If water over a dam is 1/2 inch deep,and 6 foot wide,and flow at 20 feet in 8 seconds how much water is it?

Question

I also have a drop of 10 feet. Will this power an 12 volt alt for a small power supply?

Answer 1

The previous answer was correct except for a swapping of the 8 second for the 20 feet. The amount of water that flows over the dam in 8 seconds is:

6 feet x 0.5 inch x (1 foot/ 12 inch) x 20 feet = 5 cubic feet.

The flow rate is:

rate = 5 cubic feet / 8 seconds x 60 seconds/ minute = 37.5 cubic feet per minute = 37.5 cfm

Many small cooling fans move air at this rate and require 12VDC to power them. If they were spun by air (or water) at this rate, they would generate a similar amount of electricity as is required to move them. 12VDC at around 200 mA is probably reasonable.

The weight of the water falling 10 feet will add velocity and kinetic energy which should allow it to provide more power. For DC motors (generators), the speed they turn is proportional to the voltage. A motor that provides requires 12VDC at 37.5 cfm should provide approximately 12VDC when spun at 37.5 cfm, and probably higher if spun faster by water that has accelerated by falling 10 feet.

Answer 2

The area is 6 x .5/12 = .25 sq ft. In 8 seconds, a volume of 8 x .25 = 2 cu ft passes by. The flow is then 15 cfm. The capability to drive the alternator does not depend on the voltage, but the power it needs to deliver.