Paper II — Q8
(a) The following data refers to a single-stage vapour compression refrigeration system : Refrigerant = R134a Condenser…
The following data refers to a single-stage vapour compression refrigeration system :
Refrigerant = R134a Condenser temperature = 35 °C Evaporator temperature = – 10 °C Compressor motor speed = 2800 r.p.m. Clearance ratio = 0·03 Swept volume = 269·4 cm³ Expansion index = 1·12 Compression isentropic efficiency = 75% Condensate subcooling in condenser = 5 °C
Draw P-h diagram and determine the following :
The capacity of the plant in TR
The power required in kW
The COP
The heat rejection to condenser
The second law efficiency
The properties of R134a are given in the table :
| T (°C) | P (bar) | Specific volume of saturated vapour v_g (m³/kg) | Enthalpy (kJ/kg) h_f | h_g | Entropy (kJ/kg-K) s_f | s_g |
|---|---|---|---|---|---|---|
| – 10 | 2·014 | 0·0994 | 186·7 | 392·4 | 0·9512 | 1·733 |
| 35 | 8·870 | — | 249·1 | 417·6 | 1·1680 | 1·715 |
Assume specific heat of liquid and vapour at 8·87 bar as 1·458 kJ/kg-K and 1·1 kJ/kg-K, respectively. The refrigerant at entry to compressor is in dry saturated state. 20 marks
What are the important properties of lubricants used in IC engines? Discuss their significance. 10 marks
What are the advantages and disadvantages of the Indirect Injection (IDI) swirl chamber over the open-type Direct Injection (DI) combustion chamber in CI engines? 10 marks
Calculate the critical pressure and throat area per unit mass flow rate of a convergent-divergent nozzle, expanding steam from 10 bar and dry saturated, down to atmospheric pressure of 1 bar. Assume that the inlet velocity is negligible and that the expansion is isentropic.
Use Steam tables given at the end to get the steam data.
The value of isentropic expansion index may be taken as 1·135. 10 marks
हिंदी में प्रश्न पढ़ें
निम्नलिखित आँकड़े एक एकल-चरणीय वाष्प संपीडन प्रशीतन प्रणाली से संबंधित हैं :
प्रशीतक द्रव्य = R134a संघनक तापमान = 35 °C वाष्पक तापमान = – 10 °C संपीडक (कंप्रेसर) मोटर गति = 2800 r.p.m. मुक्तांतर (क्लियरेंस) अनुपात = 0·03 प्रस्पिष्ट (स्वेप्ट) आयतन = 269·4 cm³ विस्तार सूचकांक = 1·12 संपीडन की आइसेंट्रॉपिक दक्षता (कंप्रेशन आइसेंट्रॉपिक एफिशिएंसी) = 75% संघनक में संघनित द्रव का अवशीतलन (कंडेंसेट सबकूलिंग इन कंडेंसर) = 5 °C
P-h आरेख बनाइये तथा निम्नलिखित को निर्धारित कीजिये :
संयंत्र की क्षमता, TR में
आवश्यक शक्ति, kW में
सी० ओ० पी०
संघनक को अपवाहित उष्मा (हीट रिजेक्शन टु कंडेंसर)
द्वितीय नियम दक्षता
R134a के गुणधर्म तालिका में दिये गये हैं :
| T (°C) | P (bar) | संतृप्त वाष्प का विशिष्ट आयतन v_g (m³/kg) | एन्थैल्पी (kJ/kg) | एन्ट्रॉपी (kJ/kg-K) | ||
|---|---|---|---|---|---|---|
| h_f | h_g | s_f | s_g | |||
| – 10 | 2·014 | 0·0994 | 186·7 | 392·4 | 0·9512 | 1·733 |
| 35 | 8·870 | — | 249·1 | 417·6 | 1·1680 | 1·715 |
मान लीजिये कि 8·87 bar पर तरल तथा वाष्प की विशिष्ट उष्माएँ क्रमशः 1·458 kJ/kg-K तथा 1·1 kJ/kg-K हैं। संपीडक में प्रवेश के समय प्रशीतक द्रव्य (रेफ्रिजरेंट) शुष्क संतृप्त अवस्था में होता है। (20 अंक)
आइ० सी० इंजनों में प्रयुक्त स्नेहकों के महत्त्वपूर्ण गुण कौन-कौन से होते हैं? उनके महत्व की चर्चा कीजिये। (10 अंक)
सी० आइ० इंजनों में अप्रत्यक्ष इंजेक्शन (आइ० डी० आइ०) स्वर्ल चैंबर की ओपन-टाइप प्रत्यक्ष इंजेक्शन (डी० आइ०) दहन कक्ष की तुलना में क्या-क्या लाभ और हानियाँ होती हैं? (10 अंक)
एक अभिसारी-अपसारी नुंड (कनवर्जेंट-डाइवर्जेंट नोजल) के लिये, जो कि 10 bar दाब और शुष्क संतृप्त अवस्था की भाप को वायुमंडलीय दाब 1 bar तक विस्तारित करता है, क्रांतिक दाब तथा प्रति इकाई द्रव्यमान प्रवाह दर के लिये गला (थ्रोट) क्षेत्रफल की गणना कीजिये। यह मानिये कि प्रवेश वेग नगण्य है तथा विस्तार आइसेंट्रॉपिक है।
भाप-संबंधी आँकड़ों के लिये अंत में दी गई भाप तालिकाओं का प्रयोग कीजिये।
आइसेंट्रॉपिक विस्तार सूचकांक (आइसेंट्रॉपिक एक्सपैंशन इंडेक्स) का मान 1·135 लिया जा सकता है। (10 अंक)
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.
(a) Table of properties of R134a with columns: T (°C), P (bar), Specific volume of saturated vapour v_g (m³/kg), Enthalpy (kJ/kg) h_f, h_g, Entropy (kJ/kg-K) s_f, s_g. Row 1: -10, 2.014, 0.0994, 186.7, 392.4, 0.9512, 1.733. Row 2: 35, 8.870, —, 249.1, 417.6, 1.1680, 1.715.
(c) Table: Steam Table [Saturated Water : Pressure] Columns: Pressure (kPa or MPa) | Temp. T (°C) | Specific Volume v_f (m^3/kg) | Specific Volume v_g (m^3/kg) | Internal Energy u_f (kJ/kg) | Internal Energy u_fg (kJ/kg) | Internal Energy u_g (kJ/kg) | Enthalpy h_f (kJ/kg) | Enthalpy h_fg (kJ/kg) | Enthalpy h_g (kJ/kg) | Entropy s_f (kJ/kg·K) | Entropy s_fg (kJ/kg·K) | Entropy s_g (kJ/kg·K) 0.6113 kPa | 0.01 | 0.001000 | 206.132 | 0 | 2375.3 | 2375.3 | 0.00 | 2501.3 | 2501.3 | 0 | 9.1562 | 9.1562 1.0 kPa | 6.98 | 0.001000 | 129.208 | 29.29 | 2355.7 | 2385.0 | 29.29 | 2484.9 | 2514.2 | 0.1059 | 8.8697 | 8.9756 1.5 kPa | 13.03 | 0.001001 | 87.980 | 54.70 | 2338.6 | 2393.3 | 54.70 | 2470.6 | 2525.3 | 0.1956 | 8.6322 | 8.8278 2.0 kPa | 17.50 | 0.001001 | 67.004 | 73.47 | 2326.0 | 2399.5 | 73.47 | 2460.0 | 2533.5 | 0.2607 | 8.4629 | 8.7236 2.5 kPa | 21.08 | 0.001002 | 54.254 | 88.47 | 2315.9 | 2404.4 | 88.47 | 2451.6 | 2540.0 | 0.3120 | 8.3311 | 8.6431 3.0 kPa | 24.08 | 0.001003 | 45.665 | 101.03 | 2307.5 | 2408.5 | 101.03 | 2444.5 | 2545.5 | 0.3545 | 8.2231 | 8.5775 4.0 kPa | 28.96 | 0.001004 | 34.800 | 121.44 | 2293.7 | 2415.2 | 121.44 | 2432.9 | 2554.4 | 0.4226 | 8.0520 | 8.4746 5.0 kPa | 32.88 | 0.001005 | 28.193 | 137.79 | 2282.7 | 2420.5 | 137.79 | 2423.7 | 2561.4 | 0.4763 | 7.9187 | 8.3950 7.5 kPa | 40.29 | 0.001008 | 19.238 | 168.76 | 2261.7 | 2430.5 | 168.77 | 2406.0 | 2574.8 | 0.5763 | 7.6751 | 8.2514 10.0 kPa | 45.81 | 0.001010 | 14.674 | 191.79 | 2246.1 | 2437.9 | 191.81 | 2392.8 | 2584.6 | 0.6492 | 7.5010 | 8.1501 15.0 kPa | 53.97 | 0.001014 | 10.022 | 225.90 | 2222.8 | 2448.7 | 225.91 | 2373.1 | 2599.1 | 0.7548 | 7.2536 | 8.0084 20.0 kPa | 60.06 | 0.001017 | 7.649 | 251.35 | 2205.4 | 2456.7 | 251.38 | 2358.3 | 2609.7 | 0.8319 | 7.0766 | 7.9085 25.0 kPa | 64.97 | 0.001020 | 6.204 | 271.88 | 2191.2 | 2463.1 | 271.90 | 2346.3 | 2618.2 | 0.8930 | 6.9383 | 7.8313 30.0 kPa | 69.10 | 0.001022 | 5.229 | 289.18 | 2179.2 | 2468.4 | 289.21 | 2336.1 | 2625.3 | 0.9439 | 6.8247 | 7.7686 40.0 kPa | 75.87 | 0.001026 | 3.993 | 317.51 | 2159.5 | 2477.0 | 317.55 | 2319.2 | 2636.7 | 1.0258 | 6.6441 | 7.6700 50.0 kPa | 81.33 | 0.001030 | 3.240 | 340.42 | 2143.4 | 2483.8 | 340.47 | 2305.4 | 2645.9 | 1.0910 | 6.5029 | 7.5939 75.0 kPa | 91.77 | 0.001037 | 2.217 | 384.29 | 2112.4 | 2496.7 | 384.36 | 2278.6 | 2663.0 | 1.2129 | 6.2434 | 7.4563 0.100 MPa | 99.62 | 0.001043 | 1.6940 | 417.33 | 2088.7 | 2506.1 | 417.44 | 2258.0 | 2675.5 | 1.3025 | 6.0568 | 7.3593 0.125 MPa | 105.99 | 0.001048 | 1.3749 | 444.16 | 2069.3 | 2513.5 | 444.30 | 2241.1 | 2685.3 | 1.3739 | 5.9104 | 7.2843 0.150 MPa | 111.37 | 0.001053 | 1.1593 | 466.92 | 2052.7 | 2519.6 | 467.08 | 2226.5 | 2693.5 | 1.4335 | 5.7897 | 7.2232 0.175 MPa | 116.06 | 0.001057 | 1.0036 | 486.78 | 2038.1 | 2524.9 | 486.97 | 2213.6 | 2700.5 | 1.4848 | 5.6868 | 7.1717 0.200 MPa | 120.23 | 0.001061 | 0.8857 | 504.47 | 2025.0 | 2529.5 | 504.68 | 2202.0 | 2706.6 | 1.5300 | 5.5970 | 7.1271 0.225 MPa | 124.00 | 0.001064 | 0.7933 | 520.45 | 2013.1 | 2533.6 | 520.69 | 2191.3 | 2712.0 | 1.5705 | 5.5173 | 7.0878 0.250 MPa | 127.43 | 0.001067 | 0.7187 | 535.08 | 2002.1 | 2537.2 | 535.34 | 2181.5 | 2716.9 | 1.6072 | 5.4455 | 7.0526
Model answer
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(a)(i) – (v)
P–h diagram (drawn below): P vertical (log scale), h horizontal. States: 1 dry saturated vapour at 2.014 bar; 2s isentropic compression at 8.870 bar; 2 actual compression; 3 after condensation and 5 °C subcooling; 4 after isenthalpic throttling at 2.014 bar. Process lines: 4–1 evaporation, 1–2 actual compression (1–2s dashed isentropic), 2–3 condensation/subcooling, 3–4 throttling.
State 1: h₁ = 392.4 kJ/kg, s₁ = 1.733 kJ/kg·K, v₁ = 0.0994 m³/kg.
Volumetric efficiency: p₂/p₁ = 8.870/2.014 = 4.4042, so η_v = 1 + C − C(p₂/p₁)^(1/n) = 1.03 − 0.03 × (4.4042)^(1/1.12) = 1.03 − 0.03 × 3.7573 = 0.9173.
Mass flow: swept-volume rate = 269.4 × 10⁻⁶ × 2800/60 = 0.012572 m³/s, so ṁ = η_v V̇_s/v₁ = 0.9173 × 0.012572/0.0994 = 0.1160 kg/s.
Isentropic compression: s₂ₛ = s₁ = 1.733 > s_g(8.87 bar) = 1.715, so state 2s is superheated. With c_p,v = 1.1 kJ/kg·K, T₂ₛ = 308.15 × exp[(1.733 − 1.715)/1.1] = 313.23 K, and h₂ₛ = 417.6 + 1.1 × (313.23 − 308.15) = 423.19 kJ/kg.
Actual compression: h₂ = h₁ + (h₂ₛ − h₁)/η_c = 392.4 + 30.79/0.75 = 433.45 kJ/kg, so w_c = 41.05 kJ/kg (discharge temperature ≈ 49.4 °C).
Subcooling and throttling: h₃ = h_f(35 °C) − c_p,l × 5 = 249.1 − 1.458 × 5 = 241.81 kJ/kg = h₄, so q_e = h₁ − h₄ = 150.59 kJ/kg.
(i) Capacity: Q_e = ṁq_e = 0.1160 × 150.59 = 17.47 kW = 4.97 TR (1 TR = 3.5167 kW).
(ii) Power: W = ṁw_c = 0.1160 × 41.05 = 4.76 kW.
(iii) COP = q_e/w_c = 150.59/41.05 = 3.67.
(iv) Heat rejected in the condenser: Q_c = ṁ(h₂ − h₃) = 0.1160 × (433.45 − 241.81) = 22.23 kW, which equals Q_e + W as a check.
(v) Second-law efficiency: COP_Carnot = T_L/(T_H − T_L) = 263.15/45 = 5.848, so η_II = 3.67/5.848 = 0.627 (62.7%).
(b)(i)
Important lubricant properties in IC engines:
- Viscosity: maintains oil film; too high causes pumping loss, too low causes wear.
- Viscosity index: low change of viscosity with temperature; ensures cold starting and hot protection.
- Pour point: must be low for cold-weather flow.
- Flash and fire point: high for safety and low evaporation loss.
- Oxidation and thermal stability: resist sludge, varnish and thickening at high temperature.
- Detergency and dispersancy: keep soot and deposits suspended, keeping engine clean.
- Anti-wear and extreme-pressure properties: protect cams, bearings and piston rings in boundary lubrication.
- Corrosion and rust inhibition: neutralise acidic combustion products.
- Foam resistance: prevents air entrainment and oil-film breakdown.
- Seal compatibility: prevents swelling, shrinkage and leakage.
- Alkalinity/TBN: neutralises acids and extends drain interval. Significance: together these ensure lubrication, cooling, sealing, cleaning, corrosion protection and longer engine life.
(b)(ii)
Advantages of IDI swirl chamber over open DI: lower injection pressure is needed; better fuel-air mixing at low speed and part load; smoother pressure rise, lower combustion noise; lower NOx due to lower peak temperature; less sensitive to fuel quality; wider speed range for small engines. Disadvantages: higher specific fuel consumption, often 10–20% penalty, because of heat loss and throttling between chambers; lower thermal efficiency and power output; more complex cylinder head and higher manufacturing cost; poorer cold starting, needing glow plugs; greater heat loss and thermal loading; limited to smaller engines; poorer full-load economy than open DI.
(c)
For steam, use isentropic expansion index n = 1.135. Critical pressure ratio: p*/p₁ = (2/(n+1))^(n/(n−1)) = (2/2.135)^(1.135/0.135) = 0.5774. p₁ = 10 bar, so p* = 10 × 0.5774 = 5.774 bar. Critical pressure = 5.77 bar.
From saturated steam tables at 10 bar dry saturated: v₁ = v_g = 0.1944 m³/kg. Critical velocity: c* = √[2n/(n+1) p₁v₁] = √[(2×1.135/2.135) × 10×10⁵ × 0.1944] = √(1.0632 × 1.944×10⁵) = 454.6 m/s.
Specific volume at throat: v* = v₁(p₁/p*)^(1/n) = 0.1944(1/0.5774)^(1/1.135) = 0.3155 m³/kg. Throat area per unit mass flow rate: A*/ṁ = v*/c* = 0.3155/454.6 = 6.94×10⁻⁴ m² per (kg/s). **A*/ṁ = 6.94 cm² per kg/s.**
How this answer will be evaluated
Framework: null
What earns full marks: All parts fully answered with correct calculations, clear diagrams, and detailed explanations.
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a) Determine plant capacity, power, COP, heat rejection, and second law efficiency for the R134a system. 20 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- P–h diagram with states 1, 2s, 2, 3, 4 (figure)
- η_v = 0.9173; ṁ = 0.1160 kg/s
- h2s = 423.19, h2 = 433.45 ⇒ wc = 41.05 kJ/kg; h3 = h4 = 241.81; qe = 150.59 kJ/kg
- Qe = 17.47 kW = 4.97 TR; W = 4.76 kW; COP = 3.67; Qc = 22.23 kW; η_II = 62.7%
Loses marks
- Missing P-h diagram or unmarked states
- Ignoring clearance ratio in mass flow
- Plugging numbers without governing equations
Earns more
- Correct unit conversion for TR and kW
- Explicit calculation of second law efficiency
- Consistent use of R134a property table
Extra mark
- T-s diagram provided alongside P-h
- (b)(i) Discuss important properties of IC engine lubricants and their significance. 10 marks
discuss— intro → 3-4 dimensions → example → balanced close
Must cover
- Key properties: viscosity, viscosity index, pour point, flash/fire point
- Oxidation/thermal stability, detergency/dispersancy, anti-wear/EP additives
- Corrosion inhibition, foam resistance, seal compatibility, TBN
- Significance: lubrication, cooling, sealing, cleaning, protection and engine life
Loses marks
- Listing properties without explaining significance
- Vague or generic statements
Earns more
- Mention specific lubricant types
- Discuss impact on engine wear
Extra mark
- Reference to specific lubricant standards
- (b)(ii) Compare advantages and disadvantages of IDI swirl chamber vs open-type DI chamber. 10 marks
compare— paired headings or point-wise contrast → key differences → significance → conclusion
Must cover
- IDI advantages: lower injection pressure, better mixing at low speed/part load
- Smoother pressure rise, lower noise and NOx; less fuel-quality sensitivity
- Disadvantages: higher SFC (10–20%), lower efficiency and power
- Complex head/cost, cold-start difficulty (glow plugs), heat loss; contrast with open DI
Loses marks
- Listing only advantages or only disadvantages
- Failing to contrast with DI chamber
Earns more
- Mention specific performance metrics
- Discuss impact on emissions
Extra mark
- Schematic diagrams of both chambers
- (c) Calculate critical pressure and throat area for the convergent-divergent nozzle. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- n = 1.135; p*/p1 = (2/2.135)^(1.135/0.135) = 0.5774
- p* = 5.774 bar; v1 = 0.1944 m³/kg
- c* = √[2n/(n+1)·p1v1] = 454.6 m/s
- v* = 0.3155 m³/kg ⇒ A*/ṁ = 6.94×10⁻⁴ m²/(kg/s) = 6.94 cm²/(kg/s)
Loses marks
- Ignoring isentropic expansion index
- Incorrect use of steam table data
Earns more
- Correct application of critical pressure formula
- Accurate calculation of throat area
Extra mark
- Mention of choked flow conditions
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