(a) (i) Briefly discuss HC, CO and NOₓ emission formation in SI engine. Explain the dependence of these emissions on equivalence…

(a)(i) HC, CO and NOₓ formation in SI engines

• HC (unburnt hydrocarbons): flame quenching in the thin cold layer at the walls, mixture forced into crevices (piston top-land, ring gaps) during compression and released unburnt during expansion, fuel absorbed in oil films and deposits and later desorbed, and partial burning or misfire with very lean or diluted mixtures.
• CO: incomplete oxidation of carbon when oxygen is insufficient (rich mixtures), plus CO from dissociation of CO₂ at high temperature that freezes during expansion.
• NOₓ (mainly NO): formed from atmospheric N₂ and O₂ in the hot burned gas by the thermal (extended Zeldovich) mechanism, rising exponentially with temperature and requiring free oxygen, and frozen during expansion.

Dependence on φ:

• CO is very low for lean mixtures and rises almost linearly with φ on the rich side.
• HC is lowest near φ ≈ 0.9, rises on the rich side for lack of oxygen and rises sharply for very lean mixtures due to slow burning, quenching and misfire.
• NOₓ peaks slightly lean of stoichiometric (φ ≈ 0.9–0.95), where high flame temperature and excess oxygen coexist, and falls on the rich side (oxygen short) and on the lean side (lower temperature).

(a)(ii) Maximum temperature and pressure of the Otto cycle

r = 9, p₁ = 1 bar, T₁ = 300 K, n = 1.33, AFR = 15, CV = 43000 kJ/kg, Cv = 0.717 kJ/kg·K. Working fluid = 16 kg mixture per kg fuel.

Compression 1–2: T₂ = T₁rⁿ⁻¹ = 300 × 9^0.33 = 619.5 K, p₂ = p₁rⁿ = 9^1.33 = 18.58 bar. Heat added per kg mixture: q = 43000/16 = 2687.5 kJ/kg = Cv(T₃ − T₂). T₃ = 619.5 + 2687.5/0.717 = 4368 K (maximum temperature). p₃ = p₂(T₃/T₂) = 18.58 × 4368/619.5 = 131 bar (maximum pressure). Real peaks are lower because of variable specific heats, dissociation and heat loss.

(b) Air-conditioning of the food processing room

State points: outdoor O: ω_o = 0.0209 kg/kg, h_o = 99.2 kJ/kg. Room R: ω_r = 0.00877 kg/kg, h_r = 42.4 kJ/kg.

(i) Ventilation load (standard air): OASH = 0.0204 × 90 × (45 − 20) = 45.9 kW OALH = 50 × 90 × (0.02087 − 0.00877) = 54.4 kW Ventilation load ≈ 100.3 kW (about 91.5 kW if the outdoor specific volume 0.931 m³/kg is used).

(ii) Sensible heat factors: RSHF = 35/55 = 0.636. ERSH = 35 + 0.05 × 45.9 = 37.3 kW, ERLH = 20 + 0.05 × 54.4 = 22.7 kW, ESHF = 37.3/60.0 = 0.621.

(iii) ADP and reheat: the ESHF line from R is so steep that it meets the saturation curve only near 0 °C, which would frost the coil. For an economical design take the lowest practical ADP, 5 °C (ω = 0.00542 kg/kg). The line R–ADP has ESHF′ = 1.0216 × 15/[1.0216 × 15 + 2500 × (0.00877 − 0.00542)] = 0.647. Required ERSH′ = ERLH × 0.647/0.353 = 41.6 kW, so reheat = 41.6 − 37.3 ≈ 4.3 kW. A higher ADP needs more reheat (6.7 kW at 6 °C, 10.3 kW at 7 °C), so the lowest non-frosting ADP is most economical.

(c) Condition for maximum discharge through a chimney

Let H = height, A = area, T_a and T_g = absolute temperatures of atmosphere and gas, and m = kg of air per kg of fuel, so (m + 1) kg of gas. Gas volume is taken as that of the air at the same p and T. ρ_a = p/(RT_a), ρ_g = [(m + 1)/m] p/(RT_g). Draught Δp = gH(ρ_a − ρ_g), equivalent to a hot-gas column H′ = H[(m/(m + 1))(T_g/T_a) − 1]. Velocity V = √(2gH′), discharge ṁ = ρ_g A V ∝ (1/T_g)√[(m/(m + 1))(T_g/T_a) − 1]. Maximise f = [m/((m + 1)T_a)]T_g⁻¹ − T_g⁻²: df/dT_g = −[m/((m + 1)T_a)]T_g⁻² + 2T_g⁻³ = 0 ⟹ T_g/T_a = 2(m + 1)/m, a maximum since d²f/dT_g² is negative there. As m is large (15–25), T_g ≈ 2T_a. At this point H′ = H. Hotter gas lowers density faster than velocity rises, so discharge falls. Such stack temperatures waste heat, which is why large boilers use mechanical draught.