Complex Power and Line Inductance

The definition. With \(\mathbf{V}\) a phase voltage and \(\mathbf{I}\) the line current out of the element:

The conjugate is essential. Without it the angle would be \(\theta_V + \theta_I\), which depends on where the reference was put; with it the angle is \(\theta_V - \theta_I = \phi\), which does not.

Lagging case. Take \(\mathbf{V} = 127.02\angle0^\circ\) kV and \(\mathbf{I} = 150\angle-25^\circ\) A:

Check by the line-quantity form:

The \(\sqrt{3}\) and the 3 are the same statement written with different voltages.

Leading case. With \(\mathbf{I} = 150\angle+25^\circ\):

\(P\) is unchanged; \(Q\) has reversed sign.

What the signs mean. With this convention — current out of the element into the network:

Textbooks differ. The safest habit is to state the current's direction with every result, because the algebra is identical and only the sign convention distinguishes a generator from a load.

The physical change. Both cases move 51.80 MW and both draw 150 A — so both cause identical \(I^2R\) losses in the line feeding them. What differs is where the 24 MVAr comes from: in the lagging case the network supplies it; in the leading case the load supplies it back. That is the whole subject of the previous set expressed in one sign.

The derivation. Take \(\mathbf{V}_R\) as reference and let \(\mathbf{V}_S\) lead it by \(\delta\). From \(\mathbf{V}_S = A\mathbf{V}_R + B\mathbf{I}_R\):

Then \(\mathbf{S}_R = 3\mathbf{V}_R\mathbf{I}_R^{*}\) with \(\mathbf{V}_R = |V_R|\angle0\):

Two phasors: one of fixed length that rotates with \(\delta\), and one that is entirely fixed. That structure is the circle diagram of Problem 7, already visible.

Separating real and imaginary parts:

The two coefficients. With \(|V_S| = |V_R| = 220/\sqrt3 = 127.02\) kV:

Both have the dimensions of power. Note that the whole calculation can be done in line quantities as \(V_L^2/|B|\), which gives the same 561.73 MW — the threes cancel again.

The fixed term. \(\beta_B - \beta_A = 68.20 - 0.564 = 67.63^\circ\):

At \(\delta = 20^\circ\):

Reading \(Q_R\). The negative sign says the receiving end must supply 88.3 MVAr into the line to hold its voltage at 220 kV while taking 165.8 MW. A load absorbing vars will not do that; a capacitor bank or a synchronous condenser at the receiving busbar must. Every point on the circle of Problem 7 carries this same requirement.

The derivation. Invert the two-port to get \(\mathbf{I}_S\) in terms of \(\mathbf{V}_S\) and \(\mathbf{V}_R\). Using \(D = A\) for a symmetrical line:

Then \(\mathbf{S}_S = 3\mathbf{V}_S\mathbf{I}_S^{*}\), with \(\mathbf{V}_S = |V_S|\angle\delta\):

The differences from the receiving-end pair are worth naming, because they are the usual source of error:

Everything is reversed. That is not an accident: it is the same equation viewed from the other terminal, and \(\delta\) changes sign with the viewpoint.

At \(\delta = 20^\circ\), with \(|V_S| = |V_R|\) the two coefficients are the same 548.27 and 561.73 MW as before:

The check. The real difference is the line loss:

Verify it independently from the current. \(\mathbf{I}_R = (\mathbf{V}_S - A\mathbf{V}_R)/B\) gives \(|I_R| = 493\) A and \(|I_S| = 521\) A; taking the mean square across the line:

Within 2% of 25.2. The approximation is not exact because the nominal-\(\pi\) puts the whole resistance in one branch that carries neither terminal current, but it confirms that 25 MW is a loss and not an algebraic slip.

The reactive difference is not a loss. \(Q_S - Q_R = -54.4 - (-88.3) = +33.9\) MVAr. The line has generated 33.9 MVAr net — its charging exceeds its \(I^2X\) absorption at this loading, so the line is running below its surge impedance loading. Above SIL the sign reverses.

Using \(P_R = 561.73\cos(68.20^\circ - \delta) - 208.62\) and \(Q_R = 561.73\sin(68.20^\circ - \delta) - 507.03\):

The \(\delta = 0\) row is the check. With equal voltage magnitudes and no angle there is no real transfer, and the \(+14.5\) MVAr is the line's own charging arriving at the receiving end — the Ferranti current, seen in power terms.

The shape. Written as a single cosine, \(P_R(\delta) = 561.73\cos(\beta_B - \delta) - 208.62\) is a shifted and displaced cosine:

Both departures from \(P \propto \sin\delta\) come from the same place — the resistance, which drags \(\beta_B\) down from 90° and makes \(\cos(\beta_B-\beta_A)\) non-zero.

The curve is not symmetric about its peak either. Between 0 and 20° it gains 165.8 MW; between 48 and 68° it gains only 33 MW. Power becomes progressively more expensive in angle, which is the mechanical statement of approaching the stability limit.

The reactive column is the real warning. By \(\delta = 30^\circ\) the load bus needs 160 MVAr of support to hold 220 kV; by 50° it needs 332 MVAr — more than the 325 MW being delivered. Long before the angle limit is approached, the reactive requirement has become impossible.

The maximum. Only the first term depends on \(\delta\), and it is largest when its cosine is unity:

Substituting:

The lossless estimate. With \(X = 80\ \Omega\) and both voltages at 220 kV, \(P = V_L^2\sin\delta/X\) peaks at \(\delta = 90^\circ\):

An overstatement of 71%.

Where the discrepancy sits. Two separate effects, of comparable size:

The first step costs 43 MW; the second costs 209 MW. The dominant error in the lossless formula is not that it ignores \(r\) in the magnitude, but that it throws away a whole term.

Why the fixed term vanishes when \(r = 0\). With no resistance, \(B = jX\) gives \(\beta_B = 90^\circ\), and for a nominal-\(\pi\) line \(A = 1 + ZY/2\) is very nearly real, so \(\beta_A \approx 0\):

The 208.6 MW here is \(548.27\cos67.63^\circ\), and 67.63° is only 22° from 90° — yet that 22° is worth 209 MW.

The limit is never approached in operation. A line run at \(\delta = 68^\circ\) has no margin: a small increase in demand cannot be met by a larger angle, and the system loses synchronism. Practice keeps \(\delta\) below about 30° in normal operation and below 45° after a credible contingency, which caps this line at roughly 230 MW — two-thirds of the theoretical limit.

At the limit, \(\delta = \beta_B\), so the sine term vanishes entirely:

The receiving busbar must inject 507 MVAr into the line — 1.44 times the real power it is receiving.

The apparent power and current:

At 200 MW, from the circle relation with \(\delta = 24.87^\circ\):

The comparison:

A 77% increase in delivered power costs a 164% increase in current and a fourfold increase in reactive support.

What binds first. A single ACSR conductor of the class used at 220 kV carries about 600 A continuously:

The conductor melts, the compensator does not exist, and the operator's rule is exceeded — all three long before the mathematical maximum.

Where the vars go. The series branch carries 1653 A at this point, so it absorbs \(3I^{2}X = 3(1653)^{2}(80) = 655.7\) MVAr, against \(V_L^{2}B = 29.0\) MVAr generated by the charging. The net 626.7 MVAr is supplied 507.0 from the receiving end and 119.6 from the sending end — the balance closes exactly. The reactive demand grows as the square of the current, which is why it overtakes the real transfer so abruptly.

The construction. Return to the phasor form of Problem 2:

A fixed phasor plus one of constant length and variable angle. The locus of the sum is a circle — immediately, without any algebra.

The radius is the length of the rotating phasor:

Proportional to the product of the two voltages and inversely to \(|B|\). Raise either voltage and the circle grows.

The centre is the negative of the fixed phasor:

Down and to the left of the origin — which is why the useful part of the circle lies in the fourth quadrant, where \(P > 0\) and \(Q < 0\).

Three verifications. A point at angle \(\theta = \beta_B - \delta\) measured from the positive \(P\) axis:

The three rows are the no-load point, the working point of Problem 2 and the tangent point of Problem 5 — all on one circle.

Reading the geometry. Four features carry the whole of the line's steady-state behaviour:

Where the circle crosses \(Q = 0\) is the transfer this line will carry unaided:

Only 33 MW — under a tenth of the stability limit. Beyond that every megawatt requires reactive support at the load bus.

Step 1 — locate the operating point on the circle. The circle fixes \(Q\) once \(P\) is chosen:

Step 2 — the reactive power the line will deliver at that point:

Negative: the line takes 74.7 MVAr out of the receiving busbar rather than delivering any.

Step 3 — what the load wants:

Step 4 — the compensator makes up the difference. The busbar must balance:

On the diagram this is the vertical distance from the load point \((150, +72.6)\) down to the circle at \((150, -74.7)\) — which is the whole reason the drawing was made.

At 250 MW, repeating:

The trend across the range:

\(Q_C\) outgrows \(P\) from 100 MW onwards, and by 300 MW it exceeds it.

Two separate causes, and they compound. The load's own demand grows linearly with \(P\) at fixed power factor; the line's absorption grows as \(P^2\) through \(3I^{2}X\). Doubling the transfer from 150 to 300 MW doubles the first from 73 to 145 MVAr and more than triples the second from 75 to 269.

From the relations of Problem 3, written as a phasor sum:

The same radius, the opposite centre. With \(|V_S| = |V_R|\) here:

The sending circle is the receiving circle reflected through the origin — but only when the two voltage magnitudes are equal. In general \(\mathbf{C}_S \propto |V_S|^2\) and \(\mathbf{C}_R \propto |V_R|^2\), and the symmetry is lost.

The direction of travel is also reversed. The receiving-end point sits at angle \(\beta_B - \delta\) and the sending-end point at \(\beta_B + \delta\) measured the other way from the centre — so as the load grows the two points move in opposite senses around their circles.

Operating points at three angles:

All powers in MW and MVAr.

What the generator reads. \(Q_S\) is negative throughout this range, meaning the sending end is absorbing vars from the line — the line is generating more charging than it consumes, below SIL. The machine must run under-excited, which is the condition its capability curve most restricts.

And \(Q_S\) is not monotonic: \(-42.8\), \(-54.4\), \(-49.0\) — it deepens and then recovers. Setting \(\partial Q_S/\partial\delta = 0\) gives \(\beta_B + \delta = 90^\circ\), so the minimum is at \(\delta = 21.80^\circ\):

Beyond that angle the line's own \(I^2X\) absorption begins to overtake its charging, and the generator's under-excitation eases.

Subtract the two power relations. Write \(K = 3|V|^{2}/|B|\) and \(K_A = |A|K\) with \(|V_S| = |V_R| = |V|\):

The bracket collapses by the sum-to-product identity:

The two coefficients are equal. This is the step that makes the result clean. Since \(K_A = |A|K\), the claim is \(|A|\cos(\beta_B-\beta_A) = \cos\beta_B\), or equivalently

For the nominal-\(\pi\), \(A = 1 + \mathbf{Z}\mathbf{Y}/2\) and \(\mathbf{B} = \mathbf{Z}\), so \(A^{*} - 1 = \mathbf{Z}^{*}\mathbf{Y}^{*}/2\) and

because \(\mathbf{Y}\) is purely susceptive — the line's leakage conductance is negligible. The identity is exact, not approximate.

The result:

Verification against the values computed the long way:

Exact to every figure carried.

What it says. Expanding for small angles, \(1 - \cos\delta \approx \delta^{2}/2\):

And since \(P_R \approx K\sin\beta_B\,\delta\) for small \(\delta\), the loss grows as the square of the transfer — the \(I^2R\) law, recovered from the circle diagram without ever writing a current.

And the geometric reading: at a common \(\delta\), the sending point sits at angle \(\beta_B+\delta\) from its centre and the receiving point at \(\beta_B-\delta\) from its own — reflections of each other about the \(\delta = 0\) position. Their horizontal separation is the loss, and it opens as \(1-\cos\delta\): zero at \(\delta = 0\) and widening ever faster.

The short-line reduction. Set \(A = 1\), \(\beta_A = 0\) and \(B = \mathbf{Z}\):

Written in line quantities, which is how it is usually quoted.

The lossless reduction. Set \(r = 0\) as well, so \(\beta_Z = 90^\circ\) and \(|Z| = X\):

The familiar form — and now visibly a double approximation, not one.

The comparison. With \(\mathbf{Z} = 32 + j80 = 86.16\angle68.20^\circ\) and \(X = 80\ \Omega\):

The short-line column is identical to the exact one — to every figure. That is not luck. The nominal-\(\pi\) and the short-line model share \(B = \mathbf{Z}\) exactly, and the identity proved in Problem 10 makes \(|A|\cos(\beta_B-\beta_A) = \cos\beta_B\), so the fixed term is \(V_L^2\cos\beta_Z/|Z|\) in both cases. The shunt admittance drops out of the real power relation altogether when the terminal voltages are equal.

The reactive relation is a different matter. The same comparison for \(Q_R\):

A constant offset of 14.5 MVAr — exactly the charging \(V_L^{2}B/2\) at each end. The short-line model gets the megawatts right and the megavars wrong by a fixed amount.

So the two approximations are not comparable. Dropping the shunt admittance costs nothing in \(P\) and a constant in \(Q\); dropping the resistance costs 20–30% in \(P\) at working angles and 71% at the limit. The one universally taught is the one that does the damage.

The thermal limit depends only on the conductor, not on the length:

At 0.9 power factor that is 206 MW; at unity, 229 MW.

The three limits at 200 km:

Thermal and practical-stability limits are within 2% of each other. This line is well matched to its conductor — which is what a competent design looks like.

At other lengths, keeping the same conductor and both voltages at 220 kV. These use the exact hyperbolic constants rather than the nominal-\(\pi\), so the 200 km row runs about 1% above the figures used elsewhere in the set:

Powers in MW.

The crossover. At 50 km the conductor is the constraint and stability is irrelevant — the line could carry four times its thermal rating before angle became an issue. At 500 km the position is reversed: the stability limit of 99 MW is well under half the conductor's 229 MW, and the extra copper is wasted.

Why the boundary sits near 200 km. The thermal limit is independent of length; the stability limit falls roughly as \(1/l\). They cross at whatever length makes \(V_L^{2}/|B|\) comparable to \(\sqrt3V_LI_{\text{rated}}\) — for a 220 kV single-conductor line, about 200 km. Raising the voltage to 400 kV moves the crossover out to roughly 400 km, which is why the two voltage classes have the route lengths they do.

The geometric mean distance. For a transposed horizontal line with spacings \(D\), \(D\) and \(2D\):

The inductance from the standard result of Set 5:

The reactance:

The 0.40 Ω/km assumed since Set 9, recovered from the tower drawing.

What the answer is sensitive to. Differentiate the logarithm:

A 10% change in either geometric quantity changes \(L\) by only 1.5%. The logarithm suppresses everything.

Which is why every overhead line has almost the same reactance. Across the whole range of transmission practice:

The single-conductor lines are all near 0.4 Ω/km regardless of voltage. Only bundling moves the number appreciably — and it moves the GMR, not the GMD.

The corresponding capacitance uses the physical radius instead of the GMR:

With \(r \approx 14.3\) mm this gives 0.0089 µF/km and \(b = 2.8\) µS/km — close to the 3.0 used throughout. The two logarithms differ only in the denominator, which is why \(Z_c = \sqrt{x/b}\) is so nearly the same for all overhead lines.

The bundle GMRs from the standard expressions, with \(d = 0.45\) m and \(D_s = 0.0114\) m:

A four-bundle has seventeen times the GMR of one conductor — the geometric mean radius of an object 0.45 m across, not of a wire 23 mm across.

The reactances, with GMD unchanged at 7.560 m:

Diminishing returns, and sharply. The first sub-conductor added buys 28%; the second buys a further 10%; the third a further 6%. The logarithm again: doubling the GMR subtracts \(0.2\ln2 = 0.139\) mH/km whatever the starting point, and each added sub-conductor doubles it less than the last.

The resistance falls in proportion to \(n\) — \(n\) conductors in parallel — and this matters as much as the reactance:

For the 200 km line with both voltages at 220 kV.

The limit rises by 2.45 times while the reactance falls by only 44%. The extra gain comes from the falling \(r/x\), which pushes \(\beta_B\) towards 90° and shrinks the subtracted term of Problem 5 — exactly the mechanism that made heavy series compensation counterproductive in Set 14, running the other way.

But the thermal limit rises faster. \(n\) conductors carry \(n\) times the current:

The four-bundle has crossed over: its stability limit is now below its thermal rating, and further conductor buys nothing on a 200 km route.

The calculation, with GMD \(= D\sqrt[3]{2}\) and GMR fixed at 11.4 mm:

Tripling the spacing costs 18% in reactance and 6% in power limit. Set that against the bundling result:

Bundling is the larger lever by an order of magnitude, and it acts in the useful direction.

Why the logarithm makes this so. The GMD enters as \(\ln(\text{GMD})\) and the ratio inside is already 663, so:

Matching the table. Any change of spacing is divided by about 6 before it reaches the inductance. Changing the GMR is divided by the same 6 — but bundling changes the GMR by a factor of 17, not 3.

Why towers are not built narrower. Nothing in this table is what sets the spacing. Four separate constraints do:

At 220 kV these fix the spacing at 5–7 m regardless of what the reactance would prefer, and at 400 kV at 9–12 m.

And a narrower tower has a second cost. Reducing the GMD reduces the capacitance denominator as well, so \(b\) rises with \(x\) falling — which lowers \(Z_c\) and raises SIL. A compact line therefore genuinely does carry more natural power, and compact-line designs exist for exactly this reason; but they buy it with insulation technology, not by simply moving the phases closer on a conventional tower.

The self-GMDs. Each phase is now a group of two conductors, one on each circuit. With the reversed order, phase \(a\) sits at \((0, 14)\) and \((9, 0)\):

Reversing the second circuit is what puts \(a\) diagonally opposite \(a\). Keeping the same order would give three separations of 9 m and a higher reactance — this is the low-reactance arrangement.

The equivalent GMR of each phase group, then their geometric mean:

The mutual GMDs, each the fourth root of four separations:

The reactance per phase for the pair of circuits together:

49.2% of the single circuit's 0.4082 — slightly less than half, which two independent circuits in parallel would give. The reversed phasing turns the mutual coupling to advantage.

The capacitance rises correspondingly. Using the physical radius 14.3 mm in place of the GMR:

Against 122 MW for one circuit. The natural loading has slightly more than doubled.

The limits, at 200 km with \(r = 0.08\ \Omega\)/km for the two conductors in parallel:

And the contingency governs. A double-circuit line on one tower is exposed to common-mode failure — a tower collapse, a wide fire, a lightning flashover across both circuits. Planning practice requires the surviving circuit to carry the load:

The real capability of this double circuit is 229 MVA — the same as one circuit alone — unless load can be shed on the contingency.

The equation to solve. A constant-power load fixes \(\mathbf{S}_R\) and lets \(|V_R|\) float:

Multiplying through by \(|V_R|\) gives a quadratic in \(|V_R|^{2}\) — hence two roots for every load, not one.

The two branches. Solving at a series of loads:

The upper branch is the operating point; the lower is a mathematically valid solution at high current and low voltage that no protection would permit and no load would sustain.

The nose. The two roots merge at

Beyond it there is no solution. The load cannot be served at any voltage — the system does not settle at a lower voltage, it collapses.

Against the angle limit of 353 MW. The difference is entirely in what was assumed:

Problem 5's 353 MW required 507 MVAr at the load bus. Withdraw that assumption and the same line manages 166 MW — a factor of 2.1.

The power factor moves the nose sharply:

Correcting the load's power factor raises the collapse point by 54%. This is the same lever as Problem 20 of Set 14, seen from the stability side rather than the regulation side.

The knee is the operating constraint, not the nose. At 140 MW the voltage is already 167 kV — 76% of nominal, far below anything a network would tolerate. A 0.95 pu voltage criterion caps this line at 54 MW uncompensated, and it is compensation, not the nose, that makes the line usable at all.

The assumptions are the whole difference. Both are steady-state maxima of the same two-port:

What each measures. The angle limit measures the ability of the machines to stay in step: beyond \(\delta = \beta_B\) an increase in mechanical power cannot be exported, the rotor accelerates and synchronism is lost. The voltage limit measures the ability of the network to supply reactive power: beyond the nose, no voltage exists at which the load's demand and the line's delivery agree.

The numbers for this line:

In descending order — and the two most restrictive are both reactive in origin.

Which arrives first depends on the var plan, and only on that. With a compensator at the load bus the voltage limit disappears and the thermal or angle limit binds; without one the voltage limit binds at less than half. The transfer capability of a corridor is therefore not a property of the line. It is a property of the line plus the reactive plant at its ends.

The historical pattern follows this. Angle instability dominated the mid-20th-century record because networks were sparse, machines slow and lines long relative to their loads. As systems grew denser and loads grew — especially induction-motor loads, which restore their power draw as voltage falls and so behave as the constant-power model assumes — the binding constraint moved to reactive support. Most large blackouts since 1980 have had a voltage-collapse mechanism rather than an angle one.

The two are coupled, not independent. Every megavar of support that raises the voltage limit also holds \(|V_R|\) up, which raises the angle limit through the \(|V_S||V_R|/|B|\) term. And a falling voltage raises the current for a given power, deepening the \(I^{2}X\) absorption — the feedback that makes the collapse fast.

The reference. With \(z = 0.16 + j0.40\) and \(y = j3.0\ \mu\)S/km:

Independent of length — which is exactly what makes it a useful yardstick.

The limits, from the exact hyperbolic constants with both voltages at 220 kV:

Powers in MW.

The shape. The limit falls almost exactly as \(1/l\) over the middle of the range, because \(|B| \approx |Z|\) is nearly proportional to length. It falls faster beyond 500 km, where \(\sinh\gamma l\) begins to outgrow its argument.

Why express it per SIL. The SIL absorbs the voltage class:

The \(V_L^{2}\) cancels. Since every overhead line has \(Z_c \approx 380\ \Omega\) and \(\beta \approx 1.1\times10^{-3}\), the ratio depends on the length alone — so one curve serves every voltage class. That is the St Clair curve, drawn in 1953 and still used.

The three regimes it exposes:

Read against the practical column, a 400 km line carries about 1 SIL and a 600 km line about two-thirds — which is why long AC corridors are always compensated and very long ones are built as HVDC instead.

The 30° column is the honest one. Its ratios are 63% of the theoretical maxima throughout, because \(\cos(\beta_B - 30^\circ)\) is a nearly constant fraction of 1 over the range of \(\beta_B\) that overhead lines span. That constancy is why a single rule of thumb — about 1 SIL at 400 km — survives across the whole of transmission practice.

The corridor, from Problem 16: \(r = 0.08\), \(x = 0.2010\ \Omega\)/km, \(b = 5.66\ \mu\)S/km, \(Z_c = 195.5\ \Omega\), SIL 248 MW, \(|B| = 42.9\ \Omega\), \(|A| = 0.9774\), two 600 A conductors per phase.

Limit 1 — thermal. Independent of everything else:

Limit 2 — angle, normal state. At \(\delta \le 30^\circ\) with both voltages held at 220 kV:

320 MVAr of support at the receiving bus — which must be installed for this limit to mean anything.

Limit 3 — voltage stability, no support:

And applying a \(0.95\) pu voltage criterion instead of the nose itself brings this down to 106 MW.

Limit 4 — the N−1 contingency. With one circuit out, the surviving circuit has \(|B| = 87.0\ \Omega\):

The complete picture:

All in MW.

The answer: 217 MW with reactive support installed, 166 MW without. Both are set by the contingency, not by the normal state — and the corridor's headroom in normal operation (434 MW thermal) is twice its declared capability. That gap is not waste; it is the margin that makes the contingency survivable.

What would raise it, in order of cost:

Note that series compensation — the remedy of Set 14 — is nearly useless here. It attacks \(|B|\), and \(|B|\) is not what binds.