Tank drain-time formula

Hi guys,
Need to find time required to drain a tank by gravity. I am having problems solving this one, it's got me up at nights ( semi obsessive compulsive about this one ). I know it's just a matter of differentiating or integrating or something but I can't remember which or how.

A simplified model of a draining tank is as follows:

Water tank has:
1) A hole in the bottom with an area represented by "A_drain"
2) A horizontal surface area represented by "A_surface" ( ie if tank was cylindrical this would be pi.r^2, but tank model could be round or square so please leave as just an area )
3) A water level height represented by "h"
4) Obviously a volume at any point in time represented by "A_surface * h"
5) Gravity represented by "g"

Total drain time = volume / rate of flow from hole
Volume = A_surface * h
Rate of flow from hole = A_drain * sqrt(2gh)

BUT the water-height "h" is continuously dropping so that you cannot apply sqrt(2gh) as a constant across the whole drain period. Everything is changing simultaneously and they are all dependant on each other !

Please help me with this one and also try to explain the process by which I can solve eq's like this in the future.

Thanks in advance
Nick

Welcome to the Forums :wave:

A rate of change is represented by a differential. In your case, the rate of change of volume is the flow rate out of the tank, which is dV/dt. So you can say that dV/dt = A_drain * sqrt(2gh).
Integrate it and use the boundary condition that you know, namely that at zero time (T=0) you have full volume (V=V) and that at T=T, you have V=0, remembering to include an appropriate negative sign to indicate that water is flowing out of the tank.
Then you'll just need to rearrange for T, and hey presto.

zaza

I was looking at the leaky integrator solution: da/dt=-Aa+Bs, where A=alpha, and B = Beta are both positive constants for flow in to the system and flow out respectively (you plug in the in flow to create an outflowing system)

Perhaps a little too simple, huh?

edit and I've just noticed that g needs to be represented, too :duh:

I would've used dv/dt = dh/dt * dv/dh

but i'm not sure how you'd get dh/dt!

Here is the differentiated equation from a water operators handbook:

For Circular Tank:

t = ((3.14 x dia^2) / (C x A)) x (h / (8 x g))^0.5

Where
t = time in seconds
dia = diameter of tank in ft
C = Orface coef. (0.7 to 0.91) I use 0.81 for a pipe penitrating the tank wall.
A = aera of orface in sf
h = height of water above outlet in ft
g = gravitational constant (32.2)

If you have a retangular tank it would seem that you could replace the first part (3.14 x dia^2) with (4 x L x W)

Try this out....

This homework was due about 9 months ago...

Quote:

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This homework was due about 9 months ago...

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Ha ha ha, it's not actually homework. It was something that was personally bugging me in terms of how to figure it out.

Thanks Water Guy, I appreciate you posting this up for me :)
And thanks zaza as well, sorry I didn't get around to saying thanks when you first posted :)