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
>>multiple transformers.  Also, with a 660VAC oil cap of appropriate value,
>>any "normal PF" neon can be similarly corrected.
>>
>
>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

```