Electric Arc Furnace Transformer

Looks like homework

Smells like homework.

Tastes like homework.

I'm glad I didn't step in that.

I can not perform the calculation in any way. We have the transformer impedance data at tap-1 and tap-9 from the manufacturer name plate. But can't verify the same.

Please advice me to solve the problem.

I am puzzled.

The data you give for tap 1 and 9 is, I think, the on-load data.

But the impedance is the "secondary short circuit" test data, the primary voltage (as a percent of rated, rated = 33kV) required to give secondary rated current when secondary is short-circuited.

There is no theoretical relation between rated amps and volts and impedance - for a given rated volts/amps a wide range of short-circuit impedance can be made in practice.

Do you have any other information, for example :-

- (1) open circuit secondary voltage on each tap? - this should be measurable if you have a real, working transfo.

- (2) power factor of the load for each of the volt/amp figures you give?

- (3) no-load and on-load losses?

Why do you need the impedance on taps 1, 9 ?

67model

The voltage values are no-load voltage as it is given in name plate of the transformer.

We know the no-load and load loss as 30 kW and 175kW respectively.

Is it possible to calculate impedances at various tap position considering the above data?

If the load loss is 175 kW, then this is 175/24000 = 0.0073 per unit (0.73%) of 24000 kW.

The 8% impedance has two components resistive εr and inductive εx at 90 degrees to each other. From geometry of right angle triangle for 8% impedance εz, εz² = εr² + εx²,

hence εx² = εx² - εr² = 8² - 0.73² = 63.47 : εx = 7.97%.

Regulation of a transformer loaded at rated VA is εr x cosΦ + εx x sinΦ.

Assuming resistive load [negligible inductance in load] cosΦ = 1, sinΦ = 0, hence regulation is 0.73 x 1 = 0.73% - which is negligible.

Tap 4 is 335 V, Tap 1 is 385V, hence impedance ratio is [385/335]² = 1.32. Hence impedance on tap 1 is 8%/1.32 = 6.1 % ( on 24000 kVA base).

Similarly, Tap 9 (250V) is 19% on 24000 kVA base.

I am assuming taps are actually on the HV winding and magnetic coupling is uniform, in the absence of any specific data.

Note that the tolerance on impedance is usually +/- 20% on nominal nameplate values.

The impedance is of relevance to the HV loading, as calculated above its effect on secondary with rated resistive load is negligible.

67model

Thank for your kind reply.

For furnace application power factor of the load at Tap-1 is nearly 0.15 and at Tap-9 it will be around 0.85. Also, the transformer is not constant flux i.e. constant MVA mode of operation from TAP-1 to TAP-9.

Constant MVA is from Tap-1 to Tap-4 i.e. 24MVA. From Tap-5 to Tap-9, it is constant current mode of operation i.e. 41411.4 Amp.

Your calculation for Tap-1 is almost as the name plate value but Tap-9 is far from the declared value of 12%.

Please advice on this matter.

It is necessary to be sure "eggs" are being compared with "eggs".

If the Tap 9 impedance is declared on an 18 MVA base rating as 12%, then on 24 MVA base (as I calculated it) it would become 12% x 24/18 = 16%.

Unfortunately, engineers can get so used to putting down figures in the way they habitually use - they forget to put in the base assumptions like the kVA base for an impedance. For example, whether value was mean, root-mean-square, peak or peak to peak [an omission often made when quoting vibration amplitudes].

Myself, if a transfo had multiple ratings, I would "spell it out", word by word, what base applied to each impedance, with a preference for giving all on the same base.

Usually, a necessary step in a system analysis is to get all the impedances numbered on the same base kVA, their relative effect is better shown.

The calculation I did assumed

a) all the impedance is in the secondary.

b) a perfect transformer with zero resistance and leakage reactance is between source and secondary impedance.

In reality, the reactance will be divided between primary and secondary (make them Lp and Ls, respectively). I sketch below the effect of assuming half being in primary and half in secondary, all on 24MVA base at primary voltage.

Of course, this does not include the difference in primary leakage reactance due to the different number of primary turns with change of tap (and resistance is ignored, as insignificant).

Regards,

67model

It is difficult to appreciate the short-circuit impedance of this transformer since the impedances are small

comparative with the entire circuit impedance.

In my opinion, the system is provided with a reactor at 33 kV side of 30-40 ohm and the arc resistance -depending on current level-could be of 3-4 mohm.

A very rough calculation get the impedances of 10%,8% and 4%.