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Electrical Engineering 2021 Paper I 50 marks Solve

Paper I — Q7

The following test data are obtained for a three-phase, 195 MVA, 15 kV, 50 Hz star connected synchronous machine. Open circuit…

The following test data are obtained for a three-phase, 195 MVA, 15 kV, 50 Hz star connected synchronous machine.

Open circuit test :

I_f (A)1503004506007509001200
V_LL (kV)3·757·511·213·61515·816·5

Short circuit test : I_f = 750 A, I_a = 7000 A

The armature resistance is small.

(i)

Draw the open circuit characteristic, the short circuit characteristic, the airgap line and the modified airgap line.

(ii)

Determine the unsaturated and saturated values of the synchronous reactance in pu.

(iii)

Find the field current required, if the synchronous machine is to deliver 100 MVA at rated voltage, at 0·8 leading power factor. 20 marks

(b)

A three-phase, full-wave thyristor bridge converter is operated from a three-phase, Y-connected 220 V, 50 Hz supply and the load resistance is 20 Ω.

It is required to obtain an average output voltage of 50% of the maximum possible output voltage. Determine the following :

(i)

The delay angle α

(ii)

The rms and average output currents

(iii)

The rms and average thyristor currents

(iv)

The rectification efficiency

(v)

The input PF 20 marks

(c)

A lossless transmission line has characteristic impedance Z₀ = 50 Ω. Its length is 30 m and operates at 5 MHz. The line is terminated with a load Zₗ = 60 + j50 Ω. If the phase velocity u = 0.6c on the line, find the following :

(i)

The reflection coefficient 'Γ'

(ii)

The standing wave ratio 'S'

(iii)

The input impedance 'Zᵢₙ' 10 marks

हिंदी में प्रश्न पढ़ें

एक त्रि-कला, 195 MVA, 15 kV, 50 Hz तारा (स्टार) संयोजित तुल्यकालिक यंत्र के लिए निम्नलिखित परीक्षण आँकड़े प्राप्त हुए हैं :

खुला परिपथ परीक्षण :

I_f (A)1503004506007509001200
V_LL (kV)3·757·511·213·61515·816·5

लघु परिपथ परीक्षण : I_f = 750 A, I_a = 7000 A

आर्मेचर का प्रतिरोध थोड़ा है ।

(i)

खुला परिपथ लक्षण वक्र, लघु परिपथ लक्षण वक्र, वायु अंतराल रेखा तथा परिवर्तित वायु अंतराल रेखा आरेखण कीजिए ।

(ii)

तुल्यकालिक प्रतिघात का असंतृप्त तथा संतृप्त मान pu में ज्ञात कीजिए ।

(iii)

यदि तुल्यकालिक यंत्र को निर्धारित वोल्टता तथा 0·8 अग्रगामी शक्ति गुणक पर 100 MVA प्रदान करना हो, तो वांछित क्षेत्र धारा का मान ज्ञात कीजिए । (20 अंक)

(b)

एक त्रि-कला, पूर्ण-तरंग थाइरिस्टर सेतु परिवर्तक, एक त्रि-कला, Y-संयोजित, 220 V, 50 Hz प्रदाय से चालित है तथा भार प्रतिरोध 20 Ω है।

यदि औसत निर्गत वोल्टता, अधिकतम संभव निर्गत वोल्टता की 50% चाहिए, तो निम्नलिखित का मान ज्ञात कीजिए :

(i)

विलम्बन कोण α

(ii)

निर्गत धारा का rms तथा औसत मान

(iii)

थाइरिस्टर धारा का rms तथा औसत मान

(iv)

दिष्टकरण की दक्षता

(v)

निवेश PF (20 अंक)

(c)

एक हानिरहित संप्रेषण लाइन की लाक्षणिक प्रतिबाधा Z₀ = 50 Ω है। इसकी लम्बाई 30 m है तथा यह 5 MHz पर कार्य करती है। लाइन को एक Zₗ = 60 + j50 Ω भार के साथ अन्तस्थ किया गया है। यदि लाइन पर कला वेग u = 0.6c हो, तो निम्नलिखित का मान ज्ञात कीजिए :

(i)

परावर्तन गुणांक 'Γ'

(ii)

अप्रगामी तरंग अनुपात 'S'

(iii)

निवेश प्रतिबाधा 'Zᵢₙ' (10 अंक)

Q7 of the 2021 UPSC Mains Electrical Engineering Paper I, as printed
The question as printed in the 2021 Electrical Engineering paper
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The figure this question refers to, in words

The question paper is a scan and the diagram did not survive as text. This is the figure as read from the original page — every component, value and label — so the question can be worked from the text below.

(a) Table titled 'Open circuit test' with two rows and eight columns. The first row header is 'I_f (A)' followed by values: 150, 300, 450, 600, 750, 900, 1200. The second row header is 'V_LL (kV)' followed by values: 3.75, 7.5, 11.2, 13.6, 15, 15.8, 16.5.

(c) Table: Open circuit test data for a three-phase, 195 MVA, 15 kV, 50 Hz star connected synchronous machine. Columns: I_f (A), V_LL (kV). Rows: 150, 3.75; 300, 7.5; 450, 11.2; 600, 13.6; 750, 15; 900, 15.8; 1200, 16.5. Short circuit test data: I_f = 750 A, I_a = 7000 A. Note: The armature resistance is small.

Table 1: Open circuit test data. Columns: I_f (A), V_LL (kV). Rows: (150, 3.75), (300, 7.5), (450, 11.2), (600, 13.6), (750, 15), (900, 15.8), (1200, 16.5). Table 2: Short circuit test data. Row: I_f = 750 A, I_a = 7000 A.

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) Synchronous machine from OCC and SCC data Base: 195 MVA, 15 kV, so I_base = 7506 A and Z_base = 15²/195 = 1.154 Ω. (i) Characteristics. OCC: the tabulated points (linear to about 300 A, then bending over). Air-gap line: the straight line through the origin and the linear part, slope 3.75 kV/150 A = 25 V/A (so 15 kV would need 600 A). SCC: a straight line through the origin and (750 A, 7000 A). Modified air-gap line: the line from the origin through the OCC point at rated voltage (750 A, 15 kV), slope 20 V/A.

Diagram: OCC, SCC, air-gap line and modified air-gap line of a 195 MVA, 15 kV machine

(ii) Synchronous reactance at I_f = 750 A (I_sc = 7000 A): Unsaturated (air-gap line voltage 25 × 750 = 18.75 kV): X_s = (18 750/√3)/7000 = 1.546 Ω = 1.34 pu. Saturated (OCC voltage 15 kV): X_s = (15 000/√3)/7000 = 1.237 Ω = 1.07 pu (= 1/SCR, with SCR = 750/804 = 0.933). (iii) Field current at 100 MVA, rated voltage, 0.8 leading. The flux is at the rated-voltage saturation level, so use X_s = 1.07 pu and the modified air-gap line. I_a = 100/195 = 0.513 pu at +36.87°. E_f = 1 + j1.072 × 0.513(0.8 + j0.6) = 0.670 + j0.440 = 0.802∠33.3° pu = 12.03 kV. I_f = 12 030/20 ≈ 601 A. The leading power factor means under-excitation (E_f < V). Reading E_f directly on the OCC would give about 500 A; the unsaturated reactance with the air-gap line gives about 483 A, which underestimates.

(b) Three-phase full-wave thyristor bridge Take 220 V as the line voltage: V_p = 127 V, V_m = 179.6 V, V_dm = 1.35 × 220 = 297.1 V. (i) V_dc = V_dm cos α = 0.5V_dm, so α = 60° (V_dc = 148.5 V; conduction with a resistive load is just continuous). (ii) I_dc = 148.5/20 = 7.43 A. V_rms = √3V_m[½ + (3√3/4π)cos 2α]^½ = 311.1 × 0.5416 = 168.5 V, so I_rms = 8.43 A. (iii) Each thyristor conducts 120° per cycle: I_T(avg) = I_dc/3 = 2.48 A; I_T(rms) = I_rms/√3 = 4.86 A. (iv) η = P_dc/P_ac = (148.5 × 7.43)/(168.5 × 8.43) = 1104/1420 = 77.7%. (v) Line current I_s = √(2/3) I_rms = 6.88 A; P = I_rms²R = 1420 W. PF = P/(3V_pI_s) = 1420/(3 × 127 × 6.88) = 0.54 lagging.

(c) Lossless transmission line (i) Γ = (Z_L − Z₀)/(Z_L + Z₀) = (10 + j50)/(110 + j50) = 0.247 + j0.342 = 0.422∠54.2°. (ii) S = (1 + |Γ|)/(1 − |Γ|) = 1.422/0.578 = 2.46. (iii) λ = u/f = 1.8 × 10⁸/5 × 10⁶ = 36 m, so βℓ = 2π × 30/36 = 300° and tan βℓ = −√3. Z_in = Z₀(Z_L + jZ₀ tan βℓ)/(Z₀ + jZ_L tan βℓ) = 50(60 − j36.6)/(136.6 − j103.9) = 20.4 + j2.1 Ω (≈ 20.5∠5.9° Ω), which lies between Z₀/S = 20.3 Ω and Z₀S = 123 Ω, as it must.

How this answer will be evaluated

Framework: null

What earns full marks: Correct method, clear diagrams, accurate calculations with units.

Evaluation rubric

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

  1. (a) Characteristics of the synchronous machine from OCC and SCC data, unsaturated and saturated X_s in pu, and the field current at 100 MVA, rated voltage, 0.8 leading. 20 marks

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

    Must cover

    • Draw the OCC, SCC, air-gap line (25 V/A) and modified air-gap line (20 V/A through the rated-voltage point).
    • Find X_s(unsaturated) = 1.34 pu and X_s(saturated) = 1.07 pu using base Z = 1.154 Ω.
    • Find E_f = 0.802∠33.3° pu (12.03 kV) for I_a = 0.513 pu at 0.8 leading and read I_f ≈ 601 A from the modified air-gap line.

    Loses marks

    • Missing per-unit base calculation.
    • Using X_s(unsaturated) with the air-gap line for operation at rated voltage.

    Earns more

    • Per-unit bases computed.
    • Correct angle (+36.87°) for a leading power factor.

    Extra mark

    • Phasor diagram and note that leading pf means under-excitation (E_f < V).
  2. (b) Delay angle, output and thyristor currents, efficiency and input power factor of a three-phase full-wave bridge giving 50% of the maximum output voltage. 20 marks

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

    Must cover

    • State the supply voltage taken (220 V line), V_dm = 1.35 × 220 = 297.1 V, and α = 60° from cos α = 0.5.
    • Compute I_dc = 7.43 A, V_rms = 168.5 V, I_rms = 8.43 A, and thyristor currents 2.48 A (avg) and 4.86 A (rms).
    • Compute efficiency 77.7% and input PF = 1420/(3 × 127 × 6.88) = 0.54 lagging.

    Loses marks

    • Confusing line and phase voltages.
    • Wrong efficiency formula (using P_out/P_in with different quantities).

    Earns more

    • Continuous-conduction condition at α = 60° stated.
    • Line current I_s = √(2/3) I_rms used for the PF.

    Extra mark

    • Output-voltage waveform at α = 60° sketched.
  3. (c) Reflection coefficient, standing wave ratio and input impedance of a lossless line terminated in 60 + j50 Ω. 10 marks

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

    Must cover

    • Find Γ = 0.422∠54.2° and S = 2.46.
    • Use u = 0.6c to get λ = 36 m and βℓ = 300°.
    • Find Z_in = 20.4 + j2.1 Ω using the line-input formula with tan 300° = −√3.

    Loses marks

    • Ignoring the phase velocity in the length calculation.
    • Sign error in the input-impedance formula.

    Earns more

    • Electrical length in degrees.
    • Check that |Z_in| lies between Z₀/S and Z₀S.

    Extra mark

    • Smith-chart verification.
What "Solve" is asking you to do

Choose the method, then carry it through to a final answer. Identifying what kind of problem this is and why that method applies is the first thing marked; a correct figure arrived at invisibly earns almost nothing.

Structure that answers it

Given data and what is required → method chosen, with the reason it applies → set-up (equation, circuit, free body, trial balance) → working, step by step → answer with units and any condition of validity

Where marks are lost

Doing the middle steps mentally and writing only the result. In mathematics papers, a further loss comes from giving a decimal where the exact value in surds or fractions was wanted, or from skipping the justification a part explicitly asks for.

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