Author Topic: WFC Not Over Unity  (Read 21587 times)

0 Members and 1 Guest are viewing this topic.

Offline Login to see usernames

  • Moderator
  • Sr. member
  • ***
  • Posts: 460
    • Global Kast : Water Fuel Cell Research
Re: WFC Not Over Unity
« Reply #8 on: December 13, 2012, 01:16:48 am »
I was using 1.5 inches squared which = 0.0381 meters squared

Offline Login to see usernames

  • Global Moderator
  • Hero member
  • ****
  • Posts: 4215
Re: WFC Not Over Unity
« Reply #9 on: December 13, 2012, 01:22:43 am »
You must think in meters than it will work i'm sure. it worked for me... mind that 1cm is 0,01meter so 1cm2 = 0,0001 m2

In you first calculus 0,038m = 381cm2 way a lot =)

Offline Login to see usernames

  • Global Moderator
  • Hero member
  • ****
  • Posts: 4215
Re: WFC Not Over Unity
« Reply #10 on: December 13, 2012, 01:27:26 am »
It was such a mystery to me but only now I had the clarity of mind to understand it..

Offline Login to see usernames

  • Moderator
  • Sr. member
  • ***
  • Posts: 460
    • Global Kast : Water Fuel Cell Research
Re: WFC Not Over Unity
« Reply #11 on: December 13, 2012, 02:07:48 am »
ok so like this example for Stan's VIC Transformer core?
tw = 1.09728 cm
hw = 4.72186 cm
bf = 6.25348 cm
tw = 1.09728 cm

A = ( hw * tw ) + (2( bf * tf))
A = ( 4.72186 * 1.09728 ) + (2( 6.25348 * 1.09728 ))
A = ( 5.181203 ) + (2( 6.86182 ))
A = ( 5.181203 ) + ( 13.72364 )
A = 18.904843 cm2

2 * 18.904843 = 37.809686 .....I multiplied by 2 since there are 2 core halves

37.809686 * 0.0001 = 0.0037809686 meter2

Flux = 0.3 * 0.0037809686
Flux = 0.0011343

Primary Turns = ((12V / 0.0011343) / 5000)
Primary Turns =   10579.21185 / 5000
Primary Turns = 2.1158 ???

Offline Login to see usernames

  • Global Moderator
  • Hero member
  • ****
  • Posts: 4215
Re: WFC Not Over Unity
« Reply #12 on: December 13, 2012, 02:49:47 am »
No tony sorry you maybe misread it, i meant cross-sectional area not the total area of the core... only the cross-sectional area is considered here.

http://wiki.answers.com/Q/How_do_you_calculate_the_cross_section_area_of_cylinder

for example:
typic flyback core like... 1,8cm diameter
cross-sectional area 2,54x10^-4 m2
max flux= 7,63x10-5
12v 5khz = 31 turns considering bmax = to 0,3tesla which should holds true for up to 10-20khz

Offline Login to see usernames

  • Moderator
  • Sr. member
  • ***
  • Posts: 460
    • Global Kast : Water Fuel Cell Research
Re: WFC Not Over Unity
« Reply #13 on: December 13, 2012, 02:52:53 am »
so what would it be using the measurements I gave for Stan's transformer?

Offline Login to see usernames

  • Global Moderator
  • Hero member
  • ****
  • Posts: 4215
Re: WFC Not Over Unity
« Reply #14 on: December 13, 2012, 06:20:36 am »
well the cross-sectional area is tf multiplied by the component length in the z direction which you did not provided...

the area of interest is only the area of the wire loops you see? i mean the area of the core cut.

If meyer was using ferrite, this would mean for 2cm wide maybe 1cm high is 0.0002m2

so maximum flux would be 0,00006 webber considering bmax = 0,3 tesla

So turns number considering 12v 5 khz

40 turns...

If he was using a core that has half this size in high for example than it would be a flux of 0,00003

so turns = 80

so is directly proportional to the core size... I guess his core was smaller...

« Last Edit: December 13, 2012, 07:48:08 am by sebosfato »

Offline Login to see usernames

  • Global Moderator
  • Hero member
  • ****
  • Posts: 4215
Re: WFC Not Over Unity
« Reply #15 on: December 13, 2012, 07:52:15 am »
He clearly than modulated the voltage at the input of the transformer as function of frequency applied so he could aways input power at maximum level just below saturation. Probably lowering voltage applied at lower frequencies of course.