a) How much electric energy can be generated by a hydroelectric generator from t

Question

a) How much electric energy can be generated by a hydroelectric generator from the water in a lake that is 2000 meters wide, 100 meters deep, and 8000 meters long if all the water falls through a vertical distance of 500 meters? Assume the generator is 90% efficient at converting potential energy into electrical energy. Express your answer in kWh.  

b) What is the value of the electrical energy produced from draining this volume of water if electricity rates are $0.10 per kWh

c) If the lake is drained over a period of 1 year, what is the average electrical power output in MW

Explanation / Answer

Hi Dear,

(a) volume of water in lake= length*width*depth= 8000x2000x100= 16x108 m3

density of water= 1000 kg/m3

so, mass= density x volume= 1000x 16x108= 16x1011 kg

height=500 m

g= 9.81 m/s2

potential energy= mxgxh= 16x1011x 9.81x 500= 784.8x 1013 Joules

But, 1 kWh= 3.6x 106 Joules

So, potential energy= 784.8x 1013 / (3.6x 106) = 2.18x 109 kWh

Efficiency is 90%.

So, Electrical energy= 0.9x (2.18x 109) = 1.962 x 109 kWh

(b) Value= Rate x Electrical Energy= 0.1x 1.962 x 109 = $ 196,200,000

(c)

Electrical energy= 1.962 x 109 kWh= 1962000 MWh

time= 1 year= 365 days= 365*24 = 8760h

Average electrical power= electrical energy/time= 1962000 MWh/ 8760h= 223.97 MW

See how easy it was :)

Good day!

And, don't forget to rate my answer. Cheers!

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