Paper II — Q7
(a) A two-stage turbine receives steam at 50 bar and 350 °C. At 1·5 bar, the high-pressure stage exhausts and 12000 kg of steam…
A two-stage turbine receives steam at 50 bar and 350 °C. At 1·5 bar, the high-pressure stage exhausts and 12000 kg of steam per hour is taken at this stage for process heating purposes. The remainder steam is reheated to 250 °C at 1·5 bar and then expanded through the low-pressure turbine to a condensate pressure of 0·05 bar. The power output from the turbine unit is to be 3750 kW. The isentropic efficiency of high-pressure stage is 0·84 and that of the low-pressure stage is 0·81. Calculate the rate of steam being generated per hour by the boiler.
Steam properties : 50 bar, 350 °C : h = 3068·4 kJ/kg, s = 6·4492 kJ/kg-K 1·5 bar, 250 °C : h = 2972·65 kJ/kg, s = 7·8709 kJ/kg-K
Use Steam tables given at the end to get other properties. 20 marks
What is meant by effective temperature? List and explain briefly the factors which affect optimum effective temperature in air-conditioning. 10 marks
Explain psychrometric process of adiabatic mixing of two air streams. 10 marks
Why is it necessary to cool IC engines? What are the disadvantages of overcooling an IC engine? 10 marks
हिंदी में प्रश्न पढ़ें
एक द्वि-चरणीय टरबाइन (टू-स्टेज टरबाइन) 50 bar और 350 °C पर भाप प्राप्त करता है। उच्च-दाब चरण (हाई-प्रेशर स्टेज) से 1·5 bar पर भाप निष्कासित होती है तथा इस चरण से प्रति घंटा 12000 kg भाप को प्रसंस्करण तापन (प्रोसेस हीटिंग) के लिये निकाला जाता है। शेष भाप को 1·5 bar पर 250 °C तक पुनः गर्म किया जाता है और फिर निम्न-दाब टरबाइन (लो-प्रेशर टरबाइन) से होकर 0·05 bar कंडेंसेट प्रेशर तक विस्तारित किया जाता है। टरबाइन इकाई से शक्ति उत्पादन 3750 kW है। उच्च-दाब चरण की आइसेन्ट्रॉपिक दक्षता 0·84 है तथा निम्न-दाब चरण की आइसेन्ट्रॉपिक दक्षता 0·81 है। प्रति घंटा बॉयलर द्वारा उत्पन्न भाप की दर की गणना कीजिये।
भाप के गुणधर्म : 50 bar, 350 °C : h = 3068·4 kJ/kg, s = 6·4492 kJ/kg-K 1·5 bar, 250 °C : h = 2972·65 kJ/kg, s = 7·8709 kJ/kg-K अन्य अवस्थागत गुण प्राप्त करने हेतु अंत में दी गई भाप तालिकाओं का प्रयोग कीजिये। (20 अंक)
प्रभावी तापमान से क्या तात्पर्य है? वातानुकूलन में आदर्श प्रभावी तापमान को प्रभावित करने वाले कारकों की सूची बनाइये और उन्हें संक्षेप में समझाइये। (10 अंक)
दो वायु प्रवाहों के रूद्रोष्म (एडियाबैटिक) मिश्रण की आर्द्रतामितीय (साइक्रोमैट्रिक) प्रक्रिया की व्याख्या कीजिये। (10 अंक)
आइ० सी० इंजनों को ठंडा करना क्यों आवश्यक होता है? आइ० सी० इंजन के अत्यधिक शीतलन से क्या-क्या नुकसान होते हैं? (10 अंक)
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)
State 1 (Inlet): 50 bar, 350 °C. Given: h₁ = 3068.4 kJ/kg, s₁ = 6.4492 kJ/kg-K.
HP Stage (1 → 2): The steam expands to 1.5 bar. For an isentropic expansion, s₂s = s₁ = 6.4492 kJ/kg-K. From steam tables at 1.5 bar: s_f = 1.4337, s_fg = 5.7897. Quality x₂s = (6.4492 - 1.4337) / 5.7897 = 0.866. Enthalpy h₂s = h_f + x₂s · h_fg = 467.11 + (0.866 × 2226.5) = 2395.2 kJ/kg. Isentropic work wₛ,HP = 3068.4 - 2395.2 = 673.2 kJ/kg. Actual work w_HP = η_HP × wₛ,HP = 0.84 × 673.2 = 565.5 kJ/kg. Actual exhaust enthalpy h₂ = 3068.4 - 565.5 = 2502.9 kJ/kg.
LP Stage (3 → 4): The remaining steam is reheated to 250 °C at 1.5 bar. Given: h₃ = 2972.65 kJ/kg, s₃ = 7.8709 kJ/kg-K. Expansion to 0.05 bar. Isentropic s₄s = 7.8709 kJ/kg-K. At 0.05 bar: s_f = 0.4762, s_fg = 7.9187. Quality x₄s = (7.8709 - 0.4762) / 7.9187 = 0.9338. Enthalpy h₄s = 137.75 + (0.9338 × 2423.7) = 2401.0 kJ/kg. Isentropic work wₛ,LP = 2972.65 - 2401.0 = 571.65 kJ/kg. Actual work w_LP = 0.81 × 571.65 = 463.0 kJ/kg.
Total Power Balance: Let m be the total steam rate (kg/h). Power = (Mass flow to HP × w_HP + Mass flow to LP × w_LP) / 3600. 3750 = [ (m × 565.5) + ((m - 12000) × 463.0) ] / 3600. 13,500,000 = 565.5m + 463.0m - 5,556,000. 19,056,000 = 1028.5m. m = 18,528 kg/h.
Conclusion: The rate of steam generated by the boiler is 18,528 kg/h.
(b)(i)
Effective temperature is the temperature of still, saturated air that would produce the same thermal sensation as the actual air conditions. It accounts for the combined effects of dry-bulb temperature, humidity, air velocity, and radiation.
Factors affecting optimum effective temperature:
- Humidity: High humidity reduces the body's ability to cool via evaporation, requiring a lower effective temperature for comfort.
- Air Velocity: Increased air speed enhances convective and evaporative heat loss, allowing for a higher effective temperature.
- Radiation: High radiant heat from walls or the sun increases the thermal load, necessitating a lower effective temperature.
- Activity (metabolic rate): Higher physical activity generates more internal heat, shifting the optimum lower.
- Clothing: heavier winter clothing lowers the optimum ET; light summer clothing raises it.
- Climate and season: people acclimatised to hot climates prefer a higher ET; the optimum is higher in summer than in winter.
- Age and sex: older people and women generally prefer an ET about 0.5–1 °C higher.
- Duration of stay: for short stays (shops, banks) the optimum is kept closer to outdoor conditions to avoid thermal shock; for long stays (offices, homes) a lower, steadier ET is chosen.
- Density of occupants: crowded spaces need a lower ET because of body heat and moisture. The usual comfort range is about 21–23 °C ET in winter and 22–25 °C ET in summer.
(b)(ii)
Adiabatic mixing is a psychrometric process where two air streams at different states are mixed without any heat exchange with the surroundings.
Process Characteristics:
- Conservation of Mass: The total mass of dry air in the mixture is the sum of the dry air masses of the individual streams (m₁ + m₂ = m₃).
- Conservation of Energy: The total enthalpy of the mixture equals the sum of the enthalpies of the incoming streams (m₁ h₁ + m₂ h₂ = m₃ h₃).
- Humidity Ratio: The humidity ratio of the mixture is the weighted average of the individual humidity ratios.
Chart Representation: On a psychrometric chart, the state of the mixture (Point 3) lies on the straight line connecting the initial states (Points 1 and 2). The position of Point 3 is determined by the mass ratio of the two streams.
From the two balances, (h₃ − h₁)/(h₂ − h₃) = (W₃ − W₁)/(W₂ − W₃) = m₂/m₁, so point 3 divides line 1–2 in the inverse ratio of the masses (nearer the larger stream). The dry-bulb temperature of the mixture is only approximately the mass-weighted average. If the line 1–2 crosses the saturation curve (mixing very cold and very humid air), condensation occurs and fog forms. In air-conditioning, this process is used to blend recirculated air with fresh outdoor air.
(c)
Necessity of Cooling: Internal combustion engines must be cooled because a significant portion of the heat generated by fuel combustion (approximately 30%) is not converted into mechanical work. If this heat is not removed, the engine temperature would rise to a point where metal components lose their strength, lubricating oil would break down (coking), and the engine would suffer catastrophic failure.
Disadvantages of Overcooling:
- Reduced Efficiency: More heat flows from the working gases to the cold cylinder walls, so less of the fuel energy is converted into work and thermal efficiency falls.
- Increased Friction: Overcooling increases the viscosity of the engine oil, which increases internal friction and power loss.
- Poor Combustion: A cold engine leads to incomplete combustion and a rich fuel mixture, resulting in higher fuel consumption and increased emissions.
- Corrosion: Overcooling can cause condensation of water and acids inside the engine, leading to internal corrosion.
- Wear and sludge: Unburnt fuel washes oil off the cylinder walls and dilutes the crankcase oil; condensed water forms sludge, increasing wear of rings and liners.
- Hard starting and poor vaporisation of fuel in SI engines; longer ignition delay and diesel knock in CI engines.
How this answer will be evaluated
What earns full marks: Complete working with all states identified and correct final values
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a) Determine the total steam generation rate for the boiler. 20 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- State 1: h = 3068.4 kJ/kg, s = 6.4492 kJ/kg·K (given)
- HP: x2s = 0.866, h2s = 2395.2; ws = 673.2; w_HP = 565.5 kJ/kg; h2 = 2502.9
- LP: x4s = 0.9338, h4s = 2401.0; ws = 571.65; w_LP = 463.0 kJ/kg
- Balance 3750 = [m×565.5 + (m−12000)×463.0]/3600 ⇒ m = 18,528 kg/h
Loses marks
- Using isentropic work instead of actual work
- Ignoring the 12000 kg/h extraction flow
- Unit errors in power or mass flow
Earns more
- Correctly determine enthalpy at 1.5 bar, 250°C
- Apply LP efficiency to find final exhaust enthalpy
- Sum work outputs to match 3750 kW power
- Solve for total mass flow rate in kg/h
Extra mark
- Labelled T-s diagram showing both stages
- Explicit statement of steady-flow energy equation
- (b)(i) Define effective temperature and list factors affecting it. 10 marks
explain— definition/context → points in order → small example → short close
Must cover
- Definition: temperature of still saturated air giving the same thermal sensation (DBT + humidity + velocity + radiation)
- Factors affecting the optimum: activity, clothing, climate/season, age and sex, duration of stay, density of occupants
- Humidity, air velocity and radiation define ET itself (still saturated air reference)
- Metabolic rate effect; optimum typically 21–26 °C
Loses marks
- Confusing effective with dry-bulb temperature
- Listing factors without explaining their effect
- Ignoring the 'optimum' aspect of the question
Earns more
- Mention dry-bulb temperature as baseline
- Note that optimum varies with season
- Reference standard comfort ranges (e.g., 21-26°C)
- Distinguish between summer and winter factors
Extra mark
- Reference ASHRAE or ISO comfort standards
- Mention metabolic rate as a variable
- (b)(ii) Describe the psychrometric process of adiabatic mixing. 10 marks
explain— definition/context → points in order → small example → short close
Must cover
- Adiabatic mixing: no heat exchange with surroundings
- Mass conservation: m1 + m2 = m3 (dry air)
- Energy: m1h1 + m2h2 = m3h3; humidity ratio as weighted average
- Mixture state lies on the line joining states 1 and 2, positioned by the mass ratio
Loses marks
- Assuming temperature is weighted average
- Ignoring the enthalpy balance equation
- Confusing adiabatic mixing with cooling
Earns more
- Show that mixture point lies on straight line
- Mention mass ratio determines position
- Note that humidity ratio is weighted average
- State that no condensation occurs in mixing
Extra mark
- Sketch of psychrometric chart with mixing line
- Mention application in air-conditioning systems
- (c) Explain why IC engines need cooling and overcooling effects. 10 marks
explain— definition/context → points in order → small example → short close
Must cover
- ~30% of combustion heat must be removed; else strength loss, oil coking, failure
- Overcooling: more heat loss to walls ⇒ lower thermal efficiency; higher oil viscosity and friction; poor vaporisation/combustion and emissions
- Corrosion from condensation of water and acids
- Wear and sludge from fuel wash-down and oil dilution; hard starting / CI knock
Loses marks
- Saying cooling is for power output only
- Ignoring the 'disadvantages' part of the question
- Confusing cooling with lubrication functions
Earns more
- Note that ~30% of heat is rejected
- Explain that overcooling increases fuel consumption
- Mention that cold engine runs rich mixture
- State that thermal efficiency decreases with cooling
Extra mark
- Mention specific cooling system types (air/water)
- Reference thermal stress as a failure mode
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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