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Author Topic: static or rotative DC/Pulsed DC and/or DC/AC power converter (misleadings ?)  (Read 324 times)

Offline lancaIV

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1 Hertzian cycle = 1 rotation per second

50 Hz                = 50 rotations per sec/ 3000 RPM

simple calculation !

Later on, we’ll prove experimentally that power pulse (Fig. 3) has not 15000 W and 144.80 W, but only 1.40 W

[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%.

Les dessins annexés (qui ne sont qu'un exemple d'exécution de la présente inven tion) représentent une petite machine-type pour courant monophasé à 20 -périodes en viron, d'une puissance- -ëfficace de 300 watts, alimentée par une faible pile de 6 volts 5 ampères (c'est-à-dire une puis sance moyenne de 30-watts, c'est-à-dire une multiplication par 10 qui peut être aug mentée davantage.

The accompanying drawings (which are only an exemplary embodiment of the present invention) show a typical small machine for single-phase current with approximately 20 periods(= cycles), with an effective power of 300 watts, supplied by a low battery 6 volts 5 amps (that is, an average power of 30-watts, that is, a 10-fold increase that can be increased further.

Elle porte 6 couches de spires de fil de 0.005 m/ de diamètre, c'est-à-dire environ 100 spires contre 50 à l'inducteur, ce qui permet avec une pile de 5 volts 6 ampères, d'atteindre une tension efficace de 110 volts et simultanément une intensité efficace de 3 ampères environ. chiffres calculés pour réaliser féciairage électrique d'une dizaine de lampes (110 volts-32 bougies).

[0199] Net Power Output after Feedback of 13.32 Kva

[0200] The net output at 400 Hz is 16.871 Kva (22.6 Hp); at 800 Hz, 33.7 Kva (45.2 Hp); at 3 KHz=128 Kva (171.5 Hp); at 6 KHz=256 Kva (343 Hp); at 60 Khz=2.560 Kva; at 600 KHz=25.60 Kva; at 780 KHz=33.254 Kva.


Goldbaum generator output 50 Hz grid-like

400 Hz 16,871 Kva to 50 Hz = 16,871 Kva/8  = 2,108 Kva  + 13,32 Kva input = 15,428 total output/13,32 Kva input= C.O.P 1,54

Goldbaum generator compared Meredieu generator 20 Hz output performance

400 Hz 16,871 Kva to 20 Hz = 16,871 Kva/20= 0,843  Kva + 13,32 Kva input = 14,163 total output/13,32 Kva input = C.O.P. 1,06

not to underestimate the C.O.P. 1 and C.O.P. 1,06 difference :

the surplus " ,06 " represents  x 3600 = 3 KVAh x 24h = up to 72 KVAh surplus power per day !

when I take a conventional , by quality : 90% efficiency, DC/AC inverter and coat the secondary coil

130 Electrical Energy Innovations - Our Energy Policy › 130-electrical-...

05.12.1996 — Induction Coil Coating Increases Generator Output by One-Third´

would it not reach :                     0,9 x 1,33 =  C.O.P. 1,2  ?

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