(a) What do you understand by the term EGR? Explain how EGR reduces NOₓ emission in CI engines. (10 marks) (b) The flue gas…

(a) EGR and NOₓ in CI engines

Exhaust gas recirculation (EGR) returns a controlled fraction of cooled exhaust gas to the intake charge. NOₓ forms mainly by the thermal (Zeldovich) mechanism, which needs both high local temperature (above about 1800 K) and free oxygen. EGR cuts NOₓ in two ways. The recirculated CO₂, H₂O and N₂ are inert and have a high heat capacity, so the same heat release gives a lower peak flame temperature. The dilution also lowers the oxygen concentration in the burning zone. Both effects slow NO formation sharply. Cooling the EGR gas adds to the benefit, but too much EGR raises soot, CO and fuel consumption, so the rate is controlled with load.

(b) Orsat flue-gas analysis

Take 100 kmol of dry flue gas: CO₂ 8.0, CO 0.9, O₂ 8.8, N₂ 82.3 kmol.

Oxygen supplied with air = 82.3 × 21/79 = 21.88 kmol. Carbon in fuel = 8.0 + 0.9 = 8.9 kmol.

Oxygen balance: 21.88 = 8.0 + 0.45 + 8.8 + n(H₂O)/2, so n(H₂O) = 9.25 kmol and hydrogen = 18.51 kmol.

(i) Fuel: C₈.₉H₁₈.₅₁ per 100 kmol of dry gas, that is C : H = 1 : 2.08, or CH₂.₀₈ (an olefinic type, CₓH₂ₓ roughly).

(ii) Air supplied = 21.88 + 82.3 = 104.18 kmol. Molar air–fuel ratio = 104.18/8.9 = 11.7 kmol air per kmol of carbon (CH₂.₀₈). Fuel mass = 8.9 × 12 + 18.51 = 125.3 kg, and air mass = 104.18 × 28.97 = 3018 kg, so the mass air–fuel ratio is 24.1.

(iii) Theoretical O₂ = 8.9 + 18.51/4 = 13.53 kmol. Excess air = (21.88 − 13.53)/13.53 = 61.7%.

(c) Steam nozzle efficiency and velocity coefficient

(i) Nozzle efficiency is the ratio of the actual enthalpy drop (kinetic energy gained) to the isentropic enthalpy drop between the same inlet state and exit pressure: η_N = (h₁ − h₂)/(h₁ − h₂ₛ). It measures the frictional and other irreversibility losses in the nozzle.

(ii) The velocity coefficient is the ratio of actual to isentropic exit velocity, C_v = V₂/V₂ₛ. Because kinetic energy varies as velocity squared, η_N = C_v², so C_v = √η_N. Well-made nozzles have C_v of about 0.95–0.99.

(d) Balance point of compressor and capillary tube

In a refrigeration system with a capillary tube, the operating point is where the refrigerant mass flow through the capillary tube equals the mass flow pumped by the compressor. As the evaporator pressure falls, the capillary flow rises (a larger pressure drop) but the compressor flow falls (a lower suction density and a higher pressure ratio); their intersection fixes the evaporating and condensing pressures, mass flow and subcooling. If conditions drift, the flow mismatch moves the pressures back to this point, so the system is self-regulating. A change in ambient temperature or load shifts the balance point, so the capillary length and bore are chosen to give the right capacity at design conditions.

(e) Equal friction method of duct design

In this method the frictional pressure drop per metre of duct length is kept the same in every section. The designer finds the airflow in each duct from the room loads and chooses a friction rate (commonly of the order of 1 Pa/m for comfort systems) together with a suitable main-duct velocity. For each section, the friction chart gives the equivalent diameter at that rate and the flow, and rectangular sizes are chosen to match. Fittings, grilles and coil losses are added to find the fan static pressure. It is simple and gives automatic pressure reduction as flow falls along the duct, but the pressure drop in each branch differs, so balancing dampers are needed in short branches.