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Registered Member #72
Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659
Hazmatt_(The Underdog) wrote ...
My prediction with a 240V lamp in series with the transformer:
24V across the transformer, the remainder voltage across the bulb.
I'd like to know what is measured, I am curious.
The transformer will look like a dead short on the low side.
Just curious as why you predict a voltage of less than 120v across the low side. Neglecting the magnetising current, a transformer with no load on the secondary is open circuit up to its rated voltage, Sulaiman's comment shows he gets this.
Now there are some unstated assumptions.
a) I don't think 'no load' has been mentioned yet, but when testing a transformer for magnetising current, the no load condition is obvious.
b) Similarly, we haven't discussed the rated power of the filament bulb. When you use such a bulb for protecting a device under test from excess current, it's obvious we want it to remain cold for 'normal' operating current, and get hot for fault currents. This means the bulb should be rated to pass the magnetising current with low drop. We don't know what the magnetising current is yet, so we have to measure it. We *do* know that whatever we measure up to 120v across the low side is *within the rating* of the transformer and therefore OK. I would guess that the magnetising current would fully turn on a 5w 'slumberlight', but would probably pass through a 500w halogen flood with little drop, but that's what measurement is for.
Registered Member #162
Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3140
The nameplate ratings are clear: 25 kV rated system, 21 kV nominal, rated for 50 kV for 1 minute ... so YES you can use the primary (what was the secondary) DIRECTLY connected to 240 Vac but it will be at the top end of its ratings, and how long it can be used, at what power output before overheating, I do not know
so when you are confident that the transformer is ok on 240Vac, in series with the primary, add something like this attached to the transformer.
My only concern with directly connecting to 240 Vac is that this is a fairly large transformer, quite likely that the mains breaker will trip occasionally - depending upon where in the ac cycle the transformer is powered on, zero-crossing is the worst case.
and as above, a series filament lamp would need to be high wattage. but in this specific case it is not required, other than for preventing occasional mcb trips.
CORRECTION: My main concern is your safety ... I suspect that you have negligible experience with transformers like this, if you have an eht short-circuit the transformer will provide very little current limiting, so almost 1/175 th of whatever the mains can supply before disconnect is available to kill or destroy, e.g. 13 A fuse, 6000 A rating , so less than 30 A at your eht secondary. less than 1% of that can kill you! ... You would be lucky to survive ONE mistake. Consider it as MOTs, with the shunts removed ... and on steroids. it even looks scary, and ss a general rule, I prefer live members to dead ones :)
Registered Member #72
Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659
I missed the '50kV for 1 minute' bit, well spotted.
The fact that it's time limited presumably means it would be drawing heavy magnetising current, and so dissipating hard, and well into the saturation knee.
It's 50kV at 60Hz, so that would be only 41.7kV at 50Hz. Where is redruM69?
Registered Member #135
Joined: Sat Feb 11 2006, 12:06AM
Location: Anywhere is fine
Posts: 1735
Guys, this is a STEP DOWN transformer.
50 KV for one minute is the input surge that the transformer will handle for one minute without going tits up.
21kv is its continuous input.
Just because it can tolerate 50kv into its primary for short duration, does not mean it can handle 240V into it's secondary.
If I have to, I will pull my PT out and measure the primary, it's probably .5 ohms.
It's impedance might be 2 ohms, I can measure that with my impedance meter, it doesn't matter to me.
The impedance must, MUST be very low, these are designed as small power sources for metering and instruments on the pole.
If I have to I can do a full workup on my transformer, short circuit and open circuit parameters, IF I can find the time, it is time consuming.
Honestly you guys, you should realize that a low impedance secondary (which should look like 0 ohms as a source ideally) to any load presented to it, is basically a short circuit.
Let's guess a 240V bulb is 250W, or say 1 ohm. (approximations, I will get something wrong, but this is not life or death so I don't really care, this is just a pissing contest at this point.)
the transformer probably as a .85 power factor, so P = IV * .85, 2000VA = I * 120 * .85, so I = 20A roughly
Now I = E/Z secondary, or Z secondary = E/I = 120 / 20 = 6 ohms (so I thought it was going to be much much lower than that)
The 240V lamp, on the 240V line will restrict current to 1 Amp drawn (approximations)
So with a primary current of 1 amp, and impedance of 6 ohms, E = IZ = 6 VAC on the secondary terminals ( I wagered 24 V, but these are just "napkin" calculations)
21000:120 is 175:1, so 6VAC * 175 = 1050 VAC on the primary side terminals.
I am curious, let's get it measured and see what we get.
Registered Member #72
Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659
The first transformer I wound, as a callow schoolboy, who knew just enough to get himself into trouble but not much more, used resistance wire as the primary, because 'I didn't want too much current to flow through the near zero resistance of copper wire'. Would you believe it, it was an abject failure.
A real transformer will have a very low resistance to its windings. An ideal transformer would have zero resistance, and would work slightly better than a real one.
Any transformer winding is an inductor. Inductors have a finite impedance at the operating frequency of the transformer. In a well designed transformer, operated within its ratings, that impedance is large enough to limit the no load current to a small fraction of the rated current. This small current is called the magnetising current, as it's what's required to generate the H field that pushes a B field through the core, whose changes generate the back EMF by Faraday's Law that opposes the input voltage.
What happens if you operate a transformer at too high a voltage? The B field must still change fast enough to oppose the input voltage. However, now it's got to swing to a value higher than the saturation level, which means the permeability falls, which means an H field one or more orders of magnitude is now required to create the B field. An H field orders of magnitude greater means an input current orders of magnitude greater. That increase of magnetising current is why you shouldn't operate a transformer in saturation.
With a turns ratio of 175:1, that 50kV 1 minute input becomes nominally 285v at the secondary. Which appears to allow some scope for driving the secondary at 240v, just from the ratings plate. I would still feel more comfortable measuring the magnetising current at voltages between 120v and 240v inclusive.
Hazmat, you state that the stepdown output must be very low impedance, because it's designed to drive stuff. That is true, it is. However, it only is that when the primary is being driven from a low impedance. If we look what happens when we drive the secondary with 120v and the primary is open circuit, it will see a high impedance because it's only 'seeing' the secondary inductance. Now, if you were to short circuit the 21kV winding, this short *would* be reflected back through the transformer as a very low impedance, that's what transformers do.
Registered Member #11591
Joined: Wed Mar 20 2013, 08:20PM
Location: UK
Posts: 556
Just to expand on Sulaiman's rather inflammatory post, where you went wrong is the 6 ohms. Yes the impedance of the secondary would be 6 ohms if 20 A was the short circuit current, but it isn't. 20 A is the maximum it can supply, the short circuit current will be much greater than that. 6 ohms is the load you can put on the the secondary when it is working at 120 V.
A transformer like this can be modelled with an inductance in series with the primary (leakage inductance) and an inductance in parallel with the primary (magnetising inductance). When there is no load on the transformer the current flowing will only be the (low) reactive current through the (high) magnetising inductance. If the secondary is shorted out the only current limiting will be leakage inductance which will be much lower, therefore the current will be much higher (I = V/(j*2*pi*f*L)).
There is also a series resistance which represents the resistance of the windings. This also acts to limit the maximum current.
If the transformer primary now becomes the secondary and vice versa, the transformer can be drawn with the leakage and magnetising inductance on the secondary instead, but for convention we transform them through the transformer and put them with the appropriate values on the new primary.
In short, a transformer works the same forwards as well as backwards.
Registered Member #135
Joined: Sat Feb 11 2006, 12:06AM
Location: Anywhere is fine
Posts: 1735
They work the same forward and backward ONLY when the windings are equally distributed, which is called interleaving.
If these transformers were interleaved, the reluctance plays a lesser role and yes they would be easier to energize. But the windings on these guys is primary and secondary, which makes energizing them "backward" difficult and current hungry, again a short circuit.
I had a 3180 to 120 interleaved transformer, in pieces now to make use of its core someday, if you need proof I still have one of its 2 winding cores.
It will be some time before I can do the measurements, I have all the stuff, but I have to jerry rig the power cord and variac, and 1256D's are heavy.
I gave an approximation, that's what was meant and explained.
MY PT will likely need 2000VA input in step up to deliver the same power at the HT terminals, as it will require 1500VA in step down to deliver at the LT terminals, so no it's not the same, but I will measure it.
What's your approximation then?
Assume the secondary inductance is 8H, since that's the puzzle piece you need.
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