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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