Electrical Engineering 2024 Paper II 50 marks Calculate

Paper II — Q7

(a) Calculate the real and reactive power at sending end of a transmission line while delivering 10 MVA load at 0·85 lagging…

(a)

Calculate the real and reactive power at sending end of a transmission line while delivering 10 MVA load at 0·85 lagging power factor at receiving end of line. The line parameters are A = 1, B = 12·12 ∠64·64° Ω, D = 1 and receiving end voltage of line is 33 kV. 20 marks

(b)
(i)

A binary transmission system with a transmitted power of 300 mW uses a channel with zero-mean AWGN of two-sided PSD equal to 10⁻¹⁵ W/Hz and a total transmission loss of 80 dB. If the probability of error, Pₑ is not to exceed 10⁻⁴, calculate the maximum allowable bit rate using non-coherent ASK. 10 marks

(ii)

A 2Vₚₚ audio frequency signal band-limited to 8 kHz is to be transmitted using a PCM system. If the quantization error of any sample is to be at the most ±1% of the dynamic range of the message signal, determine the minimum value of n, the minimum sampling rate and corresponding bit rate of transmission. 10 marks

(c)
(i)

Mention the techniques of increasing the voltage and current rating of converter station of HVDC transmission system. 5 marks

(ii)

Write the requirements of valves used in HVDC transmission system. 5 marks

हिंदी में प्रश्न पढ़ें
(a)

एक संचरण लाइन के प्रेषण छोर पर वास्तविक एवं प्रतिघाती शक्तियों की गणना कीजिए जबकि संचरण लाइन अपने ग्रहण छोर पर 10 MVA, 0·85 पश्चगामी शक्ति गुणांक का भार प्रदान करती है । लाइन के प्राचल A = 1, B = 12·12 ∠64·64° Ω, D = 1 और ग्रहण छोर पर लाइन की वोल्टता 33 kV है । (20 अंक)

(b)
(i)

300 mW संचरित शक्ति के साथ एक द्विआधारी संचरण तंत्र 10⁻¹⁵ W/Hz के बराबर द्विशोर PSD के शून्य माध्य AWGN तथा 80 dB कुल संचरण ह्रास वाले चैनल का उपयोग करता है । यदि त्रुटि की संभावना Pₑ, 10⁻⁴ से अधिक नहीं होनी है, तो असुसंगत ASK का प्रयोग करते हुए अधिकतम स्वीकार्य बिट दर की गणना कीजिए । (10 अंक)

(ii)

एक 2Vₚₚ ध्वनि आवृत्ति संकेत जो कि 8 kHz तक बैंड-लिमिटेड है, को एक PCM तंत्र के माध्यम से प्रेषित किया जाना है । यदि किसी नमूने की अधिकतम क्वांटाइजेशन त्रुटि सूचना संकेत की गतिक सीमा की ±1% होनी है, तो n का न्यूनतम मान, न्यूनतम प्रतिचयन (सैंपलिंग) दर तथा तत्संगत प्रेषण की बिट दर का निर्धारण कीजिए । (10 अंक)

(c)
(i)

HVDC संचरण प्रणाली के कनवर्टर स्टेशन की वोल्टता तथा धारा की दर निर्धारण (रेटिंग) को बढ़ाने वाली तकनीकों का उल्लेख कीजिए । (5 अंक)

(ii)

HVDC संचरण प्रणाली में प्रयुक्त वाल्वों की अपेक्षाओं का उल्लेख कीजिए । (5 अंक)

Q7 of the 2024 UPSC Mains Electrical Engineering Paper II, as printed
The question as printed in the 2024 Electrical Engineering paper

Model answer

Drafted with AI by UPSC Answer Check against this question's marking rubric, to the expected length. UPSC does not publish answers for Mains — this is one way to score well, not an official key. How we build rubrics and answers.

(a) Sending-end real and reactive power

Nature of the line. For a passive reciprocal line, AD − BC = 1. With A = D = 1 this gives BC = 0, so C = 0. The line is a short line with only the series impedance B, and I_S = I_R.

Receiving end. S_R = 10 MVA at 0.85 lagging: P_R = 8.5 MW; Q_R = 10 × sin(cos⁻¹0.85) = 10 × 0.5268 = 5.268 MVAr V_R = 33/√3 = 19.053 kV per phase (reference) I_R = 10×10⁶/(√3 × 33×10³) = 174.96 A, at −31.79°

Line impedance. B = 12.12∠64.64° Ω, so R = 12.12 cos 64.64° = 5.191 Ω and X = 12.12 sin 64.64° = 10.952 Ω.

Line losses (three-phase). 3I²R = 3 × 174.96² × 5.191 = 0.477 MW 3I²X = 3 × 174.96² × 10.952 = 1.006 MVAr

Sending end. P_S = 8.5 + 0.477 = 8.977 MW Q_S = 5.268 + 1.006 = 6.274 MVAr (lagging)

Check through V_S. V_S = V_R + I_R R + jI_R X = 19 053 + (772 − j478) + (1009 + j1629) = 20 834 + j1150 V = 20.87 kV per phase at δ = 3.16°, which is 36.14 kV line. S_S = 3V_SI_S* = 8.977 + j6.274 MVA, confirming the result. The sending-end power factor is 8.977/10.95 = 0.82 lagging, and the voltage regulation is (20.87 − 19.05)/19.05 = 9.5%.

Diagram: Short-line phasor diagram: V_S = V_R + I(R + jX) for the 10 MVA, 0.85 pf lagging load

(b)(i) Maximum bit rate for non-coherent ASK

Received power. 80 dB loss is a factor of 10⁸: P_R = 0.3/10⁸ = 3×10⁻⁹ W

Noise. Two-sided PSD N₀/2 = 10⁻¹⁵ W/Hz, so N₀ = 2×10⁻¹⁵ W/Hz.

Error probability. Take 300 mW as the average transmitted power, so E_b = P_RT_b is the average energy per bit. For on-off keying with envelope detection at high SNR: P_e ≈ ½ exp(−E_b/2N₀)

Required energy. ½ exp(−E_b/2N₀) ≤ 10⁻⁴ ⇒ E_b/2N₀ ≥ ln(5000) = 8.517 E_b ≥ 17.03 × 2×10⁻¹⁵ = 3.41×10⁻¹⁴ J

Maximum bit rate. R_b = P_R/E_b = 3×10⁻⁹/3.41×10⁻¹⁴ = 8.8×10⁴ bit/s ≈ 88 kbit/s If 300 mW were instead the power during a "1" pulse, the average energy would be half as large, and the rate would fall to about 44 kbit/s. Coherent ASK would allow a somewhat higher rate for the same P_e.

(b)(ii) PCM: minimum n, sampling rate and bit rate

Step size. The dynamic range is 2 V (peak-to-peak). The maximum quantisation error is half a step, so Δ/2 ≤ 0.01 × 2 V = 0.02 V ⇒ Δ ≤ 0.04 V

Number of bits. L = 2/Δ ≥ 50 levels. 2⁵ = 32 is too few, so n = 6 (64 levels). Check: Δ = 2/64 = 0.03125 V, and the maximum error is 0.0156 V = 0.78% of the range, within 1%.

Sampling rate. f_s,min = 2 × 8 kHz = 16 kHz (16 000 samples/s).

Bit rate. R_b = n f_s = 6 × 16 000 = 96 kbit/s. The minimum transmission bandwidth is R_b/2 = 48 kHz, and the SQNR for a full-scale sinusoid is about 6.02 × 6 + 1.76 ≈ 38 dB.

(c)(i) Raising the voltage and current ratings of an HVDC converter station

Voltage rating

  • Connect many thyristors in series in each valve, with RC damping circuits and grading capacitors for equal static and transient voltage sharing, plus a few redundant thyristor levels.
  • Use higher-voltage thyristors (up to about 8.5 kV per device) and multi-tier valve stacks with proper insulation coordination, bushings and clearances.
  • Connect 6-pulse bridges in series to form 12-pulse groups, and series groups per pole. Operate bipolar (±V_d) to double the line voltage for a given insulation level to earth.
  • In VSC (MMC) stations, increase the number of series-connected submodules per arm.

Current rating

  • Use larger-diameter thyristor wafers (5- and 6-inch devices carrying about 4–5 kA) instead of paralleling devices.
  • Connect valves or complete converter groups in parallel where a single device rating is not enough.
  • Use efficient de-ionised water cooling of the valves to raise the continuous and overload current.
  • In VSC stations, use larger or paralleled press-pack IGBT modules. Light-triggered thyristors are widely used so that long series strings fire simultaneously and reliably.

(c)(ii) Requirements of HVDC valves

  • Voltage withstand: high forward and reverse blocking voltage, including switching and lightning transients and high dv/dt, with uniform voltage sharing along the series string.
  • Current capability: rated continuous current, specified overloads, and fault surge currents (commutation failure, short circuits) without damage; limited di/dt at turn-on, using saturable valve reactors.
  • Firing: all series thyristors must fire simultaneously and reliably, through fibre-optic electrical or light triggering, with firing possible even at low voltage.
  • Protection: forward-overvoltage protection (breakover-diode firing), surge arresters across valves, and a few redundant levels so one failed thyristor does not trip the pole.
  • Losses and cooling: low conduction and switching losses, low reverse-recovery charge, and effective (water) cooling.
  • Reliability and maintenance: monitoring of each thyristor level, fire-safe materials, seismic and mechanical strength, low electromagnetic interference, and easy replacement of components.

How this answer will be evaluated

What earns full marks: All parts show complete method with correct phasor/circuit setup, proper unit handling, and final values with interpretation; theoretical parts name specific techniques and standards.

Evaluation rubric

Each sub-part is marked on its own, against the marks and word limit printed on the paper.

  1. (a) Find the sending-end real and reactive power for a 10 MVA, 0.85 lagging load at 33 kV with A = D = 1, B = 12.12∠64.64° Ω. 20 marks

    calculate— given → formula → substitution → result with units → interpretation

    Must cover

    • AD − BC = 1 ⇒ C = 0 (short line), so I_S = I_R
    • I_R = 10×10⁶/(√3 × 33×10³) = 174.96 A ∠−31.79°; P_R = 8.5 MW, Q_R = 5.27 MVAr
    • R = 5.191 Ω, X = 10.952 Ω; losses 3I²R = 0.477 MW, 3I²X = 1.006 MVAr
    • P_S = 8.977 MW, Q_S = 6.274 MVAr (lagging)

    Loses marks

    • Using single-phase power with line voltage
    • Adding |I|²|Z| as real loss

    Earns more

    • V_S = V_R + BI_R = 20.87 kV/phase (36.14 kV line), δ = 3.16°
    • Check by S_S = 3V_S I_S*; sending pf 0.82; regulation 9.5%

    Extra mark

    • Phasor diagram
  2. (b)(i) Find the maximum bit rate for non-coherent ASK with 300 mW transmitted, 80 dB loss, two-sided PSD 10⁻¹⁵ W/Hz and P_e ≤ 10⁻⁴. 10 marks

    calculate— given → formula → substitution → result with units → interpretation

    Must cover

    • Received power = 0.3 × 10⁻⁸ = 3×10⁻⁹ W; N₀ = 2 × 10⁻¹⁵ W/Hz
    • Non-coherent ASK (average bit energy E_b): P_e ≈ ½ exp(−E_b/2N₀)
    • E_b/2N₀ ≥ ln(1/(2×10⁻⁴)) = 8.517 ⇒ E_b ≥ 3.41×10⁻¹⁴ J
    • R_b = P_R/E_b ≈ 8.8×10⁴ bit/s (88 kbit/s)

    Loses marks

    • Taking N₀ = 10⁻¹⁵ as one-sided
    • Ignoring the 80 dB loss

    Earns more

    • States energy convention; if 300 mW is on-state power, R_b ≈ 44 kbit/s
    • Comparison with coherent ASK

    Extra mark

    • Note on optimum threshold at high SNR
  3. (b)(ii) For a 2 V p-p signal band-limited to 8 kHz, find the minimum n, sampling rate and bit rate for PCM with quantisation error ≤ ±1% of the dynamic range. 10 marks

    calculate— given → formula → substitution → result with units → interpretation

    Must cover

    • Maximum error Δ/2 ≤ 0.01 × 2 V = 0.02 V ⇒ Δ ≤ 0.04 V
    • L = 2/Δ ≥ 50 ⇒ n = 6 (64 levels)
    • f_s,min = 2 × 8 kHz = 16 kHz
    • R_b = 6 × 16 000 = 96 kbit/s

    Loses marks

    • Using Δ ≤ 0.02 V (full step instead of half step)
    • Sampling at 8 kHz

    Earns more

    • Check: with n = 6, max error = 2/128 = 0.0156 V = 0.78%
    • Minimum transmission bandwidth R_b/2 = 48 kHz

    Extra mark

    • SQNR ≈ 6.02n + 1.76 dB for a full-scale sinusoid
  4. (c)(i) State the techniques for raising the voltage and current ratings of an HVDC converter station. 5 marks

    describe— define → structure or process in order → labelled diagram → significance

    Must cover

    • Voltage: many thyristors in series in each valve with RC grading and redundant levels; series 6-pulse bridges forming 12-pulse groups per pole; bipolar operation (±V)
    • Higher-voltage thyristor wafers (up to about 8.5 kV per device) and multi-tier valve stacks with insulation coordination
    • Current: larger-diameter wafers (5–6 inch, 4–5 kA), parallel valves or converter groups, efficient de-ionised water cooling
    • For VSC/MMC: more series submodules for voltage; paralleled or larger press-pack IGBT modules for current

    Loses marks

    • Only generic statements without specific means

    Earns more

    • Light-triggered thyristors for simultaneous firing of long strings

    Extra mark

    • Example ratings of recent UHVDC links (±800 kV class)
  5. (c)(ii) State the requirements of valves used in HVDC transmission. 5 marks

    describe— define → structure or process in order → labelled diagram → significance

    Must cover

    • High forward and reverse blocking capability with transient overvoltage, dv/dt and uniform voltage sharing across the series string
    • Continuous, overload and fault surge current capability; di/dt limitation at turn-on
    • Simultaneous reliable firing of all series thyristors; forward-overvoltage protection (e.g. BOD) and redundancy
    • Low conduction and switching losses, adequate cooling, low recovery charge, high reliability and maintainability

    Loses marks

    • Listing AC breaker requirements instead

    Earns more

    • Monitoring of each thyristor level
    • Fire safety and seismic design

    Extra mark

    • Distinction between LCC thyristor valves and VSC IGBT valves
What "Calculate" is asking you to do

Apply the standard formula or schedule to data the question has already supplied — a table of readings, cost records, a balance sheet — and produce the number. The method is rarely in doubt; the marks sit in the named intermediate quantities, each of which has to appear as a labelled line.

Structure that answers it

Data as given → formula or standard treatment, named → substitution → each intermediate, labelled → result with units

Where marks are lost

Omitting an intermediate the marking scheme pays for separately, or rounding at an intermediate line so the final figure drifts. In commerce and accountancy, any figure in a statement that no numbered working note supports is treated as unearned.

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