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Re: NST Power Factor?




From: 	John H. Couture[SMTP:couturejh-at-worldnet.att-dot-net]
Sent: 	Saturday, January 10, 1998 10:11 PM
To: 	Tesla List
Subject: 	Re: NST Power Factor?

At 01:55 AM 1/9/98 +0000, you wrote:
>
>From: 	Thornton, Russ #CSR2000[SMTP:ThorntoR-at-rc.pafb.af.mil]
>Sent: 	Thursday, January 08, 1998 6:03 AM
>To: 	'Tesla discussion Group'
>Subject: 	RE: NST Power Factor?
>
>>Subject: 	Re: NST Power Factor?
>>
>>
>>Using these will not save you any money, but they will draw substantially 
>>less current and help you extend your variac's capability to drive 
>>multiple transformers.  Also, with a 660VAC oil cap of appropriate value, 
>>any "normal PF" neon can be similarly corrected.  
>>
>>-Adam
>
>Adam,
>How would one calculate the appropriate value for this correction cap?
>
>Thanks,
>
>Russ Thornton
>CSR 2040, 
>Building 989, Rm.  A1-N20
>Phone: (407) 494-6430 
>Email: thorntor-at-rc.pafb.af.mil

---------------------------------------------------------

  Russ -

  Uncorrected neon transformers are usually 50% power factor. To correct
them for 90% power factor add a capacitor calculated as follows:

    For 120 volts    c uf = .08 V A
    For 240 volts     c uf = .02 V A
    V = neon secondary volts   A = neon secondary amps

  Example  Neon  15000 volts  60 ma  60 hz  50% PF

  C = .08 x 15000 x .06 = 72 uf     For 120 volts

  C = .02 x 15000 x .06 = 18 uf      For 240 volts

------------------------

  If you are math minded and want to calculate for other conditions use the
following:

  K1 = sin(arc cos (LPF) - sin(arc cos(HPF))
  K2 = (K1 x 10^6) / (6.283 x F x Vp^2)
  C = K2 x V x A uf

  Example for above neon

  K1 = sin(arc cos(.50) - sin(arc cos(.90)) = .435
  K2 = (.435 x 10^6) / (6.283 x 60 x 120^2) = .08  (For 120 volts)
  K2 = (.435 x 10^6) / (6.283 x 60 x 240^2) = .02  (For 240 volts)  

  C = K2 x V A  = .08 x 15000 x .06 = 72 uf    For 120 volts
  C = K2 x V A  = .02 x 15000 x .06 = 18 uf    For 240 volts

  John couture