Per-Unit System and Instrument Transformers

Write down what a per-unit reactance actually says. "0.20 pu on 25 MVA, 11 kV" means the reactance is 0.20 of the impedance that would draw rated current at rated voltage:

The line-voltage form \(V_L^2/S_{3\phi}\) is used throughout; it already contains the \(\sqrt3\) factors and gives the per-phase star-equivalent ohms directly.

Change the MVA base only. Base impedance falls as \(S_{\text{base}}\) rises, so the same ohms become a larger per-unit number:

Nothing physical has changed. The machine is 20% impedant against its own rating and 80% impedant against a rating four times larger.

Change the voltage base as well. A system whose nominal voltage is taken as 10.5 kV gives

Check the ohms from the new base, which is the only reliable way to be sure a base change has been done correctly:

The same 0.968 Ω as before. Per-unit values move with the base; ohms do not.

The transformer, whose 12% is on 50 MVA:

No voltage correction is needed provided the base voltages on the two sides are in the transformer's own ratio, 11 : 132. If the 11 kV zone is on an 11 kV base, the 132 kV zone must be on a 132 kV base — and then 0.24 pu is the answer looking in from either side.

Confirm that claim in ohms, since it is the whole point of the method:

The ohms differ by \((132/11)^2 = 144\), and so do the base impedances, so the ratio survives. This is why a per-unit single-line diagram carries no transformers at all — only impedances in series.

The bases come straight from the nameplate, with \(S_{\text{base}} = 100\) kVA common to both sides:

Note that the short-circuit test was run at 45.45 A — exactly rated HV current. That is what makes the shortcuts in the next step legitimate.

Work the short-circuit test in ohms first, referred to the HV side where it was made:

Divide by the base impedance:

The same three numbers without touching an ohm. Because the test was at rated current, the applied voltage is the impedance drop and the input power is the full-load copper loss:

The second is worth dwelling on: per-unit resistance and per-unit copper loss are the same number, because \(P_{cu}/S = I^2R/(VI) = R/(V/I) = R_{\text{pu}}\) at rated current. The percentage impedance of Sets 19 to 21 is \(100Z_{\text{pu}}\), here 4.18%.

Refer the impedance to the low-voltage side and repeat. With \(a = 2200/220 = 10\):

Referring divided the ohms by 100 and the base impedance is also 100 times smaller, so the quotient is untouched. A per-unit impedance has no side.

The open-circuit test in per unit, using the LV bases:

1.1% exciting current and 0.8% iron loss — both immediately comparable with any other transformer of any rating, which is exactly what per unit is for.

Regulation becomes a one-line calculation. At full load the per-unit current is 1.0, so the approximate drop is just \(R\cos\phi \pm X\sin\phi\):

No voltages, no currents, no choice of side — and the answer is a percentage on any base.

Efficiency, likewise. Output at full load and 0.8 power factor is 0.8 pu; the losses are the two per-unit numbers already found:

Mark out the zones. A transformer separates two zones, and the base voltage changes across it in the ratio of its nameplate voltages while \(S_{\text{base}} = 25\) MVA holds everywhere:

The base voltages 11, 132 and 33 kV are in exactly the transformers' own ratios, which is the condition for the transformers to disappear from the per-unit diagram.

Convert each reactance to the 25 MVA base. The generator is already there; T1 is already there; T2 and the line are not:

The line reactance is divided by the base impedance of its own zone — 696.96 Ω, not 4.84 Ω. Using the wrong zone's base is the single commonest error in this calculation.

The elements are now in series and simply add:

The fault. With 1.0 pu voltage behind the reactance and the fault a dead short:

The short-circuit level is the base MVA divided by the per-unit reactance — a number a switchgear engineer can use directly, with no reference to voltage level at all.

Convert to amperes in each zone, using that zone's base current:

One per-unit number, three currents. As a check, 886 A at 33 kV should appear as \(886\times33/132 = 222\) A on the 132 kV side, and it does.

What the calculation would have looked like in ohms. Every impedance would have to be referred to one voltage level: the generator's \(0.15\times4.84 = 0.726\ \Omega\) becomes \(0.726\times144 = 104.5\ \Omega\) at 132 kV, T1's \(0.10\times696.96 = 69.70\ \Omega\), the line's 100 Ω, and T2's \(0.10\times696.96 = 69.70\ \Omega\) — total 343.9 Ω, giving \(76210/343.9 = 222\) A at 132 kV. The same answer, after four referrals that per unit did for free.

The leads are part of the burden, and the current travels out and back, so the length that counts is twice the run:

Convert that resistance into volt-amperes at rated secondary current, which is how CT burdens are always quoted:

The 5 A squared is what makes CT leads so expensive. The same run on a 1 A CT would impose \(1^2\times0.550 = 0.55\) VA, twenty-five times less — which is the entire reason 1 A secondaries are used where cable runs are long.

Add the instruments:

The CT is over-burdened by 16%. Beyond its rated burden a CT's exciting current rises sharply, its ratio and phase errors leave the accuracy class, and under fault conditions the core saturates and the secondary current collapses just when the protection needs it.

Work backwards for the conductor. The instruments are fixed, so the leads may take only

Choose the next standard size, 4 mm², and confirm:

4 mm² leaves 18% margin, which is sensible practice because a wattmeter may later be replaced by a transducer of higher burden.

The same burden as an impedance, which is the form Problem 5 needs:

Taking all three burdens as resistive lets them be added arithmetically. If the instruments were quoted at a power factor, the impedances would have to be added as phasors and the total would be a little smaller.

Where the error comes from. The primary ampere-turns are used for two things: driving the secondary current and magnetising the core. Only the first reaches the meter, so

Resolving \(\mathbf{I}_0\) along and across the reversed secondary current, and keeping only the component that lies along \(\mathbf{I}_p\), gives the actual ratio; the component across it gives the phase angle.

The actual ratio. With the burden at \(\delta = 30^\circ\) and the secondary current at its nominal 5 A:

The turns ratio is 100; the current ratio is 100.51. The extra half is the exciting current, which the primary must supply and the secondary never sees.

Ratio error and the actual secondary current:

An ammeter scaled 0–500 A would read \(100\times4.975 = 497.5\) A when 500 A is flowing. A CT always reads low, because the error term is always added to \(n\).

The phase angle, in radians and then in the minutes that instrument specifications use:

Positive, meaning the reversed secondary current leads the primary current. For an ammeter this is irrelevant; for the current coil of a wattmeter or a directional relay it is the error that matters, since 13.7 minutes of angle at a power factor of 0.1 shifts the reading by about 4%.

Now a purely resistive burden, \(\delta = 0\):

The burden's power factor trades one error against the other: making the burden more inductive worsens the ratio and improves the angle. This is why a CT's accuracy class is always quoted with a stated burden and burden power factor.

Turns compensation. Since the exciting current adds 0.508 to the ratio, removing 0.508 from the turns cancels it:

This is genuine practice — CTs are routinely wound with fewer secondary turns than the nominal ratio suggests. On a 100-turn secondary a whole turn is 1% of the ratio, so the correction here is coarser than the error; on a 1000-turn winding it is precise. The compensation is also exact at only one burden and one current, which is why the class limits still apply everywhere else.

A PT is a transformer running at almost no load, so its error is nothing but its voltage regulation — the drop taken by the burden current in the winding impedance. Start with that current:

The in-phase drop is what shifts the ratio, exactly as in a regulation calculation:

The quadrature drop is what turns the secondary phasor:

The exciting current contributes as well, because it flows through the primary winding whether a burden is connected or not. Only the primary half of the impedance is involved, and the exciting current is nearly in quadrature with the voltage:

Small, but not zero — and it is the reason a PT has an error even with nothing connected to it.

Combine and form the actual ratio. The primary must exceed \(n\) times the secondary by the total drop:

Negative again, and for the same reason as the CT: the instrument reads low, because part of what the primary supplies is lost inside the transformer.

The secondary voltage at exactly 11000 V primary:

A voltmeter scaled \(\times100\) would read 10942 V — 58 V low on 11 kV.

Phase angle and the zero-burden case. Adding the two contributions:

The phase angle is comfortably inside the 20-minute limit, but the ratio error of −0.53% just breaks the ±0.5% limit. The transformer is a class 1.0 unit at this burden. Reducing the burden to about 45 VA, or improving its power factor, would bring it back into class 0.5 — which is why a PT's class is meaningless unless the burden is quoted with it.