A Darrieus Rotor Has an Area Of$$1500 \mathrm {~m} ^ { 2 } \text { and operates with the optimal tip }$$

Question

Examlex · Accepted Answer

The Answer is From equation

( 10.5 )

the power is

P = \left( 0.602 \mathrm {~kg} / \mathrm { m } ^ { 3 } ight) \eta A v ^ { 3 }

Using

\mathrm { A } = 1500 \mathrm {~m} ^ { 2 }

and

\mathrm { v } = 20 \mathrm {~m} / \mathrm { s }

, and estimating the efficiency of the Darrieus rotor from Figure

10.12

at about

40 \%

the above gives the power output as

P = \left( 0.602 \mathrm {~kg} / \mathrm { m } ^ { 3 } ight) imes ( 0.40 ) imes \left( 1500 \mathrm {~m} ^ { 2 } ight) imes ( 20 \mathrm {~m} / \mathrm { s } ) ^ { 3 } = 2.89 \mathrm { MW }

A Darrieus Rotor Has an Area Of $1500 \mathrm {~m} ^ { 2 } \text { and operates with the optimal tip }$