(a) An SI engine working on the Otto cycle has cylinder bore of 210 mm and stroke length of 240 mm. The clearance volume is 1550 cc. The pressure and temperature at the beginning of compression are 1 bar and 17 °C respectively. The maximum pressure o

Question

(a)

An SI engine working on the Otto cycle has cylinder bore of 210 mm and stroke length of 240 mm. The clearance volume is 1550 cc. The pressure and temperature at the beginning of compression are 1 bar and 17 °C respectively. The maximum pressure of the cycle is 50 bar. Determine the pressure and temperature at the salient points in the cycle, the air-standard efficiency, the work done and the mean effective pressure. Show the cycle on P-v and T-s diagrams. Evaluate the fuel consumption in kg/kWh, if the calorific value of the fuel is 40 MJ/kg.

Take Cp and Cv of air as 1·005 kJ/kg-K and 0·718 kJ/kg-K respectively.

(b)

Air flowing at the rate of 100 m³/min at 40 °C DBT and 50% RH is mixed with another stream of air flowing at the rate of 20 m³/min at 26 °C DBT and 50% RH. The mixture flows over a cooling coil whose ADP temperature is 10 °C and bypass factor is 0·2. Find the DBT and RH of air leaving the coil. If this air is supplied to an air-conditioned room, where DBT of 26 °C and RH of 50% are maintained, then calculate the (i) room sensible heat factor and (ii) coil cooling capacity in tons of refrigeration.

Draw a schematic diagram of the system and show all the processes on a skeleton psychrometric chart.

Psychrometric chart is given.

Draw a schematic diagram of the system and show all the processes on a skeleton psychrometric chart.

Psychrometric chart is given.

(c) A process industry employs a medium pressure boiler to produce steam. The mass flow rate of fuel consumed is 0·847 kg/s and c.v. of the fuel is 44 MJ/kg. For efficient combustion, 16 kg of air per kg of fuel is required, for which a draught of 30 mm of the water column is required at the base of the chimney. The flue gases leave the boiler at 350 °C. The average temperature of gases in the stack may be taken as 300 °C. The atmosphere is at 20 °C.

Assuming the velocity of gases at the stack exit to be negligible, determine the height of the stack and the diameter at its base.

Also, calculate the mass flow rate of the gases.

Take Patmosphere = 101·3 kPa, Rair = Rgases = 0·287 kJ/kg-K, g = 9·81 m/s²,

ρwater = 1000 kg/m³.

UPSC Answer Check · Accepted Answer

(a) Air-standard Otto cycle: salient points, efficiency, work, mep and fuel consumption

Assumptions: air-standard Otto cycle, air as an ideal gas with constant specific heats, reversible adiabatic compression and expansion, constant-volume heat addition and rejection. γ = 1.005/0.718 = 1.40 and R = 0.287 kJ/kg·K.

Geometry: swept volume V_s = (π/4)(0.21)²(0.24) = 8.313 × 10⁻³ m³ (8313 cc); V₁ = 8313 + 1550 = 9863 cc; compression ratio r = 9863/1550 = 6.363.

Salient points (p₁ = 1 bar, T₁ = 290.15 K):
1–2 isentropic: T₂ = T₁r^(γ−1) = 290.15 × 6.363^0.3997 = 607.9 K; p₂ = p₁r^γ = 13.33 bar.
2–3 constant volume to p₃ = 50 bar: T₃ = T₂(p₃/p₂) = 607.9 × 50/13.33 = 2280 K.
3–4 isentropic: T₄ = T₃/r^(γ−1) = 1088 K; p₄ = p₃/r^γ = 3.75 bar.
Check: p₄/p₁ = T₄/T₁ = 3.75.

Air-standard efficiency: η = 1 − 1/r^(γ−1) = 1 − 1/2.095 = 52.3%.

Work per cycle: mass of air m = p₁V₁/(RT₁) = 100 × 9.863 × 10⁻³/(0.287 × 290.15) = 0.01184 kg.
Heat supplied Q₁ = mc_v(T₃ − T₂) = 0.01184 × 0.718 × 1672 = 14.22 kJ.
Heat rejected Q₂ = mc_v(T₄ − T₁) = 0.01184 × 0.718 × 798 = 6.79 kJ.
W = Q₁ − Q₂ = 7.43 kJ per cycle (check: ηQ₁ = 0.5227 × 14.22 = 7.43 kJ).

Mean effective pressure: p_m = W/V_s = 7.433/(8.313 × 10⁻³) = 894 kPa (8.94 bar).

Fuel consumption: with 52.3% of the fuel energy converted to work,
sfc = 3600/(η × CV) = 3600/(0.5227 × 40 000) = 0.172 kg/kWh.
This is the ideal air-standard value; a real engine uses roughly 1.5 to 2 times as much.

(b) Mixing of two air streams and cooling coil

Assumptions: steady flow, adiabatic mixing, barometric pressure 1.01325 bar. Properties are computed from the psychrometric relations, which agree with the chart readings to its accuracy.

State 1 (fresh air): 40°C, 50% RH, ω₁ = 0.0235 kg/kg, h₁ = 100.8 kJ/kg, v₁ = 0.9205 m³/kg.
State 2 (also the room state R): 26°C, 50% RH, ω₂ = 0.0105 kg/kg, h₂ = 52.9 kJ/kg, v₂ = 0.8616 m³/kg.
ADP: 10°C saturated, ω = 0.00763 kg/kg, h = 29.3 kJ/kg.

Mixing. Dry-air flows: ṁ₁ = 100/0.9205 = 108.6 kg/min and ṁ₂ = 20/0.8616 = 23.2 kg/min, so ṁ = 131.8 kg/min = 2.197 kg/s.
ω_M = (108.6 × 0.0235 + 23.2 × 0.0105)/131.8 = 0.0212 kg/kg.
h_M = (108.6 × 100.8 + 23.2 × 52.9)/131.8 = 92.3 kJ/kg, which corresponds to a mixture temperature t_M = 37.6°C (RH about 52%).

Coil exit S with bypass factor 0.2: the state lies 20% of the way from the ADP to M.
t_S = 10 + 0.2(37.6 − 10) = 15.5°C.
ω_S = 0.00763 + 0.2(0.0212 − 0.00763) = 0.01035 kg/kg, giving h_S = 41.8 kJ/kg and RH ≈ 94%.

(i) Room sensible heat factor. The supply air S absorbs the room load and leaves at R (26°C, 50%). At 26°C and ω_S, h = 1.005 × 26 + 0.01035 × 2549 = 52.5 kJ/kg.
Sensible part = 52.5 − 41.8 = 10.7 kJ/kg; latent part = 52.9 − 52.5 = 0.4 kJ/kg.
RSHF = 10.7/(10.7 + 0.4) ≈ 0.97. The room load is almost entirely sensible (about 23.6 kW sensible, 0.8 kW latent).

(ii) Coil cooling capacity: Q = ṁ(h_M − h_S) = 2.197 × (92.3 − 41.8) = 111 kW = 111/3.5167 = 31.6 TR (about 32 TR). Condensate enthalpy is neglected; including it lowers the figure by under 1%.

(c) Chimney height and base diameter

Given: fuel flow 0.847 kg/s with 16 kg of air per kg of fuel; draught 30 mm of water; T_a = 293.15 K; mean stack gas 573.15 K; gas at the base 623.15 K; p = 101.3 kPa.

Gas mass flow: ṁ_g = ṁ_f(1 + A/F) = 0.847 × 17 = 14.4 kg/s.
Draught pressure: Δp = ρ_w g h_w = 1000 × 9.81 × 0.030 = 294.3 Pa.
Densities: ρ_a = 101 300/(287 × 293.15) = 1.204 kg/m³; ρ_g (mean, 300°C) = 101 300/(287 × 573.15) = 0.616 kg/m³.

Height: Δp = gH(ρ_a − ρ_g), so H = 294.3/[9.81 × (1.204 − 0.616)] = 51.0 m.

Diameter at the base. With the exit velocity neglected, the draught pressure produces the gas velocity at the base, where the gas is at 350°C:
ρ_b = 101 300/(287 × 623.15) = 0.566 kg/m³.
V = √(2Δp/ρ_b) = √(2 × 294.3/0.566) = 32.2 m/s.
Volume flow = 14.4/0.566 = 25.4 m³/s, so A = 25.4/32.2 = 0.789 m² and d = √(4A/π) ≈ 1.0 m.
Using the mean density 0.616 kg/m³ instead gives d = 0.98 m, so the base diameter is about 1.0 m either way.

These are theoretical values; friction and exit losses would call for a somewhat taller or wider stack in practice.

(a) An SI engine working on the Otto cycle has cylinder bore of 210 mm and stroke length of 240 mm. The clearance volume is 1550…

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(a) Air-standard Otto cycle: salient points, efficiency, work, mep and fuel consumption

(b) Mixing of two air streams and cooling coil

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