A 10 m diameter tidal turbine is placed in an inlet and operates with a coefficient of performance of $0.3$. During a tidal period of $12.4$ hours, the tidal current in the inlet varies from $3.5 \mathrm {~m} / \mathrm { s }$ in one direction to $3.5 \mathrm {~m} / \mathrm { s }$ in the opposite direction. If it is assumed that the change in tidal current velocity is linear in time, what is the total energy output of the turbine during one tidal period.

Question

Quizplus · Accepted Answer

The Answer is Since the velocity and hence the power varies over the tidal cycle, it is necessary to write the energy as an integral over the tidal period. As the magnitude of the velocity is the same during the raising tide and the falling tide we write the energy as twice the integral over half a cycle as

E = 2 \int _ { 0 } ^ { T / 2 } P ( t ) d t

where

T

is the total tidal period. The power as a function of time is

P ( t ) = \frac { 1 } { 2 } A C ho v ( t ) ^ { 3 }

where

C

is the coefficient of performance. Since the velocity varies linearly, we write

v ( t ) = 2 v _ { 0 } t / T

Converting time to seconds (so the answer will be in joules) gives the integral

E = \frac { 8 A C ho v _ { 0 } ^ { 3 } } { ( T / 2 ) ^ { 3 } } \int _ { 0 } ^ { T / 2 } t ^ { 3 } d t = \frac { A C ho v _ { 0 } ^ { 3 } } { 8 } T

Using the values from the problem

\begin{array} { l } A = ( 3.14 ) imes ( 5 \mathrm {~m} ) ^ { 2 } = 78.5 \mathrm {~m} ^ { 2 } \C = 0.3\end{array}

\begin{array} { l } ho = 1025 \mathrm {~kg} / \mathrm { m } ^ { 3 } \v _ { 0 } = 3.5 \mathrm {~m} / \mathrm { s } \T / 2 = 6.2 \mathrm {~h} = 22320 \mathrm {~s}\end{array}

gives the energy as

E = ( 1 / 4 ) imes \left( 78.5 \mathrm {~m} ^ { 2 } ight) imes ( 0.3 ) imes \left( 1025 \mathrm {~kg} / \mathrm { m } ^ { 3 } ight) imes ( 3.5 \mathrm {~m} / \mathrm { s } ) ^ { 3 } imes ( 22320 \mathrm {~s} ) = 5.78 imes 10 ^ { 9 } \mathrm {~J}

This may be expressed in

\mathrm { kWh }

as

\left( 5.78 imes 10 ^ { 9 } \mathrm {~J} ight) / \left( 3.6 imes 10 ^ { 6 } \mathrm {~J} / \mathrm { kWh } ight) = 1.60 imes 10 ^ { 3 } \mathrm { kWh }

A 10 M Diameter Tidal Turbine Is Placed in an Inlet