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Mechanical Engineering 2021 Paper II 50 marks Derive

Paper II — Q4

(a) Two walls A and B are maintained at temperatures T_A and T_B, respectively. One end of a metal rod of length l is embedded in…

(a)

Two walls A and B are maintained at temperatures T_A and T_B, respectively. One end of a metal rod of length l is embedded in the wall A, while the other end is fixed to wall B, the rod loses heat by convection to the environment at T_∞. Derive an expression to determine (i) the temperature distribution in the rod (ii) the total heat lost by the rod (iii) the heat transferred from the wall A 20 marks

(b)

Air enters a constant-area duct at p_1 = 90 kPa, V_1 = 520 m/s and T_1 = 558°C. It is then cooled with negligible friction until it exists at p_2 = 160 kPa. Estimate : (i) V_2 (ii) T_2 and (iii) the total enthalpy of cooling in kJ/kg. Use attached chart. 20 marks

(c)

Why is it more difficult to turbocharge spark ignition engines than compression ignition engines ? Under what circumstances might supercharger be more appropriate ? 10 marks

हिंदी में प्रश्न पढ़ें
(a)

दो दीवारों A और B को क्रमशः T_A और T_B तापमानों पर बनाए रखा जाता है। l लंबाई वाली धातु की छड़ का एक सिरा दीवार A में अंतःस्थापित है, जब कि दूसरा सिरा दीवार B में अंतःस्थापित है। छड़ T_∞ तापमान पर पर्यावरण में संवहन द्वारा ऊष्मा का ह्रास करता है। निम्नलिखित को निर्धारित करने के लिए एक व्यंजक की व्युत्पत्ति करें : (i) छड़ में तापमान वितरण (ii) छड़ द्वारा समग्र ऊष्मा ह्रास (iii) दीवार A से स्थानांतरित ऊष्मा (20 अंक)

(b)

वायु p₁ = 90 kPa, V₁ = 520 m/s और T₁ = 558°C के एक नियत क्षेत्रफल वाली बाहिनी में प्रवेश करती है। तब इसे नगण्य घर्षण के साथ ठंडा किया जाता है, जब तक कि यह p₂ = 160 kPa पर निर्गत न हो जाए। आकलन करें : (i) V₂ (ii) T₂ (iii) शीतलन की समग्र पूर्ण-ऊष्मा (एन्थैल्पी) kJ/kg में। संलग्न तालिका का उपयोग करें। (20 अंक)

(c)

संपीड़न प्रज्वलन इंजनों की तुलना में स्पार्किंग प्रज्वलन इंजनों को टर्बोचार्ज करना अधिक कठिन क्यों है ? किन परिस्थितियों में उच्चदाबी निवेशक (सुपरचार्जर) अधिक उपयुक्त हो सकता है। (10 अंक)

Q4 of the 2021 UPSC Mains Mechanical Engineering Paper II, as printed
The question as printed in the 2021 Mechanical Engineering paper
Evaluate your answer to this question → See all 2021 Mechanical Engineering questions

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) Table titled 'Frictionless Duct Flow with Heat Transfer for k = 1.4' with columns: Ma, T0/T0*, p/p*, T/T*, rho*/rho = V/V*, p0/p0*: Ma = 0.0: T0/T0* = 0.0, p/p* = 2.4000, T/T* = 0.0, rho*/rho = V/V* = 0.0, p0/p0* = 1.2679 Ma = 0.02: T0/T0* = 0.0019, p/p* = 2.3987, T/T* = 0.0023, rho*/rho = V/V* = 0.0010, p0/p0* = 1.2675 Ma = 0.04: T0/T0* = 0.0076, p/p* = 2.3946, T/T* = 0.0092, rho*/rho = V/V* = 0.0038, p0/p0* = 1.2665 Ma = 0.06: T0/T0* = 0.0171, p/p* = 2.3800, T/T* = 0.0205, rho*/rho = V/V* = 0.0086, p0/p0* = 1.2647 Ma = 0.08: T0/T0* = 0.0302, p/p* = 2.3787, T/T* = 0.0362, rho*/rho = V/V* = 0.0152, p0/p0* = 1.2623 Ma = 0.1: T0/T0* = 0.0468, p/p* = 2.3669, T/T* = 0.0560, rho*/rho = V/V* = 0.0237, p0/p0* = 1.2591 Ma = 0.12: T0/T0* = 0.0666, p/p* = 2.3526, T/T* = 0.0797, rho*/rho = V/V* = 0.0339, p0/p0* = 1.2554 Ma = 0.14: T0/T0* = 0.0895, p/p* = 2.3359, T/T* = 0.1069, rho*/rho = V/V* = 0.0458, p0/p0* = 1.2510 Ma = 0.16: T0/T0* = 0.1151, p/p* = 2.3170, T/T* = 0.1374, rho*/rho = V/V* = 0.0593, p0/p0* = 1.2461 Ma = 0.18: T0/T0* = 0.1432, p/p* = 2.2959, T/T* = 0.1708, rho*/rho = V/V* = 0.0744, p0/p0* = 1.2406 Ma = 0.2: T0/T0* = 0.1736, p/p* = 2.2727, T/T* = 0.2066, rho*/rho = V/V* = 0.0909, p0/p0* = 1.2346 Ma = 0.22: T0/T0* = 0.2057, p/p* = 2.2477, T/T* = 0.2445, rho*/rho = V/V* = 0.1088, p0/p0* = 1.2281 Ma = 0.24: T0/T0* = 0.2395, p/p* = 2.2209, T/T* = 0.2841, rho*/rho = V/V* = 0.1279, p0/p0* = 1.2213 Ma = 0.26: T0/T0* = 0.2745, p/p* = 2.1925, T/T* = 0.3250, rho*/rho = V/V* = 0.1482, p0/p0* = 1.2140 Ma = 0.28: T0/T0* = 0.3104, p/p* = 2.1626, T/T* = 0.3667, rho*/rho = V/V* = 0.1696, p0/p0* = 1.2064 Ma = 0.3: T0/T0* = 0.3469, p/p* = 2.1314, T/T* = 0.4089, rho*/rho = V/V* = 0.1918, p0/p0* = 1.1985 Ma = 0.32: T0/T0* = 0.3837, p/p* = 2.0991, T/T* = 0.4512, rho*/rho = V/V* = 0.2149, p0/p0* = 1.1904 Ma = 0.34: T0/T0* = 0.4206, p/p* = 2.0657, T/T* = 0.4933, rho*/rho = V/V* = 0.2388, p0/p0* = 1.1822 Ma = 0.36: T0/T0* = 0.4572, p/p* = 2.0314, T/T* = 0.5348, rho*/rho = V/V* = 0.2633, p0/p0* = 1.1737 Ma = 0.38: T0/T0* = 0.4935, p/p* = 1.9964, T/T* = 0.5755, rho*/rho = V/V* = 0.2883, p0/p0* = 1.1652 Ma = 0.4: T0/T0* = 0.5290, p/p* = 1.9608, T/T* = 0.6151, rho*/rho = V/V* = 0.3137, p0/p0* = 1.1566 Ma = 0.42: T0/T0* = 0.5638, p/p* = 1.9247, T/T* = 0.6535, rho*/rho = V/V* = 0.3395, p0/p0* = 1.1480 Ma = 0.44: T0/T0* = 0.5975, p/p* = 1.8882, T/T* = 0.6903, rho*/rho = V/V* = 0.3656, p0/p0* = 1.1394 Ma = 0.46: T0/T0* = 0.6301, p/p* = 1.8515, T/T* = 0.7254, rho*/rho = V/V* = 0.3918, p0/p0* = 1.1308 Ma = 0.48: T0/T0* = 0.6614, p/p* = 1.8147, T/T* = 0.7587, rho*/rho = V/V* = 0.4181, p0/p0* = 1.1224 Ma = 0.5: T0/T0* = 0.6914, p/p* = 1.7778, T/T* = 0.7901, rho*/rho = V/V* = 0.4444, p0/p0* = 1.1141 Ma = 0.52: T0/T0* = 0.7199, p/p* = 1.7409, T/T* = 0.8196, rho*/rho = V/V* = 0.4708, p0/p0* = 1.1059 Ma = 0.54: T0/T0* = 0.7470, p/p* = 1.7043, T/T* = 0.8469, rho*/rho = V/V* = 0.4970, p0/p0* = 1.0979 Ma = 0.56: T0/T0* = 0.7725, p/p* = 1.6678, T/T* = 0.8723, rho*/rho = V/V* = 0.5230, p0/p0* = 1.0901 Ma = 0.58: T0/T0* = 0.7965, p/p* = 1.6316, T/T* = 0.8955, rho*/rho = V/V* = 0.5489, p0/p0* = 1.0826 Ma = 0.6: T0/T0* = 0.8189, p/p* = 1.5957, T/T* = 0.9167, rho*/rho = V/V* = 0.5745, p0/p0* = 1.0753

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) Rod between two walls with convection

Let θ = T − T∞, with constant k, h, cross-section A_c and perimeter P, steady one-dimensional conduction and no radiation. The fin equation is

d²θ/dx² − m²θ = 0, where m² = hP/(kA_c),

with θ(0) = θ_A = T_A − T∞ and θ(l) = θ_B = T_B − T∞.

(i) Temperature distribution:

T(x) = T∞ + [(T_A − T∞) sinh m(l − x) + (T_B − T∞) sinh mx]/sinh ml.

(ii) Total heat lost by convection:

Q = ∫₀ˡ hPθ dx = hP(θ_A + θ_B)(cosh ml − 1)/(m sinh ml). Using hP = kA_c m²,

Q = kA_c m tanh(ml/2)(θ_A + θ_B).

(iii) Heat from wall A (Fourier's law at x = 0):

q_A = −kA_c(dT/dx)₀ = kA_c m[θ_A cosh ml − θ_B]/sinh ml.

Similarly q_B = kA_c m[θ_B cosh ml − θ_A]/sinh ml, and Q = q_A + q_B.

(b) Rayleigh flow (frictionless duct with cooling)

For air, k = 1.4 and R = 287 J/kg K. T₁ = 831.15 K, so a₁ = √(kRT₁) = 577.9 m/s and M₁ = 520/577.9 = 0.900.

From the Rayleigh chart or relations, (p/p*)₁ = (k + 1)/(1 + kM₁²) = 1.125 and (T/T*)₁ = 1.0245. Then

(p/p*)₂ = 1.125 × (160/90) = 2.000.

The chart gives M₂ = 0.378 on the subsonic branch, with (T/T*)₂ = 0.5717.

(ii) T₂ = T₁(T/T*)₂/(T/T*)₁ = 831.15 × 0.5717/1.0245 = 463.8 K (190.6°C).

(i) From continuity, V₂ = V₁(p₁/p₂)(T₂/T₁) = 520 × (90/160) × 0.558 = 163 m/s. Check: M₂a₂ = 0.378 × 431.7 = 163 m/s.

(iii) Heat removed per kg is the fall in stagnation enthalpy: T₀₁ = T₁(1 + 0.2M₁²) = 965.7 K and T₀₂ = T₂(1 + 0.2M₂²) = 477.0 K. With c_p = 1.0045 kJ/kg K,

q = c_p(T₀₁ − T₀₂) = 491 kJ/kg removed.

(c) Turbocharging SI and CI engines

An SI engine is harder to turbocharge because it burns a premixed charge that is already in the cylinder during compression. Boost raises charge pressure and temperature, which makes knock far more likely, so compression ratio, boost and spark advance must be cut. The mixture must also stay near stoichiometric for the three-way catalyst, so the fuel enrichment that would cool the charge is limited, and the throttle gives a wide load range with a poor exhaust-energy match. A CI engine compresses only air, injects fuel late, has no knock limit of this kind, and runs lean and unthrottled, so it accepts much higher boost.

A supercharger is more appropriate where quick response is needed, since it is driven mechanically and has no turbo lag, where boost is needed at low speed or low load when exhaust energy is small, and in small engines, racing, aircraft or two-stroke scavenging. Its cost is a parasitic power drain.

How this answer will be evaluated

What earns full marks: Rigorous derivation with clear assumptions; correct application of Rayleigh flow tables; clear distinction between SI and CI engine limitations.

Evaluation rubric

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

  1. (a) Derive the temperature distribution, total heat lost and heat from wall A for the rod between two walls 20 marks

    derive— given → assumptions → stepwise derivation → result → check

    Must cover

    • Fin equation d²θ/dx² = m²θ with θ(0) = θ_A and θ(l) = θ_B
    • T(x) from the sinh solution
    • Q = kA_c m tanh(ml/2)(θ_A + θ_B) and q_A = kA_c m[θ_A cosh ml − θ_B]/sinh ml

    Loses marks

    • Using an insulated-tip boundary condition
    • Wrong boundary condition at x = l

    Earns more

    • m = √(hP/kA_c) defined with assumptions
    • Heat loss by integrating hPθ
    • q_A by Fourier's law at x = 0

    Extra mark

    • Checks Q = q_A + q_B
  2. (b) Find V₂, T₂ and the heat removed in Rayleigh flow of air with cooling 20 marks

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

    Must cover

    • M₁ = 0.900 and (p/p*)₂ = 2.000 giving M₂ = 0.378
    • T₂ = 463.8 K and V₂ = 163 m/s
    • Heat removed ≈ 491 kJ/kg from stagnation temperature change

    Loses marks

    • Using static instead of stagnation temperatures for heat
    • Using the supersonic branch

    Earns more

    • Rayleigh chart or relations used correctly
    • Subsonic branch chosen

    Extra mark

    • Cross-checks V₂ with M₂a₂
  3. (c) Explain why SI engines are harder to turbocharge than CI engines and when a supercharger is better 10 marks

    explain— definition/context → points in order → small example → short close

    Must cover

    • Knock limits boost and compression ratio in SI engines
    • CI engines compress only air and are not knock-limited
    • Supercharger suits quick response and low-speed boost

    Loses marks

    • Ignoring knock
    • Saying a supercharger uses exhaust energy

    Earns more

    • Stoichiometric operation for the catalyst limits cooling by enrichment
    • Exhaust energy low at light load

    Extra mark

    • States the parasitic power cost of superchargers
What "Derive" is asking you to do

Reach the stated expression from a starting relation, justifying every step. The destination is printed in the question, so only the route earns marks, and the assumptions you work under are part of that route.

Structure that answers it

Assumptions and notation defined → starting relation or governing equation → each step with its justification → the required expression → limiting case or boundary check

Where marks are lost

Writing the standard result first and fitting three lines to it, which an examiner reads at a glance. Marks also go on assumptions left unstated — lossless medium, small amplitude, errors independent with zero mean — and on symbols used before they are defined, even when the question says usual notations.

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