I haven't answered a physics question in almost ten years! I don’t remember exactly how I started doing it! But I know it can be calculated using the law of conservation of energy! First, consider the weight of 1,370,000 square meters * 5 meters of water. The energy generated by this weight rising 5 meters plus the energy generated by falling 5 meters equals 22%, or the electrical energy generated by a rising tide! Efficiency is simply divided by time 13
Suppose the flow rate of a hydroelectric reservoir is 9,000 cubic meters/s, the height is 40 m, and the production efficiency electricity is 75%. generated every day?
~~~~~~~~~~~~P=9.8gQH
P is the output power, the unit is kilowatt;< p> 9.8 is the acceleration of gravity;
g is the efficiency, which can simply be taken between 0.8 and 0.9 for generated power plantsales
Q is flow rate, unit: cubic meter/second;
H is the height of water, also known as Drop,
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 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 one hour
The work done by water per drywave W=mgH=ρgVH=1000*10*9000*40=
Converted into electrical energy in one day=75% W *24*60*60= The unit is joule then converted to kilowatt hours
Please do the calculation yourself