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Mechanical Engineering 2023 Paper II 50 marks Solve

Paper II — Q2

(a) A 15 m high cylinder with a cross-sectional area of 0.6 m² contains 3 m³ of liquid water at 25 °C on the top of a thin…

(a)

A 15 m high cylinder with a cross-sectional area of 0.6 m² contains 3 m³ of liquid water at 25 °C on the top of a thin insulated piston of mass 20 kg. Below the piston, argon gas is at 15 °C with a volume of 3 m³, as shown in the figure. Heat is supplied to argon such that the piston rises and pushes the water out over the top edge. Find the (i) work done (kJ) to remove the whole water from the top of the piston and (ii) heat transferred (kJ) to argon during the process. (iii) Plot the process on P-v diagram for argon. Assume atmospheric pressure (P₀) as 101 kPa, Cᵥ and R for argon as 0·312 kJ/kg-K and 0·2081 kJ/kg-K respectively. The specific volume of water at 25 °C is 0·001003 m³/kg. Neglect piston thickness. 20 marks

(b)

Air at 100 kPa and 290 K enters a gas turbine cycle with two stages of compression and two stages of expansion. This system uses ideal regenerator, reheater and intercooler. The pressure ratio across each stage is 4. 300 kJ/kg of heat is added in combustion chamber and reheater each. The regenerator increases the air temperature by 20 °C. Draw T-s plot and determine the (i) total heat rejected (kJ/kg), (ii) net work output (kJ/kg) and (iii) thermal efficiency of the system. Assume isentropic operation for all compressors and turbines. Take Cₚ of air = 1·005 kJ/kg-K and γ = 1·4. 20 marks

(c)

A convergent-divergent nozzle has a throat area of 250 mm² and an exit area of 500 mm². Air enters the nozzle with a stagnation temperature of 350 K and stagnation pressure of 1 MPa. Determine the maximum flow rate of air through the nozzle and the static pressure, static temperature, Mach number and velocity at the exit from the nozzle. Given γ = 1·4, R = 0·287 kJ/kg-K. Use Gas Table to solve the problem. 10 marks

हिंदी में प्रश्न पढ़ें
(a)

0.6 m² के अनुप्रस्थ-काट क्षेत्रफल वाले एक 15 m ऊँचे सिलिंडर में 20 kg द्रव्यमान के पतले उष्मारोधित पिस्टन के शीर्ष पर 25 °C पर 3 m³ तरल पानी निहित है। पिस्टन के नीचे, 15 °C पर 3 m³ आयतन के साथ आर्गन गैस है, जैसा कि चित्र में दर्शाया गया है। आर्गन को उष्मा की आपूर्ति इस प्रकार की जाती है कि पिस्टन ऊपर उठता है और पानी को ऊपरी किनारे से बाहर धकेलता है। ज्ञात कीजिए (i) पिस्टन के ऊपर से पूरा पानी निकालने के लिए किया गया कार्य (kJ) और (ii) प्रक्रम के दौरान आर्गन को उष्मा अंतरण (kJ). (iii) आर्गन के लिए प्रक्रम का P-v आरेख खींचिए। वायुमंडलीय दाब (P₀) को 101 kPa, आर्गन के लिए Cᵥ और R क्रमशः: 0.312 kJ/kg-K और 0.2081 kJ/kg-K मान लीजिए। 25 °C पर पानी का विशिष्ट आयतन 0.001003 m³/kg है। पिस्टन की मोटाई की उपेक्षा कीजिए। (20 अंक)

(b)

100 kPa और 290 K पर वायु संपीडन के दो चरणों और प्रसरण के दो चरणों के साथ एक गैस टरबाइन चक्र में प्रवेश करती है। यह निकाय आदर्श पुन्योजक, पुनस्तापक तथा अंतराशीतक का उपयोग करता है। प्रत्येक चरण में दाब अनुपात 4 है। दहन कक्ष और पुनस्तापक प्रत्येक में 300 kJ/kg ऊष्मा डाली जाती है। पुन्योजक वायु का तापमान 20 °C से बढ़ा देता है। T-s आरेख खींचिए और ज्ञात कीजिए (i) कुल परित्यक्त ऊष्मा (kJ/kg), (ii) शुद्ध उत्पादित कार्य (kJ/kg) तथा (iii) निकाय की तापीय दक्षता। सभी संपीडकों और टरबाइनों के लिए समएंट्रॉपी प्रचालन मान लीजिए। वायु के लिए Cₚ = 1·005 kJ/kg-K तथा γ = 1·4 लीजिए। (20 अंक)

(c)

एक अभिसारी-अपसारी टुंड का कंठ क्षेत्र 250 mm² तथा निकास क्षेत्र 500 mm² है। वायु 350 K के स्थिर तापमान तथा 1 MPa के स्थिर दाब के साथ टुंड में प्रवेश करती है। टुंड के माध्यम से वायु की अधिकतम प्रवाह दर तथा स्थैतिक दाब, स्थैतिक तापमान, मैक संख्या और वेग, टुंड के निकास पर, ज्ञात कीजिए। γ = 1·4, R = 0·287 kJ/kg-K दिया गया है। प्रश्न के हल हेतु गैस तालिका का उपयोग कीजिए। (10 अंक)

Q2 of the 2023 UPSC Mains Mechanical Engineering Paper II, as printed
The question as printed in the 2023 Mechanical 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) A vertical cross-section of a cylindrical container. Inside, a horizontal piston separates the volume into two sections. The top section contains a fluid labeled 'Water'. The bottom section contains a gas labeled 'Argon'. An arrow labeled 'P0' points vertically downward onto the top surface of the water. An arrow labeled 'g' points vertically downward from the right side of the piston, indicating gravity.

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) Work done and heat supplied to argon as the piston pushes the water out

Assumptions: argon is an ideal gas with constant specific heats; the process is quasi-static with a frictionless piston and incompressible water; g = 9.81 m/s². Only the argon is heated, since the piston is insulated.

Geometry. The water column is 3/0.6 = 5 m high and the argon column is 3/0.6 = 5 m high, so the piston starts 5 m above the base and the water surface is at 10 m. Stage 1–2: the piston rises 5 m (5 m to 10 m) until the water reaches the 15 m rim. The load is unchanged, so the pressure is constant. Stage 2–3: the piston rises from 10 m to 15 m while water spills. The water height falls linearly from 5 m to 0, so pressure falls linearly with volume.

Pressures: Water mass m_w = 3/0.001003 = 2991 kg, giving 2991 × 9.81/0.6 = 48.90 kPa. Piston: 20 × 9.81/0.6 = 0.327 kPa. P₁ = P₂ = 101 + 48.90 + 0.327 = 150.23 kPa; P₃ = 101 + 0.327 = 101.33 kPa. Argon mass: m = P₁V₁/(RT₁) = 150.23 × 3/(0.2081 × 288.15) = 7.516 kg. Volumes: V₁ = 3, V₂ = 6, V₃ = 9 m³.

(i) Work done by the argon: W₁₋₂ = P₁(V₂ − V₁) = 150.23 × 3 = 450.7 kJ. W₂₋₃ = ½(P₂ + P₃)(V₃ − V₂) = ½(150.23 + 101.33) × 3 = 377.3 kJ. W = 828 kJ. Check by energy: pushing back the atmosphere P₀ΔV = 101 × 6 = 606 kJ; lifting the piston 10 m = 1.96 kJ; lifting all the water to the rim, m_w g(15 − 7.5) = 220.1 kJ. These sum to 828 kJ.

(ii) Heat transferred. T₂ = 150.23 × 6/(7.516 × 0.2081) = 576.3 K and T₃ = 101.33 × 9/(7.516 × 0.2081) = 583.1 K. ΔU = mC_v(T₃ − T₁) = 7.516 × 0.312 × (583.1 − 288.15) = 691.5 kJ. Q = ΔU + W = 691.5 + 828.0 = 1520 kJ supplied to the argon (about 1126 kJ in 1–2 and 393 kJ in 2–3).

(iii) P–v diagram. The specific volumes are v₁ = 0.399, v₂ = 0.798 and v₃ = 1.197 m³/kg.

Diagram: P–v diagram of the piston–cylinder process

(b) Gas turbine cycle with intercooling, reheat and regeneration

Assumptions: perfect intercooling to 290 K, isentropic compressors and turbines, no pressure losses, constant c_p = 1.005 kJ/kg·K. The regenerator raises the compressed air by the stated 20 K, so the exhaust falls by 20 K. Temperature ratio per stage: τ = 4^(0.4/1.4) = 1.486.

State 1: LP compressor inlet, 290 K (100 kPa). State 2: LP compressor exit, 290 × 1.486 = 430.9 K (400 kPa). State 3: after the intercooler, 290 K. State 4: HP compressor exit, 430.9 K (1600 kPa). State 5: after the regenerator, 450.9 K. State 6: after the combustor, 450.9 + 300/1.005 = 749.4 K. State 7: HP turbine exit, 749.4/1.486 = 504.3 K (400 kPa). State 8: after the reheater, 504.3 + 298.5 = 802.8 K. State 9: LP turbine exit, 802.8/1.486 = 540.3 K (100 kPa). State 10: exhaust after the regenerator, 520.3 K.

Diagram: T–s diagram: gas turbine with intercooling, reheat and regeneration

Works: Compressors: 2 × 1.005 × (430.9 − 290) = 283.3 kJ/kg. Turbines: 1.005[(749.4 − 504.3) + (802.8 − 540.3)] = 510.2 kJ/kg.

(i) Total heat rejected = intercooler 1.005(430.9 − 290) = 141.6 plus exhaust 1.005(520.3 − 290) = 231.4, so q_rej = 373.1 kJ/kg. (ii) Net work = 510.2 − 283.3 = 226.9 kJ/kg. Check: 600 − 373.1 = 226.9. (iii) Thermal efficiency = 226.9/600 = 37.8%.

(c) Convergent-divergent nozzle: maximum flow and exit conditions

Given: T₀ = 350 K, p₀ = 1 MPa, A* = 250 mm², A_e = 500 mm², γ = 1.4, R = 287 J/kg·K.

Maximum flow (choked throat, M = 1). From the table at M = 1: T*/T₀ = 0.8333 and p*/p₀ = 0.5283, so T* = 291.7 K and p* = 0.5283 MPa. ρ* = p*/(RT*) = 528 300/(287 × 291.7) = 6.31 kg/m³. V* = √(γRT*) = √(1.4 × 287 × 291.7) = 342.3 m/s. **ṁ_max = ρ*A*V* = 6.31 × 250 × 10⁻⁶ × 342.3 = 0.540 kg/s.**

Exit (supersonic, design operation). A_e/A* = 2.0. The table gives A/A* = 1.8369 at M = 2.1 and 2.0050 at M = 2.2, so interpolating, M_e = 2.1 + 0.1 × (2.0 − 1.8369)/(2.0050 − 1.8369) = 2.197. p/p₀ = 0.10935 − 0.97 × (0.10935 − 0.09352) = 0.0940, so p_e = 94.0 kPa. T/T₀ = 0.53135 − 0.97 × (0.53135 − 0.50813) = 0.5088, so T_e = 178.1 K. V_e = M_e√(γRT_e) = 2.197 × √(1.4 × 287 × 178.1) = 588 m/s.

Continuity check: ρ_eA_eV_e = [94 000/(287 × 178.1)] × 500 × 10⁻⁶ × 587.7 = 0.540 kg/s. The same area ratio also has a subsonic solution (M ≈ 0.31, p ≈ 0.937 MPa), which occurs only if the back pressure is high enough to keep the diverging section subsonic.

How this answer will be evaluated

What earns full marks: Complete solution with all required calculations, proper diagrams, and clear physical interpretation of results

Evaluation rubric

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

  1. (a) Find the work done to remove the water, the heat transferred to argon, and plot the P-v diagram. 20 marks

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

    Must cover

    • Water column 5 m, argon 5 m; constant pressure then linear fall
    • P1 150.23 kPa and P3 101.33 kPa; argon mass 7.516 kg
    • Work 828 kJ
    • Q = 1520 kJ with P-v plot

    Loses marks

    • Taking the piston stroke as 15 m for constant pressure.

    Earns more

    • Marks for geometry, pressures, work, heat and the diagram.

    Extra mark

    • Energy check 606 + 1.96 + 220.1 kJ.
  2. (b) Draw the T-s plot of the two-stage compression, two-stage expansion cycle and find heat rejected, net work and efficiency. 20 marks

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

    Must cover

    • Stage temperature ratio 1.486
    • Temperatures: 430.9, 450.9, 749.4, 504.3, 802.8, 540.3 K
    • Heat rejected 373.1 kJ/kg and net work 226.9 kJ/kg
    • Efficiency 37.8 percent

    Loses marks

    • Using the full 120 K regeneration instead of the stated 20 K.

    Earns more

    • Marks for T-s plot, state temperatures, heat rejected, work and efficiency.

    Extra mark

    • Check q_in - q_rej = w_net.
  3. (c) Determine the maximum flow rate and exit static pressure, temperature, Mach number and velocity of the nozzle using the gas table. 10 marks

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

    Must cover

    • Choked throat T* 291.7 K, p* 0.5283 MPa
    • Maximum flow 0.540 kg/s
    • A/A* = 2 gives M = 2.197
    • p 94.0 kPa, T 178.1 K, V 588 m/s

    Loses marks

    • Using stagnation instead of static values at the exit.

    Earns more

    • Marks for choked flow, area ratio, table interpolation and exit properties.

    Extra mark

    • Continuity check and note on the subsonic root.
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