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eeseely · Jun 14, 2007, 03:48 PM

Here's a problem from the differential equations text I used some 50+ years ago.
For some reason, I found it interesting.
It's both math- and physics-related.

A square tank is filled with water and has the following dimensions:

sides = 4 ft.
depth = 6 ft

Find the time for it to empty through an inch circular hole at the tank's bottom.
Also, assume the flow factor equals for ordinary small orifices with sharp edges is 0.6

Once you have solved this problem, try calculating the time for your bathtub to empty from a given depth.

I'll confirm your answer if its correct.

galactus · Jun 14, 2007, 04:56 PM

Let's see. I will give it a go and try to remember Toricelli's Law.

The sides are 4 feet, so the area of the base is $A(h)=16 ft^{2}$ .

Torricelli's Law(A(h) in square feet, h in feet, t in seconds):

$A(h)\frac{dh}{dt}=-k\sqrt{h}$

$16\frac{dh}{dt}=\frac{-3}{5}\sqrt{h}$

Separate variables:

$16\frac{dh}{\sqrt{h}}=\frac{-3}{5}dt$

Integrate:

$16\int\frac{dh}{\sqrt{h}}=\int\frac{-3}{5}dt$

$32\sqrt{h}=\frac{-3}{5}t+C$

Assuming the tank is full, h=6 when t=0:

So, $C=32\sqrt{6}$

We have:

$32\sqrt{h}=\frac{-3}{5}t+32\sqrt{6}$

Solve for h and the height at anytime t is given by:

$h=\frac{(3t-160\sqrt{6})^{2}}{25600}$

Set h=6 and solve for t we find it takes 261.28 seconds or about 4.35 minutes to empty.

EDIT: Oops, I forgot to take into account the size of the hole so I am probably in error.

eeseely · Jun 14, 2007, 05:42 PM

Yes, you are in error... you must plug the orfice's size into the equation.

Try again...

galactus · Jun 15, 2007, 04:19 AM

I thought so. Learn me something. How does one incorporate the orifice size into the formula? The area of the 1" diameter hole in square feet would be $\frac{\pi}{576} \;\ ft^{2}$ . This is a cool problem. More interesting than most we get.

Would we use:

$\frac{dV}{dt}=A_{e}\sqrt{2gy}$ ?

Where $A_{e}=\text{area of orifice}, \;\ \sqrt{2gy}=\text{velocity}$

eeseely · Jun 15, 2007, 05:46 AM

Try this...

The average velocity of the water passing through the orifice is:

v = sqrt (2gh)

Incorporating the flow factor yield:

v = 4.8 sqrt (h)

Let h be the height of the water in the tank at time t.

The water which issues from the orifice in time dt generates a cylinder of altitude
v dt, and volume = -16 dh (cubic feet).

Hence:

V = PI(1/24)squared * v dt

After some manipulation, you'll end up with:

dt = -611(dh/sqrt h)

Take it from there...

galactus · Jun 15, 2007, 08:44 AM

So, just integrate the last equation you gave to find $t=-1222\sqrt{h}+C$

That way at t=0 and h=6, $C=1222\sqrt{6}$

We have $t=-1222\sqrt{h}+1222\sqrt{6}$

Solve for $h=\frac{(t-1222\sqrt{6})^{2}}{1493284}$

Set h=6 and solve for t to find the time(in seconds) to empty the full tank.

That gives t=5986.55 seconds or 99.78 minutes.

That seems like a long time. Oh well, if I flubbed up please straighten me out.

eeseely · Jun 17, 2007, 05:26 AM

You're getting closer...

The expression for integration is:

the integral of dt (from 6 to 0) = -611* the integral of dh/sqrt h

You're off by a factor of 2 (when you performed the integration, but your good sense told you that 99.78 minutes was too much.

Answer: 2993 seconds = 49.9 minutes