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Mechanical Engineering 2021 Paper II 50 marks Calculate

Paper II — Q7

(a) (i) A single stage impulse steam turbine rotor has a diameter of 1·2 m and runs at 3000 rpm. The nozzle angle is 18°. The…

(a)
(i)

A single stage impulse steam turbine rotor has a diameter of 1·2 m and runs at 3000 rpm. The nozzle angle is 18°. The blade speed ratio is 0·42. The relative velocity at the outlet to the relative velocity at inlet is 0·9. The outlet angle of the blade is 3° smaller than the inlet angle. For a steam flow rate of 10 kg/s find Blade angles at inlet and outlet, Axial thrust on the bearing and Power developed. 20 marks

(ii)

Describe the phenomenon of super saturated flow observed in steam nozzle using T-s diagram. How does it influence the mass flow rate through the nozzle ? 20 marks

(b)
(i)

An air-conditioned space is maintained at 27°C DBT and 50% relative humidity. The ambient conditions are 40°C DBT and 27°C WBT. The space has a sensible heat gain of 14 kW. Air is supplied to the space at 7°C saturated. Determine the following : Mass of moist air supplied to the space

(ii)

Latent heat gain of space

(iii)

Cooling load of air washer if 30% of the air supplied to the space is fresh, the remainder being recirculated. Assume humid specific heat = 1·022 kJ/kg K. Psychrometric chart is given. 20 marks

(c)

A six-cylinder four-stroke diesel engine develops a power of 250 kW at 1500 rpm. The brake specific fuel consumption is 0·3 kg/kWh. The pressures of air in the cylinder at the beginning of injection and at the end of injection are 30 bar and 60 bar respectively. The fuel injection pressures at the beginning and end of injection are 220 bar and 550 bar respectively. Assume the coefficient of discharge for the injector to be 0·65, specific gravity of fuel to be 0·85 and the atmospheric pressure to be 1·013 bar. Also assume the effective pressure difference to be the average pressure difference over the injection period. Determine the nozzle area required per injection if the injection takes place over 15° crank angle. If the number of orifices used in the nozzle are 4, find the diameter of the orifice. (Conversion 1 bar = 10⁵ Pascal) 10 marks

हिंदी में प्रश्न पढ़ें
(a)
(i)

एक एकल पद आवेग भाप टर्बाइन के चुंबक का व्यास 1·2 m है और वह 3000 rpm पर चलता है। नोजल कोण 18° है। फलक वेग अनुपात 0·42 है। निर्गम पर सापेक्ष वेग का प्रवेश पर सापेक्ष वेग से अनुपात 0·9 है। फलक का बहिर्गम कोण अंतर्गम कोण से 3° छोटा है। 10 kg/s की भाप प्रवाह दर के लिए अंतर्गम और बहिर्गम पर फलक कोणों का मान, विचरिंग पर अक्षीय प्रणोद और विकसित शक्ति का मान ज्ञात कीजिए। (20 अंक)

(ii)

T-s आरेख का उपयोग करते हुए भाप नोजल में अवलोकित अति-संतृप्त प्रवाह की घटना का वर्णन करें। यह नोजल में द्रव्यमान प्रवाह दर को कैसे प्रभावित करता है ? (20 अंक)

(b)
(i)

एक वातानुकूलित स्थान 27°C DBT और 50% (प्रतिशत) RH पर बनाए रखा जाता है। परिवेश की स्थिति 40°C DBT और 27°C WBT है। स्थान में 14 kW का संवेद उष्मा लाभ है। 7°C पर संतृप्त वायु की आपूर्ति इस स्थान में की जाती है। निम्नलिखित निर्धारित करें : स्थान में आपूर्ति की गई नम हवा का द्रव्यमान

(ii)

स्थान में गुप्त उष्मा लाभ

(iii)

धावक (वॉशर) का शीतलन भार, यदि स्थान में 30 प्रतिशत ताजी हवा की आपूर्ति की जाती है, शेष हवा का पुनःप्रचालन किया जाता है। आर्द्र विशिष्ट उष्मा = 1·022 kJ/kg K मान लें। आर्द्रतामितीय लेखाचित्र संलग्न है। (20 अंक)

(c)

एक छः सिलेंडर चार स्ट्रोक डीजल इंजन 1500 rpm पर 250 kW की शक्ति विकसित करता है। ब्रेक विशिष्ट ईंधन की खपत 0·3 kg/kWh है। अंतःक्षेपण की शुरुआत में और अंतःक्षेपण के अंत में सिलेंडर में हवा का दाब क्रमशः: 30 बार और 60 बार है। शुरुआत में और अंतःक्षेपण के अंत में ईंधन अंतःक्षेपण का दाब क्रमशः: 220 बार और 550 बार है। अंतःक्षेपक के लिए विसर्जन गुणांक 0·65, ईंधन विशिष्ट घनत्व 0·85 और वायुमंडलीय दाब 1·013 बार मान लें। प्रभावी दाब अंतर को अंतःक्षेपण अवधि पर औसत दाब अंतर के रूप में लें। प्रति अंतःक्षेपण के लिए, अपेक्षित तुंड का क्षेत्रफल निर्धारित करें, यदि अंतःक्षेपण 15° क्रैंक कोण से अधिक पर है। यदि तुंड में प्रयुक्त ऑरिफिसों की संख्या 4 है, तो ऑरिफिस का व्यास ज्ञात करें। (संपरिवर्तन 1 बार = 10⁵ पास्कल) (10 अंक)

Q7 of the 2021 UPSC Mains Mechanical Engineering Paper II, as printed
The question as printed in the 2021 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.

(b) Psychrometric Chart for barometric pressure 1.01325 bar (Sea Level). The chart plots Moisture Content (kg/kg dry air) on the x-axis (range 0.000 to 0.035) and Dry Bulb Temperature (°C) on the y-axis (range -10 to 55). Curves for Relative Humidity (10% to 90%) are shown. Diagonal lines represent Specific Enthalpy at Saturation (kJ/kg air) ranging from 0 to 145. Other lines indicate Wet Bulb or Saturation Temperature (°C), Specific Volume (m³/kg), and Enthalpy Deviation (kJ/kg dry air). A scale for Sensible Heat Factor (SHF) is provided at the bottom, with a reference point noted as 25°C, 50% RH.

Table: Saturated Water Pressure Entry. Columns: Press. (kPa), Temp. (°C), Specific Volume m3/kg (Sat. Liquid vf, Evap. vfg, Sat. Vapor vg), Internal Energy kJ/kg (Sat. Liquid uf, Evap. ufg, Sat. Vapor ug). Rows: 0.6113, 1, 1.5, 2, 2.5, 3, 4, 5, 7.5, 10, 15, 20, 25, 30, 40, 50, 75, 100, 125, 150, 175, 200, 225, 250, 275, 300, 325, 350, 375, 400, 450, 500, 550, 600, 650, 700, 750, 800.

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) Impulse turbine and supersaturated flow

(i) Blade speed u = πDN/60 = π × 1.2 × 3000/60 = 188.5 m/s. With the blade speed ratio 0.42, C₁ = u/0.42 = 448.8 m/s. For α = 18°: C_w1 = 426.8 m/s and C_f1 = 138.7 m/s.

Inlet relative velocity V_r1 = √[(426.8 − 188.5)² + 138.7²] = 275.8 m/s, and tan β₁ = 138.7/238.3, so β₁ = 30.2°. The outlet blade angle is β₂ = 27.2°.

V_r2 = 0.9 × 275.8 = 248.2 m/s. Axial component at exit C_f2 = V_r2 sin 27.2° = 113.4 m/s; relative whirl at exit = V_r2 cos 27.2° = 220.6 m/s.

Axial thrust = ṁ(C_f1 − C_f2) = 10(138.7 − 113.4) = 253 N.

Power = ṁu(ΔC_w) = 10 × 188.5 × (238.3 + 220.6) = 865 kW.

Diagram: Velocity triangles of the impulse stage and T-s diagram of supersaturated nozzle flow

(ii) In a rapidly expanding steam nozzle, condensation lags behind equilibrium. On the T–s diagram the expansion crosses the saturation line, but the vapour stays dry, as a metastable supersaturated (undercooled) vapour at a temperature below saturation, until the Wilson line, where condensation suddenly occurs. The steam is then restored to the equilibrium state in a condensation shock, with an irreversible rise in entropy and some loss of available energy.

Effect on flow: the supersaturated vapour is denser than equilibrium wet steam at the same pressure. Because the nozzle mass flow ṁ = AC/v, the lower specific volume raises the mass flow by a few per cent, and the discharge coefficient is greater than that for equilibrium flow.

(b) Air-conditioning load

From the psychrometric chart: room (27°C, 50%): h₁ = 55.6 kJ/kg and W₁ = 0.0111 kg/kg; ambient (40°C DBT, 27°C WBT): h₂ ≈ 84.5 kJ/kg and W₂ ≈ 0.0172; supply (7°C saturated): h₃ = 22.7 kJ/kg and W₃ = 0.0062.

(i) Sensible heat gain 14 = ṁ × 1.022 × (27 − 7), so ṁ = 0.685 kg/s.

(ii) Latent heat gain = ṁ(W₁ − W₃)h_fg = 0.685 × (0.0111 − 0.0062) × 2454 = 8.3 kW.

(iii) The mixture entering the washer is 30% fresh and 70% recirculated air: h_m = 0.3(84.5) + 0.7(55.6) = 64.3 kJ/kg. Washer cooling load = ṁ(h_m − h₃) = 0.685(64.3 − 22.7) = 28.5 kW.

(Chart reading error is about ±1 kJ/kg.)

(c) Fuel injector nozzle

Fuel flow = 0.3 × 250 = 75 kg/h = 0.02083 kg/s. A six-cylinder four-stroke engine injects 6 × 1500/(2 × 60) = 75 times per second, so fuel per injection = 2.778 × 10⁻⁴ kg.

Injection time = (15/360)(60/1500) = 1.667 × 10⁻³ s.

Mean pressure difference = ½[(220 − 30) + (550 − 60)] = 340 bar = 3.4 × 10⁷ Pa.

Fuel velocity through the nozzle C_d√(2Δp/ρ) = 0.65√(2 × 3.4 × 10⁷/850) = 183.9 m/s.

Nozzle area A = m/(ρ × velocity × t) = 2.778 × 10⁻⁴/(850 × 183.9 × 1.667 × 10⁻³) = 1.067 × 10⁻⁶ m² (1.067 mm²).

For four orifices, each has area 0.267 mm², so d = √(4 × 0.267/π) = 0.583 mm.

How this answer will be evaluated

What earns full marks: Complete working with all governing equations, correct units, and physically interpreted results.

Evaluation rubric

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

  1. (a) Find blade angles, axial thrust and power of the impulse turbine, and describe supersaturated flow with its effect on mass flow 20 marks

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

    Must cover

    • Blade angles β₁ = 30.2° and β₂ = 27.2° from the velocity triangles
    • Axial thrust 253 N and power 865 kW
    • Supersaturated flow on the T–s diagram up to the Wilson line and the increase in mass flow

    Loses marks

    • Taking the nozzle angle as the blade angle
    • Saying supersaturation reduces mass flow

    Earns more

    • Blade speed 188.5 m/s and C₁ = u/0.42
    • Relative velocity ratio 0.9 applied
    • Condensation shock described

    Extra mark

    • Mentions the higher discharge coefficient
  2. (b) Find air supplied, latent heat gain and air-washer cooling load 20 marks

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

    Must cover

    • Mass of air 0.685 kg/s from the sensible heat gain
    • Latent heat gain about 8.3 kW
    • Washer load about 28.5 kW with 30% fresh air

    Loses marks

    • Using the ambient state as the washer inlet
    • Using the room-ambient difference for the sensible load

    Earns more

    • State properties read from the psychrometric chart
    • Mixing enthalpy from 30% fresh and 70% recirculated air

    Extra mark

    • States the chart-reading tolerance
  3. (c) Find the nozzle area per injection and orifice diameter 10 marks

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

    Must cover

    • Fuel per injection 2.778 × 10⁻⁴ kg and time 1.667 × 10⁻³ s
    • Mean pressure difference 340 bar
    • Area 1.067 mm² and orifice diameter 0.583 mm

    Loses marks

    • Using one injection per revolution
    • Using the wrong injection duration

    Earns more

    • Injections per second from six cylinders and four-stroke operation
    • Use of C_d and fuel density

    Extra mark

    • Checks the fuel velocity of about 184 m/s
What "Calculate" is asking you to do

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Where marks are lost

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