Subject:- Renewable Energy Sources

Topic:- Wind Energy

Difficulty:- Medium

Determine the wind mill rotor diameter to operate a centrifugal pump which will have a discharge of 40000 litres/day with a total head of 10m. The pump operates for 10 hours in a day. The rated speed of wind is 6m/s. The power coefficient is 0.3. Density of air is $1.2 kg/m^3$.Assume transmission efficiency 95% , pump efficiency as 35%.

Solution :-

Pump discharge, Q=40000 litrers/day

Head, H=10m

Pump operate for 10hrs / day i.e.t=10hrs

Wind speed ,Ci=6m/s

Power coefficient. η =0.3

Density of air, $ϱ = 1.2Kg/m^3$

Transmission efficiency , ηtr=0.95

Pump efficiency ηp =0.35

i) Diameter of rotor of wind mill D:

Power available from wind to pump

P=η×Pmax × ηtr

=η x1/2 ϱACi^2×ηtr

$= 0.3 ×1/2×1.2×A×6^3×0.95$

= 36.936 × A Watts.

Power required by pump

P1= ϱ.g.QHηp

But,

Q=40000 liters/day

$= 40000/10×10^{-3}m^3/hr$

$= 4m^3/hr =4/3600 m^3/S$

∴ P1=1000 x 9.81 x 4/3600 x 10 x 0.35

= 38.15 N m/s or W

Power Available from wind = power required to run the pump

36.936 A=38.15

$36.936 ×ττ/4 D^2=38.15$

D= 1.1468m