Paper II — Q5
(a) Show in the form of a table, how the increase in the following variables affects (increase or decrease) the ignition delay…
Show in the form of a table, how the increase in the following variables affects (increase or decrease) the ignition delay period of a compression ignition (CI) engine: (i) Self-ignition temperature, (ii) Cetane number, (iii) Compression ratio, (iv) Intake pressure, (v) Intake temperature, (vi) Air-fuel ratio, (vii) Exhaust gas recirculation. 10 marks
What are the desirable characteristics of an ideal working fluid for vapour power cycle? 10 marks
What is reheat factor of a steam turbine? Derive an expression to show that the reheat factor is always greater than unity. 10 marks
Compare vapour compression and vapour absorption refrigeration systems. 10 marks
Without using psychrometric chart, calculate (i) relative humidity, (ii) humidity ratio, (iii) dew point temperature and (iv) enthalpy of moist air, when DBT is 35 °C and WBT is 23 °C. The barometer reads 755 mm of Hg. Use modified Apjohn equation (take values of pressure in bar): p_v = p'_v - (1.8p(t-t'))/2700 where, p_v = partial pressure of water vapour (w.v.) corresponding to DPT, p'_v = partial pressure of w.v. corresponding to WBT, t = DBT, t' = WBT, p = barometric pressure. Use the properties of water vapour given below: t (°C) | Vapour pressure (bar) ---|--- 10 | 0·012272, 12 | 0·014017, 14 | 0·015977, 16 | 0·018173, 18 | 0·020630, 20 | 0·023373, 22 | 0·026431, 24 | 0·029832, 32 | 0·047552, 34 | 0·053201, 36 | 0·059423. 10 marks
हिंदी में प्रश्न पढ़ें
एक तालिका के रूप में दर्शाइए कि निम्नलिखित चरों में वृद्धि, संपीडन प्रज्वलन (सी. ओ.) इंजन के प्रज्वलन विलम्ब काल को कैसे प्रभावित (वृद्धि या कमी) करती है: (i) स्वतः प्रज्वलन तापमान, (ii) सीटेन संख्या, (iii) संपीडन अनुपात, (iv) अंतर्ग्राही दाब, (v) अंतर्ग्राही तापमान, (vi) वायु-ईंधन अनुपात, (vii) रेचन गैस पुनःसंचारण। (10 अंक)
वाष्प शक्ति चक्र हेतु एक आदर्श कार्यकारी तरल की वांछनीय विशेषताएं क्या हैं? (10 अंक)
भाप-टरबाइन का पुनःस्थाप गुणक क्या है? पुनःस्थाप गुणक हमेशा एक से बड़ा होता है, यह दर्शाने हेतु एक व्यंजक की व्युत्पत्ति कीजिए। (10 अंक)
वाष्प संपीडन तथा वाष्प अवशोषण प्रशीतन प्रणालियों की तुलना कीजिए। (10 अंक)
साइक्रोमीट्रिक चार्ट का उपयोग किए बिना गणना कीजिए (i) आपेक्षिक आर्द्रता, (ii) आर्द्रता अनुपात, (iii) ओसांक तापमान तथा (iv) आर्द्र वायु की एन्थैल्पी, जब डी० बी० टी० 35 °C और डब्ल्यू० बी० टी० 23 °C है। बैरोमीटर 755 mm Hg बता रहा है। संशोधित एपीजॉन समीकरण का प्रयोग कीजिए (bar में दाब का मान लीजिए): p_v = p'_v - (1.8p(t-t'))/2700 जहाँ, p_v = डी० पी० टी० के अनुरूप जलवाष्प का आंशिक दाब, p'_v = डब्ल्यू० बी० टी० के अनुरूप जलवाष्प का आंशिक दाब, t = डी० बी० टी०, t' = डब्ल्यू० बी० टी०, p = बैरोमीटरी दाब। निचे दिए गए जलवाष्प के गुणों का उपयोग कीजिए: t (°C) | वाष्प-दाब (bar) ---|--- 10 | 0·012272, 12 | 0·014017, 14 | 0·015977, 16 | 0·018173, 18 | 0·020630, 20 | 0·023373, 22 | 0·026431, 24 | 0·029832, 32 | 0·047552, 34 | 0·053201, 36 | 0·059423। (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.
(e) Table of water vapour properties: t (°C) Vapour pressure (bar) 10 0.012272 12 0.014017 14 0.015977 16 0.018173 18 0.020630 20 0.023373 22 0.026431 24 0.029832 32 0.047552 34 0.053201 36 0.059423
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) Effect of variables on the ignition delay of a CI engine
Each variable is increased with all others held constant.
Self-ignition temperature of the fuel up: the delay increases, since the charge must reach a higher temperature before the fuel ignites. Cetane number up: the delay decreases, since a higher cetane fuel has a lower self-ignition temperature and breaks down faster. Compression ratio up: the delay decreases, as the air temperature and pressure at injection are higher. Intake pressure up: the delay decreases, since higher density speeds the pre-flame reactions. Intake temperature up: the delay decreases, as the end-of-compression temperature is higher. Air-fuel ratio up (leaner mixture): the delay increases slightly, as the cooler, less fuel-rich charge reacts more slowly. Exhaust gas recirculation up: the delay increases, since the charge has less oxygen and a lower temperature.
(b) Characteristics of an ideal working fluid for a vapour power cycle
The fluid should have a high latent heat of vaporisation and a high critical temperature, so that much energy is carried per kg and a high boiler temperature is possible without going supercritical. Its saturation pressure should be moderate at the boiler temperature, and not so low at the condenser temperature that deep vacuum and air leakage become problems. A low liquid specific volume keeps pump work small, a low liquid specific heat reduces the heat needed to reach boiling, and a low vapour specific volume keeps equipment compact. It should have good thermal conductivity and low viscosity for heat transfer, and a saturated-vapour line that gives dry or only slightly wet expansion in the turbine. Finally it should be chemically stable, non-corrosive, non-toxic, non-flammable, cheap and readily available. Water meets most of these, which is why it is used.
(c) Reheat factor of a steam turbine
Reheat factor = (sum of the isentropic enthalpy drops of the individual stages)/(isentropic enthalpy drop of the whole turbine, from the initial to the final pressure) = ΣH_i/H.
Proof that it exceeds 1. For an isentropic change, dh = v dp, so the isentropic drop between pressures p₁ and p₂ at entropy s is ∫v(p, s)dp from p₂ to p₁. The overall drop uses the initial entropy s₀: H = ∫v(p, s₀)dp. In a real multistage turbine, irreversibilities in each stage raise the entropy, so stage i begins at s_i ≥ s₀. Its isentropic drop is H_i = ∫v(p, s_i)dp over its own pressure interval. At constant pressure the specific volume of steam increases with entropy, (∂v/∂s)_p = (∂T/∂p)_s > 0. Hence v(p, s_i) > v(p, s₀) at every pressure in the stage, and so H_i exceeds the part of H lying in that pressure interval. Summing over all stages gives ΣH_i > H, so R.F. > 1.
On the h–s diagram this is because constant-pressure lines diverge towards higher entropy. Typical values are 1.02 to 1.06, and the excess drop is the heat reclaimed by the later stages from the friction losses of the earlier ones.
(d) Vapour compression versus vapour absorption refrigeration
Energy input: VCR uses mechanical work in a compressor; VAR uses heat in a generator, with only a small pump work. Refrigerant: VCR uses one pure refrigerant; VAR uses a refrigerant-absorbent pair such as NH₃-H₂O or LiBr-H₂O. Main parts: VCR has a compressor, condenser, expansion valve and evaporator; VAR replaces the compressor by absorber, pump, generator and often a rectifier. COP: VCR = Q_evap/W_comp, usually 3 to 5 or more; VAR = Q_evap/(Q_gen + W_pump), usually below 1. Moving parts and noise: VCR has a compressor, so more noise, vibration and maintenance; VAR has almost no moving parts and runs quietly. Size and cost: VCR is compact and cheap for small and medium units; VAR is bulky and costly to install but economic at large capacity when waste or solar heat is available. Temperature range: VCR reaches very low temperatures; LiBr-H₂O VAR cannot go below 0°C, and NH₃-H₂O VAR can but uses toxic ammonia. Uses: VCR in domestic refrigerators, air conditioners and cold stores; VAR in industrial plants, waste-heat recovery and solar cooling.
(e) Psychrometric properties from DBT and WBT by calculation
Given t = 35°C, t′ = 23°C, barometer 755 mm Hg: p = (755/760) × 1.01325 = 1.0066 bar. Vapour pressure at WBT, interpolating between 22°C and 24°C: p′_v = 0.026431 + 0.5 × (0.029832 − 0.026431) = 0.028132 bar. Apjohn equation: p_v = p′_v − 1.8p(t − t′)/2700 = 0.028132 − 1.8 × 1.0066 × 12/2700 = 0.028132 − 0.008053 = 0.020079 bar.
(i) Relative humidity. The saturation pressure at 35°C is (0.053201 + 0.059423)/2 = 0.056312 bar, so φ = 0.020079/0.056312 = 35.7%. (ii) Humidity ratio: ω = 0.622p_v/(p − p_v) = 0.622 × 0.020079/0.98650 = 0.01266 kg/kg dry air. (iii) Dew point: the saturation temperature at 0.020079 bar lies between 16°C (0.018173) and 18°C (0.020630), so t_dp = 16 + 2 × (0.020079 − 0.018173)/(0.020630 − 0.018173) = 17.55°C. (iv) Enthalpy: h = 1.005t + ω(2501 + 1.86t) = 35.18 + 0.01266 × 2566.1 = 67.7 kJ/kg dry air.
How this answer will be evaluated
What earns full marks: All parts answered with correct method, clear derivations, and accurate calculations.
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a) Show how an increase in each of seven variables affects the ignition delay of a CI engine. 10 marks
derive— given → assumptions → stepwise derivation → result → check
Must cover
- Self-ignition temperature up: delay up
- Cetane number, compression ratio, intake pressure and temperature up: delay down
- Leaner air-fuel ratio: delay up
- More EGR: delay up
Loses marks
- Reversing the effect of cetane number.
Earns more
- Marks for the direction of each variable with a short reason.
Extra mark
- Chemical or physical reason for each effect.
- (b) State the desirable characteristics of an ideal working fluid for a vapour power cycle. 10 marks
write short notes— define → 3-4 key features → one example → one-line significance
Must cover
- High latent heat and critical temperature
- Moderate saturation pressures at boiler and condenser
- Low liquid specific volume and specific heat
- Stable, non-corrosive, non-toxic, cheap
Loses marks
- Listing without reasons.
Earns more
- Marks for thermodynamic, heat-transfer and practical characteristics.
Extra mark
- Remark that water satisfies most of them.
- (c) Define the reheat factor and show that it is always greater than unity. 10 marks
derive— given → assumptions → stepwise derivation → result → check
Must cover
- Definition sum of stage drops over overall drop
- Isentropic drop dh = v dp
- Entropy rises through the turbine, v increases with s at constant p
- Sum of stage drops exceeds overall drop
Loses marks
- Stating the result without derivation.
Earns more
- Marks for the definition, derivation and conclusion.
Extra mark
- Divergence of isobars on h-s chart; values 1.02 to 1.06.
- (d) Compare vapour compression and vapour absorption refrigeration systems. 10 marks
compare— paired headings or point-wise contrast → key differences → significance → conclusion
Must cover
- Energy input: work versus heat
- Refrigerant versus refrigerant-absorbent pair
- COP and moving parts
- Size, cost and applications
Loses marks
- One-sided description.
Earns more
- Marks for comparison along at least five aspects.
Extra mark
- Examples of NH3-H2O and LiBr-H2O.
- (e) Calculate relative humidity, humidity ratio, dew point and enthalpy at DBT 35 C and WBT 23 C using the Apjohn equation. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- p = 1.0066 bar and p'_v = 0.028132 bar
- p_v = 0.020079 bar
- RH 35.7 percent and omega 0.01266
- Dew point 17.55 C and h 67.7 kJ/kg
Loses marks
- Reading values from a chart.
Earns more
- Marks for pressure conversion, Apjohn step, and the four properties.
Extra mark
- Interpolation shown for table values.
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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