Mechanical Engineering 2023 Paper II 50 marks Calculate

Q7

(a) A convergent-divergent nozzle receives steam at 5 bar, 250 °C and expands it isentropically into a space at 1 bar. Neglecting the inlet velocity, calculate the exit area required for a mass flow of 0·5 kg/s for the following cases: (i) When the flow is in equilibrium (ii) When the flow is supersaturated with pv^1.3 = constant Given, at 5 bar, 250 °C v = 0·4744 m³/kg, s = 7·2709 kJ/kg-K, h = 2960·7 kJ/kg and at 1 bar v_f = 0·001044 m³/kg, v_g = 1·6729 m³/kg h_f = 419·04 kJ/kg, h_g = 2676·1 kJ/kg s_f = 1·3069 kJ/kg-K, s_g = 7·3549 kJ/kg-K (b) A 1 TR refrigeration plant works on R134a simple saturated vapour compression refrigeration cycle. The evaporator and condenser temperatures are –10 °C and 44 °C respectively. Determine the (i) mass flow rate of the refrigerant, (ii) compressor power, (iii) volumetric cooling capacity and (iv) COP. Also, calculate the (v) increase in the specific compressor work due to superheat horn and (vi) throttling loss, in comparison to reversed Carnot cycle operating between the same temperature limits. Consider the entry to compressor as saturated vapour and saturated liquid refrigerant is leaving the condenser in the reversed Carnot cycle. The properties of R134a are given in the table: Take specific heat of vapour refrigerant as 1·26 kJ/kg-K. (c) Explain how the following characteristics of the lubricating oils affect the operation of an internal combustion (IC) engine: (i) Viscosity (ii) Viscosity index (iii) Pour point (iv) Flash point and fire point

हिंदी में प्रश्न पढ़ें

(a) एक अभिसारी-अपसारी नोजल 5 bar, 250 °C पर भाप प्राप्त करता है और उसे 1 bar पर समएन्ट्रॉपी विधि से किसी क्षेत्र में प्रसारित करता है। अंतरिम वेग की उपेक्षा करते हुए निम्नलिखित मामलों के लिए 0·5 kg/s के द्रव्यमान प्रवाह के लिए आवश्यक निर्गम क्षेत्र की गणना कीजिए : (i) जब प्रवाह संतुलन में हो (ii) जब प्रवाह pv^1.3 = स्थिरांक से अतिसंतृप्त हो दिया गया है, 5 bar, 250 °C पर v = 0·4744 m³/kg, s = 7·2709 kJ/kg-K, h = 2960·7 kJ/kg और 1 bar पर v_f = 0·001044 m³/kg, v_g = 1·6729 m³/kg h_f = 419·04 kJ/kg, h_g = 2676·1 kJ/kg s_f = 1·3069 kJ/kg-K, s_g = 7·3549 kJ/kg-K (b) एक 1 TR प्रशीतन संयंत्र R134a सरल संतृप्त वाष्प संपीडन प्रशीतन चक्र पर कार्य करता है। वाष्पीकरण और संघनन के तापमान क्रमशः: –10 °C और 44 °C हैं। ज्ञात कीजिए (i) प्रशीतक की द्रव्यमान प्रवाह दर, (ii) संपीडक शक्ति, (iii) आयतनिक शीतलन क्षमता तथा (iv) निष्पादन गुणांक (सी.ओ. पी.)। यह भी ज्ञात कीजिए (v) अतिताप श्रंग के कारण संपीडक के विशिष्ट कार्य में वृद्धि तथा (vi) उपरोधन हानि, समान तापमान सीमाओं के बीच चलने वाले व्युत्क्रम कार्नो चक्र की तुलना में। संपीडन में प्रवेश को संतृप्त वाष्प के रूप में मानिए और व्युत्क्रम कार्नो चक्र में संघनित्र से संतृप्त तरल प्रशीतक बाहर आ रहा है। R134a के गुण तालिका में दिए गए हैं : (c) व्याख्या कीजिए कि स्नेहक तेल की निम्नलिखित विशेषताएँ अंतर्दहन (आई. सी.) इंजन के प्रचालन को कैसे प्रभावित करती हैं : (i) श्यानता (ii) श्यानता सूचकांक (iii) बहाव बिंदु (iv) स्फुरक एवं अग्नि तापांक

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Approach

Calculate numerical solutions for all sub-parts with systematic working. For (a), apply isentropic flow relations for both equilibrium and supersaturated steam, determining exit area using continuity equation. For (b), use given R134a properties to compute cycle parameters and compare with reversed Carnot cycle. For (c), explain lubricant properties with clear IC engine operational implications. Allocate approximately 40% time to part (a) due to dual cases, 35% to part (b) for six numerical outputs, and 25% to part (c) for conceptual explanations.

Key points expected

  • Part (a)(i): Determine exit enthalpy at 1 bar using s_1 = s_2 = 7.2709 kJ/kg-K, find quality x_2, then velocity from h_0 = h_2 + V_2^2/2000, and exit area A_2 = m_dot*v_2/V_2
  • Part (a)(ii): Apply pv^1.3 = constant for supersaturated flow, calculate T_2 from isentropic relation, find v_2, h_2 using c_p for superheated steam, then velocity and area
  • Part (b): Use 1 TR = 3.517 kW, extract h_f, h_g at -10°C and 44°C from tables, compute refrigerant effect, mass flow rate, compressor work using h_2 = h_g at 44°C + c_p(T_2' - T_2), COP = RE/W
  • Part (b)(v)-(vi): Calculate Carnot COP and work, then find superheat horn loss (area under superheat on T-s diagram) and throttling loss (h_4 - h_4' on enthalpy basis)
  • Part (c): Viscosity affects film strength and pumping losses; VI ensures performance across temperature; pour point determines cold-start capability; flash/fire points indicate safety against crankcase ignition
  • Steam tables interpolation skills demonstrated where needed; all units consistent (kJ/kg, m/s, m², kg/s, kW, kJ/m³)
  • Comparison between actual VCRS and reversed Carnot cycle clearly quantified with percentage or absolute differences

Evaluation rubric

DimensionWeightMax marksExcellentAveragePoor
Concept correctness20%2Correctly distinguishes between equilibrium and supersaturated (metastable) steam flow in (a); applies Wilson line concept implicitly; uses correct VCRS cycle analysis with superheat horn and throttling loss identification in (b); accurately links each lubricant property to specific IC engine phenomena like cold-start, oil consumption, and crankcase fires.Identifies supersaturated flow but applies wrong index or confuses isentropic relations; VCRS calculations correct but Carnot comparison incomplete or conceptual errors in loss definitions; lubricant explanations generic without specific engine connections.Treats supersaturated flow as equilibrium or uses gamma=1.4 for steam; confuses VCRS with Carnot cycle or uses wrong enthalpy values; lubricant properties described without engine relevance.
Numerical accuracy25%2.5Part (a)(i): Exit area ≈ 6.8-7.2 cm²; (a)(ii): Exit area ≈ 5.5-6.0 cm² showing supersaturated reduction; Part (b): mass flow ≈ 0.024-0.026 kg/s, power ≈ 0.9-1.1 kW, COP ≈ 3.4-3.8, volumetric cooling capacity ≈ 1800-2200 kJ/m³, superheat horn ≈ 8-12 kJ/kg, throttling loss ≈ 25-35 kJ/kg; all intermediate values shown with proper significant figures.Correct methodology but arithmetic errors in 1-2 sub-parts; final answers within 10% of expected; units mostly correct but occasional mixing (bar/Pa, kW/TR).Order of magnitude errors in area (mm² vs cm²) or mass flow; COP calculated as >5 or <2; no unit consistency; key values like h_fg misread from tables.
Diagram quality15%1.5T-s diagram for steam nozzle showing both equilibrium and supersaturated expansion paths with Wilson line indicated; P-h or T-s diagram for VCRS clearly showing superheat horn (shaded) and throttling process 3-4; state points numbered; isentropic compression shown dashed vs actual; Carnot cycle overlaid or referenced for comparison.One clear diagram for either (a) or (b) with correct shape; missing superheat horn visualization or state points unlabelled; hand-sketched but recognizable.No diagrams despite question demanding visual understanding; or incorrect diagrams (e.g., Rankine cycle instead of VCRS; convergent-only nozzle sketched).
Step-by-step derivation25%2.5Explicit statement of governing equations: continuity (m_dot = rho*A*V), energy (h_0 = h + V²/2000), isentropic relations; clear quality calculation x = (s - s_f)/s_fg; supersaturated derivation with pv^n = C and T_2/T_1 = (P_2/P_1)^((n-1)/n); VCRS energy balance at each component with mass and energy equations written out; Carnot COP derivation from T_L/(T_H - T_L).Key equations stated but some steps combined or skipped; final formulas used without showing rearrangement; thermodynamic first law applied correctly but notation inconsistent.No equations stated; numbers appear without context; incorrect formula used (e.g., Bernoulli for compressible flow; ideal gas law for wet steam).
Practical interpretation15%1.5Explains why supersaturated flow gives smaller area (higher mass flow capacity) relevant to steam turbine design; discusses why actual VCRS COP is lower than Carnot and implications for Indian refrigeration energy consumption; connects lubricant VI to multi-grade oils (SAE 10W-30) for Indian climatic variation, pour point to Ladakh/Kashmir operations, flash point to turbocharged diesel safety.Brief mention of practical relevance without specific Indian context; states that metastable flow is 'faster' or 'more efficient' without explanation; generic statement about energy efficiency.No interpretation; treats as pure mathematics; or incorrect practical conclusions (e.g., supersaturated flow is undesirable without qualification; higher flash point always better without trade-offs).

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