Question

At a particular steady-state operating condition, the exhaust from a lean-burn GDI engine contains CO2: 11.5% by volume (wet measurement)
Calculate:
a. the equivalence ratio if the fuel is represented by C1H1.78.
b. the composition of the exhaust (percentage of H2O, O2, N2)

Answer 1

Given, It isu Lean mixture CO2 = 11.5% by volume (et measurement) As the amount of fuel burnt is unknown so the resulting reaction is se C, H 1.78 + y 2 O₂ + 79 Na} → H₂O +1 1.5 CO₂ + O₂ + 29 Na Conservation of carbon.- RxI= 11:5 x = 11.5 The reaction is 11. 5C, t CH178* 3523 202 + 79 N22 → 10:23540+11:5002 + O2 + 35.23X 79 M2 Divide by 1115 CH1.78 +3.06% O2 + 79 Na}>0.89 H₂O + CQ+0:086 O₂ + 3*06x79 Hz mass of air (ma) = 3.067 32+79 x 29% = 432017 kg mass of fuel (mp) = (12+32) = 44 kg Actual Air Fuel Ratio ( Alf Jactual = ma - 432.17 - 9.82 me - 44 = Now, Stoichiometric combustion - Ci H178 + 2.445 202 + 79 N2470-89H4D +CO2 +02 +2.445474 N2 One mole of fuel has a mass(mp) = (12+1.78x1) = 13-781& mass of air (ma) = (2.445832 +2:445 x 79 x 28) = 335.78 ka Stoichiometric (AVA) ratioma. 335.78 - 24,36 Equivalence ratio (O) = (A/Fje (A/F Jactual - 9.82- 21 21 mp 73:78 - ne ratio co= (Alf) Stoich - 24.36 -2.48

Now By wet analysis, Total number of moles of product = 0.89 +1 +0.086 + 11.522 - 13.49 for H₂O 0.89 6100 - 6.59% 13.49 for O₂ → 0.0868100 - 0.637% 13.49 for N f 11.522 8100 13.49 - 85.41% -