Hi Jim
On Wednesday 01 April 2009 07:49:29 pm Jim Mora wrote:
Impedance is reflected through a transformer so that say you are looking into
the primary you see the load that is on the secondary divided (in the case of a
step up transformer) by the square of the turns ratio. In your case the turns
ratio is 14400/240 or 60/1, 60^2 = 3600. So if you have a load of 3600 Ohms on
the secondary you will see 1 Ohm looking into the primary.
It works the other way in reverse, let us say (just for grins) you have some
inductive ballast that has an inductive reactance of 11.5 Ohms (at 60Hz) in
series with the primary. Looking into the secondary you will see 11.5 * 3600 =
41400 Ohms.
Lets say you want a capacitor that is resonant with that ballast at 60Hz.
1/(41400*377) = 64nF. (BTW 377 is omega which is 2*pi*f where f=60Hz)
Xl on the primary side = Xc on the secondary side. Resonance. Power factor is
unity.
Do some of these numbers look familiar?
Lets look at JavaTC with the inputs you give. It says your transformer impedance
is 41499 Ohms. (giving rated mA as 347, 5KVA/14.4KV) If it was a Neon Sign
Transformer with a nameplate rating of 14400 Volts open circuit and 347mA short
circuit it would be. But you have a Potential Transformer. Apples and oranges.
The impedance of your transformer is nowhere near that. 347mA is not the short
circuit current. It is the "rated" full load current. You transformer has a
nameplate that says something like primary 14400V secondary 120/240V 5KVA and
some percent impedance. Since I do not know the percent impedance rating of
your transformer, I will use my 5KVA Pole Pig which has a nameplate rating of
1.6 percent impedance as an example.
Calculating short circuit current:
First we need to find the full load amp rating.
5000VA/14400V = 347.2mA. Then divide FLA by unit impedance (unit impedance is
percent impedance divided by 100): 0.3472A/0.016 = 21.7A short circuit. Now
find transformer impedance: 14400V/21.7A = 663.6 Ohms.
BTW 21.7A * 60 = 1302A at 240V, with no drop (assuming an infinite buss) which
is a little bit more than my 150A service can handle.
Now lets take those "real" numbers and put them into JavaTC to see what it gives
us for a resonant size cap and for PFC capacitance.
It tells us about 4uF for resonant tank cap and 14390uF for the PFC.
Now lets "fudge" JavaTC a bit. Just put 1mA in the box where it says "Rated mA"
just so the JavaScript won't throw an error, and go over to the box where it
says "Ballast Current Amps (optional)" and put something reasonable in there,
say 20A. Then look at the output called "Transformer Output Current" which is
given in Amps, convert that to mA, in this case 500, and stick that number into
the box called "Rated mA", hit the calculate button again, and see what
happens. Resonant cap size 61nF, PFC, 221uF. Now lets try 30A. Resonant cap
size 92nF, PFC 332uF. So adjusting the ballast will not only change the
resonant point, but the PFC as well.
Another experiment we can do is to change the breakrate. 60Hz equates to a
breakrate of 120BPS. Lets change that to 500BPS by changing the 60Hz to 250Hz.
Everything else stays the same. Now we get 22nF resonant cap size and 80uF PFC.
So changing the break rate changes both the resonant point and the PFC.
In your case, you have a saturable reactor, so you can adjust the ballast on the
fly, and if I remember correctly a variable speed ASRSG, so you can vary the
breakrate on the fly. That will allow you to change both parameters at will,
run resonant or LTR or STR however you want all with a fixed tank cap size.
Cool.
Now as far as PFC goes, weeelllll---
Here is what "I" suggest. Since you are not going to power factor correct
the "natural impedance of the transformer" anyway, put the PFC caps where they
might do you some good, ahead of the ballast. Assuming all your caps are 100uF
and you have enough of them, build strings of different values and put a 20A
toggle switch in series with each string. And don't forget to put a 5meg or
10meg bleeder across each one. (cap that is) Make the first string 8 caps.
12.5uF. The next string 4 caps, 25uF. Third string 2 caps, 50uF. Fourth string
1 cap, 100uF. Fifth string, two caps in parallel for 200uF. Thus with 17 caps
you can have a range of 0 to 387.5uF in 12.5uF steps at the flip of a few
switches, like binary addition. You see where I am going with this?
BTW, take a look again at my page at http://deanostoybox.com/pfc I might have
added something since you were there last. Or not.
later
deano
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