Rectangular tank drainage time 4

[vonlueke]

Could someone please advise me on how to calculate tank drainage time for a tank-full of still water in a large, relatively shallow, rectangular, open tank having a circular drain hole in the bottom (or direct me to the correct forum or thread)? (The eng-tips search function doesn't work, in case this was already covered.)

The water falls vertically downward out of the drain hole into the atmosphere, not into an outlet pipe or fitting. The relatively-large drain hole is just a circular hole cut in the 6-mm-thick, flat, metal plate tank bottom, and the hole has essentially sharp edges (or 0.4 mm edge radii, to be exact).

I need to determine a drain hole diameter D so that, starting from a full tank (with no more water being added), the tank will drain in a certain drainage time t.

I tried to derive a simplistic differential equation and got a solution of drainage time t = [2(A1)/(A2)][h/(2g)]^0.5, where A1 = rectangular tank length times width, h = tank depth (i.e., water initial depth), A2 = drain hole area = 0.25*pi*D^2, and g = 9.80665 m/s^2.

However, the above solution neglects any transition or head losses or whatever, necking of the fluid exit stream, or whatever is significant to approximate this problem. I would say, if feasible, assume no Coriolis effect since the tank is relatively shallow and very wide and long. Could anyone advise me on how to solve this problem or provide a formula, because I didn't find a problem like this yet in my fluids nor hydraulics text book? Any help would be greatly appreciated.

 

[ve7brz]

I don't have a specific answer for this, but I would intergate the formula you are using in small increments from a full to an empty tank.

 

[desertfox]

Hi vonlueke

I came across a formula in one of my text boks for a tank emptying :-

T= (2xA/(Cd x a x (2xg)^0.5))x H^0.5

where T = time for tank to empty

A = area of tank

Cd = coeff of discharge

a = area of orifice

g = gravity const

H = depth of fluid in tank

hope this helps

 

[vonlueke]

ve7brz: Thanks for your suggestion. I did that to derive the formula in my above post. But it doesn't include (and I didn't know how to include) the head losses or whatever is applicable to this problem.

desertfox: Interesting. Your formula is the same as mine (listed in my above post) except yours has a Cd in the denominator. Thanks for your post!

Does anyone have a suggestion for the value of Cd, or how to determine it? Or does anyone have any suggestion that differs from the above formula?

 

[BobPE]

vonlueke:

Cd is the coefficient of discharge for the opening assuming its orifice discharge. If there is a pipe connected to the discharge, then you will need to factor in the headloss. Look at the opening in the tank and compare it to references in say, Cameron Hyfraulic Data or a similar book. The value will be less than 1.

BobPE

 

[vonlueke]

Thanks, BobPE. Mentioned in first post, no pipe. See first post for problem description. I currently don't have access to that reference. Would anyone be kind enough to tell me what that data would be? I.e., could anyone provide an estimate of Cd for this very standard problem? It's just, how long does it take a tank of water to drain out of a hole in the bottom of the tank? (See first post for details.) It's just a plain hole in the bottom, no pipe or anything. The opening is described in first post.

Or can Coriolis effect not be neglected for downward efflux (or would it be included in the Cd value for this problem)? Thanks for any help.

 

[prex]

In the (old) book Fluid Mechanics and Hydraulics by V.Giles (McGraw Hill), besides the confirmation of the formula given by desertfox, there's a table with the discharge coefficients for water at 15°C flowing out from a tank into the atmosphere through a thin wall. The orifice is termed vertical: I suppose this refers to the orientation of the axis, so this is exactly your case!
The coefficients vary with the orifice diameter (in the range 6.25 to 100 mm) and with the height of liquid (in the range 0.24 to 18 m).
For a diameter of 6.25 mm, the coefficient ranges from 0.647 at 0.24 m to 0.605 at 18 m, for 12.5 mm from .627 to .599, for 25 mm from 0.609 to 0.594 and for 100 mm from 0.601 to 0.593. I think that if you take 0.6 you'll have a good estimate (of course the formula is based onto a constant or average coefficient).
Concerning the vortex, this will affect only the last part of the phenomenon and should not change much the total time, though I think it will be noticeable. A vortex breaker would make the discharge time more predictable. prex

http://www.xcalcs.com

Online tools for structural design

 

[vonlueke]

Thanks, prex! Your post is immensely helpful. When you say, "The coefficients vary with...the height of liquid," do you mean initial liquid height (tank depth) or instantaneous liquid height? I assume you mean initial liquid height. Please advise if incorrect.

Taking prex's data for a 25 mm diameter hole, a 240 mm initial fluid depth, and no vortex breaker, does anyone have a feel for approximately by what percentage prex's Cd = 0.609 should be decreased to account for the vortex effect for that case? Is it something like a 1% decrease in the Cd value, or more like a 5% decrease, or what? Or how could it be approximated or assumed? Thanks.

 

[prex]

The discharge coefficients have been determined for given constant heights, but of course as the height varies during the discharge, the coefficient varies too: however as you can see the values are all close to 0.6, that's why I suggested that value. Remember also that the formula for drainage time assumes a constant Cd, though this is known to be not exactly constant.
In my opinion it is unfair to correct the discharge coefficient to account for vortices, as the vortex will affect only the final part of the transient, perhaps when the height of liquid becomes smaller than one or two times the orifice diameter. I have no idea of how much the vortex affects the discharge flow rate: a gaseous vena is formed in the center of the liquid vena, and this restricts the passage of liquid, but possibly there is less contraction on the outside of the liquid vena (this is the cause for the discharge coefficient being less than one), so the final effect might be small. I wouldn't be surprised to discover that, depending on the vortex rotational speed, the discharge rate might be even increased. prex

http://www.xcalcs.com

Online tools for structural design

 

[poetix99]

In reply to your question to "prex":

The discharge coeff., Cd is varying with instantaeous velocity (or Reynolds number, to be more to the point). The liquid head is not a fundamental parameter for an orifice discharge coefficient; it implies the velocity.

I agree 100% with PREX's practical suggestions for assuming a constant Cd, and to effectively neglect both the effect of continuously changing pressure head, and the vortex effects for the last portion of the drainage. The accuracy obtained from trying to derive an instanteous Cd and to time-march throught the entire drainage process, quasi-integrating as you go, is almost certainly not worth the trouble - the aforesaid vortex effects would introduce uncertainties greater than what you'd gained from the refinement of the Cd calculation.

 

[Graham62]

Hello, vonlueke.

I have the book "Technical Formulae" 8th edition by Gieck.
Section N7 covers "Flow of liquids from containers".

For an opening in the base (vertically downwards):

V = Cc x Cv x A x sqrt(2 x g x H)

where:
V is the volume of outlet flow
Cc is the contraction coefficient, 0.62 for sharp edge aperture and 0.97 for well rounded aperture
Cv is the velocity coefficient, 0.97 for water
A is the cross-sectional area of the outlet
g is the acceleration due to gravity
H is the depth in the container

The flow rate will of course reduce as the depth reduces.

I hope this is of some assistance.

Graham