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. $\endgroup$ – grfrazee Mar 29 '16 at 13:35