15. -12 points FRKestenCP1 6.P.088. My Notes In 2006, the United States produced

Question

15. -12 points FRKestenCP1 6.P.088. My Notes In 2006, the United States produced 282 x 109 kilowatt-hours kwh of electrical energy from 4138 hydroelectric plants 90% efficient at converting mechanical energy to electrical energy, and the average dam height is 50.0 m. wh 3 x 10%, on average each plants (a) At 282 x 109 kWh of electrical energy produced in one year, what is the average power output per hydroelectric plant? MW/dam (b) What total mass of water flowed over the dams during 2006? kg (c) What was the average mass of water per dam and the average volume of water per dam that provided the mechanical energy to generate the electricity? (The density of water is 1000 kg/m3.) average mass average volume kg/dam m3/dam (d) A gallon of gasoline contains 45.0 x 106 J of energy. How many gallons of gasoline did the 4138 dams save? gal eBook

Explanation / Answer

(a) Energy produced in one year = 282 x 10^9 kWh = 282 x 10^6 MWh

No. of hours in one year = 365 x 24

So, total MW of all the hydroelectric plant = (282 x 10^6) / (365x24)

So, average power output per hydroelectric plant = (282 x 10^6) / (365x24x4138) = 7.78 MW

(b) Suppose M is the total mass of water flowed over the dam in 2006.

So, total energy produced = Mgh

Electrical energy generated = 0.90*Mgh

and this is equal to 282 x 10^9 kWh

So, 0.90*M*9.81*50 = 282x10^9 x 10^3 x 60x60 W.s

=> M = (282x10^9 x 10^3 x 60x60) / (0.90x9.81x50) = 2300 x 10^12 kg = 2.30 x 10^15 kg

(c) Average mass of water per Dam = (2.30 x 10^15) / 4138 = 5.56 x 10^11 kg

Average volume of water = (5.56 x 10^11) / 1000 = 5.56x10^8 m^3

(d) No. of gallon of gasoline = (282x10^9 x 10^3 x 60x60) / (45.0x10^6) = 22560 x 10^6 = 2.26 x 10^10 gallons.

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