~~~~~~~~~~~P=9.8gQH
P is the output power, the unit is kilowatts;
9.8 , it is the acceleration of gravity;
g is the efficiency, which can be simply regarded as 0.8 ~ 0.9 for general power plants;
Q is the flow rate, unit: cubic meter/second;
H is the water height, also called head height,
P=9.8*0.85*0 .4*100=27.2KW.
It is recommended to purchase a hydroelectric generator with a production power of 30KW. Electricity production 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 yield 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 per hour
1 degree of electricity is equivalent to 1 kilowatt hour, i.e.
W=Pt=1000x3600=3600000 Joules
The production of electricity is the conversion of the potential energy of water into
EP=mgh
3600000=mx10x10
m=36000 kilograms. But there is no consumption when mechanical energy is converted into electrical energy. The efficiency of converting water energy into electrical energy is about 70%
So, 36,000/0.7 = 51,428 kilograms.