HVDC Transmission and FACTS

Charging current per kilometre. Each phase sits at \(V_{LL}/\sqrt3\) against earth and draws \(V_{ph}\omega C\) per kilometre of shunt capacitance:

Compare an overhead line of the same voltage, whose capacitance is nearer 0.011 µF/km: the cable's charging current is about 24 times larger for the same length. Everything below follows from that one ratio, established in Chapter 9.

The length at which the cable is full of its own charging current. If the reactor is at the sending end only, the whole charging current of the route flows in the conductor there:

With equal reactors at both ends, each end supplies half the total, so the worst conductor current is \(9.43\ell\) and

Beyond that length the cable cannot carry one ampere of load however it is compensated, unless a platform is built mid-route to compensate there as well. The 50–70 km figure quoted for uncompensated 400 kV cable in Chapter 38 is the first of these two numbers.

What is left at 100 km. Charging current at each end is \(9.43\times100 = 943\) A. It is in quadrature with a unity-power-factor load current, so the two combine as the sides of a right triangle whose hypotenuse is the thermal rating:

The cable is using 57% of its current capability to charge itself. A DC cable of the same conductor carries 1250 A at any length whatever, because at \(\omega=0\) the capacitance is charged once and then forgotten.

What it settles. The AC option is not merely dearer past about 130 km — it does not exist. That is a different kind of argument from the cost comparison of Problem 2, and it is the reason every long submarine link built anywhere in the world is a DC link, including short ones where the cost crossover has not yet been reached.

Write each total as a straight line in the route length. The whole comparison is two first-degree polynomials, and they cross once because \(T_{DC}>T_{AC}\) while \(c_{DC}<c_{AC}\):

The crossing.

The converter premium is ₹7200 crore and DC repays it at ₹8.40 crore per kilometre of route. Every break-even calculation is that one sentence.

Either side of the crossing.

At 500 km DC costs 38% more; at 1200 km it costs 18% less. The margin swings quickly because the two slopes differ by a factor of three.

The effect of losses. Converter losses are a fixed charge of about 0.7% of 2000 MW at each end, so \(2\times14 = 28\) MW that AC does not pay at all. Capitalised, that adds to \(T_{DC}\) and to the numerator. Line losses, however, favour DC and grow with distance, so they subtract from \(c_{DC}\) and enlarge the denominator:

Both numerator and denominator increase, so the sign of the change is not obvious from the structure alone. For a bipole carrying the same power on two conductors instead of three, the line-loss advantage is the larger term over any distance worth considering, and \(d^{*}\) falls — but only past a few hundred kilometres, below which the fixed converter loss dominates and pushes \(d^{*}\) the other way.

Same formula, different numbers.

Why it is eight times shorter than the overhead answer. Set the two calculations side by side:

The premium halved because the scheme is smaller, but the slope gap quadrupled because submarine AC cable is dear and needs three cores where DC needs two. A four-fold repayment rate against a halved debt gives an eight-fold shorter crossover.

Whether to believe it. No — and Problem 1 says why. At 104 km a 400 kV AC cable compensated at both ends has already lost most of its rating to charging current, and at 132 km it has lost all of it. The straight line \(C_{AC}=930+50.4d\) is therefore fiction beyond about 80 km unless mid-route compensation platforms are added, each of which is a step increase in \(T_{AC}\):

Offshore practice bears this out: AC is used out to roughly 60–80 km and HVDC beyond, a boundary set by the physics of Problem 1 rather than by the arithmetic of this one.

Rated current. A bipole is two monopoles sharing an earthed neutral, each at \(500\) kV to earth:

The two pole currents are equal and opposite at the neutral, so the earth carries only the small unbalance — a few amperes, set by differences in the two poles' control settings, not the 2000 A of a monopole.

One pole out. The healthy pole keeps its own voltage and current and returns through earth:

Two thousand amperes flowing continuously through the ground is why bipoles carry a metallic return conductor or an electrode line sited kilometres from the station: at that current the electrode corrodes buried pipework and drives DC into transformer neutrals, saturating their cores.

Expected power, with the two poles independent. Three states:

Hours of total loss, which is the number that matters to the system operator:

Identical energy availability, fifty times fewer complete outages. A 2000 MW block vanishing from a system for 175 hours a year is a planning problem; the same block halving for 172 hours is not. That asymmetry, not the energy figure, is why every large link is bipolar.

The two converter constants. Averaging one \(60^\circ\) window of the line-voltage envelope gives the familiar factor, and the overlap analysis gives the equivalent resistance:

Both depend only on the transformer, not on the operating point. Compute them once and every question about this bridge is arithmetic.

The operating point at \(\alpha=17^\circ\).

The commutation drop is 27.5 kV — over 10% of \(V_{d0}\) and more than the whole effect of the \(17^\circ\) firing delay, which cost only 11.8 kV. It is not a correction term.

The inverse question. Rearranging the same equation for \(\cos\alpha\):

The same question with the drop forgotten:

An error of \(19.4^\circ\). Fired at \(27.31^\circ\) the bridge would in fact deliver \(270.09\cos27.31^\circ-27.50 = 212.5\) kV, 11% below the target. Note also that \(7.95^\circ\) is uncomfortably close to \(\alpha_{\min}\approx5^\circ\): this bridge has almost no margin left to raise its voltage further, which is a design finding, not an arithmetic one.

The commutation equation. While current transfers from the outgoing valve to the incoming one, the two phases are short-circuited through \(2X_c\) and the driving voltage is the line voltage between them:

Solving for the angle.

A quarter of the \(60^\circ\) between firings is spent with three valves conducting instead of two. Typical values are \(15^\circ\)–\(25^\circ\) at full load; an overlap approaching \(60^\circ\) means the converter has entered an abnormal mode in which commutations run into one another.

The check. The averaging form must reproduce Problem 5's answer:

The two forms are the same statement. \(V_{d0}\cos\alpha - R_cI_d\) is what you use when \(I_d\) is known; \(\tfrac12V_{d0}[\cos\alpha+\cos(\alpha+\mu)]\) is what you use when \(\mu\) is known. Eliminating \(\mu\) between them is precisely how \(R_c=3X_c/\pi\) arose.

Sensitivity of the overlap. The right-hand side is proportional to \(I_d/V_{LL}\):

Halving the current does not halve the overlap, because \(\mu\) enters through a cosine difference and not linearly. That non-linearity is why a converter's power factor improves at light load — Problem 15 uses exactly this.

The displacement factor. The same two cosines that gave the direct voltage give the phase angle of the fundamental current:

The two expressions are identical by construction. That identity is the single most useful fact about a line-commutated converter: fixing the direct voltage fixes the power factor, with nothing left to choose.

Power on the DC side.

The independent AC-side calculation. The phase current is a \(120^\circ\) block of height \(I_d\), whose fundamental is

Two routes, one answer. The DC-side route uses \(V_dI_d\) and the AC-side route uses \(\sqrt3V_{LL}I_1\cos\phi\); that they agree is a statement that the converter is lossless and that only the fundamental carries active power.

Why it cannot be reduced. The reactive demand is

a function of \(V_d/V_{d0}\) and of nothing else. To lower \(Q\), either \(\alpha\) must fall — but \(\alpha_{\min}\approx5^\circ\) is held so that the valve has a positive forward voltage to fire into — or the overlap must fall, and the overlap is set by \(X_c\) and the current the link is dispatched to carry. The converter has one control variable and it is already committed to setting the direct voltage. The reactive power must therefore be supplied locally, by the filters and banks of Problem 15.

The direct voltage. An inverter obeys the same two-element model with \(\gamma\) replacing \(\alpha\), and the commutation drop still opposes the source:

The overlap and the two angles. The commutation equation is unchanged, but it is written from \(\gamma\) forwards. With \(V_{LLi}=V_{d0i}/1.3505 = 185.12\) kV,

Check by the averaging form: \(\tfrac12(250)(0.95106+0.73104) = 210.26\) kV, as above. The firing angle is well beyond \(90^\circ\), so \(V_d\) would be negative on the rectifier convention — which is exactly what makes the bridge absorb power from the DC side and return it to the AC system.

The extinction angle as a time. One electrical degree at 50 Hz is \(20/360 = 55.6\) µs:

The whole reason \(\gamma_{\min}\) is set at \(15^\circ\)–\(18^\circ\) rather than at the valve's \(7.2^\circ\) is to buy that margin against AC voltage dips, waveform distortion and firing-instant errors. The margin costs money: every extra degree of \(\gamma\) lowers \(\cos\gamma\), lowers \(V_{di}\), and raises the inverter's reactive demand.

The dip that causes commutation failure. Within the cycle in which the dip occurs the firing instants are already committed, so \(\beta\) is fixed at \(43.03^\circ\). A dip to a fraction \(m\) of the AC voltage enlarges the commutation term by \(1/m\) and eats into \(\gamma\):

Setting \(\gamma' = 7.2^\circ\), the point at which the valve no longer recovers:

A remote fault that depresses the receiving bus by a sixth is enough. That is an ordinary event on any transmission system, which is why commutation failure is described as the characteristic HVDC fault rather than as a malfunction — and why the inverter needs a short-circuit ratio above 2 to 3, so that its own reactive demand does not itself depress the bus.

Walk the DC circuit. Going from the rectifier's internal source to the inverter's, the loop equation is

The minus sign on \(R_{ci}\) is not an error. Written with \(\cos\gamma\), the inverter's commutation drop subtracts from its counter-voltage, so it helps the current along instead of opposing it.

The current. With identical converters the two commutating resistances cancel and only the line remains:

The terminal voltages, each converter's source less its own commutation drop, \(R_cI_d = 15.28\times1.796 = 27.43\) kV:

Powers, loss and efficiency.

The inverter's reactive demand, from the same rule used at the rectifier in Problem 7:

The inverter delivers 378 MW into its AC network and simultaneously draws 243 MVAr from it. Both stations are reactive loads; neither is a reactive source.

Why the sensitivity is extreme. The numerator of the link equation is a small difference between two large numbers, and the denominator is small too:

The numerator is 3.6% of either term. Anything that shifts either term by a fraction of a per cent shifts their difference by tens of per cent.

A 1% fall. \(V_{d0r}\) becomes \(0.99\times270.09 = 267.39\) kV, so \(V_{d0r}\cos24^\circ = 244.27\) kV:

A one-per-cent disturbance has removed a quarter of the transfer. No AC line behaves like this; the DC link does because its "impedance" is 5 Ω of conductor and nothing else.

A 5% fall. \(V_{d0r}\cos24^\circ\) becomes \(0.95\times246.74 = 234.41\) kV:

A negative answer is not a reversed current — the valves conduct one way only. It means the rectifier can no longer overcome the inverter's counter-voltage at \(24^\circ\), the current falls to zero and the link stops. A five-per-cent AC disturbance would extinguish a 380 MW transfer.

What the regulator must do. To restore \(I_d = 1.796\) kA the numerator must be returned to 8.98 kV, so the rectifier's source must again reach 246.74 kV from the reduced \(V_{d0r}=256.59\) kV:

Comfortably above \(\alpha_{\min}\), so the rectifier retains control. Advancing the firing by \(8.1^\circ\) cancels a 5% voltage collapse; the price is a poorer power factor, since \(\cos\phi\) is still \(V_d/V_{d0}\) and \(V_{d0}\) has fallen.

Where the margin runs out. The rectifier can hold 1796 A only while \(V_{d0r}\cos\alpha_{\min} \ge 246.74\) kV:

Past that the rectifier saturates at \(\alpha_{\min}\) and something else must take over the current — which is Problem 11.

The two current orders.

In normal running the actual current is 1796 A, above the inverter's order, so the inverter's current regulator is saturated and does nothing. Only the extinction-angle regulator is active there.

What the depressed rectifier can do. At 88% of its AC voltage and hard against \(\alpha_{\min}\):

Against the inverter still at \(\gamma=18^\circ\) this would give \((236.78-237.76)/5 < 0\) — no current at all. The rectifier has run out of firing angle, so it can no longer set the current, and it becomes a voltage source instead.

Control transfers to the inverter. The inverter's current regulator comes out of saturation and holds \(I_d = 1530\) A by giving up its extinction angle. The rectifier, stuck at \(\alpha_{\min}\), fixes the voltage:

The extinction angle it settles at. Inverting \(V_{di}=V_{d0i}\cos\gamma - R_{ci}I_d\):

Note the direction: \(\gamma\) has risen from \(18^\circ\) to \(23.6^\circ\). Raising \(\gamma\) lowers the inverter's counter-voltage and therefore lets more current through — which is how a device whose safety limit is a minimum \(\gamma\) can regulate current upwards. Commutation-failure risk falls during the transfer, not rises.

The delivered power.

The two stations have swapped roles, with no communication between them and no operator action. The link keeps 83% of its transfer through a disturbance that would otherwise have stopped it dead — and that is entirely the work of a 270 A offset between two current orders.

The fundamental. With ideal smoothing the phase current is a \(120^\circ\) rectangular block of height \(I_d\), and Fourier analysis of that block gives

The harmonics. Only \(h=6k\pm1\) survive, and their magnitude falls as \(1/h\):

A 5th-harmonic current of 281 A on a 200 kV bus is not a trace contaminant; it is a fifth of the fundamental and it will distort every voltage in the vicinity.

Total distortion. Summing the squares of \(1/h\) over all characteristic orders:

The series converges slowly — truncating at the 25th already gives 29.0%, so more than a tenth of the distortion sits above the 25th harmonic. That upper tail is what the damped high-pass filter branch of Problem 14 exists to catch.

The DC side. Six commutations per cycle put the ripple at

The AC and DC orders are related: a DC ripple of order \(6k\) reflects to the AC side as the pair \(6k\pm1\), which is why the two lists always appear together.

Why the triplens are absent. Two separate reasons, and both must be stated:

The three phase currents of a bridge sum to zero at every instant, so no zero-sequence component can exist; and the delta winding of the converter transformer, in the twelve-pulse arrangement of Problem 13, gives any residual triplens a circulating path that keeps them off the network. Triplens do appear in practice — as non-characteristic harmonics produced by unbalanced AC voltages or unequal firing instants — which is why a small 3rd-harmonic filter branch is often fitted.

The pole. Series connection on the DC side adds the voltages at a common current:

A \(\pm500\) kV bipole is four such bridges: two poles of two bridges each.

Where the cancellation comes from. A delta secondary's line voltages lead a star secondary's by \(30^\circ\), so the two bridges fire \(30^\circ\) apart. For a harmonic of order \(h\) that is a phase difference of \(30h\) degrees at harmonic frequency:

Two shifts must be counted, not one. The \(30^\circ\) displacement in time shifts the \(h\)th harmonic by \(30h\) degrees at harmonic frequency; referring that current through the delta winding back to the primary adds a further \(\pm30^\circ\), positive for the negative-sequence orders \(6k-1\) and negative for the positive-sequence orders \(6k+1\). The surviving orders are therefore

The 5th, 7th, 17th and 19th vanish. Two whole filter branches — the expensive low-order ones — are saved, which is more than the extra transformer costs.

The DC side. The two bridges' ripples interleave:

And a smoothing reactor of given inductance has twice the impedance at 600 Hz that it has at 300 Hz, so the ripple current falls by considerably more than the ripple voltage does.

One bridge bypassed. The bypass switch short-circuits that bridge, so the pole becomes a plain six-pulse converter carrying the same current:

Half the power, but the 5th and 7th return at full six-pulse magnitude — 281 A and 200 A from Problem 12 — into a filter yard that has no branch tuned to either. Reduced-pulse operation is therefore permitted only for limited periods and often at reduced current, and it is one of the standard reasons a converter station is derated during maintenance.

The capacitor comes first, because at fundamental frequency the branch is essentially a capacitor and its reactive output is wanted:

The reactor puts the series resonance on the 11th. Setting \(h\omega L = X_C/h\):

Verification, and the true fundamental output. At \(h=11\) the two reactances are equal and opposite:

At 50 Hz the reactor is in series with the capacitor and slightly increases the net capacitive susceptance:

0.8% more than the nominal 90 MVAr — small here, but on a filter tuned to the 3rd harmonic \(X_L = X_C/9\) and the excess reaches 12%, which cannot be ignored in the reactive balance.

Loss and residual voltage. The 11th-harmonic current sees only \(R\):

A 0.29% individual harmonic voltage, comfortably inside the 1–1.5% limits a transmission utility sets. The loss is 505 kW against the 831 MW of Problem 13 — 0.06%, which is real money over 25 years but not an engineering problem.

Detuning by 2%. A 2% loss of capacitance raises \(X_C\) to 1813.3 Ω and moves the resonance up:

The branch impedance has risen 56% and the residual harmonic voltage with it, to 1052 V. That is the price of a high \(Q\): a sharply tuned filter is also a sharply detuned one. Real filters are built with tuning taps on the reactor and are re-tuned when a capacitor unit fails, and a damped high-pass branch — deliberately low \(Q\) — is used for the upper orders where tuning cannot be maintained.

Full-load demand. Each bridge draws the 252.6 MVAr of Problem 7, and the pole is two bridges:

Four switchable steps of 70 MVAr give 280 MVAr, enough with a little to spare. Switching in steps rather than continuously is deliberate: the steps are sized so that no single switching operation moves the bus by more than about 1%.

At 30% current. With \(I_d = 540\) A and \(\alpha\) still \(17^\circ\), the commutation term scales with current:

Note that the power fell to 32.5% of rating while the reactive demand fell to 21.8% of it. The power factor improved, from 0.854 to 0.926, because a smaller current commutates faster. That improvement is the source of the difficulty in the next step.

The surplus. The filters cannot be switched out — the harmonics are still there, and at 540 A the 11th is still 38 A — so 240 MVAr of capacitive plant is committed against a demand of 110 MVAr:

A 2.2% overvoltage at a 400 kV bus is a genuine operational problem: it appears at light load, when the surrounding network is already lightly loaded and its own line charging is raising the voltage further.

What the station does about it. Three measures, used together:

The second is the elegant one and it is the one actually used: the converter's reactive appetite, which is a nuisance at full load, becomes a controllable sink at light load. Raising \(\alpha\) from \(17^\circ\) to about \(35^\circ\) at 540 A takes \(\cos\phi\) down to roughly 0.80, raising \(Q\) to about 200 MVAr and nearly closing the gap; the tap changer then restores the direct voltage that the larger \(\alpha\) would otherwise have lost. This is the same trick as the coordinated tap-and-bank control of Set 34, run at converter speed.

The uncompensated line.

A 400 km line at 400 kV is angle-limited long before it is thermally limited. Its surge impedance loading is about 600 MW, so at \(30^\circ\) it is running almost exactly at SIL — the flat-profile condition of Set 31.

The compensation required. Fix the angle and solve for the reactance the target demands:

A third of the line's reactance removed buys a 48% increase in both the scheduled transfer and the stability limit — and the equal-area argument of Set 29 turns the second of those directly into a longer critical clearing time.

The bank rating. A series capacitor is rated on the current it carries, not on the bus voltage:

218 MVAr of plant for 294 MW of extra transfer. Set 31 made the same comparison against a shunt bank and found series compensation far cheaper per megawatt gained, for the reason that its output is proportional to the current and therefore largest exactly when it is needed.

The resonance. A series \(LC\) circuit at

A 21.4 Hz torque component sits in the middle of the 15–45 Hz band in which large steam turbine-generator shafts have their torsional modes. If a big thermal unit is radially connected to this line, a plain fixed capacitor is ruled out — and Problem 17 gives the alternative.

The characteristic. The reactor sits in parallel with the capacitor, and with capacitive reactance taken positive,

where \(X_L(\alpha)\) is the reactor's effective fundamental reactance, which runs from 17 Ω at full conduction to infinity when the thyristors are blocked. Everything the device can do is read off this one hyperbola.

The three reference points.

The sign change across \(X_L(\alpha)=X_C\) is not a discontinuity the controller passes through — a band of firing angles either side of it is simply blocked, and the device operates in one region or the other.

What each state does to the line. With \(X_{\text{eff}}=132-X_{TCSC}\) and \(P = (400)^2\sin30^\circ/X_{\text{eff}}\):

A continuous range from 500 MW to 1748 MW at a fixed angle, commanded in a few milliseconds. That is the whole point: not a larger transfer, but a transfer that can be modulated, which is what damps an inter-area swing.

Why a boost of 3 is unusable. It requires \(X_L(\alpha)=64.67\ \Omega\) and gives

A 400 km line reduced to 2.7 Ω is 98% compensated. Fault currents would rise by a factor of thirty, every distance relay in the corridor would see a capacitive reactance inside its measuring loop and mis-operate, and the line would be far stiffer than the equipment at its ends. The practical ceiling on \(k\) is about 70%, which is why the vernier range is specified as a boost of 1 to 2 rather than 1 to 3.

The property that matters. The TCSC's closed-loop control is synchronised to the power-frequency current, and at subsynchronous frequencies the reactor is effectively left conducting, so the device presents an inductive impedance below 50 Hz:

The series resonance of Problem 16 requires a capacitance in the loop at 28.6 Hz. If the loop is inductive there, the resonance does not exist, no subsynchronous current can build up, and no complementary rotor torque is produced. This is the sense in which a TCSC is described as SSR-neutral, and it is the reason it is chosen over a fixed bank on lines radial to large thermal units — not because it is a better compensator, which at the same \(k\) it is not.

The two elements. The fixed capacitor must supply the whole capacitive limit, and the reactor must be able to cancel it and go 100 MVAr further:

The reactor is always larger than the swing it produces, because it must first cancel the fixed capacitor. That is the FC/TCR arrangement's one inefficiency, and it is the reason thyristor-switched capacitor banks are used alongside it on large installations.

The firing angle for 80 MVAr capacitive. Work backwards from the required net susceptance:

and invert the Fourier result for a partially-conducting reactor:

The transcendental equation \(\sigma-\sin\sigma = \text{const}\) has no closed-form inverse and is solved by bisection or by a lookup table in the controller. Two iterations from \(\sigma=2.1\) get three figures.

The whole range.

The characteristic is strongly non-linear: half the reactive swing occurs between \(90^\circ\) and \(115^\circ\) and the last quarter is spread over \(150^\circ\) to \(180^\circ\). The controller therefore works in susceptance and converts to an angle at the last step, never the other way round.

What is delivered at 0.94 pu. The SVC is a susceptance, so its output follows the square of the voltage:

A 6% voltage deficit costs 12% of the output, and the loss grows as the square. Since the device exists to act when the voltage is low, this is a defect at exactly the wrong point on the characteristic — which is Problem 19.

The injection required. The device must lift the bus by 0.09 pu:

The two ratings. Ratings are quoted at 1.0 pu, so each must be scaled by how its output derates. The operating point is \(V=0.97\), because that is where the device settles once it has done its work:

The tempting alternative is to evaluate at the depressed voltage of 0.88, giving 755 MVAr and 665 MVAr — a much larger difference of 13.6%. That is wrong, because the device does not have to deliver 585 MVAr at 0.88 pu; as it injects, the voltage rises and its own output rises with it. The correct statement is a fixed point:

Substituting \(Q_r=621.7\) and \(V=0.97\) gives \(0.88+0.0900=0.97\) ✓, and likewise for the STATCOM. At this modest depth the honest difference between the two technologies is only 3.1%.

The deeper event. Now the same two devices, ratings fixed, face an uncompensated 0.70 pu. Solve each fixed point:

The STATCOM delivers 31% more reactive power and holds the bus 1.8 percentage points higher — from a rating that is 3% smaller. The SVC has lost 43% of its rating to the voltage-squared law; the STATCOM has lost 23%.

The design conclusion.

The two are indistinguishable at rated voltage and diverge steadily below it. The device is bought for what it does during a collapse, so a comparison made at 1.0 pu — or even at 0.94 pu — understates the STATCOM's advantage systematically. Add that a STATCOM responds in a quarter of a cycle against one to two cycles, occupies about half the area, and can be given a short-term overload rating that a susceptance cannot have, and the case is made on everything except capital cost.

The natural division. Two paths across the same angle share power in inverse proportion to their reactances:

The short route carries 213 MW more than it may — a 24% overload — while the long route runs at half its capability. Nothing in the AC network can be told to do otherwise; the division is a property of the impedances, and this is precisely the situation the FACTS family exists for.

Option 1 — a TCSC in the short route, inductive. Require \(P_A/P_B = X_B/X_A'\) with \(P_A=900\) and \(P_B=700\):

The device must run in its controllable-inductor region — the full-conduction branch of Problem 17 — and it must reach 32.67 Ω there. The device of Problem 17 reached only 28.07 Ω inductive, so this application needs a larger reactor relative to its capacitor.

Option 2 — a series capacitor in the long route. Now compensate \(X_B\) downwards until the division is right:

A degree of compensation of 43.8% is routine and well inside the 70% ceiling, and the rating is 22% below the TCSC's because the capacitor sits in the lightly-loaded path where the current is smaller.

The comparison that decides it. Look at what each does to the corridor as a whole:

Both fix the division; only one improves the corridor. Adding reactance to the strong path throws away 30% of the corridor's stability limit to solve a thermal problem, while removing reactance from the weak path solves the same problem and raises the limit by 24%. Compensate the path you want more flow on, not the one you want less flow on — a rule that follows from this table and is worth more than the arithmetic that produced it.

What a UPFC adds. Both options above change a reactance, which is one of the three terms in \(P = V_1V_2\sin\delta/X\). A UPFC injects a series voltage of arbitrary magnitude and arbitrary angle, so it can instead insert an effective phase shift in one path:

The third option is the interesting one: it re-divides the flow while leaving both path reactances, and therefore the corridor's stability limit, untouched. Its cost is that the series converter must exchange real power with the line, which the shunt converter draws from the bus through the common DC capacitor — the reason a UPFC has two converters and a series capacitor has none.