(a) Can alcohols be used as fuel in IC engine? Explain with advantages and disadvantages. (10 marks) (b) A water-filled reactor…

(a) Alcohols as IC engine fuels

Yes. Methanol and ethanol can be used in SI engines neat or blended with petrol (E10, E20, flex-fuel E85/E100), and in CI engines with cetane improvers, as emulsions or in dual-fuel mode with a diesel pilot.

Advantages:

• High octane number (RON about 108–110) allows higher compression ratio and advanced timing without knock, raising efficiency.
• Oxygenated fuel gives more complete combustion with lower CO, HC and particulates, and no sulphur.
• High latent heat cools the charge, raising volumetric efficiency.
• Bio-ethanol is renewable, lowers net CO₂ and reduces crude imports. India reached about 20% ethanol blending in petrol in 2025.

Disadvantages:

• Lower calorific value (ethanol about 27 MJ/kg, methanol about 20 MJ/kg against petrol about 44 MJ/kg) means higher fuel consumption and shorter range.
• High latent heat and low vapour pressure make cold starting difficult.
• Hygroscopic, with phase separation of blends in the presence of water.
• Corrosive to aluminium, zinc and some elastomers and seals, so compatible materials are needed.
• Low cetane number makes them unsuitable for CI engines without modification, methanol is toxic, and aldehyde emissions rise.

(b) Minimum volume of the containment room

The room is insulated and evacuated, so the rupture is an unrestrained expansion: Q = 0, W = 0, hence U₁ = U₂ and u₂ = u₁ for the fixed mass. State 1 (compressed liquid, 20 MPa, 360 °C): v₁ = 0.001823 m³/kg, u₁ = 1702.8 kJ/kg. m = 1/0.001823 = 548.5 kg. State 2 (200 kPa, minimum volume): u_f = 504.5, u_fg = 2024.6 kJ/kg, x₂ = (1702.8 − 504.5)/2024.6 = 0.592, v₂ = 0.001061 + 0.592 × (0.8858 − 0.0011) = 0.5248 m³/kg. V_min = m v₂ = 548.5 × 0.5248 ≈ 288 m³.

(c) Regenerative Rankine cycle approaching Carnot efficiency

In the simple cycle part of the boiler heat is added to cold feed water from the pump temperature, so the mean temperature of heat addition is well below T_max and efficiency below Carnot. Ideal (perfect) regeneration: the feed water after the pump flows in counterflow around the turbine casing and is heated by the expanding steam, reversibly, at each point with an infinitesimal temperature difference. On the T-s diagram the expansion line becomes parallel to the feed-water heating line, so the heat given up by the steam equals the sensible heat gained by the water (equal areas). External heat is then supplied only during evaporation at T_max and rejected only during condensation at T_min, both isothermal. So η = 1 − T_min/T_max, the Carnot efficiency. In practice this is approximated by bleeding steam to several feed-water heaters, which raises the feed temperature, the mean temperature of heat addition and the efficiency, and avoids the excessive wetness of the ideal scheme.

(d) Effect of operating parameters on VCR performance

Refrigerating effect = h₁ − h₄, work = h₂ − h₁, COP = (h₁ − h₄)/(h₂ − h₁). (i) Suction pressure: lowering it raises the pressure ratio and the specific volume at suction, so refrigerating effect and mass flow fall and work rises, and capacity and COP decrease. Raising it does the opposite. (ii) Delivery pressure: raising it raises h₂ and h₃ (hence h₄), so refrigerating effect falls and work rises, and COP falls with higher discharge temperature. (iii) Subcooling: point 3 shifts left, h₄ falls, and refrigerating effect rises with no extra work, so COP rises. (iv) Superheating: if superheat is produced inside the refrigerated space, refrigerating effect rises, work per kg also rises slightly, and COP may rise or fall with the refrigerant. Superheat gained outside (suction line) adds work without useful cooling and lowers COP, but some superheat protects the compressor from liquid.

(e) Air leaving the cooling coil

Taking 40 m³/min as the air flow over the coil: State 1 (32 °C DBT, 18 °C WBT): W₁ = 0.00718 kg/kg, h₁ = 50.6 kJ/kg, v₁ = 0.874 m³/kg. ṁ = 40/(60 × 0.874) = 0.763 kg/s. h₂ = h₁ − 12/0.763 = 34.8 kJ/kg. Saturated air at the coil surface (4.5 °C): h_s = 17.7 kJ/kg, W_s = 0.00524 kg/kg. Contact factor = (h₁ − h₂)/(h₁ − h_s) = 15.7/32.9 = 0.478, so bypass factor ≈ 0.52. Leaving air lies on the line 1–s: T₂ = 4.5 + 0.52 × (32 − 4.5) ≈ 18.9 °C DBT, W₂ ≈ 0.00625 kg/kg, giving WBT ≈ 12.3 °C.