A small-scale wind turbine with the rotor diameter 1.5m uses a 500-Watt DC motor as a generator. Find the capacity factor for the machine if it delivers 100 kWh in a 30-day month. Under the standard condition of 15C temperature and 1 atm pressure, how fast would the wind have to blow for the turbine to generate half of its rate power if the machine is 30% efficient at that point

Explanation:

Diameter D is given as 1.5 m

Power rated Pr is given as 500 W

(a) Capacity Factor for the machine for 100 kWh in 30 days will be

C = P/Pr

P is the supplied power

P = kWh/days*24

= (100*1000/30*24) W

= 138.89 W

C = 138.89/500

= 0.2778

i.e C = 27.78%

(b) How fast the wind have to blow under standard condition of 15 °C and I atm pressure, for the turbine to generate half of its rate power if the machine is 30% efficient at that point.

T = 15+273 k = 288 k

Pressure p = 1 atm = 101.325 kPa

efficiency η = 30%

wind velocity V is unknown

Pr/2 = 500/2 = 250 W

Density ρ = p/RT

ρ = 101.325/287*288

= 1.226 kg/m³

Pr/2 = ηρAV²

V = √Pr/ηρA

= √500/[1.226*(π/4)*1.5²*0.3]

= 27.73 m/s

answer:

explanation:

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