Why does my model of a wind turbine show high RPM with low power output?

I am designing a wind turbine to generate 1300 W.

I modeled the output voltage of an AC generator using the following specifics:

• Number coil turns = 300.
• Copper wire gauge = 1.02 mm.
• Knowing the density of copper, I calculated the weight of the coil to be 4.39 kg.
• Magnetic field strength = 1.24 T. (here are the specs for the magnet)
• Area = 1 m2.
• Rotational speed = 45 RPM.

I plugged these numbers into the output voltage equation:

$$\epsilon = NBA\omega\sin(\omega t)$$

to get a value of 1753 V. I am assuming that my generator will generate at least the 0.74 A to output a power of 1300 W.

Then, I modeled my turbine. I used the equation:

$$P_T = \frac{1}{2}\rho A v^3 C_p$$

to calculate rotor power. I assumed:

• My turbine would be 30% efficient.
• The wind velocity would be 10 m/s.
• The area swept by blades would be 2.63 m2.
• The density of air would be 1.25.

After plugging these values in, I get a power yield of 493 W. However, I find that the rotor speed is well above 45 RPM. This equation gives the tip speed ratio of the turbine:

$$\lambda = \frac{wr}{v}$$

Knowing that the optimum tip speed ratio of a three-blade turbine is 4.2, radius of the rotor blade is 0.915 m, and velocity of air is 10 m/s, that angular velocity would be 45.9 rad/s or 440 RPM.

How can I get 1300 W from this turbine when the theoretical yield from power of the rotor is only 493 W, even though my rotor should turn at the correct RPM for the generator to produce 1300 W?

• Converting this to rpm would yield 440 rpm That's quite fast for a wind turbine. It will probably shake itself to pieces. Mar 29 '16 at 13:35
• @EnergyNumbers what am I doing wrong? Mar 29 '16 at 14:01
• the kinetic energy of wind gives a power yield of 1643.89W (without multiplying Cp). However, I am assuming my turbine will be 30% efficient; thus, I am getting a rotor power of 493W. My turbine has a very small swept area, so is my assumption that my efficiency of the turbine be 30% false?? I know the power coefficient is related to tip speed ratio and pitch, but will the small area also have an effect? Mar 29 '16 at 14:47
• @user510 What's the source of energy for a wind turbine and how do you capture it? If the rotor power you calculated is too low, what design value(s) do you have the greatest ability to control and how do you change those values to capture more of the wind's energy?
– Air
Mar 30 '16 at 23:26
• It sounds like you are working backwards, you need to first design a turbine which will generate the power you need and then design a generator to suit that power requirement. The dimensions of a generator give you its maximum capacity they don't guarantee that it will generate that amount of power. You may also want to look at existing generators specs for a reality check. Mar 30 '16 at 23:27

At $C_p=0.3$ and $v=10\ \mathrm{m/s}$, you simply need an area of: $$A=\frac{2P_T}{C_pv^3\rho}=7.07\ \mathrm{m^2}$$ $$\Rightarrow R=1.5\ \mathrm{m}$$
Moreover, you say $\lambda=4.2$ is optimal for a three-blade turbine, but it actually depends on the blade profile, which is optimized for a given "design tip speed ratio". For more information about this, see chapter 3.7 of "Burton, Wind energy handbook, 2011".
• No, the generator max power should be a bit higher than the rated output of the rotor, because gusts of wind might exceed the rated wind speed ($v=10m/s$) and make the wind turbine output more than 1300W. But ultimately the whole system won't output more power than the rotor. Mar 31 '16 at 6:56