- #1
drewman13
- 9
- 0
Here are two equations showing a nearly equivelant energy output for a given volume and velocity of water. Using the formula:
EKin = M/2 x Vsquared
250/2 x 15.34m/s x 15.34m/s = 29,414 KW (requires 5 times more volume)
50/2 x 35m/s x 35m/s = 30,625 KW (requires only 2.3x more velocity)
Since the velocity is squared, isn't it better to look to use velocity over volume? IF velocity can be acheived through another means other than water pressure via water depth, wouldn't that be the most efficient way to go?
EKin = M/2 x Vsquared
250/2 x 15.34m/s x 15.34m/s = 29,414 KW (requires 5 times more volume)
50/2 x 35m/s x 35m/s = 30,625 KW (requires only 2.3x more velocity)
Since the velocity is squared, isn't it better to look to use velocity over volume? IF velocity can be acheived through another means other than water pressure via water depth, wouldn't that be the most efficient way to go?