Question 14.14: Power Generation in a Wind Farm The average wind speed at a ...
Power Generation in a Wind Farm
The average wind speed at a proposed HAWT wind farm site is 12.5 m/s (Fig. 14–112). The power coefficient of each wind turbine is predicted to be 0.41, and the combined efficiency of the gearbox and generator is 92 percent. Each wind turbine must produce 2.5 MW of electrical power when the wind blows at 12.5 m/s.
(a) Calculate the required diameter of each turbine disk. Take the average air density to be 𝜌 = 1.2 kg/m³.
(b) If 30 such turbines are built on the site and an average home in the area consumes approximately 1.5 kW of electrical power, estimate how many homes can be powered by this wind farm, assuming an additional efficiency of 96 percent to account for the powerline losses.
We are to estimate the required disk diameter of a wind turbine
and how many homes a wind farm can serve.
Assumptions 1 The power coefficient is 0.41 and the combined gearbox/generator efficiency is 0.92. The power distribution system has an efficiency of 96 percent.
Properties The air density is given as 1.2 kg/m³.
Analysis (a) From the definition of power coefficient,
\dot{W}_{\text {rotor shaft output }}=C_P \frac{1}{2} \rho V^3 A=C_P \frac{1}{2} \rho V^3\left(\pi D^2 / 4\right)
But the actual electrical power produced is lower than this because of gearbox and generator inefficiencies,
\dot{W}_{\text {electrical output }}=\eta_{\text {gearbox/generator }} \frac{C_P \pi \rho V^3 D^2}{8}
which we solve for diameter,
D=\sqrt{8 \frac{\dot{W}_{\text {electrical output }}}{\eta_{\text {generator }} C_P \pi \rho V^3}}=\sqrt{8 \frac{2.5 \times 10^6 W \left(\frac{ N \cdot m / s }{ W }\right)\left(\frac{ kg \cdot m / s ^2}{ N }\right)}{(0.41)(0.92) \pi\left(1.2 \frac{ kg }{ m ^3}\right)\left(12.5 \frac{ m }{ s }\right)^3}}
= 84.86 m ≅ 85 m
(b) For 30 machines, the total electrical power produced is
30(2.5 MW) = 75 MW
However, some of that is lost (wasted – turned into heat) due to inefficiencies in the power distribution system. The electrical power that actually makes it to people’s home is thus
\left(\eta_{\text {power distribution system }}\right)(\text { total power })=(0.96)(75 MW )=72 MW
Since an average home consumes power at a rate of 1500 W, the number of homes served by this wind farm is calculated as
\text { Number of homes }=\frac{\left(\eta_{\text {power distribution system }}\right)(\text { number of turbines })\left(\dot{W}_{\text {electrical output per turbine }}\right)}{\dot{W}_{\text {electrical usage per home }}}
=\frac{(0.96)(30 \text { turbines })\left(2.5 \times 10^6 W / \text { turbine }\right)}{1.5 \times 10^3 W / \text { home }}
=4.8 \times 10^4 \text { homes }= 4 8 , 0 0 0 \text { homes }
Discussion We give the final answers to two significant digits since we cannot expect any better than that. Wind farms of this size and larger are being constructed throughout the world.