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

• songoku
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.
haruspex said:
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

songoku said:
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.

songoku said:
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.

songoku said:
##\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
haruspex said:
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

songoku said:
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.
songoku said:
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
haruspex said:
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

songoku said:
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.
songoku said:
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
haruspex said:
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

songoku said:
##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
haruspex said:
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

Delta2
songoku said:
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
haruspex said:
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

songoku said:
How can all the terms consisting g eliminated? Thanks
I only had one equation with g:
haruspex said:
##\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.

Last edited:
songoku
Thank you very much for all the help and explanation haruspex, Delta2, Lnewqban, jbriggs444

Lnewqban and Delta2

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