Calculating Drain Time of a Tank with Closed Outlet: What is the Formula?

[Mehdi622]

Experts,

I need to calculate how long does it take to drain a tank through port in bottom of the tank, but not any other parts of this system is exposed to atmosphere.

So as the tank drains, it creates negative pressure at top of the fluid inside the tank? any formula to calculate the time required to drain it or drain a certain volume of the tank?

 

[BvU]

Hello Mehdi, welcome to PF !

Negative pressures are an unknown phenomenon in physics, but lower than atmospheric is certainly possible.

So you need an equation for the flow through the port as a function of the pressure difference between port and liquid surface. Then you integrate that wrt time and there you are.

Pressure at liquid surface depends on temperature (if the liquid evaporates) and on volume (of the gas above the liquid -- if there is any at the start of the draining). You will need initial conditions and some physical properties to deal with this. A liquid nitrogen tank with a hole behaves different from a tank with crude oil.

And the integration can be done numerically if things become too complicated.

Good luck !

--

 

[Mehdi622]

Thanks for your reply,

Assume 100m of pipe, filled with water, top of the pipe is closed (not open to atmosphere) and we are tyring to drain the pipe through 1.125" nozzle in the bottom of the pipe.

is there any equation develoed for this scenario? if yes please share

 

[sophiecentaur]

The rate at which the tank drains will depend on the volume of the air space on top of the water. AS soon as you start to lose water, the pressure on top will drop and the rate will depend upon the fractional change in air volume (plus temperature drop etc. etc., if you're being fussy)
I just spotted the 100m bit. Is that 100m of vertical run?

 

[Mehdi622]

Thanks for the rely

The pipe is initially filled 100% with water, 100 m of pipe and we try to drain it through the 1.125" at the bottom of the pipe. top f the pipe is not exposed to atmosphere. like a straw filled with water?

any formula developed for this scenario?

 

[sophiecentaur]

This link could give you a ball park figure, once you have decided on the actual pressures involved.

 

[BvU]

Assume 100m of pipe, filled with water, top of the pipe is closed (not open to atmosphere) and we are tyring to drain the pipe through 1.125" nozzle in the bottom of the pipe.

We're getting somewhere. No tank but a pipe. More than 1" diameter if I am allowed to guess. But you can enlighten us. And horizontal I suppose ? (If not, you get quite a different problem, so please don't leave us poor helpers in the dark about this and about further circumstances !)
And water, not liquid nitrogen and no crude oil either.

Anything else? Is the object of the exercise to get it empty asap or to fix a leak asap ?

Or is it homework ?

is there any equation developed for this scenario? if yes please share

You bet. Oil companies, water companies, schoolteachers, students, all have dealt with it, so: yes.

--

 

[sophiecentaur]

So the pipe was originally filled with water by being pumped upwards from the bottom (requiring about 10 Bar of pressure ?) then the pump is turned off and the water drains out again? We really do need some help with this.

 

[Mehdi622]

Thanks for the rely.

Sorry, I think I should have been more clear.

The actual case is, 100m of vertical pipe, ID 4.276 inches. The pipe is filled with water and top of the pipe is closed (not exposed to atmosphere). Bottom of the pipe there is a 1.125" opening (nozzle) exactly at the center of the pipe and it is the only way for water to drain.

So we fill the pipe with water from "Top" (bottom nozzle is closed at this stage), once the pipe filled with water we close the top, then open the bottom nozzle.

any other factors like temperature etc are to be excluded.

I was looking for an equation to find out how long it takes to drain the pipe (considering top of the pipe is closed), but i could not find any.

This is a part of a problem with one of the tools failure, and answer to the above question (sharing an formula etc.) would be a huge help.

 

[jbriggs444]

It seems clear that the pipe will drain in two stages.

In the first stage, water will squirt from the nozzle under a net pressure of 9 bars and declining to 0 bars as the pipe drains from 100 meters full to only 10 meters full. There will be a partial vacuum inside the pipe above the surface of the water filled only with water vapor. At room temperature, the vapor pressure will be negligible.

At the end of the first stage, the system is in an equilibrium. Water pressure at the bottom of the pipe is equal to atmospheric pressure. Water no longer squirts. But a 1.125" opening is large enough to allow air bubbles to enter the pipe and for water to "glug" out.

The first stage is easy to model. The exit velocity of the water stream will depend on the water pressure at the bottom of the pipe. Let P be the current pressure difference between the bottom of the pipe and atmospheric pressure. Let V be an incremental volume of water. Let $\rho$ be the density of water and v be the exit velocity. Then we can equate the kinetic energy of the water ($\frac{mv^2}{2} = \frac{V{\rho}v^2}{2}$) with the pressure energy of the water $PV$. That yields:

$PV = \frac{V{\rho}v^2}{2}$

Simplify that, solve for v and multiply by nozzle cross-section to get flow rate. Express the pressure difference P as a function of the height of water in the column. Express the change in water height in terms of flow rate and you have a differential equation for water height in terms of time. Solve that and you have the time required to get through stage one.

Stage two is tougher.

 

[Mehdi622]

It seems clear that the pipe will drain in two stages.

In the first stage, water will squirt from the nozzle under a net pressure of 9 bars and declining to 0 bars as the pipe drains from 100 meters full to only 10 meters full. There will be a partial vacuum inside the pipe above the surface of the water filled only with water vapor. At room temperature, the vapor pressure will be negligible.

At the end of the first stage, the system is in an equilibrium. Water pressure at the bottom of the pipe is equal to atmospheric pressure. Water no longer squirts. But a 1.125" opening is large enough to allow air bubbles to enter the pipe and for water to "glug" out.

The first stage is easy to model. The exit velocity of the water stream will depend on the water pressure at the bottom of the pipe. Let P be the current pressure difference between the bottom of the pipe and atmospheric pressure. Let V be an incremental volume of water. Let $\rho$ be the density of water and v be the exit velocity. Then we can equate the kinetic energy of the water ($\frac{mv^2}{2} = \frac{V{\rho}v^2}{2}$) with the pressure energy of the water $PV$. That yields:

$PV = \frac{V{\rho}v^2}{2}$

Simplify that, solve for v and multiply by nozzle cross-section to get flow rate. Express the pressure difference P as a function of the height of water in the column. Express the change in water height in terms of flow rate and you have a differential equation for water height in terms of time. Solve that and you have the time required to get through stage one.

Stage two is tougher.

jbriggs444, Thanks for the reply.

Considering two scenarios below:

1) the 100m vertical pipe is filled with water, top of the pipe is open to atmosphere, then we open the 1.125" nozzle

2) the 100m vertical pipe is filled with water, top of the pipe NOT open to atmosphere, then we open the 1.125" nozzle

Basically, the time required to drain the pipe from full to 10m full is not different for both cases above?

 

[BvU]

Time is different. Consider the end situation: pipe open means 1 atmosphere at the liquid level plus 10 m of water. Pipe closed means water vapour pressure (almost nothing) plus 10 m of water. So the driving pressures at the nozzle are different at the end. In fact hey are different from the beginning (open: 1 atm +100 m, closed: approx. 100 m )

 

[Mehdi622]

Time is different. Consider the end situation: pipe open means 1 atmosphere at the liquid level plus 10 m of water. Pipe closed means water vapour pressure (almost nothing) plus 10 m of water. So the driving pressures at the nozzle are different at the end. In fact hey are different from the beginning (open: 1 atm +100 m, closed: approx. 100 m )

ys,Guys,
I am not really good with correlating and finding right formula. What i need to have is actually the formula to calculate time required to drain the top closed pipe (filled with water) to More than 10 m filled (e.g from 100m to 30m) in top closed system. Appreciate if you share a clear formula. Thank you

 

[equi_librium]

First calculate how rapidly the water would flow if the pipe were entirely open at the top and the water is free flowing (i.e. ignoring the effect of the viscosity of water). The flow rate is proportional to the cross section of the outlet ( pi*r^2) = (3.146)(1/2*1.125)^2 falling (free flowing in 1 dimension) and the velocity at any time 't', V = 0 + gt (initial flow velocity is 0) g: acceleration due to gravity. In this case V increases until all the water is evacuated from the pipe.

In the actual case in which the top of the pipe is closed, the water "glugs" out of the pipe in a way that requires a volume of air entering the opening to equal the volume of water leaving. This condition will 'limit' V to some constant value related to the frequency of the glugs (which are abrupt interruptions of flow) Fg = const (not actually but close). So now you have a series of V = 0 + gt with t [0,1/Fg]. Integrate the flow for each series of glug flows (don't laugh). This represents the actual flow rate.

You could approximate the actual case with a free flowing pipe with a much smaller outlet using the above analysis if you like, but it's not necessary.

BTW Fg is related to how rapidly a bubble of 1/2 the cross section of pipe can propagate upward by 1 diameter of the pipe (1.125"). This is the time that it would take a column of water 1.125" tall to free-flow through a pipe of 1/2*1.25".

Good luck. I hope this is correct.

 

[equi_librium]

First calculate how rapidly the water would flow if the pipe were entirely open at the top and the water is free flowing (i.e. ignoring the effect of the viscosity of water). The flow rate is proportional to the cross section of the outlet ( pi*r^2) = (3.146)(1/2*1.125)^2 falling (free flowing in 1 dimension) and the velocity at any time 't', V = 0 + gt (initial flow velocity is 0) g: acceleration due to gravity. In this case V increases until all the water is evacuated from the pipe.

In the actual case in which the top of the pipe is closed, the water "glugs" out of the pipe in a way that requires a volume of air entering the opening to equal the volume of water leaving. This condition will 'limit' V to some constant value related to the frequency of the glugs (which are abrupt interruptions of flow) Fg = const (not actually but close). So now you have a series of V = 0 + gt with t [0,1/Fg]. Integrate the flow for each series of glug flows (don't laugh). This represents the actual flow rate.

You could approximate the actual case with a free flowing pipe with a much smaller outlet using the above analysis if you like, but it's not necessary.

BTW Fg is related to how rapidly a bubble of 1/2 the cross section of pipe can propagate upward by 1 diameter of the pipe (1.125"). This is the time that it would take a column of water 1.125" tall to free-flow through a pipe of 1/2*1.25".

Good luck. I hope this is correct.

correction: substitute the term throttle frequency f for Fg to eliminate confusion.