Q5Power System - I

Question

Q.5. (a) Explain the working of SF6 gas circuit breaker.

(b) An 11kV, 3-phase transmission line has a resistance of 1.5 ohm and reactance of 4 ohm per phase. Calculate the percentage regulation and efficiency of the line when a total load of 5000kVA at 0.8 lagging power factor is supplied at 11kV at the distant end.

Answer 1

(a) Working of SF6 Gas Circuit Breaker

An SF6 gas circuit breaker uses sulphur hexafluoride gas — an electronegative gas with very high dielectric strength (about 2.5-3 times that of air at the same pressure) and superior heat-absorption/arc-quenching capability — as both the insulating and arc-extinguishing medium, sealed within the breaker enclosure at moderate pressure (typically 3-7 bar). When the contacts separate to interrupt current, an arc is drawn between them; the surrounding SF6 gas absorbs free electrons in the arc (due to its strongly electronegative molecular property), rapidly de-ionizing the arc path and rebuilding dielectric strength across the widening gap. In the common puffer design, the breaker's own contact-separation mechanism mechanically compresses a volume of SF6 gas, which is then released as a high-velocity blast directed axially through a nozzle onto the arc at the critical moment; interruption occurs at (or close to) the natural current zero-crossing, when the gas's rapidly recovering dielectric strength exceeds the rate of rise of the transient recovery voltage appearing across the contacts. Self-blast (thermal) and rotating-arc SF6 breaker designs use alternative mechanisms (arc-generated pressure rise, or magnetically rotated arc to distribute heating) to achieve similar rapid arc extinction with reduced operating mechanism force requirements compared to a pure puffer design.

(b) Percentage Regulation and Efficiency Calculation

Given: 11 kV, 3-phase line, R = 1.5 Ω/phase, X = 4 Ω/phase, load = 5000 kVA at 0.8 lagging pf, supplied at 11 kV at the receiving (distant) end.

I = \frac{S}{3 V _{r}} = \frac{5 , 000 , 000}{3 \times 11 , 000} = 262.43 A

Using the approximate voltage-drop formula (phase basis), with cosφ=0.8, sinφ=0.6, and Vr(phase) = 11,000/√3 = 6350.85 V:

Approx. drop (phase) = I (R cos ϕ + X sin ϕ) = 262.43 \times (1.5 \times 0.8 + 4 \times 0.6) = 262.43 \times 3.6 = 944.7 V

V_{s} (phase) \approx V_{r} (phase) + drop = 6350.85 + 944.7 = 7295.6 V, V_{s} (line) = 7295.6 \times 3 = 12, 634 V

% Regulation = \frac{V _{s} - V _{r}}{V _{r}} \times 100 = \frac{12 , 634 - 11 , 000}{11 , 000} \times 100 \approx 14.85%

Note: using the more precise (exact, quadrature-inclusive) voltage drop formula, Vs(phase) = √[(Vr+IRcosφ+IXsinφ)² + (IXcosφ-IRsinφ)²], gives a slightly refined value: (IXcosφ-IRsinφ) = 262.43×(4×0.8-1.5×0.6) = 262.43×(3.2-0.9) = 262.43×2.3 = 603.6 V; Vs(phase) = √(7295.6²+603.6²) = √(53,225,780+364,334) = √53,590,114 ≈ 7320.5 V, giving Vs(line) ≈ 12,679 V and %Regulation ≈ 15.26% — both the approximate and exact methods give a voltage regulation in the range of approximately 14.9-15.3%, reflecting the fairly heavily loaded (high current relative to voltage class) nature of this short 11 kV line segment.

Transmission efficiency:

P_{r} = S cos ϕ = 5000 \times 0.8 = 4000 kW

P_{l oss} = 3 I^{2} R = 3 \times (262.43)^{2} \times 1.5 = 309, 770 W \approx 309.8 kW

η = \frac{P _{r}}{P _{r} + P _{l oss}} \times 100 = \frac{4000}{4000 + 309.8} \times 100 \approx 92.81%

Result: the line has an approximate voltage regulation of about 15% and a transmission efficiency of approximately 92.8%, reflecting the relatively high current (262 A) drawn at this modest 11 kV distribution voltage level for a 5 MVA load — illustrating why such heavier loads are normally served at a higher sub-transmission voltage in practice to keep both regulation and losses within more typical, acceptable limits (regulation under 5-10%, efficiency above 95-97%) for distribution-class equipment.

Extended Discussion: SF6 Breaker Rating Classes and Arc-Interruption Media Comparison

SF6 circuit breakers are manufactured across the full range from indoor medium-voltage switchgear (11-36 kV, typically single-pressure puffer design in a compact metal-clad enclosure) up to the largest EHV/UHV outdoor breakers (400 kV to 1200 kV, using multiple series interrupter units per pole with grading capacitors to share the recovery voltage evenly). Compared to the alternative arc-quenching media used historically and in present-day practice — oil (bulk oil and minimum-oil breakers, now largely obsolete due to fire risk and high maintenance), air-blast (requiring a large compressed-air plant and noisy operation, now rare), and vacuum (dominant at distribution/medium voltage up to about 36-40 kV, but not yet cost-effective at transmission voltages for very high system voltage ratings) — SF6 offers the best overall combination of compact size, low maintenance, and voltage-scalability from medium voltage right up to UHV, which is why it has become the dominant circuit breaker technology at transmission voltage levels worldwide, notwithstanding its potent greenhouse-gas environmental drawback that is now driving research into SF6-free alternative gas mixtures (such as fluoronitrile/CO2 blends) for new installations.

Extended Numerical Check: Exact Regulation via Complex Phasor Method

As an independent verification of the voltage-drop calculation, the sending-end phase voltage can be computed directly using complex phasor arithmetic rather than the approximate/exact scalar drop formulas used above. Taking Vr(phase) = 6350.85∠0° V as reference, and Ir = 262.43∠-36.87° A = 262.43×(0.8-j0.6) = 209.94-j157.46 A (using cosφ=0.8, sinφ=0.6 lagging):

Z = R + j X = 1.5 + j 4.0 Ω, I_{r} \cdot Z = (209.94 - j 157.46) (1.5 + j 4.0)

I_{r} Z = (209.94 \times 1.5 + 157.46 \times 4.0) + j (209.94 \times 4.0 - 157.46 \times 1.5) = (314.91 + 629.84) + j (839.76 - 236.19) = 944.75 + j 603.57 V

V_{s} (phase) = V_{r} + I_{r} Z = (6350.85 + 944.75) + j 603.57 = 7295.60 + j 603.57 V

∣ V_{s} ∣ (phase) = 7295.6 0^{2} + 603.5 7^{2} = 53, 225, 800 + 364, 298 = 53, 590, 098 \approx 7320.5 V

This matches exactly the 'exact' method value obtained earlier (7320.5 V), confirming that the full complex-phasor approach and the resolved-component (in-phase/quadrature) approximate-exact method are algebraically identical, as expected, and giving Vs(line) = 7320.5×√3 ≈ 12,679 V and %Regulation = (12,679-11,000)/11,000×100 ≈ 15.26%, reconfirming the earlier result.

Extended Discussion: Why Regulation and Efficiency Move in Opposite Directions Here

It is instructive to note that this line simultaneously exhibits a fairly high voltage regulation (~15%) alongside a very respectable transmission efficiency (~92.8% here, or up to about 97% when computed via the input-power-referenced formula rather than the receiving-power-referenced formula used above) — these two performance measures are not directly linked, since regulation depends primarily on the line's reactance (which does not dissipate real power but does cause a substantial voltage phase/magnitude shift for a lagging power factor load), while efficiency depends primarily on the line's resistance (which directly dissipates I²R real power loss). A line can therefore have low losses (high efficiency) yet still exhibit significant voltage regulation if its X/R ratio is high, exactly the situation illustrated by this problem's R=1.5Ω, X=4Ω per phase (X/R ratio of 2.67), and this decoupling of regulation from efficiency is precisely why both quantities must be calculated and checked separately when evaluating a transmission or distribution line's overall performance, rather than assuming that a low-loss line will automatically also exhibit good (low) voltage regulation.

For comparison, if this same 5000 kVA, 0.8-lagging-pf load were instead supplied at a leading power factor of 0.8 leading, the sinφ term in the drop formula would effectively change sign, reducing or even reversing the net voltage drop and potentially causing the receiving-end voltage to exceed the sending-end voltage — the voltage-rise phenomenon discussed elsewhere in this paper for leading power factor loads — while the transmission efficiency, being governed only by I²R loss (which depends on the current magnitude squared regardless of its phase angle relative to voltage), would remain essentially unchanged from the lagging-pf case for the same load kVA. This further reinforces the point that voltage regulation is fundamentally a reactive-power/phase-angle-dependent quantity, whereas transmission efficiency is fundamentally a resistive-loss/current-magnitude-dependent quantity, and the two must always be evaluated as logically separate performance criteria for any transmission or distribution line design study.

Finally, note that if power-factor-correction capacitors were installed at the receiving end to raise the load power factor from 0.8 lagging closer to unity, both regulation and efficiency for this same 11 kV line would improve simultaneously: a higher power factor reduces the reactive current component (lowering the IXsinφ term in the regulation formula) and also reduces the total current magnitude for the same real power delivered (lowering I²R loss and hence improving efficiency), which is precisely why power factor improvement at the load end is one of the most cost-effective practical measures available to a distribution utility for simultaneously improving voltage profile and reducing losses on lines exactly like the one analyzed in this problem.