By the end of this chapter you should be able to:
Explain why machine parameters are normalised.
Define the base quantities and say which two are chosen freely.
Convert voltage, current, impedance and power to per unit.
Prove that per-unit impedance is the same on either side of a transformer.
Convert an impedance from one base to another.
Relate per-unit to percentage impedance.
Apply three-phase bases using line values.
Compute a fault level from per-unit reactances.
Why Normalise?
Transformers of various sizes, ratings and voltage ratios are used throughout a power system.
The parameters of the equivalent circuits of these machines vary over a very large range.
Comparison is made far simpler if all the parameters are normalised.
If simple scaling is done, the scaling factors must be carried forward through every calculation. Expressing on a percent basis is one example of such scaling.
If instead the scaling is done on a logical basis, a simple representation results without the bother of the scaling factors.
Different systems of units are also in use in different countries (FPS, CGS, MKS), and these undergo revision over the years. If the parameters can be freed from the units, the system becomes very simple.
The per-unit system is developed with these aspects in mind. The parameters are referred to some base values and thus get scaled. A common base is adopted in view of the different ratings of the equipment; for an individual machine, its own nominal parameters are used as the base.
The Base Quantities
The base quantities are ordinary electrical quantities, related to one another by the ordinary circuit laws.
The four base quantities \(V_{base}\), \(I_{base}\), \(S_{base}\) and \(Z_{base}\) are linked by two equations, so only two may be chosen independently.
In practice the two chosen are \(S_{base}\) and \(V_{base}\), because those are what a nameplate gives. The others follow:
Normally \(S_{base}\) and \(V_{base}\) are known from nameplate details, and the other base values are derived from them.
The second form of \(Z_{base}\) is the one used in practice, since it needs only the two chosen quantities. Note that it is quadratic in voltage — the source of most base-conversion errors.
Per-Unit Values
Every quantity is then divided by its own base.
A per-unit value is a pure number — volts divided by volts, ohms divided by ohms. The units have gone, and with them the question of which system of units was used.
This is the point made in Section 51-1 about FPS, CGS and MKS. It also means that per-unit values may be multiplied and divided freely without unit bookkeeping, which is why the ordinary circuit laws hold in per unit exactly as they do in ohms and amperes.
Problem. For the 50 kVA, 2200/220 V transformer used throughout Chapters 41 to 46, whose equivalent impedance referred to the primary is \(R_{01} = 1.359~\Omega\) and \(X_{01} = 4.599~\Omega\), find the base quantities on each side and express the impedance in per unit.
Take the machine's own rating as the base: \(S_{base} = 50\) kVA.
| HV side | LV side | |
|---|---|---|
| \(V_{base}\) | 2200 V | 220 V |
| \(I_{base} = S_{base}/V_{base}\) | 22.727 A | 227.27 A |
| \(Z_{base} = V_{base}^{2}/S_{base}\) | 96.80 Ω | 0.9680 Ω |
Sample working, the HV side.
The impedance in per unit.
Comment. The per-unit impedance is 0.0495, or 4.95 % — exactly the percentage impedance measured by the short-circuit test in Chapter 46. The two are the same quantity under different names, as Section 51-7 makes explicit.
Note the ratio of the two base impedances: \(96.80/0.9680 = 100 = a^{2}\). The base impedance scales exactly as the referring factor does, which is the whole reason the next section's result holds.
Choosing the Bases
Two rules govern the choice, and the second is the one that makes the system work.
- One \(S_{base}\) for the whole system. The volt-ampere base is chosen once and used everywhere — commonly 100 MVA for a transmission study, or the rating of the largest machine.
- A different \(V_{base}\) in each voltage zone, related by the transformer turns ratios. If the base is 11 kV on the generator side of an 11/132 kV transformer, it must be 132 kV on the other side.
A network divides into zones separated by transformers. Within a zone the voltage base is constant; crossing a transformer, it changes in the ratio of the turns.
Choose the bases this way and every transformer becomes a plain series impedance with no ratio at all. Choose them any other way and the ratios reappear, and the method's principal advantage is lost.
Note that the base voltages are set by the transformer nameplate ratios, not by the actual operating voltages, and not by the tap in use. A base is a reference, not a measurement.
Per-Unit Impedance Is Side-Independent
This is the central result of the chapter, and the reason the per-unit system is used at all.
If all the equivalent-circuit parameters are referred to the secondary side, and the per-unit values are computed with the secondary voltage and current as the bases, there is no change in the per-unit values.
Proof. Let \(a = N_1/N_2\). Referring an impedance from the primary to the secondary divides it by \(a^{2}\):
The base voltage on the secondary side is smaller by \(a\), while \(S_{base}\) is unchanged:
Therefore
Problem. Verify the theorem for the transformer of Example 51.1 by computing the per-unit impedance from the secondary side.
Refer the impedance to the secondary. With \(a = 2200/220 = 10\):
Divide by the secondary base impedance.
Compare with the primary side.
Comment. Identical, as the proof requires. The ohmic values differ by a factor of 100, but the per-unit values are the same number.
The practical consequence is worth stating plainly. In ohms, one must constantly ask "referred to which side?" — the question that generated Chapter 44's \(R_{01}, X_{01}\) and \(R_{02}, X_{02}\) pairs, and every opportunity for error that came with them. In per unit the question does not arise.
It follows too that a transformer nameplate need quote only one impedance figure, with no side attached — which is exactly what nameplates do.
Changing Base
Many times, when more transformers are involved in a circuit, one is required to choose a common base value for all of them. The parameters of all the machines are then expressed on this common base, and the conversion naturally changes their per-unit values.
The rule follows from the definition. An impedance in ohms is fixed; only the base changes:
An impedance \(Z_{p.u.,old}\) on the old base of \(S_{base,old}\) and \(V_{base,old}\) is modified to the new base \(S_{base,new}\), \(V_{base,new}\) as above.
When the voltage base is unchanged — the usual case — this reduces to a simple ratio of the kVA bases.
The voltage term is squared and the power term is not. Forgetting the square is the commonest error in per-unit work, and it is worth checking dimensionally each time: \(Z_{base} = V^{2}/S\), so the correction must carry \(V^{2}\).
Problem. The two transformers of Chapter 49 — A of 200 kVA at 4 % and B of 300 kVA at 5 %, at the same voltage — are to be placed on a common 500 kVA base. Convert both, and confirm the load-sharing ratio.
Conversion. The voltage base is common, so only the kVA ratio applies:
Load sharing. From Chapter 49, the kVA divide inversely as the impedances:
Comment. The ratio 0.8333 is exactly what Chapter 49 obtained by converting to a 100 kVA base instead. The choice of common base is arbitrary and cancels out of any ratio — only consistency matters.
Notice too what the conversion reveals. On their own bases the impedances were 4 % and 5 %, which look similar; on a common base they are 10.0 % and 8.33 %, and the inequality that costs the bank capacity is now plainly visible. This is precisely the trap Chapter 49 named: equal on their own bases means unequal on a common one, in the ratio of the ratings.
Per-Unit and Percentage Impedance
The two are the same quantity, differing only by a factor of one hundred.
Defined in Chapter 43 as the fraction of rated voltage needed to circulate rated current on short circuit. Preferred on nameplates and in test reports.
The same number as a fraction of unity. Preferred in calculation, because per-unit quantities may be multiplied without factors of 100 accumulating.
The numerator is the voltage drop across the impedance at rated current, and the denominator is the rated voltage. That is exactly the definition of percentage impedance, expressed as a fraction rather than a percentage.
The same identity explains a fact used repeatedly in Part 3: since \(Z_{p.u.}\) is the fractional drop at rated current, the short-circuit current in per unit is simply \(1/Z_{p.u.}\). A machine of 5 % impedance passes 20 times rated current on a solid short circuit.
Three-Phase Bases
For three-phase work the bases are taken as the total three-phase volt-amperes and the line-to-line voltage. The convenient result is that the formulas keep their single-phase shape.
The current base carries a \(\sqrt3\), but the impedance base does not — \(Z_{base} = V_{base}^{2}/S_{base}\) is identical in form to the single-phase case.
So is the star or delta connection relevant? No. With line voltage and three-phase VA as bases, the per-unit impedance of a balanced three-phase bank is the same whether it is connected in star or in delta — one more ratio that disappears.
This is why a three-phase transformer's percentage impedance is quoted without stating the connection, and why a per-unit single-line diagram needs no \(\sqrt3\) anywhere in it. The factor is absorbed into the bases once, at the start.
Problem. For the transformer of Example 51.1, with \(R = 0.014039\) pu and \(X = 0.047510\) pu, find the voltage regulation at full load at 0.8 lagging, unity, and 0.8 leading power factor, working entirely in per unit.
The approximate regulation formula of Chapter 45, written in per unit with \(I = 1.0\) pu at full load:
At 0.8 lagging (\(\cos\phi = 0.8\), \(\sin\phi = 0.6\), plus sign):
At unity.
At 0.8 leading (minus sign):
Comment. Chapter 45 obtained \(+3.975\,\%\), \(+1.405\,\%\) and \(-1.727\,\%\) for this machine by working in volts and amperes, referring everything to a chosen side and tracking a 227.27 A secondary current throughout.
Here the same three numbers appear from two per-unit constants and a power factor — no currents, no voltages, no referring, and no side to keep track of. The tiny differences in the third decimal come only from rounding the ohmic values.
This is the practical payoff of the whole chapter. Regulation in per unit is just \(R\cos\phi \pm X\sin\phi\), and since \(R\) and \(X\) in per unit are properties of the design rather than of the size, transformers of quite different ratings that share a design have identical regulation.
Advantages and Cautions
Turns ratios disappear, so a multi-voltage network becomes one ordinary circuit.
Per-unit impedances of like machines fall in a narrow, familiar range, so an erroneous value is obvious at sight.
The units vanish, and with them the choice of unit system.
Manufacturers' data are already in this form.
The \(\sqrt3\) of three-phase working is absorbed into the bases once.
Fault levels follow immediately as \(S_{base}/Z_{p.u.}\).
Every quantity must be on the same \(S_{base}\); mixing bases is the commonest error.
Voltage bases must follow the nameplate ratios, not the operating voltages or tap positions.
The base conversion is quadratic in voltage, linear in kVA.
Physical insight is dulled — a per-unit answer must be converted back before anything is ordered or built.
Where the transformer ratio does not match the base ratio, an ideal transformer of the residual ratio remains in the circuit.
| Quantity | Typical per unit |
|---|---|
| Transformer impedance, distribution | 0.03 to 0.06 |
| Transformer impedance, large power | 0.08 to 0.20 |
| Transformer resistance | 0.005 to 0.02 |
| Exciting current | 0.02 to 0.06 |
| Synchronous machine, transient reactance | 0.15 to 0.35 |
| Overhead line, per 100 km at 132 kV | 0.1 to 0.3 |
The narrowness of these ranges is itself the argument for the system. A transformer impedance of 0.8 pu or 0.0008 pu is wrong on sight, whereas 77 Ω or 0.077 Ω requires knowing the rating before it can be judged.
Problem. A generator of 25 MVA, 11 kV with \(X = 0.15\) pu feeds a transformer T1 of 30 MVA, 11/132 kV with \(X = 0.10\) pu, then a line of 100 Ω, then a transformer T2 of 25 MVA, 132/33 kV with \(X = 0.08\) pu. Using a 50 MVA base, find the total reactance and the three-phase fault level and current at the 33 kV bus.
Base voltages by zone: 11 kV at the generator, 132 kV on the line, 33 kV at the load — set by the transformer ratios.
Converting each element to the 50 MVA base.
For the line, the impedance is in ohms, so its base impedance is needed:
Total, in series.
Fault level.
Fault current at 33 kV.
Check, directly from the fault MVA.
Comment. Four elements at three different voltages, two of them transformers, and the calculation is a single addition. In ohms this would require referring the generator, the line and both transformers to one common voltage level, with two squared turns ratios applied and every opportunity for error.
Note which element dominates. The generator's 0.300 pu is a third of the total, and the line's 0.287 pu almost as much — the transformers together contribute only 0.327 pu. The per-unit form makes such comparisons immediate, whereas in ohms the generator's reactance and the line's would be at different voltage levels and could not be compared at all until referred.
Summary and Key Formulas
Machine parameters vary over a huge range; normalising them to chosen bases removes both the scaling factors and the units.
Of the four base quantities, only two are chosen freely — normally \(S_{base}\) and \(V_{base}\) from the nameplate. The rest follow.
\(Z_{base} = V_{base}^{2}/S_{base}\) — quadratic in voltage, which governs every base conversion.
One \(S_{base}\) for the whole system; a different \(V_{base}\) in each zone, related by the nameplate turns ratios.
Per-unit impedance is the same referred to either side, because impedance and base impedance are both referred by \(1/a^{2}\), which cancels. The turns ratio vanishes from the circuit.
Base change: \(Z_{new} = Z_{old}\,(V_{old}^{2}/S_{old})(S_{new}/V_{new}^{2})\); with the voltage base unchanged, simply the ratio of kVA bases.
\(\%Z = 100\,Z_{p.u.}\) — the same quantity. Short-circuit current is \(1/Z_{p.u.}\) in per unit.
Three-phase bases use total VA and line voltage; the \(\sqrt3\) appears in \(I_{base}\) but cancels from \(Z_{base}\), and the star or delta connection becomes irrelevant.
Regulation in per unit is simply \(R\cos\phi \pm X\sin\phi\).
Fault level is \(S_{base}/Z_{p.u.}\).
| Quantity | Formula | Notes |
|---|---|---|
| Base power | \(S_{base} = V_{base}I_{base}\) | single phase |
| Base current | \(I_{base} = S_{base}/V_{base}\) | ÷\(\sqrt3\,V_{base}\) if three-phase |
| Base impedance | \(Z_{base} = V_{base}^{2}/S_{base}\) | same form in three-phase |
| Per-unit voltage | \(V_{p.u.} = V/V_{base}\) | dimensionless |
| Per-unit impedance | \(Z_{p.u.} = Z\dfrac{S_{base}}{V_{base}^{2}}\) | — |
| Side-independence | \(Z'_{p.u.} = Z_{p.u.}\) | the \(1/a^{2}\) cancels |
| Base change | \(Z_{new} = Z_{old}\dfrac{V_{old}^{2}}{S_{old}}\dfrac{S_{new}}{V_{new}^{2}}\) | quadratic in \(V\) |
| Same voltage base | \(Z_{new} = Z_{old}\,S_{new}/S_{old}\) | the usual case |
| Percentage | \(\%Z = 100\,Z_{p.u.}\) | same quantity |
| Short-circuit current | \(I_{sc} = 1/Z_{p.u.}\) pu | 20× at 5 % |
| Regulation | \(R\cos\phi \pm X\sin\phi\) pu | + lagging, − leading |
| Fault level | \(S_{fault} = S_{base}/Z_{p.u.}\) | at the fault point |
Common Mistakes
Mixing bases. Every impedance in a calculation must be on the same \(S_{base}\). Nameplate values almost never are.
Forgetting the square on voltage in the base-conversion formula.
Using operating voltages as base voltages. Bases follow the nameplate ratios and never change with the tap or the load.
Choosing voltage bases that ignore the transformer ratios. The ratios then reappear as ideal transformers in the circuit.
Applying \(\sqrt3\) to the base impedance. It cancels; only \(I_{base}\) carries it.
Asking which side a per-unit impedance refers to. The question has no meaning — that is the point of Section 51-5.
Confusing per unit with percent by a factor of 100 in the middle of a calculation.
Worrying about star or delta when the bases are line voltage and three-phase VA.
Reporting a per-unit answer as a physical one. Convert back before quoting a current or a voltage.
Treating a per-unit value as a property of the machine alone. It is a property of the machine and its base.
Chapter Review
Begin every problem by fixing \(S_{base}\) for the whole system and \(V_{base}\) for each zone. Almost every per-unit error is a base error.
P51.1 A 100 kVA, 11000/400 V transformer has an equivalent impedance of 8.0 Ω referred to the HV side. Find its per-unit impedance on its own rating.
Show answer
\[Z_{base} = \frac{(11\,000)^{2}}{100\,000} = \frac{121\times10^{6}}{10^{5}} = 1210~\Omega\]\[Z_{p.u.} = \frac{8.0}{1210} = 0.006612~\mathrm{pu} = 0.661\,\%\]P51.2 Verify P51.1 from the LV side, given that the impedance referred to LV is 0.010578 Ω.
Show answer
\[Z_{base} = \frac{(400)^{2}}{100\,000} = \frac{160\,000}{100\,000} = 1.600~\Omega\]Identical, as the side-independence theorem requires.\[Z_{p.u.} = \frac{0.010578}{1.600} = 0.006611~\mathrm{pu} \quad\checkmark\]P51.3 A 15 MVA transformer has 9 % impedance on its own rating. Express it on a 100 MVA base at the same voltage.
Show answer
A large number, but perfectly correct — it is 9 % of a base six and two-thirds times smaller.\[Z = (0.09)\frac{100}{15} = 0.6000~\mathrm{pu} = 60.0\,\%\]P51.4 A generator has \(X = 0.20\) pu on 20 MVA, 11 kV. Find its per-unit reactance on a 50 MVA, 10.5 kV base.
Show answer
Both bases change, so the full formula is needed:\[X_{new} = (0.20)\frac{(11)^{2}}{20}\frac{50}{(10.5)^{2}} = (0.20)\frac{121}{20}\frac{50}{110.25}\]Note the voltage terms are squared and the power terms are not.\[= (0.20)(6.050)(0.45351) = 0.5487~\mathrm{pu}\]P51.5 A three-phase system has \(S_{base} = 100\) MVA and \(V_{base} = 33\) kV. Find \(I_{base}\) and \(Z_{base}\).
Show answer
\[I_{base} = \frac{100\times10^{6}}{\sqrt3\,(33\,000)} = 1749.5~\mathrm{A}\]The \(\sqrt3\) appears in the current base only.\[Z_{base} = \frac{(33\,000)^{2}}{100\times10^{6}} = \frac{1.089\times10^{9}}{10^{8}} = 10.89~\Omega\]P51.6 A transformer of 5 % impedance supplies a solid short circuit at its terminals. Find the short-circuit current in per unit and as a multiple of rated.
Show answer
Twenty times rated current — which is why the short-circuit test of Chapter 46 is conducted at 5 % of rated voltage.\[I_{sc} = \frac{1}{Z_{p.u.}} = \frac{1}{0.05} = 20~\mathrm{pu}\]P51.7 Prove that the per-unit impedance of a transformer is the same referred to either side.
Show answer
Let \(a = N_1/N_2\). Referring an impedance from primary to secondary divides it by \(a^{2}\):The base voltage on the secondary side is smaller by \(a\), while \(S_{base}\) is common to both sides:\[Z'_{ohm} = \frac{1}{a^{2}}Z_{ohm}\]Dividing,\[Z'_{base} = \frac{(V_{base}/a)^{2}}{S_{base}} = \frac{1}{a^{2}}\frac{V_{base}^{2}}{S_{base}} = \frac{1}{a^{2}}Z_{base}\]\[Z'_{p.u.} = \frac{(1/a^{2})Z_{ohm}}{(1/a^{2})Z_{base}} = \frac{Z_{ohm}}{Z_{base}} = Z_{p.u.}\]The reason in words: the impedance and its base are referred by the same factor \(1/a^{2}\), so their ratio is unaltered. The turns ratio therefore disappears from the per-unit equivalent circuit, and a transformer reduces to a plain series impedance.
This holds only because \(S_{base}\) is common to the two sides and the voltage bases are in the turns ratio — which is precisely why the bases must be chosen that way.
P51.8 Explain why the per-unit system is preferred to working in ohms for power-system calculations.
Show answer
Transformers vanish. With bases in the turns ratio, every transformer becomes a series impedance. A network at five voltage levels becomes one ordinary series-parallel circuit.
Values fall in a narrow range. Transformer impedances cluster around 0.05 to 0.2 pu whatever the rating, so an erroneous figure is obvious. In ohms, 77 Ω cannot be judged without knowing the rating and voltage.
Units disappear, and with them the system of units used.
Manufacturers already publish in this form, as percentage impedance.
The \(\sqrt3\) is absorbed once into the bases, and star or delta ceases to matter.
Fault levels follow immediately as \(S_{base}/Z_{p.u.}\).
The cost is a loss of physical feel, and the constant obligation to keep every quantity on one base. Answers must be converted back to amperes and volts before anything is specified.
P51.9 A 20 MVA, 33/11 kV transformer of 10 % impedance feeds a fault on its 11 kV side. Using a 20 MVA base, find the fault MVA and fault current.
Show answer
On its own rating \(Z = 0.10\) pu, so\[S_{fault} = \frac{20}{0.10} = 200~\mathrm{MVA}\]\[I_{base} = \frac{20\times10^{6}}{\sqrt3\,(11\,000)} = 1049.7~\mathrm{A}\]This assumes an infinite source behind the transformer; a real source impedance would reduce it.\[I_{fault} = \frac{1049.7}{0.10} = 10\,497~\mathrm{A}\]P51.10 Two transformers, 1000 kVA at 5 % and 2000 kVA at 6 %, are paralleled. Put both on a 2000 kVA base and find the load sharing for a 2500 kVA load.
Show answer
\[Z_A = (0.05)\frac{2000}{1000} = 0.1000~\mathrm{pu}, \qquad Z_B = (0.06)\frac{2000}{2000} = 0.0600~\mathrm{pu}\]\[S_A = (2500)\frac{0.0600}{0.1600} = 937.5~\mathrm{kVA}, \qquad S_B = (2500)\frac{0.1000}{0.1600} = 1562.5~\mathrm{kVA}\]Loadings: \(937.5/1000 = 93.75\,\%\) and \(1562.5/2000 = 78.13\,\%\).
A overloads first, at a total load of \(1000/0.375 = 2667\) kVA, so the 3000 kVA bank is limited to 2667 kVA — 11.1 % of capacity stranded, exactly the effect of Chapter 49.
MCQ 1. How many base quantities may be chosen independently?
(a) one (b) two (c) three (d) fourShow answer
(b) two — normally \(S_{base}\) and \(V_{base}\); the rest follow.MCQ 2. The base impedance equals:
(a) \(V_{base}S_{base}\) (b) \(V_{base}^{2}/S_{base}\) (c) \(S_{base}/V_{base}^{2}\) (d) \(V_{base}/S_{base}\)Show answer
(b) \(V_{base}^{2}/S_{base}\).MCQ 3. The per-unit impedance of a transformer referred to the primary compared with the secondary is:
(a) \(a^{2}\) times (b) \(1/a^{2}\) times (c) the same (d) \(a\) timesShow answer
(c) the same — the \(1/a^{2}\) cancels between impedance and base.MCQ 4. In the base-conversion formula, voltage appears:
(a) linearly (b) squared (c) cubed (d) not at allShow answer
(b) squared, since \(Z_{base} = V^{2}/S\).MCQ 5. Percentage impedance equals:
(a) \(Z_{p.u.}\) (b) \(10\,Z_{p.u.}\) (c) \(100\,Z_{p.u.}\) (d) \(Z_{p.u.}/100\)Show answer
(c) \(100\,Z_{p.u.}\) — the same quantity.MCQ 6. In three-phase per-unit work, the \(\sqrt3\) appears in:
(a) \(Z_{base}\) only (b) \(I_{base}\) only (c) both (d) neitherShow answer
(b) \(I_{base}\) only — it cancels from the impedance base.MCQ 7. A machine of 4 % impedance passes a short-circuit current of:
(a) 4 pu (b) 0.04 pu (c) 25 pu (d) 400 puShow answer
(c) 25 pu, since \(1/0.04 = 25\).MCQ 8. Base voltages in different zones should be related by:
(a) the operating voltages (b) the transformer turns ratios (c) the tap positions (d) the load currentsShow answer
(b) the transformer turns ratios, or the ratios reappear in the circuit.MCQ 9. A typical per-unit impedance for a distribution transformer is about:
(a) 0.0005 (b) 0.05 (c) 0.5 (d) 5.0Show answer
(b) 0.05, that is 5 %.MCQ 10. The fault level at a point is:
(a) \(S_{base}Z_{p.u.}\) (b) \(S_{base}/Z_{p.u.}\) (c) \(Z_{p.u.}/S_{base}\) (d) \(S_{base} - Z_{p.u.}\)Show answer
(b) \(S_{base}/Z_{p.u.}\).
Explain why machine parameters are normalised, and what the per-unit system adds over simple scaling.
Define the base quantities and explain why only two may be chosen independently.
State the rules for choosing bases across a multi-voltage network.
Prove that per-unit impedance is the same referred to either side of a transformer.
Derive the base-conversion formula and explain why voltage enters squared.
Show that per-unit and percentage impedance are the same quantity.
Explain why the \(\sqrt3\) cancels from the three-phase base impedance.
Give the advantages and the cautions of the per-unit system.
Part 3 has treated the transformer as a single-phase device throughout. Chapter 52 turns to the three-phase transformer — its construction as three separate single-phase units or as a single three-limb core, and why the three-limb core works at all when the three fluxes must sum to zero.
Chapter 53 then takes up the winding connections: star, delta and zigzag, in the four useful combinations. Each has a proper place — delta on the side that must carry the third-harmonic magnetising current of Chapter 42, star where a neutral is needed for a four-wire supply, zigzag where earth-fault current must be provided. The vector groups that Chapter 49's fourth paralleling condition referred to are simply a notation for the phase shift each combination introduces, and two transformers of different groups cannot be paralleled however well their ratios and impedances match.
Part 3 closes with the open-delta connection, in which two single-phase transformers do the work of three at 57.7 % of the capacity — the last of the transformer's practical surprises before Part 4 begins on induction motors.