Tank draining

[Durk]

Hi all,

I have a question which has plagued me with quiet awhile.

If a tank (circular or other) has to be drained, what is the formula to calculate the total time taken to empty such a tank?

Also, if the outlet of the tank was connected to the inlet of another tank at the same level (same volume), can the time taken to bring the two tanks to the same level be calculated? (I am assuming as the other tank fills the difference in head will decrease, slowing down the flowrate)

I understand that there are a large amount of factors such as type of outlet, working fluid etc etc involved but a user friendly formula would be much appreciated, even a discussion on such a topic would be informative.

Thanks for your time

 

[BigInch]

Depends on the flowrate obtained at the outlet, a function of the pressure differential across the outlet, which is a function of the depth of liquid in the tank. Two tanks connected together must also include the inlet, outlet pressure of the connecting pipe, hence head loss of the pipe, outlet pressure being a function of inlet pressure to the second tank, which is also a function of depth in the second tank. Yes a simple system can easily be approximated with a spreadsheet calculation using time steps and keeping track of depths, inlet and outlet pressures, pipe flowrate, etc. Most any fluids mechanics text will have such examples. Look for the 3 reservoir problems and add tank pressures at the reservoirs.

I have not failed. I've just found 10,000 ways that won't work."-Edison “If Edison had a needle to find in a haystack, he would proceed at once to examine straw after straw until he found the object of his search. I was a sorry witness of such doings, knowing that a little theory and calculation would have saved 90% of his work.- Tesla

 

[bimr]

Try this formula:

http://www.efunda.com/formulae/fluids/draining_tank.cfm

 

[JStephen]

Note that the flow rate decreases as the level drops, so you get a little calculus problem out of it, but shouldn't be hard to solve. The actual flow rate comes from Bernouli's equation. For an open drain, figure zero velocity, zero pressure at the top of the tank, zero pressure at the drain outlet, and solve for velocity as a function of head difference. Include local loss coefficients where appropriate. Probably figure the time to get the level down to the top of the pipe, rather than the center, as the area starts changing then, too.

In real life, most people wouldn't drain a full water tank; they'd draw the level down as far as possible before isolating it.

 

[bimr]

The drain discharg rate varies since the water height continues to fall. The time to drain is related to the volume of water, the initial height of the water and the cross-sectional area.

A good guess for an average flow rate would be 1/2 of the discharge rate at the start.

http://www.grow.arizona.edu/Grow--GrowResources.php?ResourceId=188

 

[Durk]

Thanks for your thoughts.

I have looked at the example and formula supplied by efunda (thanks Bimr), and worked the solution for a typical size tank, 10m diameter, 10m depth. I calculated for the top level and worked a table for the falling values. I think i got a bit carried away with the 1 sec intervals but it is interesting to see all the same.
While this seems about as good an answer as one could get, i can't help but think that a integration/differentiation (as mentioned) method might be alot less work and even more accurate(if my method is correct!).
My skills in the calculus department are quiet weak, so if somebody has a strong handle on calculus it would be good to see if it can be done.
Also, following the values worked out from the tank drain, i will look at the idea of calculating the simultaneous filling of an equal tank as previously mentioned. I imagine that a few headaches will accompany this (calculus again might be the prefered option!). I have attached the workings........ appreciate any comments. Thanks

http://files.engineering.com/getfil...7c9903a5669&file=Drain_a_tank_calculation.xls

 

[BigInch]

Accuracy only depends on time step size. Using 1 second steps for tank levels will give too much accuracy in relation to the equations which predict flow out of this tank and into the other tank (Cd variations), and the pipe head loss equation, all of which could easily deviate by as much as 10% from real values.

All you can integrate is the flowrate, once it is determined for each step and all the higher math you can possibly think of won't improve the accuracy of the Cd or Hl values you need to use to get that flowrate.

I have not failed. I've just found 10,000 ways that won't work."-Edison “If Edison had a needle to find in a haystack, he would proceed at once to examine straw after straw until he found the object of his search. I was a sorry witness of such doings, knowing that a little theory and calculation would have saved 90% of his work.- Tesla

 

[bimr]

You would use calculus if you desired to calculate the volume of water discharged. However, that is something that you alreday know because the it is the same as the tank volume.

The time to drain is:

T = V/(4 A (sq rt h) )

V = Volume
A= Drain Area
h= height of liquid above drain

You also should have seen that the drain time varied with 1 over the square of the drain diameter.

 

[Durk]

I worked through your calc. Bimr and found the time to be approx. 8070 sec. If this is correct, how can there be such a difference between our answers(mine being close to 13,000 sec.). While i love a simple formula myself, i can't help but think that this leaves out some critical data such as gravity. Big Inch speaks of such things as Cd variation, i think more in respect to the 2 tank scenario but at 10% this is still quiet away off the aforementioned figures.

 

[bimr]

The equations from the two sources listed above are basically the same with the exception that the arizona.edu site does not show a discharge coefficient.

The discharge coefficient is C, whose value is typically between 0.90 and 0.98.

So, there should be little difference between the numbers. You must be making an error.

http://www.grow.arizona.edu/Grow--GrowResources.php?ResourceId=188

http://www.efunda.com/formulae/fluids/draining_tank.cfm

 

[BigInch]

Exactly what I'm talking about. That Cd value is for nicely rounded flush edges.

Cd of as low as 0.61 might be used for square flush edges through a thin walled tank, 0.52 if the exit pipe sticks into the tank past the tank wall.

I think you have square edges.

What's the value now?

8000

I have not failed. I've just found 10,000 ways that won't work."-Edison “If Edison had a needle to find in a haystack, he would proceed at once to examine straw after straw until he found the object of his search. I was a sorry witness of such doings, knowing that a little theory and calculation would have saved 90% of his work.- Tesla

 

[quark]

Durk,

Check 20 Aug post of bimr. The average flowrate is half the initial instantaneous flowrate. Actually you maximum flowrate initially and then zero in the last second. You can get the average flowrate by (initial flowrate+final flowrate)/2 or initial flowrate/2.

Now, Q = 0.98*0.007693*(2*9.81*10)[sup]0.5[/sup], which is 0.1056 cu.mtr/sec. Half of this is 0.0528cu.mtr/sec.

So, time taken will be 785/0.0528 = 14868 sec (approximately)

In your spreadsheet, you have emptied out only 779.26 cu.mtr after 13322 seconds. You have to still drain 5.74 cu.mtr. Even if we consider that flowrate at 13322nd second remains constant further, you still require about 5.74/0.00922 = 623 seconds.

 

[quark]

The constant given by bimr in the simplified form should, actually, be 4.43 which is nothing but (2g)[sup]0.5[sup]. So, critical data like gravity was not left out.

Time calculation based on average flowrate between initial and final conditions (i.e half of initial flowrate) can be established by integation as well.

If A is CS area of tank and a is CS area of nozzle, then the flowrate through nozzle will be Cdxax(2gZ)[sup]0.5[/sup].

If we consder a drop of dZ in liquid column height correspondingly, in a time dt then the volume flowrate out of the tank is AxdZ/dt.

Therefore, AxdZ/dt = Cdxax(2gZ)[sup]0.5[/sup]

By rearranging the terms,

(A/(Cdxax(2g)[sup]0.5[/sup]))dZ/Z = dt

Integrating both sides, we get

(A/(Cdxax(2g)[sup]0.5[/sup]))x2xZ[sup]1/2[/sup] = t

Applying the boundary conditions of 0 and 10 meters for Z,

t = 78.5*2*(10-0)[sup]1/2[/sup]/(Cdx0.00785x4.43)

or t = 14276.65/Cd

If our Cd is 0.98, then t = 14568 sec.

If Cd is 0.9, then t = 15863 sec.

The descripancy in my earlier post was due to the fact that I considered Cd twice. Otherwise, it is 0.0528*0.00785/0.007693 = 0.05388 cu.mtr/s

So, time will be 785/0.05388 = 14569 sec.

 

[BigInch]

No no no.

It is not correct to apply the calculated velocity to the entire cross-sectional area of the nozzle. It must be applied to the area of the vena contracta.

Rounded Sharp-edged Short tube Borda
Cc 1.0 0.62 1.0 0.52
Cv 0.98 0.98 0.8 0.98

Cd = Cc x Cv = 0.62 * 0.98 = 0.61

Q = A x Cd x (2gH)^0.5

exit velocity = Cv * (2gH)^0.5
Cv = 0.98
exit velocity = 0.98 *(2 * 9.81 * D)^0.5 = 13.727 m/s
Cc = 0.62
Area of Vena Contracta = 3.14*((0.1m)^2)/4 * Cc = 0.00486 m2
Flowrate = 0.00486 m2 * 13.727 m/s = 0.06681 m3/s
Flowrate = Cv * Cc * (2gH)^0.5
Flowrate = Cd * (2gH)^0.5
where Cd = Cv * Cc

Flowrate = 0.66809 m3/s


"I'm all in favor of keeping dangerous weapons out of the hands of fools. Let's start with typewriters."
- Frank Lloyd Wright (1868-1959)

 

[BigInch]

0.066809

"I'm all in favor of keeping dangerous weapons out of the hands of fools. Let's start with typewriters."
- Frank Lloyd Wright (1868-1959)

 

[quark]

You are perfectly right. I was just showing that the simple equation yeilds the same result as that of infinitesimal time interval calculation. That is why I wrote the final equation in terms of Cd as variable.

 

[BigInch]

OK then with Cd = 0.61, sharp edge exit, apx 6.25 hours

Ya 1 sec is a rediculous waste of computer resources.

Acceptable
Error calculated
in Q time time to reach
% step 0.1 meter depth
seconds Hours

1 60 5.85
1 120 5.83
1 240 5.8
1 300 5.5
0.05 10 5.87


"I'm all in favor of keeping dangerous weapons out of the hands of fools. Let's start with typewriters."
- Frank Lloyd Wright (1868-1959)

 

[bimr]

The orifice coefficient for a short tube, square edge with no separation is 0.82.

The orifice coefficient for a short tube, square edge with fluid separation from the walls is 0.61.

One would think that 0.82 would be more correct.

Not really sure that the original post was concerned about the minutiae of orifice coefficients. I thought he was interested in the theoretical concept of calculating the time to drain a tank.

 

[BigInch]

I'd use 0.82 if it was piped to another tank.

"I'm all in favor of keeping dangerous weapons out of the hands of fools. Let's start with typewriters."
- Frank Lloyd Wright (1868-1959)