Tidal Barrage System Generates 15 MWe from 85% Efficient

Question

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The Answer is

ext { The gravitational potential energy is } E = m g h ext { so per cubic meter of water the }

energy is If the generator produces

15 \mathrm { MWe }

at

85 \%

efficiency the total energy of the water is

\left( 15 \mathrm { MW } _ { \mathrm { e } } ight) / ( 0.85 ) = 17.65 \mathrm { MW }

The power is

P = v A E / \mathrm { m } ^ { 3 }

So the area is found to be

A = \frac { P } { v \left( E / m ^ { 3 } ight) }

E / \mathrm { m } ^ { 3 } = \left( 1025 \mathrm {~kg} / \mathrm { m } ^ { 3 } ight) imes \left( 9.8 \mathrm {~m} / \mathrm { s } ^ { 2 } ight) imes ( 5.7 \mathrm {~m} ) = 5.73 imes 10 ^ { 4 } \mathrm {~J} / \mathrm { m } ^ { 3 }

Substituting appropriate values gives

A = \frac { 1.765 imes 10 ^ { 7 } \mathrm {~W} } { ( 6 \mathrm {~m} / \mathrm { s } ) imes \left( 5.73 imes 10 ^ { 4 } \mathrm {~J} / \mathrm { m } ^ { 3 } ight) } = 51.3 \mathrm {~m} ^ { 2 }

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