Mark Hall wrote:
> Hi:
>
> I just purchased a 3.3KV @ 1000 ma, 40 pound, "never got warm @800
> ma", transformer off the web. I measured the secondary resistance
> with
> a Fluke 8024B DVM using several different ranges and in all cases it
> came to 170 ohms. There is a secondary centertap.. and each side to
> center gives about 85 ohms.. presumeably no open or hi resistance
> connections. The lowest primary resistance (lots of probe digging and
>
> twisting) reads ~1.5 ohms. He says it's a 110VAC primary which I can
> live with. Is a DVM a "legal" way to measure secondary resistance, or
>
> should I invite my buddies over and pop a 6-pack to watch thing melt
I suspect the transformer too. It sounds too light, and too resistive. I
believe the following argument proves this.
3.3 kV@ 1 A is 3300 W if run into a purely resistive load - upon which
most, but not all transformers are rated. At 115 V primary voltage,
assuming no magnetisiting current, this is 29 A primary current. With a
DC resistance of 1.5 Ohms, this is a i^2R loss of 29*29*1.5=1261 W due
to the primary. The secondry i^2R loss is 1*1*170 Ohms = 170 W. Hence
the total copper loss (primary + secondry), is 1431 W. Typically
transformers (and motors) have about equal copper and iron losses
(theoretically efficiency is maximum then) so the losses in this
transformer at 3.3 kW output would be about 2*1431=2862W. Hence the
input power would be about 3300+2862=6162 W , or a maximum of 53%
efficiency.
In other words, the highest possible efficiency for your transformer at
3.3 kW output, would be 53%, with 2.9 kW produced as heat. No
transformer, except the very smallest, would be designed to run at 53%
peak efficiency. Transformers are designed to run at peak efficiency at
maximum load.
Whether the transformer is acceptable for your likely duty cycle is
another question, and could best be determined by using lamps as loads
for test purposes.
G8WRB.
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