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Re: [Amps] Plate transformer current unknown

To: amps@contesting.com
Subject: Re: [Amps] Plate transformer current unknown
From: "Will Matney" <craxd1@ezwv.com>
Reply-to: craxd1@ezwv.com
Date: Tue, 03 May 2005 14:55:35 -0400
List-post: <mailto:amps@contesting.com>
Angel,

One last thing, since that core is that big, measure the thickness of an 
outside leg. A 4 inch core shows a 2 inch leg, and a 3 inch core shows a 1-1/2 
inch leg. We might be able to tell from it then.

Best,

Will

*********** REPLY SEPARATOR  ***********

On 5/3/05 at 1:38 PM Will Matney wrote:

>Angel,
>
>If the transformer stack is 11" wide X 9" high then about the closest
>thing to a standard EI lam is a 4(3)MH which has a 4 inch center leg and
>is 10" high X 11-1/8" wide. Now if the stack is 10" deep, that makes 4 X
>10 = 40 square inches. That sounds a little big to me though as generally
>a rectangular stack is used mostly which would have been 4 X 8 inches or
>36 square inches of core area. Anyhow, lets work this backwards;
>
>a = 0.16 X sqrt P
>
>P = a / 0.16 ^2
>
>40 sq in / 0.16 = 250 then 250 ^2 = 62,500 watts. (12 kilogauss at 60 Hz)
>36 sq in / 0.16 = 225 then 225 ^2 = 50,625 watts. (12 kilogauss at 60 Hz)
>
>One should note, I'm not sure as I didn't go back and check, I may have
>wrote 0.61 for the formula where it should be 0.16. Sorry about that if I
>did.
>
>The turns per volt will roughly be 5 / a or 5 / 40" = 1/8 TPV, and 5 / 36"
>= 3/16 TPV.
>
>Something dont sound right about those measurements for the core, double
>check those and only measure the core itself (lamination stack W X H X D)
>as the depth is what I thought about. The next stock lam is a 3" but RCA
>could have had a custom made like a 3-1/2" leg. For 3-1/2" at 10" deep
>would be 35 sq in., and for a 3" leg, 30 sq inches. 3-1/2" would be 47,850
>watts and at 3" would be 35,156 watts. An approximate weight using a 10"
>stack with the 4(3)MH lam would be 258.75 pounds. By your dimensions
>given, maybe 200 pounds. If it's that big, it may well be this size in kVA
>or in watts.
>
>3000 x 1.414 = 4242 VDC. 50,000 watts / 4242 VDC = 11.75 amperes (with no
>losses).
>
>Assuming a 5% PS loss, 11.75 X 1.05 = 12.34 amperes.
>
>Keep in mind thought that the lead size will be in conduit rating for
>amperes. Here, 18 gauge is 13 to 18 amps for chassis wiring.
>
>Remember that these figures are ballpark ones and not exact.
>
>Best,
>
>Will
>
>*********** REPLY SEPARATOR  ***********
>
>On 5/3/05 at 12:05 PM Angelo Karabetsos wrote:
>
>>I am unable to calculate anything to do with windings or core due to the
>>fact it is in a metal case encapsulated in black epoxy. I discovered this
>>after removing the lid. I'll say one thing, it's darn heavy and hard to
>>move
>>around without anywhere to grab. The secondary wire gauge looks to be
>>about
>>18ga and the dimensions are 11"x9"x10" and has a RCA nameplate. Not too
>>much
>>more information I can determine. Now I know what to look for at flea
>>markets.
>>One other question. What should the B+ voltage be after rectification.
>>Lets
>>say 3000v secondary through a fullwave bridge rectifier and a bank of ten
>>470uf 450v caps? I read somewhere to multiply by 0.9 and that should be a
>>good rule of thumb. What I am trying to determine is how to calculate the
>>proper transformer for the job and keep voltage drop under load to
>>respectable figure. Not an easy task to figure out with surplus parts.
>>
>>Angelo
>>
>>&gt;From: &quot;Will Matney&quot; &lt;craxd1@ezwv.com&gt;
>>&gt;Reply-To: craxd1@ezwv.com
>>&gt;To: amps@contesting.com
>>&gt;Subject: Re: [Amps] Plate transformer current unknown
>>&gt;Date: Fri, 29 Apr 2005 11:28:25 -0400
>>&gt;
>>&gt;Angelo,
>>&gt;
>>&gt;Really you cant tell that way (using the resistance) unless you know
>>the
>>mean lenght of turn, how many turns per layer, and how many layers they
>>are
>>in the winding. This then would give you how long the wire is and you
>>could
>>then use the resistance and the length to determine the wire size. If
>it's
>>possible to use a micrometer or a wire gauge to see what the secondary
>>wire
>>size is, then one can find out what its rating is. In amateur service,
>>ICAS
>>used anywhere from 700 to 900 circular mils per ampere. CCS uses anyhwere
>>from say 1000 to 1200 cir. mil. per amp. If you can find the wire size,
>go
>>look at a wire chart for how many circular mils it has and divide by the
>>above numbers to determine its current carrying ability. IE, 10,000
>>circular
>>mils running at 1000 cir mil per amp = 10 amperes.
>>&gt;
>>&gt;Best,
>>&gt;
>>&gt;Will
>>&gt;
>>&gt;
>>&gt;*********** REPLY SEPARATOR  ***********
>>&gt;
>>&gt;On 4/29/05 at 3:01 PM Angelo Karabetsos wrote:
>>&gt;
>>&gt; &gt;I would like to find out the current rating of a transformer.
>>Primary
>>&gt; &gt;voltage taps 0v-110v-115v-208v-230v and secondary taps
>>3450v-0v-3450v and
>>&gt; &gt;secondary resistance of 53.5ohms from 0v to 3450v tap and
>107ohms
>>from
>>&gt; &gt;3450v
>>&gt; &gt;to 3450v tap. All measurements done with a simple DMM. The
>>commercial
>>&gt; &gt;amplifier it came out of has a plate current meter with a max
>>reading of
>>&gt; &gt;600ma. How could I find out the CCS or ICAS rating?
>>&gt; &gt;Cheers Angelo
>>&gt; &gt;
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