Per-Unit Quantities — Solved Problems

Set 3 — Per-Unit Quantities

Twenty worked problems on the change of variable that makes power system analysis tractable. Express every quantity as a fraction of a base value chosen per voltage zone, and three things happen at once: transformer turns ratios disappear from the impedance diagram, the \(\sqrt{3}\) factors of Set 1 cancel identically, and every machine impedance falls into a narrow, memorable range regardless of its size. The price is one piece of discipline — the bases must be chosen consistently — and almost every error in this topic is a violation of it.

Textbook Chapter 4 · 20 solved · 12 practice · 3 challenge · 12 MCQs

i Method Recap

Two bases are chosen; the rest follow. Pick \(\text{MVA}_B\) — one value for the entire system — and \(\text{kV}_B\) for one zone. Everything else is derived, never chosen: \(Z_B = \text{kV}_B^2/\text{MVA}_B\) and \(I_B = \text{MVA}_B/(\sqrt{3}\,\text{kV}_B)\).
The base MVA is common to the whole network; the base kV changes at every transformer, in the ratio of its nameplate. A network with three voltage levels has three base voltages and one base MVA. Getting this backwards is the single most common structural error in the subject.
Base change: \(Z_{pu,\text{new}} = Z_{pu,\text{old}}\left(\dfrac{\text{MVA}_{B,\text{new}}}{\text{MVA}_{B,\text{old}}}\right)\left(\dfrac{\text{kV}_{B,\text{old}}}{\text{kV}_{B,\text{new}}}\right)^{2}\). Note the directions — MVA the right way up, kV inverted and squared. Manufacturers quote impedance on the equipment's own rating, so this conversion is needed for almost every component in a real study.
A transformer's per-unit impedance is the same referred to either winding, provided the base voltages on the two sides are in the transformer's own ratio. This is the whole point of the method: it is what removes the ideal transformer from the impedance diagram.
Percent impedance is per-unit \(\times\,100\). Nothing more. A "5% transformer" has \(Z_{pu} = 0.05\) on its own rating.
Short-circuit MVA \(= \text{MVA}_B/Z_{pu}\). The reciprocal relationship means a low per-unit impedance implies a high fault level — which is why transformer impedance is deliberately specified, not merely tolerated.
Three-phase quantities and per-phase quantities give the same base impedance. \(Z_B = \text{kV}_{LL}^2/\text{MVA}_{3\phi} = \text{kV}_{LN}^2/\text{MVA}_{1\phi}\) identically. This is why per-unit calculations never carry an explicit \(\sqrt{3}\) or a factor of 3 — they cancel by construction.

VideoWalkthrough

Problem 1Warm-upBase Quantities

A three-phase system is to be analysed on a base of 100 MVA and 132 kV. Find the base impedance, the base current and the base phase voltage, and verify that they are mutually consistent.

Solution

Base impedance, from the two chosen bases:

\[ Z_B = \frac{(\text{kV}_B)^2}{\text{MVA}_B} = \frac{(132)^2}{100} = 174.24\ \Omega \]

Base current, from three-phase apparent power:

\[ I_B = \frac{\text{MVA}_B \times 10^{6}}{\sqrt{3}\,\text{kV}_B \times 10^{3}} = \frac{100\times10^{6}}{\sqrt{3}(132\,000)} = 437.4\ \text{A} \]

Base phase voltage:

\[ V_{B,\text{ph}} = \frac{132}{\sqrt{3}} = 76.21\ \text{kV} \]

The consistency check — Ohm's law must hold among the bases themselves:

\[ \frac{V_{B,\text{ph}}}{I_B} = \frac{76\,210}{437.4} = 174.2\ \Omega = Z_B\ \checkmark \]

This is not a coincidence but a design requirement of the per-unit system. If the bases did not satisfy Ohm's law among themselves, then \(V_{pu} = I_{pu}Z_{pu}\) would not hold and the whole scheme would collapse.

Only two bases are ever chosen; the other two are consequences. Choosing \(Z_B\) or \(I_B\) independently is not a shortcut, it is an error — it breaks the Ohm's-law consistency and every subsequent per-unit result silently inherits the inconsistency. Choose MVA and kV, derive the rest, and never revisit the decision.

Answer\(Z_B = 174.24\,\Omega\), \(I_B = 437.4\) A, \(V_{B,\text{ph}} = 76.21\) kV

Problem 2Warm-upMachine p.u. Impedance

A three-phase generator rated 1000 kVA, 33 kV has an armature resistance of 20 \(\Omega\)/phase and a synchronous reactance of 70 \(\Omega\)/phase. Calculate the per-unit impedance of the generator.

Solution

With no other system specified, the machine's own ratings are the natural choice of base:

\[ \text{kV}_B = 33\ \text{kV},\qquad \text{MVA}_B = 1.0\ \text{MVA} \]

Base impedance:

\[ Z_B = \frac{(33)^2}{1.0} = 1089\ \Omega \]

Per-unit impedance is simply the ohmic value divided by the base:

\[ Z_{pu} = \frac{Z}{Z_B} = \frac{20 + j70}{1089} = 0.0184 + j0.0643\ \text{p.u.} \]

In magnitude and angle:

\[ Z_{pu} = 0.0669\angle 74.05^\circ\ \text{p.u.} \]

The numbers themselves carry a diagnosis. A synchronous reactance of 0.064 p.u. is implausibly small — real synchronous reactances run from 1.0 to 2.0 p.u., and even a sub-transient reactance is around 0.15. What the 70 \(\Omega\) and 33 kV really describe is a machine of far larger MVA than 1000 kVA. That instinct — that a per-unit value which falls outside the familiar band signals bad data rather than an unusual machine — is one of the most practically useful things the per-unit system provides.

Answer\(Z_{pu} = 0.0184 + j0.0643 = 0.0669\angle74.05^\circ\) p.u.

Problem 3Exam levelBase Change

A generator has a sub-transient reactance \(X'' = 0.25\) p.u. on its own rating of 25 MVA, 11 kV. Find \(X''\) on (a) a 100 MVA, 11 kV base and (b) a 100 MVA, 13.8 kV base.

Solution

The base-change formula follows directly from the fact that the ohmic impedance is a property of the machine and does not change:

\[ X_{pu,\text{new}} = X_{pu,\text{old}}\left(\frac{\text{MVA}_{B,\text{new}}}{\text{MVA}_{B,\text{old}}}\right)\left(\frac{\text{kV}_{B,\text{old}}}{\text{kV}_{B,\text{new}}}\right)^{2} \]

aThe voltage base is unchanged, so only the MVA ratio acts:

\[ X'' = 0.25\left(\frac{100}{25}\right)\left(\frac{11}{11}\right)^{2} = 0.25(4) = 1.00\ \text{p.u.} \]

bNow both ratios act, and the voltage ratio is inverted and squared:

\[ X'' = 0.25\left(\frac{100}{25}\right)\left(\frac{11}{13.8}\right)^{2} = 1.00 \times 0.6354 = 0.635\ \text{p.u.} \]

Sanity check via ohms, which should be invariant. On the machine's own base, \(Z_B = 11^2/25 = 4.84\ \Omega\), so \(X'' = 0.25(4.84) = 1.21\ \Omega\). On the 100 MVA, 13.8 kV base, \(Z_B = 13.8^2/100 = 1.9044\ \Omega\):

\[ X''_{pu} = \frac{1.21}{1.9044} = 0.635\ \text{p.u.}\ \checkmark \]

Case (b) is a warning, not an exercise. Changing the voltage base of a machine while keeping its physical identity means the machine is no longer being described at its rated voltage — legitimate only when the machine genuinely sits in a zone whose base voltage was set by some other transformer's ratio. If your base kV differs from a machine's rated kV, stop and check whether the base was chosen correctly, because more often than not it was not.

Answer(a) \(X'' = 1.00\) p.u. · (b) \(X'' = 0.635\) p.u.

Problem 4Exam levelTransformer p.u.

A three-phase \(\Delta\)–Y transformer rated 100 kVA, 11 kV/400 V has primary and secondary leakage reactances of 12 \(\Omega\)/phase and 0.05 \(\Omega\)/phase respectively. Calculate the per-unit reactance referred to each side and comment on the result.

Solution

The line-voltage ratio, which is what must be used for a three-phase bank irrespective of the winding connections:

\[ K = \frac{400}{11\,000} = 0.03636 \]

Case 1 — HV side as base. \(\text{kV}_B = 11\), \(\text{MVA}_B = 0.1\):

\[ Z_B = \frac{(11)^2}{0.1} = 1210\ \Omega \]

\[ X_{01} = X_1 + \frac{X_2}{K^2} = 12 + \frac{0.05}{(0.03636)^2} = 12 + 37.81 = 49.81\ \Omega/\text{phase} \]

\[ X_{pu} = \frac{49.81}{1210} = 0.0412\ \text{p.u.} \]

Case 2 — LV side as base. \(\text{kV}_B = 0.4\), \(\text{MVA}_B = 0.1\):

\[ Z_B = \frac{(0.4)^2}{0.1} = 1.6\ \Omega \]

\[ X_{02} = K^2X_1 + X_2 = (0.03636)^2(12) + 0.05 = 0.0159 + 0.05 = 0.0659\ \Omega/\text{phase} \]

\[ X_{pu} = \frac{0.0659}{1.6} = 0.0412\ \text{p.u.} \]

The two are identical. In ohms the reactance differs by a factor of 756; in per-unit it does not differ at all.

This identity is the entire justification for the per-unit system. Because the transformer's impedance is the same number on both sides, the ideal transformer that separates them contributes nothing and can be deleted from the impedance diagram — leaving a single connected network with no turns ratios in it. Everything from Set 16 onwards, every \(Y\)-bus and every fault study, depends on this deletion being legitimate. It is legitimate only when the base voltages on the two sides stand in the transformer's own ratio, which is why base selection is treated as a discipline rather than a preference.

Answer\(X_{pu} = 0.0412\) p.u. from either side — identical

Problem 5Warm-upPercent Impedance

A 500 kVA, 11 kV/415 V distribution transformer is marked "impedance 5%". Find its impedance in ohms referred to the HV side and to the LV side, and verify the relation between them.

Solution

Percent impedance is nothing more than per-unit impedance multiplied by 100, always on the equipment's own rating:

\[ Z_{pu} = \frac{5}{100} = 0.05\ \text{p.u. on }0.5\ \text{MVA} \]

HV side:

\[ Z_B = \frac{(11)^2}{0.5} = 242\ \Omega,\qquad Z = 0.05(242) = 12.10\ \Omega \]

LV side:

\[ Z_B = \frac{(0.415)^2}{0.5} = 0.3445\ \Omega,\qquad Z = 0.05(0.3445) = 0.01722\ \Omega \]

The two ohmic values must be related by the square of the turns ratio:

\[ \left(\frac{11\,000}{415}\right)^{2} = (26.51)^2 = 702.8, \qquad \frac{12.10}{0.01722} = 702.7\ \checkmark \]

The nameplate percentage means something physical: 5% is the fraction of rated voltage that must be applied to the primary, with the secondary short-circuited, to drive rated current. It is measured in exactly that way. It also states directly that a short circuit at the secondary terminals will draw \(1/0.05 = 20\) times rated current — which is why the same number that a designer treats as a regulation figure, a protection engineer treats as a fault-level figure.

Answer\(Z_{HV} = 12.10\,\Omega\), \(Z_{LV} = 0.01722\,\Omega\), ratio \(= 702.7 = (11\,000/415)^2\)

Problem 6Exam levelShort-Circuit MVA

The 500 kVA, 11 kV/415 V, 5% transformer of Problem 5 is supplied from a source of effectively infinite fault level. Find the short-circuit MVA and the symmetrical fault current at its LV terminals.

Solution

With an infinite source the pre-fault voltage stays at 1.0 p.u. and the only impedance in the loop is the transformer's own:

\[ I_{f,pu} = \frac{V_{pu}}{Z_{pu}} = \frac{1.0}{0.05} = 20\ \text{p.u.} \]

Short-circuit MVA follows immediately:

\[ \text{MVA}_{sc} = \frac{\text{MVA}_B}{Z_{pu}} = \frac{0.5}{0.05} = 10\ \text{MVA} \]

Base current on the LV side:

\[ I_B = \frac{500\times10^{3}}{\sqrt{3}(415)} = 695.6\ \text{A} \]

Hence the fault current:

\[ I_f = 20 \times 695.6 = 13\,912\ \text{A} \approx 13.9\ \text{kA} \]

\[ \text{Check: } I_f = \frac{\text{MVA}_{sc}\times10^{6}}{\sqrt{3}(415)} = \frac{10\times10^{6}}{718.8} = 13\,912\ \text{A}\ \checkmark \]

Transformer impedance is specified, not merely tolerated. A 2.5% transformer of the same rating would double the fault level to 27.8 kA and would demand switchgear of twice the breaking capacity — while a 10% unit would halve the fault current but double the voltage regulation. The impedance a purchaser specifies is therefore a direct trade between voltage quality on one side and fault duty and equipment cost on the other, and it is settled at the point of purchase rather than in the design office afterwards.

Answer\(\text{MVA}_{sc} = 10\) MVA, \(I_f = 13.9\) kA at 415 V

Problem 7Exam levelParallel Machines

Two generators operate in parallel on the same 11 kV bus:

\(G_1\): 50 MVA, 11 kV, \(X'' = 0.20\) p.u.
\(G_2\): 75 MVA, 11 kV, \(X'' = 0.15\) p.u.

Using a common base of 100 MVA, 11 kV, find each reactance on the common base, the equivalent reactance of the pair, and the fault MVA at the bus.

Solution

The two reactances are quoted on different bases and cannot be combined until they are on the same one. This is the step most often skipped, and skipping it is not a small error — here it would misstate the fault level by more than 30%.

\[ X_1 = 0.20\left(\frac{100}{50}\right) = 0.40\ \text{p.u.},\qquad X_2 = 0.15\left(\frac{100}{75}\right) = 0.20\ \text{p.u.} \]

The voltage bases are already equal at 11 kV, so only the MVA ratio acts.

The two machines are in parallel as seen from the bus:

\[ X_{eq} = \frac{X_1X_2}{X_1 + X_2} = \frac{(0.40)(0.20)}{0.60} = 0.1333\ \text{p.u.} \]

Fault MVA at the bus:

\[ \text{MVA}_{sc} = \frac{100}{0.1333} = 750\ \text{MVA} \]

Check by adding the individual contributions, each computed on its own rating:

\[ \frac{50}{0.20} + \frac{75}{0.15} = 250 + 500 = 750\ \text{MVA}\ \checkmark \]

Fault MVAs of parallel sources add; per-unit reactances do not. The check above is worth internalising as an independent route: because \(\text{MVA}_{sc} = \text{MVA}_B/X_{pu}\), combining reactances in parallel is arithmetically identical to adding fault levels in series. Utilities quote system strength as a fault level in MVA precisely because it is the additive quantity, and a network planner can total the contributions of several in-feeds without ever choosing a base.

Answer\(X_1 = 0.40\), \(X_2 = 0.20\), \(X_{eq} = 0.1333\) p.u., \(\text{MVA}_{sc} = 750\) MVA

Problem 8Challenge-liteBase Selection

For the single-line diagram shown, take a base of 50 MVA and 11 kV in the generator zone. Express every series reactance in per-unit on that base.

\(G\): 25 MVA, 11 kV, \(X = 0.20\) p.u.
\(T_1\): 30 MVA, 11/132 kV, \(X = 0.10\) p.u.
Line: \(100\ \Omega\) reactance at 132 kV
\(T_2\): 25 MVA, 132/33 kV, \(X = 0.08\) p.u.

Three voltage zones, one base MVA, three base voltages

Solution

Step 1 — mark the zones. Transformers divide the system into three zones. The base MVA of 50 applies to all of them; the base voltage changes at each transformer in that transformer's nameplate ratio:

\[ \text{Zone 1: } 11\ \text{kV},\qquad \text{Zone 2: } 11\times\frac{132}{11} = 132\ \text{kV},\qquad \text{Zone 3: } 132\times\frac{33}{132} = 33\ \text{kV} \]

The base voltages happen to coincide with the nameplate voltages here because the transformer ratios are consistent. When they are not — a 11/130 kV transformer feeding a 132 kV line, say — the base follows the transformer ratio, not the line's nominal voltage.

Step 2 — generator. Same voltage base, different MVA base:

\[ X_G = 0.20\left(\frac{50}{25}\right)\left(\frac{11}{11}\right)^{2} = 0.400\ \text{p.u.} \]

Step 3 — transformer \(T_1\). Its rated voltages match the zone bases on both sides, so only the MVA ratio acts:

\[ X_{T1} = 0.10\left(\frac{50}{30}\right) = 0.1667\ \text{p.u.} \]

Step 4 — the line. Its ohms must be divided by the base impedance of zone 2:

\[ Z_{B2} = \frac{(132)^2}{50} = 348.48\ \Omega, \qquad X_{\text{line}} = \frac{100}{348.48} = 0.287\ \text{p.u.} \]

Step 5 — transformer \(T_2\):

\[ X_{T2} = 0.08\left(\frac{50}{25}\right) = 0.160\ \text{p.u.} \]

Step 6 — the series total. Because every ideal transformer has vanished, these four reactances are simply in series:

\[ X_{\text{total}} = 0.400 + 0.1667 + 0.287 + 0.160 = 1.014\ \text{p.u.} \]

Note what did not have to be done. No impedance was referred through a turns ratio, no \(\sqrt3\) appeared, and the four elements — a machine, two transformers and a line, spanning three voltage levels — added like four resistors in a first-year circuit. That is the entire return on the discipline of base selection, and it is why the method survives into every commercial power-system package in use today.

Answer\(X_G = 0.400\), \(X_{T1} = 0.167\), \(X_{\text{line}} = 0.287\), \(X_{T2} = 0.160\); total \(= 1.014\) p.u.

Problem 9Warm-upZone Base Currents

For the system of Problem 8, find the base current in each of the three zones and verify that the ratios match the transformer ratios.

Solution

Base current uses the same base MVA everywhere and the zone's own base voltage:

\[ I_{B1} = \frac{50\times10^{6}}{\sqrt{3}(11\,000)} = 2624\ \text{A} \]

\[ I_{B2} = \frac{50\times10^{6}}{\sqrt{3}(132\,000)} = 218.7\ \text{A} \]

\[ I_{B3} = \frac{50\times10^{6}}{\sqrt{3}(33\,000)} = 874.8\ \text{A} \]

Checking the ratios against the transformers:

\[ \frac{I_{B1}}{I_{B2}} = \frac{2624}{218.7} = 12 = \frac{132}{11}\ \checkmark, \qquad \frac{I_{B3}}{I_{B2}} = \frac{874.8}{218.7} = 4 = \frac{132}{33}\ \checkmark \]

The consequence is worth stating explicitly: a current of, say, 0.9 p.u. flowing through the whole radial system is 2362 A at 11 kV, 196.8 A at 132 kV and 787.3 A at 33 kV — one per-unit number, three ammeter readings.

Base currents transform inversely with base voltages, exactly as physical currents do through a transformer. That is not decorative: it is the reason a single per-unit current can describe a series path spanning several voltage levels at once. Converting back to amperes is therefore always the last step, done once per zone, and never in the middle of a calculation.

Answer\(I_{B1} = 2624\) A, \(I_{B2} = 218.7\) A, \(I_{B3} = 874.8\) A

Problem 10Exam levelLoad in p.u.

A load of 40 MW at 0.8 power factor lagging is supplied at 33 kV. Using a base of 50 MVA, 33 kV, express the load as a per-unit complex power, a per-unit current and a per-unit impedance, assuming the bus voltage is at 1.0 p.u. Convert the impedance back to ohms.

Solution

Apparent power of the load:

\[ S = \frac{40}{0.8} = 50\ \text{MVA at } \cos^{-1}(0.8) = 36.87^\circ \]

In per-unit on a 50 MVA base:

\[ S_{pu} = \frac{50\angle 36.87^\circ}{50} = 1.0\angle 36.87^\circ = 0.8 + j0.6\ \text{p.u.} \]

Note that the real part is numerically the power factor and the imaginary part is \(\sin\phi\) — a pleasant consequence of the base having been chosen equal to the load's own apparent power.

Per-unit current, from \(S = VI^{*}\) with \(V = 1.0\angle0^\circ\):

\[ I_{pu} = \left(\frac{S_{pu}}{V_{pu}}\right)^{*} = (1.0\angle 36.87^\circ)^{*} = 1.0\angle -36.87^\circ\ \text{p.u.} \]

Per-unit impedance:

\[ Z_{pu} = \frac{V_{pu}}{I_{pu}} = \frac{1.0\angle 0^\circ}{1.0\angle -36.87^\circ} = 1.0\angle 36.87^\circ = 0.8 + j0.6\ \text{p.u.} \]

Back to ohms:

\[ Z_B = \frac{(33)^2}{50} = 21.78\ \Omega, \qquad Z = 21.78\angle 36.87^\circ = (17.42 + j13.07)\ \Omega \]

\[ \text{Check: } P = \frac{3(33\,000/\sqrt3)^2(17.42)}{17.42^2 + 13.07^2} = 40\ \text{MW}\ \checkmark \]

A load has an impedance only at one particular voltage. The \(0.8 + j0.6\) p.u. computed here is the impedance that draws 40 MW when the bus sits at 1.0 p.u. If the voltage falls to 0.9 p.u., a constant-impedance model would deliver only \(0.81 \times 40 = 32.4\) MW, while a real motor load would draw almost the same power at a higher current. Which model is appropriate is a modelling decision, and load-flow programs offer all three — constant impedance, constant current and constant power — because none of them is universally right.

Answer\(S = 1.0\angle36.87^\circ\), \(I = 1.0\angle-36.87^\circ\), \(Z = 0.8+j0.6\) p.u. \(= (17.42 + j13.07)\,\Omega\)

Problem 11Exam levelRegulation in p.u.

A 33 kV feeder of impedance \((5 + j12)\,\Omega\) supplies a load of 8 MVA at 0.85 power factor lagging. Using a base of 10 MVA, 33 kV and taking the receiving-end voltage as 1.0 p.u., find the sending-end voltage in per-unit and in kV, and the voltage regulation.

Solution

Base impedance and the feeder in per-unit:

\[ Z_B = \frac{(33)^2}{10} = 108.9\ \Omega, \qquad Z_{pu} = \frac{5 + j12}{108.9} = 0.0459 + j0.1102\ \text{p.u.} \]

Load current in per-unit, with \(\mathbf{V}_R = 1.0\angle0^\circ\) as reference:

\[ S_{pu} = \frac{8}{10} = 0.8\ \text{p.u.}, \qquad \phi = \cos^{-1}(0.85) = 31.79^\circ \]

\[ \mathbf{I}_{pu} = 0.8\angle -31.79^\circ = 0.680 - j0.421\ \text{p.u.} \]

Voltage drop along the feeder:

\[ \Delta \mathbf{V} = \mathbf{I}Z = (0.680 - j0.421)(0.0459 + j0.1102) = 0.0777 + j0.0556\ \text{p.u.} \]

Sending-end voltage:

\[ \mathbf{V}_S = 1.0 + 0.0777 + j0.0556 = 1.0777 + j0.0556 = 1.0791\angle 2.95^\circ\ \text{p.u.} \]

\[ V_S = 1.0791 \times 33 = 35.61\ \text{kV} \]

Voltage regulation:

\[ \text{Reg} = \frac{|V_S| - |V_R|}{|V_R|} \times 100 = \frac{1.0791 - 1.0}{1.0}\times100 = 7.91\% \]

The real part of the drop is far larger than the imaginary part, even though the feeder is mostly reactive. That inversion is characteristic and useful: with a lagging load the reactive component of current works against the reactance to produce an in-phase drop, which is what actually lowers the magnitude at the receiving end. The quadrature component merely shifts the angle. This is why compensating reactive power — not resistance — is the lever that fixes a voltage problem, and it is the seed of the whole of Set 34 on voltage control.

Answer\(V_S = 1.079\angle2.95^\circ\) p.u. \(= 35.61\) kV, regulation \(= 7.91\%\)

Problem 12Challenge-liteThree-Winding Transformer

A three-winding transformer has ratings: primary 66 kV, 15 MVA; secondary 13.2 kV, 10 MVA; tertiary 2.3 kV, 5 MVA. Short-circuit tests give

\[ Z_{ps} = 7\%\ \text{on }15\ \text{MVA},\qquad Z_{pt} = 9\%\ \text{on }15\ \text{MVA},\qquad Z_{st} = 8\%\ \text{on }10\ \text{MVA} \]

Find the star-equivalent impedances \(Z_p\), \(Z_s\), \(Z_t\) on a common base of 15 MVA, 66 kV.

Solution

The three measurements are on two different MVA bases. Bring \(Z_{st}\) onto the common 15 MVA base first — nothing can be combined until they agree:

\[ Z_{st} = 0.08\left(\frac{15}{10}\right) = 0.12\ \text{p.u.} \]

Each measured impedance is the sum of two arms of the star equivalent, since a short-circuit test energises two windings and shorts one:

\[ Z_{ps} = Z_p + Z_s,\qquad Z_{pt} = Z_p + Z_t,\qquad Z_{st} = Z_s + Z_t \]

Three equations, three unknowns. Adding all three and halving gives \(Z_p + Z_s + Z_t\), and subtracting each measurement in turn isolates one arm:

\[ Z_p = \tfrac{1}{2}\big(Z_{ps} + Z_{pt} - Z_{st}\big) = \tfrac{1}{2}(0.07 + 0.09 - 0.12) = 0.02\ \text{p.u.} \]

\[ Z_s = \tfrac{1}{2}\big(Z_{ps} + Z_{st} - Z_{pt}\big) = \tfrac{1}{2}(0.07 + 0.12 - 0.09) = 0.05\ \text{p.u.} \]

\[ Z_t = \tfrac{1}{2}\big(Z_{pt} + Z_{st} - Z_{ps}\big) = \tfrac{1}{2}(0.09 + 0.12 - 0.07) = 0.07\ \text{p.u.} \]

Verifying against the measurements:

\[ Z_p + Z_s = 0.07\ \checkmark,\qquad Z_p + Z_t = 0.09\ \checkmark,\qquad Z_s + Z_t = 0.12\ \checkmark \]

The star arms are a computational device, not physical windings. Nothing inside the tank corresponds to \(Z_p = 0.02\) p.u., and one arm frequently comes out negative — perfectly legitimate, since only the pairwise sums are measurable and only those are constrained to be positive. A negative arm in a three-winding equivalent is a common sight in real data and is never, by itself, evidence of an error.

Answer\(Z_p = 0.02\), \(Z_s = 0.05\), \(Z_t = 0.07\) p.u. on 15 MVA, 66 kV

Problem 13Exam levelMotor on a System Base

A 5000 hp, 6.6 kV synchronous motor operates at 0.85 power factor with an efficiency of 92% and has \(X'' = 0.20\) p.u. on its own rating. Find its rating in kVA and its sub-transient reactance on a 25 MVA, 6.6 kV system base.

Solution

The horsepower figure is mechanical output. Convert to kilowatts:

\[ P_{\text{out}} = 5000 \times 0.746 = 3730\ \text{kW} \]

Electrical input, using the efficiency:

\[ P_{\text{in}} = \frac{3730}{0.92} = 4054\ \text{kW} \]

Apparent power — this is the machine's own MVA base, and it is what the manufacturer's 0.20 p.u. refers to:

\[ S_{\text{rated}} = \frac{4054}{0.85} = 4770\ \text{kVA} = 4.77\ \text{MVA} \]

Base change to the system base. The voltage bases agree at 6.6 kV, so only the MVA ratio acts:

\[ X'' = 0.20\left(\frac{25}{4.77}\right) = 1.048\ \text{p.u.} \]

Cross-check in ohms. On its own base, \(Z_B = 6.6^2/4.77 = 9.132\ \Omega\), so \(X'' = 1.826\ \Omega\). On the system base, \(Z_B = 6.6^2/25 = 1.742\ \Omega\):

\[ X''_{pu} = \frac{1.826}{1.742} = 1.048\ \text{p.u.}\ \checkmark \]

Two conversions, in the right order, and neither optional. Horsepower to kilowatts, then kilowatts to kVA through both efficiency and power factor — and it is the resulting kVA, not the horsepower and not the kilowatts, that is the machine's base. A motor's contribution to a fault is calculated from this reactance, so a motor whose rating was taken as "5 MVA because 5000 hp sounds like 5 MVA" would misstate the fault contribution by about 5%, which is exactly the sort of quiet error that survives a review.

Answer\(S_{\text{rated}} = 4.77\) MVA, \(X'' = 1.048\) p.u. on 25 MVA

Problem 14Challenge-liteBase Errors

A transmission line of 80 \(\Omega\) reactance operates at 132 kV. On a 100 MVA base, find its per-unit reactance. Then compute what would result if the base voltage of the 33 kV zone were mistakenly used, and quantify the consequence for a fault calculation.

Solution

The correct calculation, using the base voltage of the zone the line actually occupies:

\[ Z_B = \frac{(132)^2}{100} = 174.24\ \Omega, \qquad X_{pu} = \frac{80}{174.24} = 0.459\ \text{p.u.} \]

The erroneous calculation, using 33 kV:

\[ Z_B = \frac{(33)^2}{100} = 10.89\ \Omega, \qquad X_{pu} = \frac{80}{10.89} = 7.346\ \text{p.u.} \]

The error factor is the square of the voltage ratio:

\[ \frac{7.346}{0.459} = 16 = \left(\frac{132}{33}\right)^{2} \]

The consequence for a fault study. If this line were the only impedance in the loop, the correct fault level is

\[ \text{MVA}_{sc} = \frac{100}{0.459} = 218\ \text{MVA}, \qquad\text{against}\qquad \frac{100}{7.346} = 13.6\ \text{MVA} \]

A sixteen-fold understatement of the fault level. Switchgear selected on the wrong figure would be catastrophically under-rated, and the error would not reveal itself until the first fault.

Base errors are silent, and they scale as the square. Nothing in the arithmetic looks wrong — 7.346 p.u. is a perfectly well-formed number. The only defence is the sanity band: transmission-line reactances run from roughly 0.05 to 0.5 p.u. on a 100 MVA base, transformers 0.05 to 0.15, machines 0.1 to 0.3 sub-transient. A value an order of magnitude outside its band is a base error until proved otherwise, and checking every element against that band before proceeding costs a minute and catches nearly all of them.

AnswerCorrect \(X_{pu} = 0.459\); wrong base gives \(7.346\) — a factor of 16, understating fault level 218 MVA as 13.6 MVA

Problem 15Warm-upBack to Real Units

In a 100 MVA, 11 kV zone a machine carries a current of \(0.85\angle -25^\circ\) p.u. at a terminal voltage of \(1.0\angle 0^\circ\) p.u. Find the actual current in amperes, the actual terminal voltage, and the real and reactive power in MW and Mvar.

Solution

Base current for the zone:

\[ I_B = \frac{100\times10^{6}}{\sqrt{3}(11\,000)} = 5249\ \text{A} \]

Actual current — magnitude scaled by the base, angle carried straight through:

\[ \mathbf{I} = 0.85(5249)\angle -25^\circ = 4462\angle -25^\circ\ \text{A} \]

Actual terminal voltage, a line value since the base is a line value:

\[ V = 1.0 \times 11 = 11\ \text{kV (line)} \]

Complex power in per-unit:

\[ S_{pu} = \mathbf{V}\mathbf{I}^{*} = (1.0\angle 0^\circ)(0.85\angle 25^\circ) = 0.85\angle 25^\circ\ \text{p.u.} \]

And in real units, scaling by the base MVA:

\[ S = 0.85(100)\angle 25^\circ = 85\angle 25^\circ\ \text{MVA} = 77.0 + j35.9 \]

\[ P = 77.0\ \text{MW},\qquad Q = 35.9\ \text{Mvar} \]

\[ \text{Check: } S = \sqrt{3}(11\,000)(4462) = 85.0\ \text{MVA}\ \checkmark \]

Angles never scale; only magnitudes do. The per-unit transformation multiplies every quantity by a positive real number, so phase relationships pass through untouched — which is exactly why a per-unit phasor diagram is a faithful picture of the real one. It also means the power factor, being the cosine of an angle, is the same number in per-unit as in real units, and needs no conversion at any stage.

Answer\(I = 4462\angle-25^\circ\) A, \(V = 11\) kV, \(P = 77.0\) MW, \(Q = 35.9\) Mvar

Problem 16Challenge-liteWhy No √3 Appears

Show that the base impedance computed from three-phase quantities and from per-phase quantities is identical, and verify numerically for a 30 MVA, 33 kV system.

Solution

Start from the per-phase definition, which is the physically meaningful one — impedance is a per-phase quantity:

\[ Z_B = \frac{V_{B,LN}}{I_B} \]

Substitute the line-to-neutral voltage and the base current in terms of three-phase quantities:

\[ V_{B,LN} = \frac{\text{kV}_{LL}}{\sqrt3}\times10^{3}, \qquad I_B = \frac{\text{MVA}_{3\phi}\times10^{6}}{\sqrt3\,\text{kV}_{LL}\times10^{3}} \]

Dividing, the two \(\sqrt3\) factors cancel exactly:

\[ Z_B = \frac{\text{kV}_{LL}\times10^{3}/\sqrt3}{\text{MVA}_{3\phi}\times10^{6}/(\sqrt3\,\text{kV}_{LL}\times10^{3})} = \frac{(\text{kV}_{LL})^{2}}{\text{MVA}_{3\phi}} \]

Now the per-phase route. Here \(\text{MVA}_{1\phi} = \text{MVA}_{3\phi}/3\) and \(\text{kV}_{LN} = \text{kV}_{LL}/\sqrt3\):

\[ \frac{(\text{kV}_{LN})^{2}}{\text{MVA}_{1\phi}} = \frac{(\text{kV}_{LL})^{2}/3}{\text{MVA}_{3\phi}/3} = \frac{(\text{kV}_{LL})^{2}}{\text{MVA}_{3\phi}} \]

The factor of 3 in the numerator and the factor of 3 in the denominator are the same factor of 3, and they cancel. The two definitions are one definition.

Numerically, for 30 MVA and 33 kV:

\[ \frac{(33)^2}{30} = 36.30\ \Omega \qquad\text{and}\qquad \frac{(33/\sqrt3)^2}{30/3} = \frac{(19.053)^2}{10} = \frac{363.0}{10} = 36.30\ \Omega\ \checkmark \]

This identity is why per-unit calculations are so unforgiving of a mixed convention and so forgiving of everything else. Because the \(\sqrt3\) and the 3 cancel by construction, a correct per-unit calculation never contains either — so the appearance of a stray \(\sqrt3\) anywhere in per-unit working is a reliable signal that line and phase quantities have been mixed. Treat it as an error indicator rather than a step to be checked.

Answer\(Z_B = \text{kV}_{LL}^2/\text{MVA}_{3\phi} = \text{kV}_{LN}^2/\text{MVA}_{1\phi} = 36.30\,\Omega\)

Problem 17Exam levelSystem Fault Level

For the system of Problem 8, with a total series reactance of 1.014 p.u. on a 50 MVA base, find the per-unit fault current, the fault MVA and the actual fault current for a three-phase fault at the 33 kV bus. Assume 1.0 p.u. pre-fault voltage.

Solution

Per-unit fault current, from a single application of Ohm's law to the whole three-voltage-level system:

\[ I_{f,pu} = \frac{1.0}{1.014} = 0.986\ \text{p.u.} \]

Fault MVA:

\[ \text{MVA}_{sc} = \frac{50}{1.014} = 49.3\ \text{MVA} \]

Actual fault current at the 33 kV bus, using the base current of zone 3 found in Problem 9:

\[ I_f = 0.986 \times 874.8 = 863\ \text{A} \]

\[ \text{Check: } I_f = \frac{49.3\times10^{6}}{\sqrt3(33\,000)} = 863\ \text{A}\ \checkmark \]

The same fault expressed in the other zones — one per-unit current, three ammeter readings:

\[ \text{at }132\ \text{kV}: 0.986(218.7) = 215.6\ \text{A}, \qquad \text{at }11\ \text{kV}: 0.986(2624) = 2588\ \text{A} \]

Note how modest this fault level is. At 49.3 MVA the 33 kV bus is weak, because the 0.287 p.u. line reactance and the 0.40 p.u. generator reactance dominate the loop.

A single per-unit division has replaced a chain of impedance referrals across three voltage levels. Done in ohms, this problem would require referring the generator and both transformer impedances through two turns ratios before any addition was possible, with two opportunities to square a ratio the wrong way up. The per-unit route has neither. That saving, multiplied by a few hundred buses, is why no serious fault study has been done in ohms since the 1930s.

Answer\(I_f = 0.986\) p.u., \(\text{MVA}_{sc} = 49.3\) MVA, \(I_f = 863\) A at 33 kV

Problem 18Exam levelFault at Two Locations

A 30 MVA, 11 kV generator with \(X'' = 0.15\) p.u. supplies a 30 MVA, 11/132 kV transformer with \(X = 0.10\) p.u. Find the fault MVA and fault current for a three-phase fault (a) at the generator terminals and (b) on the 132 kV side of the transformer.

Solution

Both items are already on a 30 MVA base and their rated voltages match the zone bases, so no base change is needed. Take \(\text{MVA}_B = 30\).

aFault at the generator terminals. The transformer is on the far side of the fault and carries no fault current, so only the generator reactance is in the loop:

\[ \text{MVA}_{sc} = \frac{30}{0.15} = 200\ \text{MVA} \]

\[ I_B = \frac{30\times10^{6}}{\sqrt3(11\,000)} = 1575\ \text{A}, \qquad I_f = \frac{1575}{0.15} = 10\,498\ \text{A} \approx 10.5\ \text{kA} \]

bFault on the 132 kV side. Now both reactances are in series:

\[ X_{\text{total}} = 0.15 + 0.10 = 0.25\ \text{p.u.}, \qquad \text{MVA}_{sc} = \frac{30}{0.25} = 120\ \text{MVA} \]

\[ I_B = \frac{30\times10^{6}}{\sqrt3(132\,000)} = 131.2\ \text{A}, \qquad I_f = \frac{131.2}{0.25} = 525\ \text{A} \]

Both results in one table:

\[ \begin{array}{lccc} \text{Fault location} & X_{pu} & \text{MVA}_{sc} & I_f \\ \hline \text{11 kV terminals} & 0.15 & 200 & 10.5\ \text{kA} \\ \text{132 kV side} & 0.25 & 120 & 525\ \text{A} \end{array} \]

Transformer impedance is the cheapest fault limiter in the system. Adding 0.10 p.u. cut the fault level from 200 to 120 MVA. The same principle, applied deliberately, is why generator step-up transformers are specified with relatively high impedance and why series reactors are installed on heavily interconnected busbars — the fault level at a strong bus can otherwise outgrow the switchgear installed to interrupt it, and replacing switchgear is far more expensive than adding reactance.

Answer(a) 200 MVA, 10.5 kA at 11 kV · (b) 120 MVA, 525 A at 132 kV

Problem 19Challenge-liteSingle-Phase Bank

Three identical single-phase transformers, each rated 20 MVA, 66 kV/11.55 kV with \(X = 0.08\) p.u. on its own rating, are connected as a bank with the HV windings in delta and the LV windings in star. Find the three-phase rating, the line-voltage ratio of the bank, and the per-unit reactance on the bank's own three-phase base.

Solution

Three-phase rating. Each unit contributes its full rating:

\[ S_{3\phi} = 3 \times 20 = 60\ \text{MVA} \]

HV side, delta. Each winding stands across a line pair, so the line voltage equals the winding voltage:

\[ V_{L,HV} = 66\ \text{kV} \]

LV side, star. Each winding stands between a line and the neutral, so the line voltage is \(\sqrt3\) times the winding voltage:

\[ V_{L,LV} = \sqrt3 \times 11.55 = 20.0\ \text{kV} \]

\[ \text{Bank ratio} = 66/20\ \text{kV} \]

Per-unit reactance. Take the LV side. The winding reactance in ohms, on the unit's own single-phase base:

\[ Z_{B,\text{unit}} = \frac{(11.55)^2}{20} = 6.670\ \Omega, \qquad X = 0.08(6.670) = 0.5336\ \Omega\ \text{per winding} \]

Now the bank's own three-phase base on the LV side:

\[ Z_{B,\text{bank}} = \frac{(20.0)^2}{60} = 6.667\ \Omega \]

\[ X_{pu} = \frac{0.5336}{6.667} = 0.080\ \text{p.u.} \]

Unchanged. The \(\sqrt3\) on the voltage was squared to 3 in the base impedance, and the factor of 3 on the MVA cancelled it exactly.

The per-unit impedance of a bank equals that of one unit, whatever the connection. Δ–Y, Y–Δ, Y–Y or Δ–Δ — the cancellation above happens identically in every case, because the connection scales the voltage base by some factor and the MVA base by the square of nothing at all, leaving the ratio fixed. This is why data sheets quote a single per-unit impedance for a bank without stating the connection, and why replacing a Y–Y bank with a Δ–Y bank changes the phase shift and the zero-sequence path but not one number in the positive-sequence impedance diagram.

Answer\(S_{3\phi} = 60\) MVA, ratio \(66/20\) kV, \(X_{pu} = 0.08\) p.u. — unchanged

Problem 20Challenge-liteComplete System

Draw the per-unit impedance diagram for the system below on a base of 10 MVA, 13.8 kV in the generator zone, and find the generator internal voltage required to hold 1.0 p.u. at the load.

\(G\): 10 MVA, 13.8 kV, \(X = 0.15\) p.u.
\(T_1\): 10 MVA, 13.8/138 kV, \(X = 0.10\) p.u.
Line: 20 km at \(j0.5\ \Omega/\text{km}\)
\(T_2\): 5 MVA, 138/13.8 kV, \(X = 0.08\) p.u.
Load: 4 MW at 0.9 p.f. lagging, 13.8 kV

Solution

Zone bases. One base MVA of 10 throughout; base voltages follow the transformer ratios:

\[ \text{Zone 1: }13.8\ \text{kV},\qquad \text{Zone 2: }138\ \text{kV},\qquad \text{Zone 3: }13.8\ \text{kV} \]

Generator and \(T_1\). Both are already on 10 MVA with matching voltage bases, so they transfer unchanged:

\[ X_G = 0.15\ \text{p.u.},\qquad X_{T1} = 0.10\ \text{p.u.} \]

Line. Total reactance \(20 \times 0.5 = 10\ \Omega\), divided by the zone-2 base impedance:

\[ Z_{B2} = \frac{(138)^2}{10} = 1904.4\ \Omega, \qquad X_{\text{line}} = \frac{10}{1904.4} = 0.00525\ \text{p.u.} \]

Negligible against everything else — a short line at a high voltage is electrically almost nothing, which is precisely the argument for transmitting at 138 kV rather than 13.8 kV.

Transformer \(T_2\). Rated 5 MVA, so a base change is required:

\[ X_{T2} = 0.08\left(\frac{10}{5}\right) = 0.160\ \text{p.u.} \]

Total series reactance:

\[ X_{\text{total}} = 0.15 + 0.10 + 0.00525 + 0.160 = 0.4153\ \text{p.u.} \]

Load current. With \(\mathbf{V}_{\text{load}} = 1.0\angle0^\circ\):

\[ S = \frac{4}{0.9} = 4.444\ \text{MVA} \Rightarrow S_{pu} = 0.4444\angle 25.84^\circ, \qquad \mathbf{I} = 0.4444\angle -25.84^\circ\ \text{p.u.} \]

Generator internal voltage:

\[ \mathbf{E} = 1.0\angle0^\circ + (0.4444\angle-25.84^\circ)(j0.4153) = 1.0 + 0.1846\angle 64.16^\circ \]

\[ = 1.0805 + j0.1661 = 1.093\angle 8.74^\circ\ \text{p.u.} = 15.09\ \text{kV} \]

Actual currents, using the zone base currents \(I_{B1} = I_{B3} = 418.4\) A and \(I_{B2} = 41.84\) A:

\[ I_{\text{load}} = 0.4444(418.4) = 186.0\ \text{A}, \qquad I_{\text{line}} = 0.4444(41.84) = 18.60\ \text{A} \]

The angle of \(\mathbf{E}\) is the load angle, and it is the quantity that governs stability. Here \(\delta = 8.74^\circ\) — comfortable, since the steady-state limit is \(90^\circ\) and practice keeps well below it. The same series reactance that produced this angle reappears as the denominator of \(P = EV\sin\delta/X\) in Set 24, so the impedance diagram assembled here is not merely a fault-study tool; it is the same diagram the stability analysis will use, and this is why the effort of building it correctly is repaid several times over.

Answer\(X_{\text{total}} = 0.415\) p.u., \(\mathbf{E} = 1.093\angle8.74^\circ\) p.u. \(= 15.09\) kV

Practice

Practice Problems

Work each on paper before opening the answer. The answer is given so you can check yourself; the method is deliberately not.

P1. Find the base impedance and base current for a base of 200 MVA, 220 kV.

Show answer
\(Z_B = 220^2/200 = \mathbf{242}\ \Omega\); \(I_B = 200\times10^6/(\sqrt3\cdot220\,000) = \mathbf{524.9}\) A.
P2. A transformer is 8% on 20 MVA. What is its per-unit impedance on a 100 MVA base?

Show answer
\(0.08(100/20) = \mathbf{0.40}\) p.u. A common slip is to divide instead of multiply — remember that a smaller machine looks more impedant on a larger base.
P3. A generator is \(X'' = 0.12\) p.u. on 60 MVA, 13.8 kV. Find its reactance in ohms.

Show answer
\(Z_B = 13.8^2/60 = 3.174\ \Omega\), so \(X'' = 0.12(3.174) = \mathbf{0.381}\ \Omega\).
P4. A 25 MVA transformer of 6% impedance feeds a bus from an infinite source. Find the fault MVA and comment on the switchgear rating needed at 11 kV.

Show answer
\(25/0.06 = \mathbf{417}\) MVA. At 11 kV that is \(417\times10^6/(\sqrt3\cdot11\,000) = \mathbf{21.9}\) kA, so 25 kA switchgear would be the next standard rating up.
P5. A 132 kV line has 60 \(\Omega\) reactance. Find its per-unit value on 100 MVA, and check it lies in the usual band.

Show answer
\(60/174.24 = \mathbf{0.344}\) p.u. — within the 0.05–0.5 band expected of a transmission line on a 100 MVA base, so no base error is indicated.
P6. Two transformers, 10 MVA at 5% and 15 MVA at 6%, operate in parallel. Find the equivalent per-unit impedance on a 25 MVA base.

Show answer
On 25 MVA: \(0.125\) and \(0.10\). Parallel: \((0.125)(0.10)/0.225 = \mathbf{0.0556}\) p.u. Check by fault level: \(200 + 250 = 450\) MVA, and \(25/0.0556 = 450\) MVA ✓
P7. A base of 50 MVA, 11 kV is chosen in the generator zone of a system whose step-up transformer is 11/220 kV. What are the base voltage and base current in the transmission zone?

Show answer
\(\mathbf{220}\) kV and \(50\times10^6/(\sqrt3\cdot220\,000) = \mathbf{131.2}\) A. The base MVA stays at 50 — it never changes across a transformer.
P8. A 15 MVA, 11 kV machine has \(X'' = 0.18\) p.u. What is the symmetrical short-circuit current at its terminals?

Show answer
\(I_B = 787.3\) A, so \(I_f = 787.3/0.18 = \mathbf{4374}\) A. Equivalently \(15/0.18 = 83.3\) MVA.
P9. A load draws 0.6 p.u. current at 0.9 p.u. voltage with a 0.8 lagging power factor. Find \(P\) and \(Q\) in per-unit.

Show answer
\(S = (0.9)(0.6) = 0.54\) p.u., so \(P = \mathbf{0.432}\) and \(Q = \mathbf{0.324}\) p.u.
P10. Three single-phase units of 10 MVA, 33/6.35 kV, 7% are connected Y–Y. Find the bank's three-phase rating, line-voltage ratio and per-unit impedance.

Show answer
\(\mathbf{30}\) MVA; ratio \(\sqrt3(33)/\sqrt3(6.35) = \mathbf{57.2/11}\) kV; impedance \(\mathbf{7\%}\) — unchanged, as always.
P11. On a 100 MVA base, a system has 0.05 p.u. source reactance and a 0.10 p.u. transformer. Find the fault MVA at the transformer's LV terminals.

Show answer
\(100/0.15 = \mathbf{667}\) MVA. Note the source is much the stronger of the two, so the transformer sets the fault level.
P12. A 400 kV line of 90 \(\Omega\) is computed as 0.9 p.u. on a 100 MVA base. Is this plausible? If not, what base voltage was probably used?

Show answer
Not plausible — the correct value is \(90/(400^2/100) = 90/1600 = \mathbf{0.056}\) p.u. A result of 0.9 implies \(Z_B = 100\ \Omega\), i.e. a base of \(\mathbf{100}\) kV was used instead of 400 kV.

Challenge

Challenge Problems

Each of these needs an idea rather than a formula. Decide what the governing principle is before opening the answer.

C1. Prove that a transformer's per-unit impedance is identical referred to either winding, and identify the precise condition on the base voltages under which the proof holds. What goes wrong in the impedance diagram if that condition is violated — for example, if a designer picks 11 kV and 132 kV as bases for a transformer actually wound 11/138 kV?

Show answer
Let the transformer have turns ratio \(a = V_1/V_2\) and impedance \(Z_2\) referred to side 2. Referred to side 1 it is \(Z_1 = a^2Z_2\). With base voltages \(V_{B1}, V_{B2}\) and common \(S_B\):
\[ \frac{Z_1}{Z_{B1}} = \frac{a^2Z_2}{V_{B1}^2/S_B} = \frac{a^2 Z_2 S_B}{V_{B1}^2}, \qquad \frac{Z_2}{Z_{B2}} = \frac{Z_2S_B}{V_{B2}^2} \]
These are equal if and only if \(a^2/V_{B1}^2 = 1/V_{B2}^2\), that is \(V_{B1}/V_{B2} = a\).

The condition: the base voltages must stand in exactly the transformer's own turns ratio. Nothing else is required — the absolute values are free.

If violated: with bases 11/132 on a 11/138 kV transformer, \(V_{B1}/V_{B2} = 12\) but \(a = 12.545\). The two per-unit impedances now differ by \((12.545/12)^2 = 1.093\), so the impedance is 9.3% larger seen from one side than the other. The ideal transformer can no longer be deleted: what remains is an off-nominal-ratio ideal transformer of ratio \(1{:}1.045\) that must be retained in the model.

This is not a pathology to be avoided — it is exactly how tap-changing transformers are represented in load flow, as a per-unit impedance plus a residual ideal transformer of ratio \(1{:}t\). What the per-unit system removes is the nominal ratio; the deviation from nominal is real, controllable, and is the whole mechanism of voltage control in Set 34.
C2. Machine reactances cluster in narrow per-unit bands almost independently of rating — sub-transient reactance is near 0.15 p.u. for a 10 MVA machine and for a 1000 MVA machine alike. Explain physically why this should be so, and what it implies about the fault current a machine can deliver relative to its own rating.

Show answer
Per-unit reactance is \(X_{pu} = X\,S_{\text{rated}}/V_{\text{rated}}^2\). Now \(X\) itself scales with the machine's geometry: reactance \(\propto N^2\mu A/\ell\) and rated voltage \(\propto N\phi f\), while rated MVA \(\propto\) the machine's active volume. Working the scaling through, the geometric factors cancel to leading order and \(X_{pu}\) depends on proportions — airgap length relative to bore, winding depth relative to pole pitch — not on absolute size.

Those proportions are fixed by the same physics for every machine: the airgap must be small enough that the excitation is economical and large enough that mechanical clearance and armature reaction are tolerable. Designers of a 10 MVA and a 1000 MVA machine face the same optimum and arrive at the same ratios, so both land near \(X'' \approx 0.15\) p.u.

Implication: every synchronous machine delivers roughly \(1/0.15 \approx 6\text{--}7\) times its own rated current into a terminal fault, regardless of size. This is why fault levels track installed capacity so closely, why a network planner can estimate a fault level from a one-line diagram before any data arrives, and why a per-unit value far outside the band is a data error rather than an unusual machine — the argument of Problems 2 and 14.
C3. The per-unit system is often described as "dimensionless". Examine that claim carefully. Is a per-unit voltage the same kind of object as a per-unit impedance? What breaks if one uses two different base MVAs in different parts of the same network, and why is this different from using different base kV?

Show answer
Dimensionless, but not interchangeable. Each per-unit quantity is a pure number, but it is a ratio to a different base, and those bases carry the dimensions. A per-unit voltage of 1.05 and a per-unit impedance of 1.05 are both dimensionless and mean entirely unrelated things. What the system guarantees is that the relations between them survive: \(V_{pu} = I_{pu}Z_{pu}\) and \(S_{pu} = V_{pu}I_{pu}^*\) hold only because the four bases were chosen to satisfy those same relations, as verified in Problem 1.

Two base MVAs: catastrophic. Suppose zone 1 uses 100 MVA and zone 2 uses 50. A current of 1.0 p.u. flowing from zone 1 into zone 2 becomes 2.0 p.u. on crossing the boundary — but no transformer is present to justify a change, and KCL is violated at the boundary node. The network is no longer a connected circuit; series impedances can no longer simply be added, and every fault and load-flow result is wrong.

Two base kVs: not merely allowed but required. The base voltage must change at every transformer, in the transformer's ratio — and it does so precisely because a real transformer is there to change the real voltage. The base change mirrors a physical change; that is why it is legitimate. There is no physical device that changes MVA, so there is no justification for a base MVA that changes.

The asymmetry, then, is not a convention: base voltage tracks a physical transformation that actually occurs, and base MVA tracks a conserved quantity that does not.

Self-Test

Multiple-Choice Questions

MCQ 1. The base impedance of a three-phase system is:
(a) \(\text{kV}_{LL}^2/\text{MVA}\) (b) \(\sqrt3\,\text{kV}_{LL}^2/\text{MVA}\) (c) \(\text{kV}_{LL}^2/(3\,\text{MVA})\) (d) \(\text{kV}_{LL}/\text{MVA}\)

Show answer
(a). No \(\sqrt3\) or 3 appears — they cancel identically, as shown in Problem 16.
MCQ 2. In a network with four voltage levels, the number of base MVA values is:
(a) 1 (b) 2 (c) 4 (d) depends on the transformers

Show answer
(a) one. The base MVA is common to the entire network; only the base kV changes, once per transformer.
MCQ 3. A transformer's per-unit impedance referred to the primary compared with that referred to the secondary is:
(a) larger by \(a^2\) (b) smaller by \(a^2\) (c) equal (d) larger by \(a\)

Show answer
(c) equal — provided the base voltages stand in the transformer's ratio. This is the property that removes the ideal transformer from the diagram.
MCQ 4. An impedance of \(Z_{pu}\) on \(\text{MVA}_1\) becomes, on \(\text{MVA}_2\) at the same voltage base:
(a) \(Z_{pu}\,\text{MVA}_1/\text{MVA}_2\) (b) \(Z_{pu}\,\text{MVA}_2/\text{MVA}_1\) (c) \(Z_{pu}(\text{MVA}_2/\text{MVA}_1)^2\) (d) unchanged

Show answer
(b). MVA goes the right way up, unsquared; only the kV ratio is inverted and squared.
MCQ 5. A transformer marked 6% has a per-unit impedance, on its own rating, of:
(a) 6 (b) 0.6 (c) 0.06 (d) 0.006

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(c) 0.06. Percent impedance is per-unit \(\times\,100\), nothing more.
MCQ 6. The short-circuit MVA of a system of per-unit impedance \(Z_{pu}\) on base \(\text{MVA}_B\) is:
(a) \(\text{MVA}_B\,Z_{pu}\) (b) \(\text{MVA}_B/Z_{pu}\) (c) \(Z_{pu}/\text{MVA}_B\) (d) \(\sqrt3\,\text{MVA}_B/Z_{pu}\)

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(b). Low impedance means a high fault level — the two are reciprocal.
MCQ 7. Three single-phase transformers of per-unit impedance \(z\) each are connected as a Δ–Y bank. The bank's per-unit impedance is:
(a) \(z/3\) (b) \(3z\) (c) \(\sqrt3 z\) (d) \(z\)

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(d) \(z\). The voltage-base and MVA-base changes cancel exactly, whatever the connection — Problem 19.
MCQ 8. Base current and base voltage across a transformer are related by:
(a) both scale with the turns ratio (b) base current scales inversely with base voltage (c) base current is unchanged (d) base voltage is unchanged

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(b). Their product, the base MVA, is common to the whole network, so one must fall as the other rises.
MCQ 9. A sub-transient reactance computed as 8 p.u. on a 100 MVA base most likely indicates:
(a) a very small machine (b) a base error (c) a saturated machine (d) a correct value for a small generator

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(b) a base error. Even a very small machine on a 100 MVA base rarely exceeds 2–3 p.u. A value an order of magnitude out is a wrong base until proved otherwise.
MCQ 10. In per-unit, the power factor of a load is:
(a) divided by the base (b) multiplied by \(\sqrt3\) (c) unchanged (d) referred to the base MVA

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(c) unchanged. The transformation scales magnitudes by positive reals and leaves every angle alone, so any cosine of an angle is invariant.
MCQ 11. Two areas of one network are analysed with base MVAs of 100 and 50 respectively. The result is:
(a) valid, if the base kVs are right (b) valid, but currents must be scaled (c) invalid — KCL is violated at the boundary (d) valid only for balanced systems

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(c). A per-unit current would change value at a boundary where no transformer exists. See Challenge C3.
MCQ 12. One arm of a three-winding transformer's star equivalent comes out negative. This means:
(a) a measurement error (b) an arithmetic error (c) nothing is wrong — only the pairwise sums are physical (d) the transformer is faulty

Show answer
(c). The star arms are a computational device; only \(Z_{ps}\), \(Z_{pt}\) and \(Z_{st}\) are measurable, and a negative arm is common in real data.

Reference

Key Formulas

Quantity	Relation	Notes
Per-unit definition	\(X_{pu} = X_{\text{actual}}/X_{\text{base}}\)	Applies to every quantity alike
Base impedance	\(Z_B = \text{kV}_B^2/\text{MVA}_B\)	Line kV and three-phase MVA
Base impedance (per phase)	\(Z_B = \text{kV}_{LN}^2/\text{MVA}_{1\phi}\)	Identical value — the 3s cancel
Base current	\(I_B = \dfrac{\text{MVA}_B\times10^6}{\sqrt3\,\text{kV}_B\times10^3}\)	Amperes
Base consistency	\(Z_B = V_{B,LN}/I_B\)	Choose two bases only; derive the rest
Base change	\(Z_{pu}^{\text{new}} = Z_{pu}^{\text{old}}\dfrac{\text{MVA}_{new}}{\text{MVA}_{old}}\left(\dfrac{\text{kV}_{old}}{\text{kV}_{new}}\right)^{2}\)	MVA upright, kV inverted and squared
Percent impedance	\(\%Z = 100\,Z_{pu}\)	Always on the equipment's own rating
Short-circuit MVA	\(\text{MVA}_{sc} = \text{MVA}_B/Z_{pu}\)	Fault levels of parallel sources add
Fault current	\(I_f = I_B/Z_{pu}\)	Use the base current of the faulted zone
Base kV across a transformer	\(\text{kV}_{B2} = \text{kV}_{B1}\times\) nameplate ratio	MVA base never changes
Transformer p.u.	same from either side	Only if base kVs are in the turns ratio
Off-nominal ratio	residual ideal transformer \(1{:}t\)	What remains when bases ≠ turns ratio
Single-phase bank	\(S_{3\phi} = 3S_{1\phi}\), \(Z_{pu}\) unchanged	Any connection: Y–Y, Δ–Y, Y–Δ, Δ–Δ
Three-winding star arms	\(Z_p = \tfrac12(Z_{ps}+Z_{pt}-Z_{st})\)	Cyclic for \(Z_s\) and \(Z_t\); may be negative
Complex power	\(S_{pu} = V_{pu}I_{pu}^{*}\)	Same form as in real units
Angles	unchanged by the transformation	Power factor is base-independent
Sanity band — lines	\(0.05\)–\(0.5\) p.u.	On a 100 MVA base
Sanity band — transformers	\(0.05\)–\(0.15\) p.u.	On the transformer's own rating
Sanity band — machines	\(X'' \approx 0.15\) p.u.	Nearly independent of rating

Diagnostics

Common Mistakes

Changing the base MVA somewhere in the network. It is one value for the whole system. Changing it breaks KCL at the boundary and invalidates everything downstream — Challenge C3.
Inverting the base-change formula. MVA the right way up, kV inverted and squared. Getting the MVA ratio backwards is the single most frequent slip in the whole topic.
Using a zone's nominal voltage instead of the base derived from the transformer ratio. They coincide only when the transformer ratios are consistent. When they are not, the base follows the transformer and an off-nominal ratio remains in the model.
Dividing a line's ohms by the wrong zone's base impedance. Problem 14 shows the error is a factor of 16 for a 132/33 kV mix-up, and it is completely silent.
Forgetting to convert a manufacturer's impedance to the study base. Nameplate values are on the equipment's own rating. Problem 7 misstates the fault level by a third if this is skipped.
Introducing a \(\sqrt3\) into a per-unit calculation. There is never one. Its appearance means line and phase quantities have been mixed — treat it as an error indicator.
Taking a motor's horsepower as its MVA base. Convert through 0.746, then through efficiency, then through power factor — Problem 13.
Converting to amperes mid-calculation. Stay in per-unit until the final step, then convert once per zone using that zone's base current.
Assuming a per-unit impedance is dimensionless in the sense of being interchangeable. A 1.05 p.u. voltage and a 1.05 p.u. impedance are ratios to different bases and have nothing to do with one another.
Choosing \(Z_B\) or \(I_B\) independently. Only two bases are free. Choosing a third breaks the Ohm's-law consistency demonstrated in Problem 1, and every later result inherits the inconsistency silently.
Rejecting a negative arm in a three-winding equivalent. It is normal. Only the pairwise sums are physical quantities — Problem 12.
Not checking results against the sanity bands. A one-minute check of every element against its expected band catches nearly every base error before it propagates into a fault study.

Looking Ahead

The impedance diagram assembled in Problem 20 is a small thing — four reactances in series — but it is the same object, built the same way, that every remaining set in this book operates on. Set 4 develops it into a full reactance diagram for a multi-machine system, and from Set 16 onwards it is no longer drawn at all but stored as a matrix.

What the per-unit system bought was the right to treat that diagram as a single connected circuit. Transformers are gone; \(\sqrt3\) factors are gone; a machine, a line and a transformer add like resistors. Everything that follows — the \(Y\)-bus of Set 16, the Gauss–Seidel and Newton–Raphson iterations of Sets 19 and 20, the sequence networks of Set 22, the swing equation of Set 24 — is written in per-unit and would be almost unmanageable without it. The discipline of base selection practised here is therefore not an exercise in bookkeeping; it is the price of admission to the rest of the subject.