, $H$, is irrelevant.

@haruspex has provided a big hint. You know the pressure at the apex.

 

[haruspex]

So I take position 1 as the left end pipe in the pool and position 2 as the apex of the tube.

$P_1+ \frac{1}{2} \rho {{v_1}}^2 + \rho gh_1=P_2 + \frac{1}{2} \rho{{v_2}}^2 + \rho gh_2$

$P_{\text{atm}}+\rho gh+ \frac{1}{2} \rho {{v_1}}^2 + \rho g(H-h)=P_2 + \frac{1}{2} \rho{{v_2}}^2 + \rho gH$

$h$ is function of time , $P_2=0$

Can I take $v_1$ as 0?

Thanks

Much better.
Yes, v₁=0.
But on the left you have both ρgh and ρg(H-h), which makes no sense because that would reduce to ρgH, canceling the term on the right.

Next, you need an equation relating h to v₂.

 

[songoku]

I am not getting it. If you are considering the left hand portion of the pipe then the right hand side and, in particular, $H$, is irrelevant.

@haruspex has provided a big hint. You know the pressure at the apex.

Much better.
Yes, v₁=0.
But on the left you have both ρgh and ρg(H-h), which makes no sense because that would reduce to ρgH, canceling the term on the right.

Next, you need an equation relating h to v₂.

Let me try again

$P_1+\frac{1}{2} \rho {v_1}^2 + \rho gh_1 = P_2 + \frac{1}{2} \rho {v_2}^2 + \rho gh_2$

Taking the bottom of the pool as reference, $h_1=0$ and $h_2=h$

$P_{\text{atm}} + \rho gh'= \frac{1}{2} \rho {v_2}^2 + \rho gh$, where $h'$ is the depth of water in the pool

---------------------------------------------------------------------------------------------------------------------------------------------------

For equation relating $h'$ to $v_2$:

$\frac{d \text{(volume)}}{dt}=-a.v_2$ where $a$ is area of pipe and the negative sign because the volume is decreasing

$A.\frac{dh'}{dt}=-a.v_2$ , where $A$ is area of poolI need to use integration?

$\int dh'=-\frac{a}{A} .v_2 \int dt$

$h'=-k.v_2.t+c$ , where $k=\frac{a}{A}$

Taking $h'=h$ when $t=0$, I get:

$h'=-k.v_2.t+h$

-------------------------------------------------------------------------------------------------------------------------------------------------------
So:
$P_{\text{atm}} + \rho gh'= \frac{1}{2} \rho {v_2}^2 + \rho gh$

$P_{\text{atm}} + \rho g(-k.v_2.t+h)=\frac{1}{2} \rho {v_2}^2 + \rho gh$The term $\rho gh$ will cancel out?

$P_{\text{atm}} - \rho gk.v_2.t=\frac{1}{2} \rho {v_2}^2$To move all the water to the apex of the tube $\rightarrow h'=0$
$h'=-k.v_2.t+h$

$t=\frac{h}{k.v_2}$Then:
$P_{\text{atm}} - \rho gk.v_2.t=\frac{1}{2} \rho {v_2}^2$

$P_{\text{atm}} - \rho gk.v_2 \frac{h}{k.v_2}=\frac{1}{2} \rho {v_2}^2$

$P_{\text{atm}} - \rho gh=\frac{1}{2} \rho {v_2}^2$Is this even correct? Thanks

 

[haruspex]

I need to use integration?

Yes, but v₂ is a variable, so it is not as simple as you have treated it.
Go back to your differential equation:

$A.\frac{dh'}{dt}=-a.v_2$ , where $A$ is area of pool

Substitute for v₂ what you get from Bernoulli.

 

[songoku]

Yes, but v₂ is a variable, so it is not as simple as you have treated it.
Go back to your differential equation:

Substitute for v₂ what you get from Bernoulli.

$P_{\text{atm}}+\rho gh'=\frac{1}{2} \rho{v_2}^2 + \rho gh$

$v_2=\sqrt{\frac{2}{\rho} (P_{\text{atm}}+\rho gh' - \rho gh)}$

----------------------------------------------------------------------------------------------
$A.\frac{dh'}{dt}=-a.v_2$

$A.\frac{dh'}{dt}=-a.\sqrt{\frac{2}{\rho} (P_{\text{atm}}+\rho gh' - \rho gh)}$

$\int \frac{1}{\sqrt{(P_{\text{atm}}+\rho gh' - \rho gh)}}dh'=\int -\frac{a}{A} \sqrt{\frac{2}{\rho}}dt$

$\frac{2\sqrt{P_{\text{atm}}+\rho gh' - \rho gh}}{\rho g}=-\frac{a}{A} \sqrt{\frac{2}{\rho}} t+c$

Taking $h'=h$ when $t=0$, I get:
$\frac{2\sqrt{P_{\text{atm}}+\rho gh' - \rho gh}}{\rho g}=-\frac{a}{A} \sqrt{\frac{2}{\rho}} ~t+\frac{2 \sqrt{P_{\text{atm}}}}{\rho g}$

Is this correct? If yes, the next step I have in my mind is:
1. Put $h'=0$ to find the time taken for the water to reach the apex of the pipe

2. Find $v_2$ using bernoulli equation by setting $h'=0$

3. Using the formula $\Delta y=u.t+\frac{1}{2} gt^2$ to find the time taken by the water to fall from apex to container, where $\Delta y=-20~\text{m}$ and $u=-v_2$

Is my idea correct? Thanks

 

[haruspex]

$P_{\text{atm}}+\rho gh'=\frac{1}{2} \rho{v_2}^2 + \rho gh$

$v_2=\sqrt{\frac{2}{\rho} (P_{\text{atm}}+\rho gh' - \rho gh)}$

----------------------------------------------------------------------------------------------
$A.\frac{dh'}{dt}=-a.v_2$

$A.\frac{dh'}{dt}=-a.\sqrt{\frac{2}{\rho} (P_{\text{atm}}+\rho gh' - \rho gh)}$

$\int \frac{1}{\sqrt{(P_{\text{atm}}+\rho gh' - \rho gh)}}dh'=\int -\frac{a}{A} \sqrt{\frac{2}{\rho}}dt$

$\frac{2\sqrt{P_{\text{atm}}+\rho gh' - \rho gh}}{\rho g}=-\frac{a}{A} \sqrt{\frac{2}{\rho}} t+c$

Taking $h'=h$ when $t=0$, I get:
$\frac{2\sqrt{P_{\text{atm}}+\rho gh' - \rho gh}}{\rho g}=-\frac{a}{A} \sqrt{\frac{2}{\rho}} ~t+\frac{2 \sqrt{P_{\text{atm}}}}{\rho g}$

Is this correct? If yes, the next step I have in my mind is:
1. Put $h'=0$ to find the time taken for the water to reach the apex of the pipe

2. Find $v_2$ using bernoulli equation by setting $h'=0$

3. Using the formula $\Delta y=u.t+\frac{1}{2} gt^2$ to find the time taken by the water to fall from apex to container, where $\Delta y=-20~\text{m}$ and $u=-v_2$

Is my idea correct? Thanks

Setting h'=0 to find t, yes. not sure whether you are expected to worry about what happens thereafter.
numerically, what do you get?

If you do want to consider the next stage, it is different. We now have an ascending column of water in the pipe. It all moves at the same speed, has diminishing mass, but subject to a constant pressure difference. So it will accelerate at an increasing rate. Bernoulli does not apply.
This gives you a new differential equation to solve.

 

[songoku]

Setting h'=0 to find t, yes. not sure whether you are expected to worry about what happens thereafter.
numerically, what do you get?

$\frac{2\sqrt{P_{\text{atm}}+\rho gh' - \rho gh}}{\rho g}=-\frac{a}{A} \sqrt{\frac{2}{\rho}} ~t+\frac{2 \sqrt{P_{\text{atm}}}}{\rho g}$

Taking $P_{\text{atm}}=1 \times 10^5 ~\text{Pa} , \rho = 1000 ~kg/m^3, g=10~m/s^2, a=5~cm^2, A=50~m^2, h'=0, h= 2~m$, I get $t=1.5 \times 10^4$ seconds

I think this is just the time needed for the water to move from the bottom of the pool to the apex of the pipe so that's why I think I need to calculate the time taken to go down 20 m to the container by using kinematics formula and taking the acceleration to be acceleration of free fall

If you do want to consider the next stage, it is different. We now have an ascending column of water in the pipe. It all moves at the same speed, has diminishing mass, but subject to a constant pressure difference. So it will accelerate at an increasing rate. Bernoulli does not apply.
This gives you a new differential equation to solve.

1. By "has diminishing mass", is it because some of the mass of water has turned to vapour?

2. Constant pressure difference is equal to $P_{\text{atm}} - P_{\text{top}} = P_{\text{atm}}-0=P_{\text{atm}}$ ?

3. Why will the water accelerate at increasing rate? Can't we say the water will move with constant speed because from flow rate continuity equation: $a_1.v_x=a_2.v_y$ the value of $v_x$ will be the same as $v_y$ since the area of pipe is constant?

Thanks

 

[haruspex]

I get t=1.5×104 seconds

I got 64,000s before, much closer to the given answer. I'll try to find my scribbles.

1. By "has diminishing mass", is it because some of the mass of water has turned to vapour?

No, it's because the column is rising in the pipe, no more water is coming in at the bottom, but it is spilling over at the top.
AS I posted, you can ignore that some will turn to vapour. That happens just once at the beginning.

2. Constant pressure difference is equal to Patm−Ptop=Patm−0=Patm ?

Yes.

3. Why will the water accelerate at increasing rate? Can't we say the water will move with constant speed because from flow rate continuity equation: a1.v1=a2.v2 the value of v1 will be the same as v2 since the area of pipe is constant?

At any instant the whole column will move at the same speed, but it has diminishing mass and a constant force, so not only will the speed increase, the acceleration will increase.

 

[songoku]

I got 64,000s before, much closer to the given answer. I'll try to find my scribbles.

The given answer is 600 seconds. So I guess my working and equation is already correct so it is just a matter of plugging value

The time I need to calculate is only the time taken to reach the apex of the pipe? No need to calculate the time taken by the water to fall from apex to the container and add them to get total time?

At any instant the whole column will move at the same speed, but it has diminishing mass and a constant force, so not only will the speed increase, the acceleration will increase.

The constant force comes from: $F=P_{\text{atm}} \times \text{area of pipe}$ ?

If, let say, I want to continue to the vertical pipe on the container's side, is this how I do it:

$F=P.a$

$\frac{d(mv)}{dt}=P.a$

$v.\frac{dm}{dt}=P.a$

$v.\frac{d(\rho V)}{dt}=P.a$

$v.\rho \frac{d(a.H)}{dt}=P.a$

$v.\rho .a \frac{dH}{dt}=P.a$

$v.\rho \frac{dH}{dt}=P$

Then integrating to get $H$ in term of $t$, find the constant of integration by putting $H=20~m$ for $t=0$ and finally find the time taken by all water to go in container by setting $H=0$?

Thanks

 

[haruspex]

The given answer is 600 seconds.

Yes, sorry, I have been assuming that is just an error in taking the ratio of the areas and it should have said 60,000.

The time I need to calculate is only the time taken to reach the apex of the pipe? No need to calculate the time taken by the water to fall from apex to the container and add them to get total time?

The question is unclear. As I wrote, you are probably only meant to find the time to empty the pool. Finding the time to empty the left side of the pipe is an interesting exercise, but will be insignificant in comparison. The time for it all then to fall into the container is surely irrelevant.

$\frac{d(mv)}{dt}=P.a$

$v.\frac{dm}{dt}=P.a$

No, v is changing too. And I strongly discourage the use of $\frac{d(mv)}{dt}=F$. Where mass is being gained or lost, it ignores the momentum lost with departing mass or gained with arriving mass.
In the present case, water spilling over the top had speed v, so carries momentum out of the system consisting of the column.
You can write one equation relating the rate of change of mass to the speed of the water, and another relating the rate of change of speed to the current mass.

 

[songoku]

No, v is changing too. And I strongly discourage the use of $\frac{d(mv)}{dt}=F$. Where mass is being gained or lost, it ignores the momentum lost with departing mass or gained with arriving mass.

So $\frac{d(mv)}{dt}=F$ can be used if $m$ is constant and not a good one if $m$ changes?

In the present case, water spilling over the top had speed v, so carries momentum out of the system consisting of the column.
You can write one equation relating the rate of change of mass to the speed of the water, and another relating the rate of change of speed to the current mass.

I am not sure how to proceed.

At the top of the pipe, the speed of the water is $v_2$ (as obtained by bernoulli equation). Let say the mass of water is $m$, then the momentum of water is $mv_2$

After time $\Delta t$, water of mass $\Delta m$ leaves the pipe (into the container) so the mass of water in the pipe will be $m-\Delta m$ and the velocity will be $v_2+\Delta v$

The new momentum will be:
$p_2=(m-\Delta m)(v_2+\Delta v)=mv_2+m\Delta v-v_2\Delta m-\Delta m \Delta v\approx mv_2+m\Delta v-v_2\Delta m$

$\Delta p=p_2-p_1=mv_2+m\Delta v-v_2\Delta m-mv_2=m\Delta v-v_2\Delta m$
-----------------------------------------------------------------------------------------------------------------

$\frac{\Delta p}{\Delta t}=P.a$

$m\frac{dv}{dt}-v_2\frac{dm}{dt}=P.a$

Is this correct? Thanks

 

[haruspex]

At the top of the pipe, the speed of the water is v2 (as obtained by bernoulli equation).

Initially, yes, in fact all of the water in the left part of the pipe will be at that speed.

So $\frac{d(mv)}{dt}=F$ can be used if $m$ is constant and not a good one if $m$ changes?

$m\frac{dv}{dt}-v_2\frac{dm}{dt}=P.a$

Note what that last equation gives for P=0. It says the water continues to accelerate up the pipe!
As I posted, it is best to avoid using $\frac{d(mv)}{dt}$. I have seen many students get the wrong answer that way.
It is wrong here because it treats the spilt water as passing its momentum to the water remaining in the column.

I laid out a perfectly straightforward path: write one equation for the force on what is in the pipe at some instant, and the resulting acceleration, dv/dt, and another equation for how the velocity leads to the rate of change of mass in the column.
So you have dv/dt as a function of m, and dm/dt as a function of v.
This will lead to a second order ODE.

 

[songoku]

Note what that last equation gives for P=0. It says the water continues to accelerate up the pipe!

When $P=0\rightarrow m\frac{dv}{dt}=v_2\frac{dm}{dt}$

How to conclude from this equation that the water continues accelerating upwards?

I laid out a perfectly straightforward path: write one equation for the force on what is in the pipe at some instant, and the resulting acceleration, dv/dt, and another equation for how the velocity leads to the rate of change of mass in the column.
So you have dv/dt as a function of m, and dm/dt as a function of v.

From where should I start setting up the equation? I have no other idea besides from $F=P.a$

This will lead to a second order ODE.

I think this is beyond me.

I will see if I still can understand up only until setting up the equation. If this is also beyond my understanding, I will end this thread.

Thanks

 

[haruspex]

How to conclude from this equation that the water continues accelerating upwards?

The way you defined it, dm/dt is positive, so your equation makes dv/dt positive.

From where should I start setting up the equation?

If at some instant the mass is m and the force is P.a, what is the acceleration?
If the velocity is v, what is the rate of loss of mass?

 

[songoku]

If at some instant the mass is m and the force is P.a, what is the acceleration?

$P.a=m \times \text{acceleration}$

$P.a=m \frac{dv}{dt}$

$\frac{dv}{dt} = \frac{P.a}{m}$

Is this correct?

If the velocity is v, what is the rate of loss of mass?

Sorry, I have no idea to set this one

 

[haruspex]

$P.a=m \times \text{acceleration}$

$P.a=m \frac{dv}{dt}$

$\frac{dv}{dt} = \frac{P.a}{m}$

Is this correct?Sorry, I have no idea to set this one

If it is rising at speed v with a cross-sectional area a, what is the volumetric flow rate?
What then is the mass flow rate? That will be the rate at which water leaves the column.

Since I am increasingly sure this is beyond what you are expected to do, I am quite happy to supply the rest of the solution to this part if you do not wish to pursue it yourself.

 

[songoku]

If it is rising at speed v with a cross-sectional area a, what is the volumetric flow rate?
What then is the mass flow rate? That will be the rate at which water leaves the column.

Since I am increasingly sure this is beyond what you are expected to do, I am quite happy to supply the rest of the solution to this part if you do not wish to pursue it yourself.

Since I have very little to no idea of how to proceed, I think it is better for you to help me supplying the rest of solution.

Thanks

 

[haruspex]

Since I have very little to no idea of how to proceed, I think it is better for you to help me supplying the rest of solution.

Thanks

$\dot v= \frac{P.a}m-g$
$\dot m=-\rho a.v$
When you have simultaneous differential equations the way forwards often involves differentiation.
$\ddot v= -\frac{P.a}{m^2}\dot m$
Now we can eliminate m:
$\ddot v=\frac 1{Pa}\dot v^2\rho a.v$
$\frac{\ddot v}{\dot v}=\frac 1{P}\dot v\rho v$
Integrating:
$\ln(\dot v)=c+\frac 1{2P}\rho v^2$
$\dot v=Ce^{\frac 1{2P}\rho v^2}$
When t=0, v=v₂, m=$\rho ha$, $\dot v=\frac{P}{\rho h}-g$
$\frac{P}{\rho h}-g=Ce^{\frac 1{2P}\rho v_2^2}$
$\dot v=(\frac{P}{\rho h}-g)e^{\frac 1{2P}\rho (v^2-v_2^2)}$
$(\frac{P}{\rho h}-g)t=\int _{v_2}^{\infty}e^{\frac 1{2P}\rho (v_2^2-v^2)}dv$
$=\int _0^{\infty}e^{\frac 1{2P}\rho (-2vv_2-v^2)}dv$
Which we can look up in tables of the Gaussian distribution.

 

[songoku]

Now we can eliminate m:
$\ddot v=\frac 1{Pa}\dot v^2\rho a.v$

I don't understand this part. This is what I tried:

$\dot v= \frac{P.a}m-g \rightarrow m=\frac{P.a}{\dot v +g}$

$\frac{1}{m^2} = \frac{(\dot v +g)^2}{(P.a)^2}$
-------------------------------------------------------------------------------------------------------

$\ddot v= -\frac{P.a}{m^2}\dot m$

$=-P.a \frac{(\dot v +g)^2}{(P.a)^2} .(-\rho .a.v)$

$=\frac{\rho .a.v}{P.a}(\dot v^2 + 2 \dot v g + g^2)$

How can all the terms consisting $g$ eliminated? Thanks

 

[haruspex]

How can all the terms consisting g eliminated? Thanks

I only had one equation with g:

$\dot v= \frac{P.a}m-g$

and I differentiated it wrt t, so g disappeared for a while.
But note that it reappeared when the boundary condition was used. Most likely, your development will arrive at the same point.

 

[songoku]

Thank you very much for all the help and explanation haruspex, Delta2, Lnewqban, jbriggs444