Fluid Dynamics

The aim of the experiment was to fill up a big bucket that has a small hole drilled at the bottom of it with water to a height h of three inches and measaure the time t it takes for the bucket to empty out 16 ounces or approximately 473.18 ml of water.

	1 st	2nd run	3rd run	4th run	5th run	6th run
Time to empty (sec)	40.68	41.16	41.77	41.65	42.94	42.70

"6 t_iàt_ave= 41.81 sec

Calculating theoretical time:
 
The diameter d of the drain hole was measured to be 0.5 cm and so the radius r is .25 cm. Hence the area of the drain hole is 1.96*10^-5 cm^2.

Next,

from the consveration of mass one gets M1= M2= rA1v1t= rA2v2t= A1v1= A2v2 where A*v is equal to volume flow rate V/t .

Now looking at the diagram of the bucket and using bernoulli's equation one derives that v1--the velocity at which the water exits the small hole of the bucket--is equal to (2gh)^(1/2) where h is equal to Y_2 minus Y_1.  

Putting the equations together: V/t = A1(2gh)^(1/2) --> so t = V/((A1)(2gh)^(1/2))

Therefore t-theoretical = 21.05 sec.

Percent error is there for 49.65 percent which is regretably way,way too large.

This inevitalby suggest that the measured diameter is way off.  Therefore rearraning the equation for time and solving for r one gets 0.001772. The diameter is then 0.003544. The error in the diameter is 29.11 percent. 

Furthermore replacing the new area to find the theoretical-time it took to drain the 473.18 ml of water, one gets t-theoretical = 41.83 sec. This yields a percent error in time of 0.04 percent.