When Lake Mead is nearly full, the water at Hoover Dam effectively falls about 2

Question

When Lake Mead is nearly full, the water at Hoover Dam effectively falls about 200.0  m before encountering the hydroelectric turbines (a more accurate description would be to say that the water is "squirted" through the turbines by the water pressure at the bottom of the lake, but the "squirting" speed is the same as the speed after a free fall of 200.0  m ). After passing through the turbines, the water speed is almost zero. Assume the efficiency of the turbines for generating electric power is 93.1  percent .

For this question, use only energy considerations without resorting to a free-body diagram of the water in contact with the blades of the generator turbines. The generators convert mechanical energy to electrical energy with an efficiency of 93.1  percent , meaning only 6.9  percent of the converted mechanical energy is lost to thermal energy.

Part A

For each liter of water which passes through the turbines, what is the amount of electrical energy produced by the turbines when the lake is nearly full? One liter of water has a mass of one kg.

Part B

If the rate of water flow through the turbine intakes is 7.60×105  liters/s , what is the rate of electric energy output from the hydroelectric generators at Hoover Dam when the lake is nearly full? Give your answer in gigawatts (GW); one GW is one billion or 109 watts.

Give your answer in gigawatts (GW).

Explanation / Answer

(a)

formula for electrical energy

E = mgh = 1(9.8) (200) = 1960 J

the amount of electrical energy produced by the turbines when the lake is nearly full is

E = effciency ( E) = 0.931 ( 1960) = 1824.76 J

(b)

Power output= 1824.76 J/L ( 7.60×105  liters/s )= 1.38 GW

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