Electrical Engineering 2024 Paper II 50 marks Calculate

Q6

(a) Calculate the power loss in the transmission system given in the following figure. The numerical values of transmission system are: I₁ = 0·75 ∠0° PU, I₂ = 0·8 ∠0° PU, V₃ = 1·2 ∠0° PU, Z₁ = (0·07 + j0·15) PU, Z₂ = (0·06 + j0·20) PU, Z₃ = (0·05 + j0·06) PU 20 marks (b) The fuel input equations of two power plant operations are given as: F₁ = 0·3 P₁² + 35 P₁ + 125, ₹/hr F₂ = 0·2 P₂² + 30 P₂ + 140, ₹/hr If the maximum and minimum loading on each unit is 90 MW and 20 MW respectively and the total consumption demand is 200 MW, then calculate the economical operating schedule and corresponding cost of generation. If load is equally shared by both units, calculate the savings achieved by loading the units as per equal incremental production cost. Neglect the transmission losses. 20 marks (c) A DM transmitter with a fixed step size of 0·25 V is given a sinusoidal message signal. Determine the maximum permissible amplitude of the message signal, if slope overload is to be avoided. Assume sampling frequency ten times the Nyquist rate. 10 marks

हिंदी में प्रश्न पढ़ें

(a) निम्नलिखित चित्र में दी गई संचरण प्रणाली में शक्ति हानि की गणना कीजिए । संचरण प्रणाली के आंकिक मान निम्न प्रकार हैं : I₁ = 0·75 ∠0° PU, I₂ = 0·8 ∠0° PU, V₃ = 1·2 ∠0° PU, Z₁ = (0·07 + j0·15) PU, Z₂ = (0·06 + j0·20) PU, Z₃ = (0·05 + j0·06) PU 20 अंक (b) दो शक्ति संयंत्रों के संचालन के ईंधन निविष्ट समीकरण निम्न प्रकार हैं : F₁ = 0·3 P₁² + 35 P₁ + 125, ₹/घंटे F₂ = 0·2 P₂² + 30 P₂ + 140, ₹/घंटे यदि प्रत्येक इकाई पर अधिकतम व न्यूनतम भार क्रमशः: 90 MW और 20 MW तथा कुल खपत मांग 200 MW हो, तो मितव्ययी परिचालन अनुसूची और उससे संबंधित उत्पादन लागत की गणना कीजिए । यदि दोनों इकाइयों द्वारा कुल भार को समान रूप से सहभाजित किया जाता है, तो समान बढ़ोतरी उत्पादन लागत के अनुसार इकाइयों के भारण से प्राप्त बचत की गणना कीजिए । संचरण हानि को उपेक्षित किया गया है । 20 अंक (c) एक 0·25 V के स्थायी चरण परिमाण वाले DM प्रेषित्र (ट्रांसमीटर) को एक ज्यावक्रीय सूचना संकेत दिया जाता है । यदि प्रवणता अधिभार (स्लोप ओवरलोड) से बचना है, तो सूचना संकेत के अधिकतम अनुज्ञेय आयाम का निर्धारण कीजिए । न्यूनतम चयन आवृत्ति को नाइक्विस्ट दर से दस गुना माना गया है । 10 अंक

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Directive word: Calculate

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How this answer will be evaluated

Approach

This is a multi-part numerical problem requiring systematic calculation across three distinct domains: power system analysis, economic dispatch, and communication systems. Spend approximately 40% of time on part (a) transmission loss calculation (20 marks), 35% on part (b) economic load dispatch with cost comparison (20 marks), and 25% on part (c) delta modulation slope overload condition (10 marks). Begin each part with the relevant formula, show complete step-by-step working with proper units, and conclude with clear final answers for each sub-part.

Key points expected

  • For (a): Correct application of power loss formula P_L = |I|²R for each branch, proper handling of complex currents through each impedance, and summation of I²R losses across all three branches (Z₁, Z₂, Z₃) with correct PU to actual conversion if needed
  • For (b): Setting up equal incremental cost criterion dF₁/dP₁ = dF₂/dP₂, solving simultaneous equations with P₁ + P₂ = 200 MW constraint, checking against generator limits (20-90 MW), calculating total cost for economic schedule and equal load sharing (100 MW each), then computing savings
  • For (c): Deriving slope overload condition |dm(t)/dt|_max ≤ Δ/T_s = Δ·f_s, applying to sinusoidal signal m(t) = A_m sin(ω_m t) to get A_m ≤ Δ·f_s/(2πf_m), using f_s = 10 × 2f_m = 20f_m (Nyquist rate), and final amplitude calculation
  • Correct handling of per-unit system in (a) with proper identification of resistive components from complex impedances
  • Verification of generator operating limits in (b) and recalculation if limits are violated (though 200 MW demand with 20-90 MW limits allows feasible solution)
  • Clear presentation of cost comparison in (b) showing economic dispatch savings over simple equal loading

Evaluation rubric

DimensionWeightMax marksExcellentAveragePoor
Concept correctness20%10Correctly identifies I²R loss mechanism for (a), equal incremental cost criterion for (b), and slope overload condition relating step size to signal derivative for (c); properly distinguishes between real power loss and complex power in (a)Identifies basic formulas but confuses real power loss with |I|²|Z| in (a), or misapplies economic dispatch condition in (b), or uses wrong Nyquist rate relationship in (c)Fundamental conceptual errors such as using P = VI instead of I²R for losses, ignoring generator limits in economic dispatch, or completely wrong slope overload condition
Numerical accuracy25%12.5All three parts show precise calculations: correct branch currents and I²R summation in (a), accurate Lagrange multiplier solution yielding P₁≈72.7 MW, P₂≈127.3 MW with cost savings ~₹136/hr in (b), and exact amplitude A_m = 0.25×20/(2π) ≈ 0.796 V in (c)Minor arithmetic errors in one part (e.g., wrong decimal place in cost calculation or impedance calculation) but correct methodology throughoutMajor calculation errors in multiple parts, wrong final answers due to calculator mistakes or unit confusion, or missing cost comparison in (b)
Diagram quality10%5Reconstructs the transmission system diagram for (a) showing nodes, branches with impedances, current directions, and voltage reference; sketches DM system with integrator, comparator, and sampler for (c); clear labeling throughoutBasic diagram for (a) with missing labels or incorrect current directions, or omits diagram for (c) despite its relevanceNo diagrams provided, or completely wrong circuit topology in (a) that invalidates the analysis
Step-by-step derivation25%12.5Complete sequential working: (a) shows KCL/KVL for finding branch currents, then individual losses, then total; (b) derives dF/dP equations, sets equal, solves with constraint, verifies limits, calculates both costs; (c) starts from slope condition, substitutes sinusoidal maximum slope, applies sampling frequency relationSome steps skipped or combined (e.g., jumps to final formula without showing dF/dP derivation), or missing verification of generator limits in (b)Only final answers with no working, or incorrect sequence of steps that obscures logical flow, or missing cost comparison calculation in (b)
Practical interpretation20%10Interprets (a) results in context of transmission efficiency and Indian grid voltage levels; for (b) discusses significance of economic dispatch in reducing fuel costs for NTPC-type thermal plants; for (c) relates DM step size selection to voice signal encoding in Indian telecom systemsBrief mention of practical relevance without specific context, or generic statements about 'saving money' without grid-specific insightNo interpretation provided, or irrelevant discussion unrelated to power systems or communication engineering practice

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