I'm not a hydropower professional, so I can only tell you roughly: from what you said, the theoretical maximum value of hydropower is:
Pmax = mgh/t = ρVgh/t = (1000 x 3 x 9.8 x 10 ÷ 1) Watt ≈ 300 kilowatts.
Generally, the energy conversion efficiency of small hydropower generation units is around 0.7 (70%). Therefore, the installed power under the above conditions is approximately 200 kW.
~~~~~~~~~~~~P=9.8gQH
P is the output power in kilowatts;
9.8 , this is gravity, acceleration;
g is the efficiency, which can be simply taken as 0.8~0.9 for general power plants;
Q is flow rate, unit: cubic meter/second; p>
H is the water height, also called gap,
P=9.8*0.85*0.4*100=27.2KW.
It is recommended to purchase a hydroelectric generator with a production power of 30KW. The production of electricityelectricity is 27 kilowatt hours per hour~~~~~~~~~~
The above formula is correct, but the answer is too different
P=9.8GQH= 9.8*0.8*0.6*100=470KW. I don't know how to calculate 27.2kw per 100 meters of water height. Even a flow rate of 0.4 is not that small in terms of efficiency, small unit generators usually take 0.8, and turbines. also take 0.9. Full Take 0.8. If it is an impact type, the return is 2 points lower, so take 0.8. For mixed flow, take 0.82. For the positive solution, I choose impact 400KW or 500KW. 500 for this station. 500 degrees in an hour