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Mechanical Engineering 2021 Paper II 50 marks Compulsory Prove

Paper II — Q1

(a) Discuss briefly the functional differences between a fan, a blower and a compressor. (10 marks) (b) Prove that shock cannot…

(a)

Discuss briefly the functional differences between a fan, a blower and a compressor. 10 marks

(b)

Prove that shock cannot occur in subsonic flow. 10 marks

(c)

Consider a large plane wall of thickness L = 0·4 m, thermal conductivity k = 2·3 W/m°C and surface area A = 20 m². The left side of the wall is maintained at a constant temperature of T₁ = 80°C while the right side loses heat by convection to the surrounding air at Tₐ = 15°C with a heat transfer coefficient of h = 24 W/m²°C. Assuming constant thermal conductivity and no heat generation in the wall, (i) obtain a relation for the variation of temperature in the wall. (ii) evaluate the rate of heat transfer through the wall. 10 marks

(d)

The following equation has been proposed for the heat transfer coefficient in natural convection from long vertical cylinders to air at atmospheric pressure: h̄_c = (536.5(Tₛ - T_∞)⁰.33)/T where T = the film temperature = ((Tₛ + T_∞))/2 and T is in the range 0 to 200°C. The corresponding equation in dimensionless form is (h̄_c L)/K = C(Gr Pr)^m. Compare the two equations to determine the values of C and m such that the second equation will give the same results as the first. Use properties of dry air at 100°C and one atmosphere: K = 0·0307 W/(mk), g = 9·8 m/sec², μ = 21·673×10⁻⁶ NS/m², Cₚ = 1022 J/(kg K). The absolute pressure of one atmosphere = 101,000 N/m². The gas constant R (for air) = 287 J/kg K. Symbols have their usual meaning. 10 marks

(e)

Combustion in a diesel engine is assumed to begin at inner dead centre and to be at constant pressure. The air-fuel ratio is 27 : 1, the calorific value of the fuel is 43000 kJ/kg, and the specific heat (at constant volume) of the products of combustion is given by: Cᵥ = (0·71 + 20 × 10⁻⁵ T) kJ/(kg K). R for products = 0·287 kJ/(kg K). If the compression ratio is 15 : 1, and the temperature at the end of compression is 870 K, determine the percentage of stroke at which combustion is completed. 10 marks

हिंदी में प्रश्न पढ़ें
(a)

पंखा (फैन), आधमाता (ब्लोअर) और संपीडक के बीच प्रकार्यात्मक अंतरों को संक्षेप में समझाइए। (10 अंक)

(b)

सिद्ध कीजिए कि अवध्वनिक प्रवाह में प्रघात घटित नहीं हो सकता है। (10 अंक)

(c)

मोटाई L = 0·4 m, ऊष्मा चालकता k = 2·3 W/m°C और पृष्ठीय क्षेत्रफल A = 20 m² वाली एक बड़ी समतल दीवार पर विचार करें। दीवार के बायीं ओर का तापमान T₁ = 80°C पर स्थिर बनाए रखा जाता है, जबकि दाहिनी ओर Tₐ = 15°C तापमान पर परिवेश की हवा में संवहन द्वारा ऊष्मा-अंतरण गुणांक h = 24 W/m²°C के साथ ऊष्मा का ह्रास होता है। स्थिर ऊष्मा चालकता और दीवार में कोई भी ऊष्मा उत्पादन न होने को मानते हुए: (i) दीवार में तापमान परिवर्तन के लिए एक संबंध प्राप्त करें। (ii) दीवार के माध्यम से ऊष्मा हस्तांतरण की दर का मूल्यांकन करें। (10 अंक)

(d)

वायुमंडलीय दबाव पर लंबे उर्ध्वाधर सिलिंडर से हवा में प्राकृतिक संवहन के लिए ऊष्मा हस्तांतरण गुणांक का समीकरण h̄_c = (536 · 5(Tₛ-T_∞)⁰ · 33)/T द्वारा प्रस्तावित है। जहाँ T = फिल्म तापमान = ((Tₛ+T_∞))/2 और T, 0 से 200°C की सीमा में है। विमारहित रूप में संगत समीकरण (h̄_c L)/K = C(Gr Pr)^m है। दोनों समीकरणों की तुलना करके, दूसरे समीकरण में C और m के उपमानों को निर्धारित करें जो पहले समीकरण के एक सा परिणाम देगा। 100°C और एक वायुमंडलीय दबाव पर शुष्क हवा के निम्न गुणों का उपयोग करें: K = 0·0307 W/(mk), g = 9·8 m/sec², μ = 21·673×10⁻⁶ NS/m², Cₚ = 1022 J/(kg K), एक वायुमंडलीय यथार्थ दाब = 101,000 N/m², वायु के लिए गैस नियतांक R = 287 J/kg K। संकेताक्षरों का सामान्य अर्थ है। (10 अंक)

(e)

ऐसा माना जाता है कि डीजल इंजन में दहन आंतरिक निश्चल्य स्थिति पर शुरू होता है तथा दहन में दबाव स्थिर रहता है। वायु ईंधन अनुपात 27 : 1, ईंधन का ऊष्मीय मान 43000 kJ/kg, दहन के उत्पादों की विशिष्ट ऊष्मा (अचर आयतन पर) Cᵥ = (0·71 + 20 × 10⁻⁵ T) kJ/(kg K) एवं उत्पादों के लिए R = 0·287 kJ/(kg K) दिया गया है। यदि संपीडन अनुपात 15 : 1 है, और संपीडन के अंत में तापमान 870 K है, तो ज्ञात करें कि कितने प्रतिशत चरण (स्ट्रोक) पर दहन पूरा होता है। (10 अंक)

Q1 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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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) Fan, blower and compressor

All three raise the pressure of a gas, and they differ in how much. A fan gives a very small rise (up to about 1 kPa, pressure ratio near 1) with large volume flow; density change is negligible, so the flow is treated as incompressible, as in ventilation, cooling and furnace draught. A blower gives a moderate rise (about 1 kPa to roughly 100 kPa, pressure ratio up to about 1.1–1.2); density change is small, and it serves forced draught, exhaust and conveying. A compressor gives a high rise (pressure ratio above about 1.2, often several times that); density and temperature rise markedly, so compressibility, cooling and thermodynamic irreversibility must be handled, as in refrigeration, gas turbines and pneumatic systems.

(b) No shock in subsonic flow

For a steady normal shock in a perfect gas, continuity, momentum and energy give p₂/p₁ = 1 + [2γ/(γ + 1)](M₁² − 1) and ρ₂/ρ₁ = (γ + 1)M₁²/[(γ − 1)M₁² + 2]. The entropy rise is

Δs/R = [1/(γ − 1)] ln(p₂/p₁) − [γ/(γ − 1)] ln(ρ₂/ρ₁).

Differentiating with respect to M₁²,

d(Δs/R)/d(M₁²) = 2γ(M₁² − 1)²/{M₁²[(γ − 1)M₁² + 2][2γM₁² − (γ − 1)]} ≥ 0.

At M₁ = 1, Δs = 0. Since Δs does not decrease as M₁ rises, Δs < 0 for M₁ < 1. The second law requires Δs ≥ 0 for an adiabatic process, so a shock cannot exist in subsonic flow. It needs M₁ > 1.

(c) Plane wall with convection on one side

(i) With constant k and no generation, d²T/dx² = 0, so T = C₁x + C₂. At x = 0, T = 80°C, so C₂ = 80. At x = L, −k(dT/dx) = h(T_L − T_a), so −kC₁ = h(65 + C₁L) and

C₁ = −65h/(k + hL) = −65 × 24/(2.3 + 9.6) = −131.1°C/m.

Hence T(x) = 80 − 131.1x °C, and T(L) = 27.6°C.

(ii) Q = ΔT/(L/kA + 1/hA) = 65/(0.4/46 + 1/480) = 65/0.010779 = 6030 W ≈ 6.03 kW.

(d) Constants C and m for natural convection

Equating hL/K = C(Gr Pr)^m with h = 536.5ΔT^0.33/T: the ΔT-exponent 0.33 ≈ 1/3 fixes m = 1/3 (and makes h independent of L). Take T as the film temperature in kelvin, 373.15 K.

ρ = p/RT = 101 000/(287 × 373.15) = 0.943 kg/m³; ν = μ/ρ = 2.298 × 10⁻⁵ m²/s; Pr = μC_p/K = 0.7215; β = 1/373.15.

Gr Pr = gβPrΔT L³/ν² = 3.588 × 10⁷ ΔT L³, so (Gr Pr)^(1/3) = 329.8 ΔT^(1/3) L.

Then 536.5ΔT^(1/3)L/(373.15 × 0.0307) = C × 329.8ΔT^(1/3)L gives 46.83 = 329.8C, so

C ≈ 0.142 and m = 1/3.

This lies near the usual turbulent constant (0.10–0.13) for a one-third exponent. Reading T in °C would give C = 0.53, which is not physically consistent with m = 1/3.

(e) End of combustion in a diesel engine

Per kg of air, the fuel is 1/27 kg, so the products weigh 28/27 kg. Heat per kg of products = 43 000/28 = 1535.7 kJ/kg.

C_p = C_v + R = 0.997 + 0.0002T. At constant pressure,

1535.7 = 0.997(T₃ − 870) + 0.0001(T₃² − 870²),

giving T₃ = 2060.4 K.

V₃/V₂ = (28/27)(T₃/T₂) = (28/27)(2060.4/870) = 2.456.

Stroke volume = V₁ − V₂ = 14V₂, so combustion ends at (2.456 − 1)/14 × 100 = 10.4% of the stroke.

How this answer will be evaluated

What earns full marks: All parts answered with correct method, clear steps, and accurate calculations.

Evaluation rubric

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

  1. (a) Discuss the functional differences between a fan, a blower and a compressor 10 marks

    discuss— intro → 3-4 dimensions → example → balanced close

    Must cover

    • Fan: very small pressure rise, incompressible treatment
    • Blower: moderate pressure rise, small density change
    • Compressor: high pressure ratio with significant density and temperature change

    Loses marks

    • Calling all three the same
    • Giving no distinguishing pressure rise

    Earns more

    • Typical pressure ratios or ranges quoted
    • Typical applications of each

    Extra mark

    • Notes the compressibility and cooling consideration for compressors
  2. (b) Prove that a shock cannot occur in subsonic flow 10 marks

    justify— claim → 3-4 reasons → evidence → conclusion

    Must cover

    • Normal-shock relations from conservation laws
    • Entropy change Δs expressed in terms of M₁
    • Δs < 0 for M₁ < 1, violating the second law

    Loses marks

    • Asserting the result without entropy argument
    • Using isentropic relations across the shock

    Earns more

    • Pressure and density ratios stated correctly
    • Derivative of Δs with respect to M₁² shown non-negative

    Extra mark

    • Notes Δs = 0 at M₁ = 1 as the weak-shock limit
  3. (c) Find the temperature distribution in the wall and the heat transfer rate 10 marks

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

    Must cover

    • Linear profile T = C₁x + C₂ with T(0) = 80°C
    • Convection boundary condition at x = L giving C₁ = −131.1°C/m
    • Q ≈ 6.03 kW

    Loses marks

    • Applying T = 15°C at the right face
    • Using the wrong area or conductivity units

    Earns more

    • Right-face temperature about 27.6°C
    • Thermal-resistance cross-check

    Extra mark

    • States the assumptions of steady one-dimensional conduction
  4. (d) Find C and m so that the dimensionless correlation matches the given equation 10 marks

    analyse— intro → causes → effects → stakeholders/linkages → way forward

    Must cover

    • m = 1/3 from the ΔT exponent
    • Gr Pr evaluated from air properties at 100°C
    • C ≈ 0.142 with T taken in kelvin

    Loses marks

    • Using β in °C
    • Not stating the unit used for T

    Earns more

    • ρ, ν, Pr and β computed correctly
    • Matching of coefficients shown

    Extra mark

    • Compares C with the usual turbulent value or explains the unit choice for T
  5. (e) Find the percentage of stroke at which combustion ends 10 marks

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

    Must cover

    • Heat per kg of products 43 000/28 and C_p = C_v + R
    • T₃ ≈ 2060 K from the constant-pressure energy balance
    • V₃/V₂ = 2.456 and about 10.4% of the stroke

    Loses marks

    • Using C_v instead of C_p
    • Using V₁ − V₂ = V₂

    Earns more

    • Quadratic in T₃ solved correctly
    • Stroke volume 14V₂ used

    Extra mark

    • Accounts for the added fuel mass in the volume ratio
What "Prove" is asking you to do

Establish that the statement holds for every case it claims, not for one representative case. The argument must be closed: each line follows from a definition, a hypothesis, or a named theorem you are entitled to use.

Structure that answers it

Given and to prove, restated → theorem or construction to be used, named → the argument line by line → conclusion stated as proved

Where marks are lost

Testing one example, which illustrates but proves nothing. On an if and only if claim, proving one direction and stopping forfeits that half outright, and degenerate cases — zero, the empty set, the equality case — have to be disposed of rather than assumed away.

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