118 POWER PLANT ENGINEERINGPutting all the values, we get
Solidity = [(30 × 0.25)/( π × 10)] × 100 in %
= 24%.
Example 14. A 5 metre diameter rotor is rotating at 15 revolutions per minute (rpm) and the
wind speed is 3 m/s, calculate tip speed ratio of the rotor.
Solution. Tip speed ratio = [(π × dia of rotor × revolution per sec)/ Wind speed]
Given that;
Diameter of rotor = 5 m
Revolution = 15 RPM = 15/60 = 0.25 revolution per second
Wind speed = 3 m/sec
Tip speed ratio = [(π × 5 × 0.25)/ 3] = 1.3
Tip speed ratio = 1.3.
Example 15. Ocean wave on the coast of Tamil Nadu, India were with following data :
Amplitude 1 m, Period 6 s. Calculate the following:
Wavelength, velocity, energy density, density, power extracted from a wave of 10.0 m with a
power density, energy in 100 m wide wave. Assume density of ocean water as 1000 kg/m^3.
Solution. For ocean wave;
Wavelength (λ) = 1.56 T^2 m
In this example λ = 1.56 T^2 = 1.56 × 62 = 56.16 m
Wave velocity, c = Wavelength(λ)/Period (T) =56.6/6 = 9.36 m/s
Frequency, f = 1/T = 1/6 s–1 = 0.1667
Surface energy density
E/A = 1/2. D. a^2 .g J/m^2 = 1.2 × 1000 × 1 × 9.81 J/m^2
= 4.905 × 1000 J/m^2.
Energy in 100 m wide wave
Area A = Wavelength (λ) × Width (W) = 56.16 × 100 m.
Energy = E/A × A
= (4.905 × 1000) × (56.16 × 100)
= 275.46 × 100000 J = 27.546 J.
Power = E × f W
= 27.546 × 10^3 × (1/6) = 4.56 × 10 W.
Power density = P/A
= (4.56 × 10^3 )/(56.16 × 100)
= 256.08 × 10^5 W/m
= 25.60 kW/m^2.