Paper II — Q3
3.(a) A solidly earthed 400 KV, 3 phase busbar system is connected with two incoming and four outgoing lines (feeders). A…
3.(a) A solidly earthed 400 KV, 3 phase busbar system is connected with two incoming and four outgoing lines (feeders). A differential protection is provided with switchgear of 4000 MVA capacity having the following parameters:
CT secondary resistance = 0.8 Ω Lead wire resistance = 1.2 Ω Relay load = 1.0 Ω CT magnetization current = 0.3 mA/V Max. full load current in one feeder = 100 A Voltage setting of over current relay = 100 V
If the O.C. relay in the spill path is set at 1.0 A, find the following:
The maximum 'through fault' current up to which the protection scheme remains stable. 20 marks
Whether the switchgear is capable to handle maximum through fault current. 20 marks
The value of minimum internal fault current that can be detected by protection scheme. 20 marks
The pick-up setting for detecting minimum internal fault current of 90 Amp. 20 marks
3.(b) Consider a signal detector with an input
r = ±A + n
where +A and −A occur with equal probability and the noise variable n is characterized by the Laplacian pdf shown.
p(n) = (1/√2σ) e^(-|n|√2/σ)
Determine the probability of error as a function of the parameters A and σ. 20 marks
Determine the SNR required to achieve an error probability of 10^(-6). 20 marks
3.(c) A coil of 300 V moving iron voltmeter has a resistance of 500 ohms and an inductance of 0.8 H. The instrument reads correctly at 50 Hz AC supply and takes 100 mA at full scale deflection. What is the percentage error in the instrument reading, when it is connected to 200 V DC supply. 10 marks
हिंदी में प्रश्न पढ़ें
3.(a) दो आने वाले प्रदायकों व चार जाने वाले प्रदायकों को एक दृढ़ता से भू संपर्कित, त्रिकला, 400 KV बसबार तंत्र से जोड़ा गया है । एक 4000 MVA क्षमता वाले सिचंगियर द्वारा अंतरण संरक्षण (डिफरेंशियल प्रोटेक्शन) प्रदान कराया गया है । प्रणाली के प्राचल निम्न प्रकार हैं :
CT का द्वितीयक प्रतिरोध = 0.8 Ω चालक तार का प्रतिरोध = 1.2 Ω रिले भार का प्रतिरोध = 1.0 Ω CT की चुंबकत्व धारा = 0.3 mA/V प्रत्येक प्रदायक की अधिकतम पूर्ण भार धारा = 100 A अधिधारा रिले की बोल्टता का निर्धारण = 100 V
यदि अधिधारा रिले, जो परिपथ में लगाई गई है, को 1.0 पर निर्धारित किया गया है, तो निम्नलिखित तथ्यों का निर्धारण करें ।
अधिकतम शु-फाल्ट धारा का मान, ताकि संरक्षण प्रणाली संतुलित रहे । 20 अंक
क्या संरक्षण प्रणाली अधिकतम शु-फाल्ट धारा को सहन करने में सक्षम है ? 20 अंक
संरक्षण प्रणाली द्वारा संसूचित न्यूनतम अंतरण दोष धारा का मान ज्ञात करें । 20 अंक
न्यूनतम अंतरण दोष धारा 90 A के लिए पिकअप सेटिंग का मान ज्ञात करें । 20 अंक
3.(b) एक संकेत संसूचक का निवेश निम्नप्रकार है
r = ±A + n
+A व −A समप्रायिकता के साथ घटित होता है तथा रवर (नॉयज वेरिएबल) n की विशेषता को लाप्लासियन pdf द्वारा दर्शाया गया है ।
p(n) = (1/√2σ) e^(-|n|√2/σ)
प्राचल A व σ के फलन के रूप में त्रुटि की प्रायिकता का निर्धारण करें । 20 अंक
10^(-6) त्रुटि प्रायिकता के लिए आवश्यक SNR ज्ञात करें । 20 अंक
3.(c) एक 300 V चल लोहे वोल्टमापी की कुण्डली का प्रतिरोध 500 ohm व प्रेरकत्व 0.8 H है । मापक यंत्र 50 Hz AC आपूर्ति पर दोषरहित मापन करता है तथा पूर्ण स्केल विचेषण के समय 100 mA धारा ग्रहण करता है । जब यह मापन यंत्र 200 V DC आपूर्ति के साथ जोड़ा जाता है तो प्रतिशत त्रुटि का मान ज्ञात करें । 10 अंक
Model answer
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(a) High-impedance busbar differential protection The CT ratio is not given; with a 1 A relay setting and 100 A feeder load it is taken as 100/1. There are 6 CTs (2 incoming, 4 outgoing). (a) Maximum stable through-fault current. On an external fault the faulted feeder's CT may saturate completely, so it offers only its winding and lead resistance, R_CT + R_L = 0.8 + 1.2 = 2.0 Ω, to the current of the healthy CTs. The relay (spill) circuit sees I_f,sec(R_CT + R_L), which must not exceed the voltage setting of 100 V: I_f,sec = 100/2.0 = 50 A, so I_f(max) = 50 × 100 = 5000 A. The relay's own burden is not in this loop. (b) Switchgear. Its rated current is 4000 × 10⁶/(√3 × 400 × 10³) = 5774 A, which exceeds 5000 A (equivalently 3464 MVA of through-fault against 4000 MVA), so the switchgear can handle the maximum stable through-fault current. (c) Minimum detectable internal fault current. At the setting voltage each CT draws 0.3 mA/V × 100 V = 0.03 A, so the six CTs draw 0.18 A. With the relay set at 1.0 A: I_min,sec = 1.0 + 0.18 = 1.18 A, so I_min = 118 A primary. (d) Pick-up for a 90 A internal fault. 90 A is 0.90 A secondary, of which 0.18 A magnetises the CTs, so the relay pick-up setting is 0.90 − 0.18 = 0.72 A.
(b) Detector with Laplacian noise (i) For equiprobable ±A the optimum threshold is 0, and by symmetry P_e = P(n > A) = ∫_A^∞ (1/(√2σ)) e^(−√2 n/σ) dn = (1/(√2σ)) × (σ/√2) e^(−√2A/σ) P_e = ½ e^(−√2 A/σ). (ii) The noise variance is σ², so SNR = A²/σ² and P_e = ½ e^(−√(2·SNR)). For 10⁻⁶: e^(−√(2 SNR)) = 2 × 10⁻⁶, so √(2 SNR) = ln(5 × 10⁵) = 13.12 and SNR = 86.1 (≈ 19.35 dB), i.e. A/σ ≈ 9.28.
(c) Moving-iron voltmeter on dc At full scale on ac: Z = 300/0.1 = 3000 Ω; X_L = 2π × 50 × 0.8 = 251.3 Ω, so the total resistance (coil plus series resistor) is R = √(3000² − 251.3²) = 2989.5 Ω. On 200 V dc there is no reactance: I = 200/2989.5 = 66.9 mA. The scale is calibrated for ac with impedance 3000 Ω, so the pointer reads 66.9 mA × 3000 = 200.7 V. Error = (200.7 − 200)/200 ≈ +0.35% (reads high).
How this answer will be evaluated
What earns full marks: Complete derivations with correct formulas, clear circuit models, and proper interpretation of results.
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a) Maximum stable through-fault current, switchgear adequacy, minimum detectable internal fault current and pick-up setting of a high-impedance busbar differential scheme. 20 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Use the stability condition I_f,sec(R_CT + R_L) = 100 V with R_CT + R_L = 2.0 Ω to get 50 A secondary and 5000 A primary (CT ratio 100/1 stated as assumed).
- Compare 5000 A with the switchgear rating 5774 A (4000 MVA at 400 kV) and conclude it is adequate.
- Find I_min = 118 A (1.0 A relay + 6 × 0.03 A) and the pick-up 0.72 A for a 90 A fault.
Loses marks
- Including the relay burden in the stability loop.
- Omitting CT magnetising currents in the minimum operating current.
Earns more
- Magnetising current at the setting voltage (0.3 mA/V × 100 V) used for all six CTs.
- Clear distinction between a through fault (CT saturation) and an internal fault.
Extra mark
- Sketch of the circulating-current scheme with the spill path.
- (b) Probability of error for antipodal signalling in Laplacian noise and the SNR required for 10⁻⁶. 20 marks
derive— given → assumptions → stepwise derivation → result → check
Must cover
- Take the threshold at 0 and set up P_e = P(n > A) with the Laplacian pdf.
- Integrate to get P_e = ½ exp(−√2 A/σ).
- Use SNR = A²/σ² to obtain SNR ≈ 86.1 (19.35 dB) for P_e = 10⁻⁶.
Loses marks
- Wrong integral setup or limits.
- Not expressing P_e in terms of A and σ.
Earns more
- Absolute value in the pdf handled by symmetry.
- Clear definition of SNR in terms of A and σ.
Extra mark
- Comparison with Gaussian noise at the same SNR.
- (c) Percentage error of a moving-iron voltmeter calibrated at 50 Hz ac when used on 200 V dc. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Find the ac impedance 3000 Ω and X_L = 251.3 Ω, hence the total resistance 2989.5 Ω.
- Find the dc current 66.9 mA on 200 V.
- Find the reading 200.7 V and the error of about +0.35%.
Loses marks
- Ignoring the inductance in the ac impedance.
- Wrong percentage-error calculation.
Earns more
- Correct reactance at 50 Hz.
- Distinction between ac impedance and dc resistance.
Extra mark
- Remark that on dc the reactance vanishes, so the reading is slightly high.
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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