please help guys?

Question

The Grand Coulee Dam is 1270 m long and 170 m high. The electrical power output from generators at its base is approximately 1920 MW.
How many cubic meters of water must flow from the top of the dam per second to produce this amount of power if 93.0 \% of the work done on the water by gravity is converted to electrical energy? (Each cubic meter of water has a mass of 1000 {\rm kg}.)

Answer 1

The energy developed by a mass m moving vertically through a distance h in a gravitational acceleration g is mgh. So, each metric ton generates
1000*9.8*170*.93 = 1,549,380 J of energy. So the amount of water to generate 1,920,000,000 J is
1,920,000,000/1,549,380 = 1,239.2 cu meters. Since a watt is 1 J/sec, there must be 1,239.2 cubic meters of water/sec flowing.

Doug

Answer 2

The energy provided by water falling from the top of the dam is mass*g*h. At 90% efficiency this becomes 0.93*mass*g*h. If m is in kg and h in meters, this comes out in joules. This energy per second is one watt.

1920 * 10^6 joules/sec = g*0.93*mass/sec*170

mass/sec = 1920*10^6/(g*0.93*170) kg/sec

The mass of the water is 1000 kg/ m^3 * V(m^3), so the volume is

V = (mass/1000) = (1920*10^3/(g*0.93*170) m^3/sec

for g = 9.8 m/sec^2 this comes out 1.24*10^3 m^3/sec

Answer 3

117cf/s