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

In summary: The pressure at the top point is greater than the pressure at the bottom point because the water is pushing down on the top point.
  • #1
songoku
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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
Untitled.png

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
 
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  • #2
songoku said:
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.
 
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  • #3
haruspex said:
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...:confused:
 
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  • #4
Delta2 said:
I must be doing the same mistake then cause i am also getting 10000...:confused:
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.
 
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  • #5
haruspex said:
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?

haruspex said:
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
 
  • #6
songoku said:
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?
 
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  • #7
haruspex said:
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
 
  • #8
songoku said:
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?
 
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  • #9
haruspex said:
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
 
  • #10
songoku said:
Force from the pump?
No, the pump is above the water. In physics, there is no such thing as suction.
 
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  • #11
haruspex said:
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
 
  • #12
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?
 
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  • #13
haruspex said:
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
 
  • #14
songoku said:
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?
 
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  • #15
haruspex said:
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
 
  • #16
songoku said:
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?
 
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  • #17
jbriggs444 said:
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
 
  • #18
songoku said:
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.
songoku said:
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?
 
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  • #19
songoku said:
... 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."
 
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  • #20
songoku said:
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?
 
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  • #21
jbriggs444 said:
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?

Lnewqban said:
Did you create this sketch yourself?
No, I post the original question

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

haruspex said:
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
 
  • #22
songoku said:
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.
songoku said:
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.
songoku said:
After reaching the top part of the pipe, the water will flow down due to gravity.
Yes.
songoku said:
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.
 
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  • #23
haruspex said:
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?
 
  • #24
songoku said:
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.
 
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  • #25
haruspex said:
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
 
  • #26
songoku said:
$$\rho g(H-h)$$
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?
 
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  • #27
jbriggs444 said:
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
 
  • #28
songoku said:
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.
 
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  • #29
songoku said:
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.
 
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  • #30
jbriggs444 said:
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 said:
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
 
  • #31
songoku said:
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:
songoku said:
##A.\frac{dh'}{dt}=-a.v_2## , where ##A## is area of pool
Substitute for v2 what you get from Bernoulli.
 
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  • #32
haruspex said:
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
 
  • #33
songoku said:
##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.
 
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  • #34
haruspex said:
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
 
  • #35
songoku said:
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.
songoku said:
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.
songoku said:
2. Constant pressure difference is equal to Patm−Ptop=Patm−0=Patm ?
Yes.
songoku said:
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.
 
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