physics
posted by AE .
Water flows steadily from an open tank into a pipe. The elevation of the top of the tank is 10.2 m, and the elevation at the pipe is 3.90 m. The initial crosssectional area of the pipe is 6.40×10^−2 m; and at where the water is discharged from the pipe, it is 2.30×10^−2 m. The crosssectional area of the tank is very large compared with the crosssectional area of the pipe. Assuming that Bernoulli's equation applies, compute the volume of water that flows across the exit of the pipe in the time interval 2.40 seconds.

The average head is 10.23.9=6.3 m
g=9.8 m/s²
When the area of the orifice is small compared to the tank area, and if the tank is vented, pressure at the water surface equals that of the outlet, then the output velocity is approximately √(2gh).
Volume of discharge
=velocity * area * time
=√(2gh)*2.30×10^−2*2.4
=0.6 m³ approx.
For the derivation of the simplified formula, see:
http://www.engineeringtoolbox.com/bernouilliequationd_183.html