This is just for example.
The current at the input and the amplitude of something at the output.
This means that if the device was inefficient and then it was improved, then this does not necessarily result in more energy at the output than at the input.
Efficiency means ?
Richard /Dick Fradella example !
https://worldwide.espacenet.com/publicationDetails/biblio?CC=US&NR=2012256422A1&KC=A1&FT=D&ND=3&date=20121011&DB=EPODOC&locale=en_EP#[0090] Generator power and efficiency with wind turbine drive is computed below, for a representative example of the present invention, at maximum shaft speed, mid-speed, and minimum usable speed, using a few simplifying approximations. Shaft speed, power, and the other variables in the computations herebelow are exemplary, and not intended as limiting the present invention in any way. This will help explain FIG. 1 and FIG. 2 configuration operation, distinctions and improvements over widely used prior art generators:
[0091] Let maximum speed equal 1000 revolutions per minute (rpm), mid-speed equal 500 rpm, and minimum speed equal 100 rpm. Also, let maximum stator current Imax=10 amperes, and nominal VDC=100 volts. Further, let Q1-Q4 power MOSFET ON resistance Rdson=0.01 ohm, inductor L1-L4 series pair winding resistance RL=0.1 ohm. Also, stator winding resistance Rs=0.15 ohm, stator voltage Vmax=100 volts at 1000 rpm, and fly-back (free-wheeling) diode D1-D8 forward drop Vf=1-volt at 10 amp. These parameters are consistent with a test prototype, according to the present invention, developed to generate power from wind turbines.
[0092] At 1000 rpm, Vmax=100 volts, so PWM duty-cycle (Ton)/(Ton+Toff) is essentially zero. Therefore, losses=Imax<2>(RL+Rs)+2 VfImax=(10 amp)<2>(0.25 ohm)+(2 volt)(10 amp), amounting to 45 watts loss. Output power=(Imax)*(Vmax)=(Imax)*(VDC)=(10 amp)(100 volts)
=1000 watts. So, generator efficiency at maximum speed and maximum power is about
95% for this example of generator and integrated electronics parameters.
[0093] At 500 rpm, Imax=(10 amp)/(4)=2.5 amps; and Vmax=(100 volts)*(0.5)=50 volts. So PWM duty-cycle=1⁄2. Average pulse power generated=(Imax)*(Vmax)=(Imax)*(VDC)/2=(2.5 amp)(50 volt)
=125 watts. Losses to maintain inductor current=Imax<2>(RL+Rs+Ron)=(2.5 amp)<2>(0.26 ohm)=1.6 watts. Fly-back diode losses=2 Vf*Imax/2=(0.6 v)(2.5 amp)=1.5 watts. So total losses=3.1 watts. Therefore, mid-speed generator efficiency is about
97%.
[0094] At 100 rpm, Imax=(10 amp)/(100)=0.1 amp; and Vmax=(100 volts)/(10)=10-volts. So PWM duty-cycle= 9/10. Average pulse power generated=(Imax)*(Vmax)=(Imax)*(VDC)/10=(0.1 amp)(10 v)
=1 watt. Losses to maintain stator and inductor current=Imax<2>(RL+Rs+2*Rdson)=(0.1 amp)<2>(0.27 ohm)=0.0027-watt. Fly-back diode losses=(2*Vf)*(Imax)/10=(0.6 v)(0.1 a)/5=0.012 watts. So total losses=0.015-watt. Thus, generator efficiency at low speed is about
98%. Device nominal capacity full use efficiency ! inverse/cube law !
Above highest efficiency = lowest (uneconomical) conversion gain
We can see windgenerators with moving blades/rotor,consuming more energy than generating !
Triangulation : physical efficiency / technical efficiency / financial efficiency
wmbr
OCWL
