[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]

RE: What size PFC ...

Terry -

Note that the equation you are using is correct for PFC only when the
reactive volt amps (VAR) is used. It is a coincidence that with your problem
the reactive volt amps (991) and the watts (900) are very close. It is
obvious if the power factor was 80% instead of 67.2% you would need less
than the 182.7 uf capacitance to correct the power factor to 100%. Your test
amps would then be 9.38 amps instead of 11.16 amps. You can find the lower
capacitance by using the K factor (multiplier) I showed in 1996.
   K = sin(arc cos(Power factor))
   K = sin(arc cos(.8)) = 0.6
Without the capacitor the volt amps would be
   VA = 900/.8 = 1125
The reactive volt amps would be
   VAR = 1125 * .6 =675

The necessary PFC capacitor would then be
   C = 675/(2 x pi x F x Vi^2)
   C = 675/(6.283 x 60 x 120^2) = 124.3 uf
This is less than the 182.7 uf because the power factor correction was only
for 80%. As you can see if the 900 was used instead of the 675 the cap would
be too large and would cause the power factor to be leading.

Think of a right angle triangle. The base is 900 watts. The hypotenuse is
1125 and the vertical is 675 VAR. The power factor angle is
   Angle = arc cos(900/1125) = 36.87 degrees
The power factor is
   Power factor = cos(36.87) = 80%

Are there any other coilers who can verify the above? What would be the PFC
capacitor if Terry found the "no cap" test was 13 amps instead of 11.16

John Couture


-----Original Message-----
From: Tesla list [mailto:tesla-at-pupman-dot-com]
Sent: Monday, September 04, 2000 12:30 PM
To: tesla-at-pupman-dot-com
Subject: RE: What size PFC ...

Original poster: Terry Fritz <twftesla-at-uswest-dot-net>

Hi John,

Ok, we seem to be saying the same thing some how ;-))  You last equation
seems oddly familiar ;-))

">The PFC capacitance = 991.8/(6.283 * 60 * 120^2) = 182.7 uf"

Much like  Cpfc = Vo x Io / (2 x pi x F x Vi^2)   ;-))

The difference is that I assume "Vo x Io" Is the "VARS" to be corrected.
Since MicroSim found the more accurate data to get the real number 991.8
the true capacitance of 182.7 can be determined.

However, suppose we do not have MicroSim or $2000 worth of test
equipment...  How can we determine the VARs?  I think the point is, if you
don't know the VARs, estimate it as "Vo x Io" and you will be close
enough...  An estimate, of course, but the basic idea is to find a
capcitance value easily with the data at hand.



At 11:15 PM 9/3/00 -0700, you wrote:
>The problem you show is different from the one I showed. The problem I
>solved was for a 50% to 90% PF correction. Your problem is for a 100%
>correction and a different no cap power factor. I will use your amperes.
>The 11.16 amps at 120 volts = 1339.2 VA
>The 7.5 amps at 120 volts = 900 VA
>The power factor is 900/1339.2 = .672 = 67.2% power factor
>The power factor angle is arc cos(.672) = 47.78 degrees
>The VARs to correct to 100% = 900 * tan(47.78) = 991.8
>The PFC capacitance = 991.8/(6.283 * 60 * 120^2) = 182.7 uf
>You would need 182.7 uf to correct your NST to 100% power factor.
>Actually your NST is operating at 1339.2 Var without a cap. However, the
>VA is only a guess for the 100% power factor operation because of other
>unknowns. You can confirm this only by test with a 182.7 uf cap. Even this
>cap might give a leading power factor.
>As you can see this problem is complicated and can be done in several ways.
>Try this on MicroSim. You should get the same answer.
>John Couture
>-----Original Message-----
>From: Terry Fritz [mailto:twftesla-at-uswest-dot-net]
>Sent: Sunday, September 03, 2000 2:57 PM
>To: John H. Couture
>Subject: RE: What size PFC ...
>Hi John,
>At 12:42 PM 9/2/00 -0700, you wrote:
>>Terry -
>>The rearranged equation is still incorrect when used with active volt
>>The equation is correct only when used with reactive volt amps. To convert
>>active to reactive amps you need to use complex numbers or trig functions.
>>prefer to use trig functions as I show in my post to the Tesla List
>>"PFC for Neons". Can it be that long ago?
>>Note that using your "close enough" equation will always give you a
>>power factor which is worst that a lagging power factor. It would be
>>interesting to see what a model by microSim would show.
>>To properly meter the TC load you need at least 4 meters, volts, amps,
>>watts, power factor. The power factor meter is required to tell you if the
>>load is leading or lagging. A VAR meter would help and save you having to
>>the necessary calculations.
>>Bart I am glad to hear that you are researching the problem. As Terry
>>pointed out there may be other issues and your work may shed more light on
>>the subject.
>>John Couture
>I pulled up your old post:
>PFC for Neons
>	To: Tesla List
>	Subject: PFC for Neons
>	From: "John H. Couture"
>	Date: Sun, 14 Jul 1996 17:55:23 GMT
>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 = .079 V A
>    For 240 volts    c uf = .020 V A
>    V = neon secondary volts     A = neon secondary amps
>The factors       K1 = sin(arccos(LPF)-sin(arccos(HPF))
>                  K1 = .43 for 50% to 90% power factor
>For 120 volts     K2 = (.43 x 10^6)/(6.283 F V)           F = 60 Hz
>                  K2 = (.43 x 10^6)/(377 x 120^2) = .079
>For 240 volts     K2 = .0198 or .02
>Example:   Neon 15000 volts  60 ma  120 volts  60 Hz
>           C = .079 x 15000 x .06 = 71.1 uf
>sniped efficiency text...
>	So your saying I should have a 71.1 uF cap instead of my present 200uF on
>my 15/60 coil...
>Let's look at the RMS current draw of my 15/60 coil with various PFC caps
>using MicroSim:
>Cap size	AC line current		Notes
>0uF 		11.16 ARMS		No PFC cap
>71.1uF		9.19 ARMS		John's equation
>165.8uF	7.75 Arms		"Terry's" equation
>200uF		7.72 ARMS  		The "real" best PFC cap size
>	Terry