Power calculations

[willib]

a while back i saw some struggling with the question of power dissapated in the gen.(internal resistance) and power output for a gen.

This is how i do it.

V=I*R

For a phase voltage of 24V and a phase resistance of 0.054 ohms and a load resistance of 0.5 ohms, first calculate the total resistance R(total) = 0.054 + 0.5 = 0.554 ohms , next calculate the total current ,V / R( total) ..

I = 24/0.554 = 43.32A

the coil voltage = IR = 43.32* 0.054 = 2.34V

the load voltage = 43.32*0.5 = 21.66V

the power dissapated in the coils will be the coil Voltage (2.34) squared divided by the coil resistance 0.054 ohms.

p(coil)= (2.34)^2 / 0.054 = 101.4 W

the power dissipated in the load is the load Voltage squared divided by the load resistance.

P(load)= (21.66)^2 / 0.5 = 938 W

[scottsAI]

How about showing the gen connected to a battery?

For simplicity we will use 12v for the battery,  no rectifiers losses, OK?

Battery impedance is << gen impedance. So will ignore it also.

I = (24 - 12) / 0.054 = 222.222 amps

Power into battery is 12 * 222 = 2,667watts.

How much power is dissipated in the generator?

2,667watts.

With high winds the generators output voltage can go over 48v.

I = (48 - 12) / 0.054 = 667amps now were smoking!

Power into battery is 12 * 667 = 8kw

Power in generator is (48 - 12) * 667 = 24kw.

To keep the generator from burning up it must furl before this point.

This shows why targeting a low cut in speed will limit the high end because of excessive voltage, forcing early furling to protect the generator. Or use limiting generator impedance, which will limit generator output at all points.

The use of star / delta can help, by reducing the voltage by 1.73 in higher wind speeds. But is it enough?

Have fun,

Scott.