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# Crazy mother of a physics problem?

The Grand Coulee Dam is 1270 m long and 170 m high. The electrical power output from generators at its base is approximately 1900 MW.

How many cubic meters of water must flow from the top of the dam per second to produce this amount of power if 87% of the work done on the water by gravity is converted to electrical energy? (Each cubic meter of water has a mass of 1000 kg.) g=9.80, express answer in m^3/s

### 2 Answers

- 1 decade agoFavorite Answer
At the top of the Dam, the water has 100% of its energy <or if you prefer energy per second - that is power >

87% of power <=> 1900 MW

100% of power <=> 2184 MW

At the top of the Dam, the potential energy the water possesses is what can be ideally converted into 100% power.

Density of water = 1000 kgm^-3

Let the mass of water be y kg

Potential energy of water = y * 9.80 ms^-2 * 170 m

= 1666 y Joules

Each second it can deliver 1666y Watt of energy

That is 1666y W <=> 2, 184, 000, 000 W

y = 1310924 Kg of water each second

Using density of water, above, each second, 1311 m^3 of water must flow from the top of the dam

Your answer is ultimately 1311m^3/s

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- Anonymous1 decade ago
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