Paper II — Q2
(a)(i) 3 kg of air is compressed in a reversible steady flow polytropic process from 100 kPa, 40°C to 1000 kPa. During this…
3 kg of air is compressed in a reversible steady flow polytropic process from 100 kPa, 40°C to 1000 kPa. During this process the law of compression followed is PV^1.25 = C. Determine the shaft work, heat transferred and the change in entropy. Assume for air C_v = 0.717 kJ/kg K and R = 0.287 kJ/kg K. (ii) Distinguish between pdv work and -vdp work. 20 marks
Calculate the displacement thickness and momentum thickness of a laminar boundary layer, in terms of the nominal boundary layer thickness δ, for the following velocity distribution: u/U_0 = sin(π/2 y/δ) 20 marks
An ideal gas turbine engine operates with air as the working fluid at a pressure ratio 18 : 1 and a maximum temperature of 700°C. The air enters the compressor at 100 kPa and 20°C. Determine the thermal efficiency, the heat addition and the temperature of exhaust air. For air take C_p = 1.0035 kJ/kg K and γ = 1.4. 10 marks
हिंदी में प्रश्न पढ़ें
100 kPa, 40°C से 1000 kPa तक एक प्रतिक्रम्य अपरिवर्ती प्रवाह पॉलिट्रॉपिक प्रक्रम में 3 kg वायु संपीड़ित होती है। इस प्रक्रिया के दौरान संपीड़न नियम PV^1.25 = C का पालन होता है। शाफ्ट-कार्य, हस्तांतरित ऊष्मा तथा एन्ट्रॉपी में परिवर्तन निर्धारित करें। हवा के लिए C_v = 0.717 kJ/kg K और R = 0.287 kJ/kg K मान लें। (ii) pdv कार्य और -vdp कार्य के बीच अंतर करें। (20 अंक)
निम्नलिखित वेग वितरण के लिए अभिहित सीमांत परत मोटाई δ के संदर्भ में स्तरीय सीमांत परत की विस्थापन मोटाई और संवेग मोटाई की गणना करें: u/U_0 = sin(π/2 y/δ) (20 अंक)
एक आदर्श गैस टरबाइन इंजन दबाव अनुपात 18 : 1 और अधिकतम तापमान 700°C पर कार्यात्मक-तरल वायु से संचालित है। हवा 100 kPa और 20°C पर संपीड़क में प्रवेश करती है। ऊष्मीय दक्षता, ऊष्मा योग और रेचन हवा का तापमान निर्धारित करें। हवा के लिए C_p = 1.0035 kJ/kg K और γ = 1.4 लें। (10 अंक)
Model answer
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(a) Polytropic compression of air
(i) C_p = C_v + R = 0.717 + 0.287 = 1.004 kJ/kg K and n = 1.25.
T₂ = T₁(p₂/p₁)^((n−1)/n) = 313.15 × 10^0.2 = 496.3 K, so ΔT = 183.16 K.
Shaft work (steady flow, reversible) W_s = −m∫v dp = −m[n/(n − 1)]R(T₂ − T₁) = −3 × 5 × 0.287 × 183.16 = −788.5 kJ. The air receives 788.5 kJ of shaft work.
Heat transfer from the steady-flow energy equation: Q = ΔH + W_s = 3 × 1.004 × 183.16 − 788.5 = 551.7 − 788.5 = −236.8 kJ. About 236.8 kJ is rejected.
Entropy change ΔS = m[C_p ln(T₂/T₁) − R ln(p₂/p₁)] = 3[1.004 × 0.4605 − 0.287 × 2.3026] = −0.595 kJ/K.
(ii) The p dv work, ∫p dv, is the boundary work of a closed system and is the area under the p–v curve. The −v dp work is the reversible shaft work of a steady-flow device such as a compressor or turbine, the area to the left of the p–v curve. For the same path they differ by flow work: ∫p dv − (−∫v dp) = Δ(pv). For a polytropic process, −∫v dp = n × ∫p dv.
(b) Boundary-layer thicknesses
With η = y/δ and u/U₀ = sin(πη/2):
Displacement thickness δ* = δ∫₀¹[1 − sin(πη/2)]dη = δ[1 − 2/π] = 0.3634δ.
Momentum thickness θ = δ∫₀¹ sin(πη/2)[1 − sin(πη/2)]dη = δ[2/π − 1/2] = 0.1366δ.
Here ∫₀¹ sin(πη/2)dη = 2/π and ∫₀¹ sin²(πη/2)dη = 1/2. The shape factor is H = δ*/θ = 2.66.
(c) Brayton cycle
T₁ = 293.15 K, T₃ = 973.15 K, and r_p^((γ−1)/γ) = 18^(0.2857) = 2.2838.
T₂ = 293.15 × 2.2838 = 669.5 K.
Heat addition q_in = C_p(T₃ − T₂) = 1.0035 × 303.7 = 304.7 kJ/kg.
Exhaust temperature T₄ = T₃/2.2838 = 426.1 K = 153°C.
Thermal efficiency η = 1 − 1/2.2838 = 56.2%.
How this answer will be evaluated
What earns full marks: All parts solved with correct method, clear steps, and physical interpretation.
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a) Find shaft work, heat transfer and entropy change for polytropic compression and distinguish p dv work from −v dp work 20 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- T₂ = 496.3 K and shaft work 788.5 kJ input
- Heat rejected 236.8 kJ and ΔS = −0.595 kJ/K
- p dv is closed-system boundary work, −v dp is steady-flow shaft work
Loses marks
- Using p dv work as the shaft work
- Wrong sign for heat or entropy
Earns more
- Use of n/(n − 1) in the shaft work
- Steady-flow energy equation applied for heat
- Relation Δ(pv) between the two works
Extra mark
- States −∫v dp = n∫p dv for a polytropic process
- (b) Find displacement and momentum thickness for u/U₀ = sin(πy/2δ) 20 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Definitions of δ* and θ
- δ* = (1 − 2/π)δ = 0.3634δ
- θ = (2/π − 1/2)δ = 0.1366δ
Loses marks
- Using the momentum thickness formula for δ*
- Arithmetic error in the integrals
Earns more
- Substitution η = y/δ
- Integrals evaluated correctly
Extra mark
- Gives the shape factor H ≈ 2.66
- (c) Find thermal efficiency, heat addition and exhaust temperature of the gas turbine 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- T₂ = 669.5 K from 18^(0.2857)
- Heat addition ≈ 304.7 kJ/kg
- Exhaust 153°C and efficiency 56.2%
Loses marks
- Using °C in the isentropic relation
- Using the wrong exponent
Earns more
- Absolute temperatures used
- T₄ from isentropic expansion
Extra mark
- Notes the ideal-cycle efficiency depends only on pressure ratio
What "Calculate" is asking you to do
Apply the standard formula or schedule to data the question has already supplied — a table of readings, cost records, a balance sheet — and produce the number. The method is rarely in doubt; the marks sit in the named intermediate quantities, each of which has to appear as a labelled line.
Structure that answers it
Data as given → formula or standard treatment, named → substitution → each intermediate, labelled → result with units
Where marks are lost
Omitting an intermediate the marking scheme pays for separately, or rounding at an intermediate line so the final figure drifts. In commerce and accountancy, any figure in a statement that no numbered working note supports is treated as unearned.
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