The Grand Coulee Dam is 1270 m wide and 170 m high. The electrical power output

Question

The Grand Coulee Dam is 1270 m wide and 170 m high. The electrical power output from the generators at the base of the dam is approximately 2000 MW.

How many cubic meters per second of water must flow from the top of the dam to produce this amount of power if all the initial gravitational potential energy is converted to electrical energy? (Each cubic meter of water has a mass of 1000 kg.)
V=_____m^3/s

What would be the answer to part (a) if the conversion process to electrical energy were only 92.0%?
V=_____m^3/s

Explanation / Answer

i assume 2000 be 2000 Mwatt 2.00*10^9 watt = Q*170 m Q = 2.00*10^9 Nm/sec/170m = 1,18*10^7 N/sec = 1,20*10^6 kg/sec = 1200 m^3/sec if ? =0,92, then Q1 = Q/? = 1200 m^3/sec/0.92 = 1304.3 m^3/sec

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