Regulation and Efficiency

From Set 12, the sending-end conditions and performance:

The Ferranti rise and surge impedance loading:

The three failures, against ordinary limits:

The line runs at 300 MW against an SIL of 422 MW — that is, at 71% of its natural loading, so it is a net generator of reactive power.

Computing \(\mathbf{V}_S = A\mathbf{V}_R + B\mathbf{I}_R\) at each loading, with \(I_R\) scaling in proportion to the power:

Two things move together and one moves against them. Regulation rises steadily with loading; efficiency falls; and the sending-end power factor swings from strongly leading to lagging, crossing unity near 320 MW.

At 100 MW the sending-end power factor is 0.593 leading — the line's charging current dominates the load current almost completely.

The loss is essentially \(3I^2R\) integrated along the line, so it grows as the square of the loading while the delivered power grows only linearly:

— a hyperbolic fall, which is what the table shows.

At light load the loss does not fall to zero, because the charging current still flows and dissipates in the conductors:

What would improve it. Efficiency depends on \(r\) and on the current, so:

At \(r = 0.16\ \Omega\)/km this line carries a single unbundled conductor. A twin-bundle design of the same voltage, as in Set 7, would have \(r \approx 0.016\ \Omega\)/km and an efficiency above 98%.

Regulation is

and \(V_S\) itself contains both effects. Separate them by asking what each would give alone.

The Ferranti part. Set the load to zero: \(V_S = |A|V_R\) and the regulation would be

— which is not the answer, because the Ferranti effect only produces regulation when a fixed \(V_S\) is held. Take instead \(V_S\) at its loaded value of 460.6 kV and remove the load:

The two components of the 108.4 kV total rise:

The remedies map directly onto the split:

Reactors distributed along the line cancel a fraction \(k\) of its shunt susceptance, so the effective per-km admittance becomes

Since \(z\) is unchanged and only the magnitude of \(y\) alters:

The angles are untouched, because \(y\) remains purely imaginary.

The electrical length falls:

against 25.6° uncompensated — the line now behaves as though it were 253 km long.

The new constants:

The Ferranti rise falls accordingly:

Applying \(\gamma_{\text{eff}} = \gamma\sqrt{1-k}\) and \(Z_{c,\text{eff}} = Z_c/\sqrt{1-k}\) at each degree:

To bring the Ferranti rise within 5% the table shows \(k = 0.5\) gives 4.99% — just inside:

Exactly at the 420 kV insulation class limit, so 60% would be chosen in practice for margin.

Reading the columns against each other reveals the trade. As \(k\) rises:

The line's total shunt susceptance and its charging:

Reactors absorbing 60% of it:

Split equally between the two ends:

Reactance and inductance of the total bank, per phase:

Each end's reactor is twice this reactance and twice the inductance — 2778 \(\Omega\) and 8.84 H per phase.

These are large machines: an 8.8 henry oil-immersed reactor for 400 kV service stands several metres high and weighs tens of tonnes per phase.

Using the compensated constants from Problem 5:

What improved: the Ferranti rise, from 10.4% to 4.0% — a 62% reduction, exactly as designed.

What got worse:

What barely changed: the regulation, from 27.08% to 25.55%. The lower \(V_{R,\text{NL}}\) was almost entirely offset by the higher \(V_S\) needed to drive the load.

The line now runs at 300 MW against an SIL of 267 — above its natural loading, where before it was below. The reactive character has reversed.

Series compensation reduces the reactive part of \(z\), leaving the resistance alone:

Note that the angle of \(z\) changes now, unlike the shunt case — so \(\gamma\) and \(Z_c\) both rotate as well as scale.

Recomputing from scratch:

The new constants:

Performance at 300 MW, 0.95 lagging:

Recomputing at each degree of series compensation:

Setting the two mechanisms side by side at equal effect on the Ferranti rise (about 5%):

The same improvement in \(|A|\), and an SIL differing by a factor of 1.8.

The reason. Both reduce \(|\gamma| = \sqrt{|z||y|}\) and hence improve \(|A|\). But

The mechanisms. Both reduce \(|\gamma| = \sqrt{|z||y|}\) and so improve \(|A|\). They differ entirely in their effect on \(Z_c = \sqrt{z/y}\):

At equal effect on the Ferranti rise (about 5%), from Problems 6 and 10:

Series compensation is better on every count here — better regulation, higher SIL, smaller rating. So why are reactors used at all?

Because they solve a problem series capacitors cannot:

So the two are complementary, not alternatives. Series capacitors serve the heavily loaded condition — regulation, power transfer, stability. Shunt reactors serve the lightly loaded one — Ferranti, energisation, generator var absorption.

Both modifications applied to the per-km parameters:

Recomputing:

Performance, against the three single-remedy cases:

The combination gives the best Ferranti figure (1.94%) and a regulation of 17.78% — but the regulation is still far above the 10% target.

Note also that SIL has landed at 346 MW: the series compensation's increase and the shunt's decrease have partly cancelled, leaving it near the original.

At 300 MW, 0.95 lagging:

Every form of compensation made the efficiency slightly worse.

The reason is straightforward. The loss is \(I^2R\) and neither remedy alters \(R\). What they do alter is the current distribution:

The changes are fractions of a percentage point — small, but the sign is the point: compensation is not a loss-reduction measure.

The 300 MW load at 0.95 lagging demands

Correcting to each target requires the capacitors to supply the load's demand plus any leading component:

Evaluating, with the resulting line performance:

Correction to 0.95 leading brings the regulation to 9.05% — inside the 10% target at last, and the sending voltage down to 427.9 kV.

The efficiency, however, falls slightly to 87.4%, because the leading current adds to the conductor current without delivering power.

Computing \(Q_S = 3V_SI_S\sin\phi_S\) and \(Q_R = P_R\tan\phi_R\) at each loading, with a negative \(Q_S\) meaning the line delivers vars back to the source:

The crossover in \(Q_S\) — where the sending end changes from absorbing vars to supplying them — occurs near 340 MW.

The line's own net generation falls steadily from 168 MVAr at no load to 23 MVAr at 500 MW, as the \(I^2X_L\) absorption grows to offset the \(V^2B\) generation.

Setting the two equal gives the surge impedance loading:

Though the tabulated net generation does not reach zero until beyond 500 MW, because the load's own reactive demand at 0.95 lagging is being supplied through the line as well.

Case A — 422 MW at unity power factor. The receiving current is real:

Through the exact constants \(A = 0.9060\angle2.35^\circ\), \(B = 166.87\angle68.94^\circ\):

A rise of 61.6 kV over the line — 15.4% — at the loading that is supposed to give a flat profile.

Case B — terminated in \(Z_c\). Now the current is set by the impedance, not chosen:

The same magnitude — but leading by 10.90°, because \(Z_c\) of a real line is slightly capacitive.

Repeating the calculation gives \(V_S = 435.9\) kV line — a rise of 8.98%, not 15.4%. And the sending-end angle is:

Identical to eight figures. That is the signature of a matched line: the voltage at any point is the sending voltage retarded in phase by \(\beta x\) and attenuated by \(e^{-\alpha x}\), nothing more.

The profiles. Both cases delivered 609.4 A, so the difference is entirely in the phase of that current:

The matched profile is a pure exponential. Check it:

Exactly. The 8.98% rise is not a Ferranti effect and not a regulation problem — it is the attenuation of a wave, read backwards.

The power factor is the same at both ends. At the receiving end \(P + jQ = 414.6 - j79.84\) MVA; at the sending end \(492.4 - j94.83\). The ratios:

A matched line has the same power factor everywhere along it — \(\cos(10.90^\circ) = 0.982\) leading. It neither generates nor absorbs net vars; it merely passes them in proportion.

The efficiency, however, is poor: 414.6/492.4 = 84.20%. Nearly 78 MW lost. SIL is a reactive optimum, not a loss optimum — it is the loading at which \(I^2X_L\) absorption exactly matches \(V^2B\) generation, and \(r\) plays no part in that balance at all.

The lossless case. With \(r = g = 0\), \(\gamma = j\beta\) is purely imaginary and the hyperbolic functions become circular:

With \(Z_c = \sqrt{x/b}\) now purely real — the surge impedance proper.

At \(\beta l = \pi\) — one half wavelength — \(\cos\pi = -1\) and \(\sin\pi = 0\):

What that means. Substituting into the two-port equations:

The receiving voltage equals the sending voltage in magnitude whatever the load — including no load. Regulation is identically zero, and the sending current equals the receiving current. The line has become a perfect 1:1 transformer with a 180° phase shift.

The half-wave length for this line. From \(\beta = 1.11634\times10^{-3}\) rad/km:

Corresponding to a propagation velocity \(v = f\lambda = 50\times5628 = 2.814\times10^{5}\) km/s, or \(0.938c\).

With the resistance restored. Evaluating the exact constants at \(l = 2814\) km with \(z = 0.16 + j0.40\):

Using \(\cosh(u + j\pi) = -\cosh u\). The magnitude is above unity, not equal to it.

And \(B = Z_c\sinh(\gamma l)\) with \(\sinh(u + j\pi) = -\sinh u\):

Not zero — 243 Ω, larger than the whole 400 km line's \(|B|\) of 167 Ω.

So the ideal fails on both counts. On no load the receiving voltage is now \(V_S/|A| = 0.843V_S\) — a 16% drop, the opposite of Ferranti. And the loss over 2814 km is crippling: \(e^{-2\alpha l} = e^{-1.21} = 0.298\), so less than a third of the incident power survives the journey.

The requirement. We need \(\beta l = \pi\) over 400 km:

Seven times the phase constant the line has.

How to get it. Since \(\gamma = \sqrt{zy}\), multiplying both \(x\) and \(b\) by a factor \(m\) multiplies \(\gamma\) by \(m\) while leaving \(Z_c = \sqrt{z/y}\) nearly unchanged. That means added series inductance and added shunt capacitance — the reverse of both conventional remedies.

Because \(r\) cannot be scaled with them, the exact factor is found by solving \(\beta l = \pi\) for \(m\) in

which gives \(m = 7.167\), a little above the naive 7.035 because the resistance drags \(\beta\) down slightly.

The resulting parameters:

The attenuation over the whole line is now only 0.0876 Np — a twentieth of what 2814 km of real line would cost.

The constants:

Against 0.9060 and 166.9 Ω uncompensated. \(|B|\) has fallen by a factor of five and \(|A|\) is within 0.4% of unity — the no-load voltage now falls by 0.38% instead of rising by 10.37%.

The bill. The added shunt susceptance over 400 km is \(\Delta B = (7.167-1)\times3.0\times10^{-6}\times400 = 7.40\times10^{-3}\) S:

Six times the line's own charging of 192 MVAr, and nearly three times the SIL.

And the added series reactance:

Against the line's own 160 Ω. We would be adding six times more reactance than the line possesses, at 400 kV, carrying full load current.

And the profile is the real objection. Loading the synthesised line at 300 MW, 0.95 lagging with \(V_R = 400\) kV:

The terminals are well behaved; the interior is not. At 100 km the voltage is 461 kV and at 300 km it is 350 kV — a 111 kV swing inside a line whose two ends differ by 33 kV. Insulation must be rated for the maximum, not the terminal value.

The two-reactor case. Each end carries 57.6 MVAr, so each reactor has

The sending-end reactor sits on the source busbar and does not affect the profile at all; only the receiving-end one does.

Setting up the profile. Measure \(x\) from the receiving end. The current leaving the line there is the reactor current:

Evaluating at \(x = 400\) and scaling so that \(|V(400)| = 400\) kV:

The maximum is not at either end. Scanning the profile finely puts it at \(x = 119\) km, where \(V = 419.6\) kV — 3.6 kV above the receiving end and 19.6 kV above the sending end.

Why the bulge exists. A lumped reactor absorbs charging current only where it stands. Between the two reactors the line's distributed capacitance goes on generating vars with nothing local to absorb them, so the voltage climbs until the series reactance carrying that current away creates enough drop to arrest it. The two reactors have suppressed the terminal symptom and left an interior one.

Three units of 38.4 MVAr. Repeating with \(B_r = 2.40\times10^{-4}\) S at each end and the same at 200 km, the mid-point reactor drawing \(-jB_r\mathbf{V}_{\text{mid}}\) out of the through current:

Maximum 417.5 kV at 80 km. The kink at 200 km is the mid-point reactor taking its share.

The comparison, for identical installed MVAr:

The terminal voltage is essentially unchanged — 0.1 kV — but the interior peak falls by 2.1 kV. The compensation has been made to act where it is needed rather than where it is convenient.

The starting point. Uncompensated, at 300 MW and 0.95 lagging, from Problems 1 and 2:

Both targets are missed by a wide margin.

Step 1 — shunt reactors alone, 60%. From Problem 6, \(|A|\) rises to 0.9619:

The no-load target is nearly met; the regulation target is untouched, and SIL has been cut by 37%.

Step 2 — series capacitors alone, 50%. From Problem 9:

A large improvement in regulation, and SIL raised by 30% — but 18.29% is still not 10%.

Step 3 — both together. With \(k_{\text{sh}} = 0.60\) and \(k_{\text{se}} = 0.50\) the effective parameters become \(z' = 0.16 + j0.20\) and \(y' = j1.20\times10^{-6}\):

The no-load target is now met with room to spare — 1.94% against 3% — and SIL has settled at 346 MW, between the two single remedies.

But the regulation is 17.78%. Recomputing at 300 MW, 0.95 lagging:

Barely better than series compensation alone. Adding shunt reactors to a series-compensated line does almost nothing for regulation, because \(|A|\) was already near unity and \(|B|\) is unchanged.

Step 4 — the missing variable. Nothing done to the line will fix this, because the drop is being caused by the load's own reactive demand travelling the length of it:

Installing shunt capacitors at the receiving-end busbar to carry the load to 0.95 leading requires

and the improvement is decisive:

The three load power factors, on the compensated line:

Only the last meets the 10% target — and note the efficiency, which improves to unity power factor and then falls, because the leading current is as large as the lagging one was.

The specification. Three separate installations, each answering a different symptom:

What is not fixed. The efficiency, at 87.4%, is worse than the uncompensated line's 89.5%. No arrangement of reactors and capacitors changes \(r = 0.16\ \Omega\)/km, and the leading current needed for the regulation target actually raises \(I^2R\). Getting the efficiency up requires a second circuit, a bundled conductor, or 765 kV — a construction decision, not a compensation one.

And the reactors and capacitors must be switched. From Problem 15 the line's reactive requirement swings 300 MVAr between no load and full load and reverses sign near 340 MW. Leaving 197 MVAr of shunt capacitors connected at no load would add to a rise the reactors are there to suppress. The correct plant is switched in both directions, sequenced to the load.