Electric Power Devices study guide

What this area covers

Electric Power Devices centres on the transformer, with a growing share for the equipment that now sits beside it: battery energy storage, photovoltaic plants and wind turbines. You need to model a transformer from its test data and nameplate, predict how it behaves under load and in parallel, and read the ratings of inverter-connected sources well enough to size and compare them. This area accounts for 8–12 questions on the exam.

The ideas everything else rests on

The equivalent circuit comes from two tests. A transformer is modelled as an ideal transformer plus a series branch (winding resistance and leakage reactance) and a shunt branch (core loss and magnetising reactance). The open-circuit test, at rated voltage with one winding open, gives the shunt branch. The short-circuit test, at reduced voltage with one winding shorted and rated current flowing, gives the series branch. Know which side each test is usually run on and refer values to one side before you combine them.

Percent impedance is a short-circuit test result in disguise. It is the share of rated voltage that drives rated current through a shorted transformer. It sets fault current, voltage regulation and how parallel units share load. On the transformer's own base, percent impedance divided by 100 is the per-unit impedance, ready to drop into a system diagram after a change of base. Parallel units first need the same voltage ratio, phase shift and polarity, or circulating current flows even with no load. Once those match, load divides in inverse proportion to impedance on a common base, so the unit with the lower percent impedance on its own rating reaches full load first.

An autotransformer carries most of its power by conduction. Only the part of the power that crosses the series winding is transformed magnetically. That is why an autotransformer of a given winding rating can pass much more power when the two voltages are close:

$$S_{auto} = S_{w} \frac{V_H}{V_H - V_L}$$

The trade-off is no electrical isolation and a lower impedance, which raises fault current.

Efficiency peaks where the losses match. Core loss is roughly constant whenever the unit is energised; copper loss grows with the square of load. Efficiency is

$$\eta = \frac{P_{out}}{P_{out} + P_{core} + P_{cu}}$$

and it is highest at the load where copper loss equals core loss. All-day efficiency uses energy over 24 hours instead of power at one instant, which rewards low core loss for lightly loaded distribution units.

Storage and PV have two ratings each. A battery system has a power rating and an energy rating; their ratio is its duration, and round-trip losses mean you get back less than you put in. A PV plant has a DC array rating at standard test conditions and an AC inverter rating; when the array is larger than the inverter, output clips at the inverter limit on bright days.

How to study it

1. Reduce open-circuit and short-circuit test data to an equivalent circuit, referred first to the high side and then to the low side.
2. Compute full-load current on both sides, and fault current from percent impedance, with the Transformer current calculator.
3. Move a nameplate impedance onto a system base with the Per-unit converter, then work load sharing between two parallel units with unequal impedances.
4. Work regulation and efficiency at several load levels and power factors; find the load for peak efficiency and compute an all-day figure from a simple load profile.
5. Practise three-phase bank connections (wye–wye, delta–wye, delta–delta) and their phase shifts, and check kVA and current with the Three-phase power calculator.
6. Finish with storage and PV ratings: duration, usable energy and inverter loading.

Mistakes that cost points

• Leaving values on the wrong side. Impedances from the two tests may be measured on different windings. Refer both to one side with the square of the turns ratio before adding them.
• Sharing load by kVA rating instead of impedance. Equal ratings do not mean equal shares if the per-unit impedances differ.
• Using the winding rating as the throughput of an autotransformer. The power it passes is larger than its winding rating; know which one the problem gives you.
• Scaling core loss with load. Only copper loss changes with the square of load. Core loss stays put as long as voltage does.

References worth having

• IEEE Std C57.12.00 (general requirements for liquid-immersed transformers)
• IEEE Std 1547 (interconnection of distributed energy resources)
• Chapman, Electric Machinery Fundamentals: the transformer chapter