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Electrical Engineering 2023 Paper I 50 marks Calculate

Paper I — Q8

(a) A 415 V, 4-pole, 3-phase, 50 Hz, star-connected squirrel cage induction motor has per phase parameters of r_s = 1·1 Ω, r_r =…

(a)

A 415 V, 4-pole, 3-phase, 50 Hz, star-connected squirrel cage induction motor has per phase parameters of r_s = 1·1 Ω, r_r = 1·3 Ω, X_m = 167 Ω, and X_ls = X_lr = 3·5 Ω, all at rated frequency. The motor has a rated slip of 4%. Its speed is to be controlled by VVVF method using a VSI (Voltage Source Inverter). Find the voltage to be applied to the motor at 5 Hz operating frequency to maintain same peak torque as in 50 Hz. Also determine the speed at which rated torque appears at this frequency. (Neglect core losses in the motor and all motor parameters referred from stator side) 20 marks

(b)
(i)

A 300 W carrier is modulated to a depth of 70%. Calculate the total power transmitted in case of Vestigial Side Band (VSB) modulation. Assume 15% of the other side band is transmitted along with wanted side band. Also find the saving in power when compared to Double Side Band (DSB) transmission. 10 marks

(ii)

Find the temperature of the attenuator in the system shown in the figure below so that the overall noise figure of the system does not exceed 3·5 dB. The attenuator introduces a loss of 3 dB. The system consists of the attenuator (loss = 3 dB), an amplifier (gain = 1000, noise figure = 1.5 dB) and a mixer (gain = 100, noise figure = 8 dB) in cascade. 10 marks

(c)

A bank of three identical single-phase transformers having 11000 V/231 V voltage ratio are connected in delta-star combination with delta side connected to 11 kV, 3-phase balanced supply. The star side is supplying a balanced load of 120 kVA at 0·8 pf lag. A single-phase load of 40 kW, upf is now connected between one line and neutral of the secondary side. Calculate the input line currents at the delta side under this condition. (Neglect any magnetising currents of the transformers) 10 marks

हिंदी में प्रश्न पढ़ें
(a)

एक 415 V, 4-ध्रुव, त्रि-कला, 50 Hz, तारा-संयोजित गिलहरी पंजर प्रेरण मोटर के निर्धारित आवृत्ति पर प्रति कला प्राचल r_s = 1.1 Ω, r_r = 1.3 Ω, X_m = 167 Ω, तथा X_ls = X_lr = 3.5 Ω हैं । मोटर का निर्धारित सर्पण 4% है । मोटर की गति को एक VSI (वोल्टता स्रोत प्रतिलोमक) की सहायता से VVVF पद्धति द्वारा नियंत्रित किया जाना है । 5 Hz की संक्रियात्मक आवृत्ति पर मोटर में अनुप्रयुक्त की जाने वाली वह वोल्टता ज्ञात कीजिए जो उच्चतम बलाघूर्ण को 50 Hz की आवृत्ति पर उच्चतम बलाघूर्ण के बराबर बनाए रखे । वह गति भी ज्ञात कीजिए जो इस आवृत्ति पर निर्धारित बलाघूर्ण प्रदान करे । (मोटर के सभी प्राचल स्टेटर सिरे से संदर्भित हैं तथा मोटर के कोर क्षरण को अनदेखा कीजिए)

(b)
(i)

एक 300 W के वाहक को 70% गहराई तक मॉडुलित (प्रमाणित) किया गया है । अवशेषी पार्श्व बैंड (VSB) मॉडुलन (तारतम्य) में कुल संचारित शक्ति परिकलित कीजिए । यह मानिए कि चाहे गए पार्श्व बैंड के साथ दूसरे पार्श्व बैंड के 15% हिस्से का संचारण किया जाना है । द्वि पार्श्व बैंड (DSB) के संचारण की तुलना में शक्ति में बचत भी ज्ञात कीजिए ।

(ii)

चित्र में प्रदर्शित तंत्र में क्षीणक का वह तापमान ज्ञात कीजिए जिससे तंत्र का समग्र रवांक 3·5 dB से अधिक न हो । क्षीणक 3 dB का क्षरण उत्पन्न करता है । तंत्र में क्षीणक (क्षरण = 3 dB), प्रवर्धक (लाभांक = 1000, रवांक = 1·5 dB) तथा मिश्रित्र (लाभांक = 100, रवांक = 8 dB) श्रेणीक्रम में हैं ।

(c)

तीन एकसमान एकल-कला परिणामित्रों, जिनका बोल्टता अनुपात 11000 V/231 V है, का समूह त्रिकोण-तारा संयोजन में जोड़ा गया है जहाँ त्रिकोण सिरा 11 kV के त्रि-कला संतुलित प्रदाय से जुड़ा है । तारा सिरा 0·8 शक्ति गुणांक पश्चता पर 120 kVA का संतुलित भार प्रदाय कर रहा है । अब द्वितीयक सिरे की एक लाइन तथा न्यूट्रल के मध्य एक 40 kW, इकाई शक्ति गुणांक, का एकल-कला भार जोड़ा जाता है । इस स्थिति में त्रिकोण सिरे पर निवेश लाइन धाराएँ परिकलित कीजिए । (परिणामित्रों की चुंबकन धाराओं को अनदेखा कीजिए)

Q8 of the 2023 UPSC Mains Electrical Engineering Paper I, as printed
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Part (b)
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(b) A block diagram of a cascaded system consisting of three blocks connected in series from left to right. The input is on the far left and the output is on the far right. The first block is labeled 'Attenuator' with 'loss = 3dB'. The second block is labeled 'Amplifier' with 'Gain = 1000' and 'Noise figure = 1.5 dB'. The third block is labeled 'Mixer' with 'Gain = 100' and 'Noise figure = 8 dB'.

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(a) VVVF control: voltage at 5 Hz and the speed for rated torque

Per phase, V₅₀ = 415/√3 = 239.6 V. Leakage reactances at frequency f scale as X = X₅₀ (f/50), so X₅₀ = X_ls + X_lr = 7 Ω and X₅ = 0.7 Ω. X_m = 167 Ω is large, so the magnetising branch is moved to the terminals and ignored in the torque.

Voltage for equal peak torque. T_max = 3V²/{2ω_s[r_s + √(r_s² + X²)]} with ω_s = 2π f/2 for four poles. At 50 Hz: r_s + √(r_s² + X²) = 1.1 + √(1.21 + 49) = 8.186 Ω. At 5 Hz: 1.1 + √(1.21 + 0.49) = 2.404 Ω. Equal T_max requires V₅²/(5 × 2.404) = V₅₀²/(50 × 8.186), so V₅ = 239.6 × √(0.1 × 2.404/8.186) = 239.6 × 0.1714 = 41.1 V per phase, i.e. about 71 V line-to-line. This is more than the constant V/f value (41.5 V line), because the stator resistance drop must be compensated at low frequency.

Rated torque. At 50 Hz and s = 0.04, with X = 7 Ω: T_rated = [3V²(r_r/s)]/[ω_s((r_s + r_r/s)² + X²)] with r_r/s = 32.5 Ω and ω_s = 157.1 rad/s, giving T_rated = 3 × 57 408 × 32.5/[157.1 × (33.6² + 49)] = 30.3 N·m.

Speed at 5 Hz. Synchronous speed N_s = 120 × 5/4 = 150 rpm and ω_s = 15.71 rad/s. Set T = 30.3 N·m with V₅ = 41.1 V and X = 0.7 Ω. Writing x = r_r/s: 3 × 41.1² x/[15.71((1.1 + x)² + 0.49)] = 30.3, which reduces to 30.3x² − 255.9x + 51.5 = 0, whose stable (larger) root is x = 8.24 Ω. Then s = 1.3/8.24 = 0.158 and N = 150 × (1 − 0.158) ≈ 126 rpm.

Diagram: Torque speed curves of the induction motor at 50 Hz and at 5 Hz with the same peak torque 67 N m; rated torque 30.3 N m at 1440 rpm and 126

(b)(i) VSB transmission power

For m = 0.7 and P_c = 300 W, each side band carries P_SB = m²P_c/4 = 0.49 × 300/4 = 36.75 W.

VSB: carrier + wanted side band + 15 per cent of the other = 300 + 36.75 + 0.15 × 36.75 = 300 + 36.75 + 5.51 = 342.3 W.

DSB: 300 + 2 × 36.75 = 373.5 W.

Saving compared with DSB = 373.5 − 342.3 = 31.2 W, about 8.4 per cent of the DSB power.

(b)(ii) Attenuator temperature for the noise figure limit

The attenuator has loss L = 3 dB = 2, so its gain G₁ = 0.5. At physical temperature T its noise figure is F₁ = 1 + (L − 1)T/T₀ with T₀ = 290 K. The amplifier has F₂ = 10^0.15 = 1.413 and G₂ = 1000, and the mixer has F₃ = 10^0.8 = 6.31.

Friis: F = F₁ + (F₂ − 1)/G₁ + (F₃ − 1)/(G₁G₂) = F₁ + 0.4125/0.5 + 5.31/500 = F₁ + 0.825 + 0.0106.

The limit is F ≤ 10^0.35 = 2.239, so F₁ ≤ 2.239 − 0.836 = 1.403. Then T/290 ≤ 0.403, so T ≤ 117 K (about −156 °C). An attenuator at the usual 290 K would give F₁ = 2 and exceed the limit.

Diagram: Cascade of attenuator 3 dB loss, amplifier gain 1000 noise figure 1.5 dB, and mixer gain 100 noise figure 8 dB, with Friis formula

(c) Delta-star transformer bank with an unbalanced load

Phase ratio = 11 000/231 = 47.6, and the star line voltage is 231 × √3 = 400 V. The balanced load gives I_L = 120 000/(√3 × 400) = 173.2 A at 0.8 lagging (−36.9°). The 40 kW unity-power-factor load on phase a draws 40 000/231 = 173.2 A in phase with V_an.

Secondary phase currents: I_a = 173.2∠−36.9° + 173.2∠0° = 328.6∠−18.4° A, I_b = 173.2∠−156.9° A and I_c = 173.2∠83.1° A.

Each delta winding current equals the matching star current divided by 47.6: 6.90 A, 3.64 A and 3.64 A. The delta line currents are the differences of adjacent winding currents: I_A = I_AB − I_CA, I_B = I_BC − I_AB and I_C = I_CA − I_BC. Evaluating these gives |I_A| = 8.42 A, |I_B| = 9.92 A and |I_C| = 6.30 A.

Check: the three line currents sum to zero, and I_C is unchanged from the balanced value, 120 000/(√3 × 11 000) = 6.30 A. The zero-sequence part of the single-phase load, 57.7 A on the secondary, circulates inside the delta and does not appear in the lines.

Diagram: Delta-star bank of three 11000/231 V transformers supplying a balanced 120 kVA load and a 40 kW single-phase load on phase a, with resulting

How this answer will be evaluated

What earns full marks: Complete working with all diagrams, correct calculations, clear assumptions, and proper units throughout.

Evaluation rubric

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

  1. (a) Find the voltage at 5 Hz giving the same peak torque and the speed at which rated torque appears. 20 marks

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

    Must cover

    • X scales with frequency: 7 ohm at 50 Hz, 0.7 ohm at 5 Hz
    • T_max expression and V5 = 41.1 V per phase (about 71 V line)
    • Rated torque about 30.3 N m at 4 per cent slip
    • Slip about 0.158 and speed about 126 rpm at 5 Hz

    Loses marks

    • Applying constant V/f (41.5 V line) instead.

    Earns more

    • Marks for the peak-torque condition, rated torque and the slip solution.

    Extra mark

    • Comment that V/f must be boosted for the stator drop.
  2. (b)(i) Calculate the VSB transmitted power and the saving over DSB. 10 marks

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

    Must cover

    • Side band power m^2 Pc/4 = 36.75 W
    • VSB = 300 + 36.75 + 0.15 x 36.75 = 342.3 W
    • DSB = 373.5 W
    • Saving about 31.2 W

    Loses marks

    • Using amplitude instead of power.

    Earns more

    • Marks for each power and the saving.

    Extra mark

    • Percentage saving about 8.4.
  3. (b)(ii) Find the attenuator temperature so that the noise figure does not exceed 3.5 dB. 10 marks

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

    Must cover

    • Attenuator F1 = 1 + (L - 1) T/290 with L = 2
    • Friis formula with G1 = 0.5, G2 = 1000
    • Limit F = 2.239 gives F1 <= 1.403
    • T <= 117 K

    Loses marks

    • Treating attenuator noise figure as equal to its loss regardless of temperature.

    Earns more

    • Marks for the cascade formula and the temperature.

    Extra mark

    • Noting that 290 K would exceed the limit.
  4. (c) Find the delta-side line currents of the transformer bank with the added single-phase load. 10 marks

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

    Must cover

    • Secondary load currents 173.2 A each; phase a gets an extra 173.2 A in phase
    • Phase ratio 47.6
    • Delta winding currents and line currents as differences
    • |IA| = 8.42 A, |IB| = 9.92 A, |IC| = 6.30 A

    Loses marks

    • Dividing line currents directly by the phase ratio.

    Earns more

    • Marks for the secondary currents, turns ratio and the line-current subtraction.

    Extra mark

    • Zero-sequence circulation in the delta.
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