Language: 
To browser these website, it's necessary to store cookies on your computer.
The cookies contain no personal information, they are required for program control.
  the storage of cookies while browsing this website, on Login and Register.

GDPR and DSGVO law

Storing Cookies (See : http://ec.europa.eu/ipg/basics/legal/cookies/index_en.htm ) help us to bring you our services at overunity.com . If you use this website and our services you declare yourself okay with using cookies .More Infos here:
https://overunity.com/5553/privacy-policy/
If you do not agree with storing cookies, please LEAVE this website now. From the 25th of May 2018, every existing user has to accept the GDPR agreement at first login. If a user is unwilling to accept the GDPR, he should email us and request to erase his account. Many thanks for your understanding.
Amazon Warehouse Deals ! Now even more Deep Discounts ! Check out these great prices on slightly used or just opened once only items.I always buy my gadgets via these great Warehouse deals ! Highly recommended ! Many thanks for supporting OverUnity.com this way.

User Menu

Tesla Paper

Free Energy Book

Get paid

Donations

Please Donate for the Forum.
Many thanks.
Regards, Stefan.(Admin)

A-Ads

Powerbox

Smartbox

3D Solar

3D Solar Panels

DC2DC converter

Micro JouleThief

FireMatch

FireMatch

CCKnife

CCKnife

CCTool

CCTool

Magpi Magazine

Magpi Magazine Free Rasberry Pi Magazine

Battery Recondition

Battery Recondition

Arduino

Ultracaps

YT Subscribe

Gravity Machines

Tesla-Ebook

Magnet Secrets

Lindemann Video

Navigation

Products

Products

WaterMotor kit

Statistics

  • *Total Members: 84074
  • *Latest: qwee

  • *Total Posts: 894589
  • *Total Topics: 15733
  • *Online Today: 44
  • *Most Online: 103
(December 19, 2006, 11:27:19 PM)
  • *Users: 3
  • *Guests: 19
  • *Total: 22

Author Topic: static or rotative DC/Pulsed DC and/or DC/AC power converter (misleadings ?)  (Read 331 times)

Offline lancaIV

  • elite_member
  • Hero Member
  • ******
  • Posts: 4367
1 Hertzian cycle = 1 rotation per second


50 Hz                = 50 rotations per sec/ 3000 RPM

simple calculation !






https://fbadhusha.weebly.com/uploads/3/8/9/5/3895546/ele-pulse-power.pdf

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




https://worldwide.espacenet.com/publicationDetails/biblio?DB=EPODOC&II=1&ND=3&adjacent=true&locale=en_EP&FT=D&date=20121011&CC=US&NR=2012256422A1&KC=A1


[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%.






https://worldwide.espacenet.com/publicationDetails/biblio?DB=EPODOC&II=0&ND=3&adjacent=true&locale=en_EP&FT=D&date=19291018&CC=FR&NR=667647A&KC=A

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).




https://worldwide.espacenet.com/publicationDetails/biblio?DB=EPODOC&II=2&ND=3&adjacent=true&locale=en_EP&FT=D&date=20140213&CC=US&NR=2014043128A1&KC=A1


[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
https://www.ourenergypolicy.org › 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  ?

Free Energy | searching for free energy and discussing free energy


 

OneLink