HYDROPOWER output EQUATION

The formula used to calculate power output in the hydropower is wrong, needs corrections.

Ht X g X rate of water discharged = watts

If so

Can we get more output by supplying more water by increasing the size of water supplying pipe for a given hydropower installation? No, that means quantity of water discharged at the same velocity does not increase output.
What will happen if we move the same electromagnetic set to different places of different altitude? Will the output will be equal. No, because at the place where value of ‘g’ is greater, nearer to sea level, the velocity of water will be greater even the height of the penstock pipe is same than at a place where value of ‘g’ is less i.e. at higher altitude. Faster velocity of water will rotate the turbine faster, thus the output will be greater, more rpm more output.

Instead of using discharge rate, more correct formula could be velocity of water hitting the runner of the turbine.

What about rmp, wire turnings (more the turnings the more quantity or ampere of electricity), and number of poles (more the poles the more is the voltage) in the generator installed?

Making stronger electromagnet requires more turnings, more the number of turnings of wire the stronger the magnet is? Reverse process is electrical effect of magnet. Greater the turnings coil the same magnet will have more electrical effect than when the number of turnings is less.

So I think the formula should be expressed
Number of turnings of wire X number of poles in the generator X rpm or velocity of water

RPM or velocity of water depends upon height of the falling water column (head height) at a given place. The value of ‘g’ is constant at a given place.