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Chapter 3 Working hypotheses and preliminary analysis LHV V,syngas = LHVm,syngas = 4.6 MJ/kg = 4.26 MJ (3.15) ρsyngas 1.08 kg/Nm3 Nm3 Remembering that νst is 2.38 both for CO and H2 (see Equations 3.4 and 3.5), the stoichiometric amount of combustion air can be obstained, still performing a weighted average (in this case obviously referring to molar fractions): νst,syngas = ∑νst,i ⋅ Xi = νst,CO ⋅ XCO + νst,H2 ⋅ XH2 + νst,CH4 ⋅ XCH4 = i = 2.38 Nm3a ⋅ 0.20 + 2.38 Nm3a ⋅ 0.20 + 9.52 Nm3a ⋅ 0.01 = 1.05 Nm3a (3.16) N m 3f N m 3f N m 3f N m 3f Considering excess air ranging from 0 to 50%, the volumetric air to fuel ratio is about 1 ÷ 1.6 and thus the volumetric LHV of the syngas/air mixture is: LHVV,syngas 4.26 MJ/Nm3 MJ LHVV,mixture,syngas = ν +1 = (1÷1.6)+1 Nm3 /Nm3 ≅ 1.65 ÷ 2.1 Nm3 (3.17) gfg In this case, there is a certain difference between the two values, which essentially relates to the different densities: the methane LHV is roughly 10 ÷ 20% higher than the syngas one. This means that a given volumetric flow rate does not yield the same power input: an engine with a certain displacement fed with syngas will generate lower power than in case of methane fuelling, i.e. it is subject to derating. Indeed there are some methods to overcome this phenomenon, at least partially (increasing the engine compression ratio, performing a heavier turbo-charging, etc. [3.12]), but independently from that, this is a practical aspect that does not affect energy balances. For the purposes of this work, it was sufficient to verify that in both of the two different feeding cases the same power output corresponds to the same gas mass flow rate, as modelled by ThermoflexTM. Then, it is not a problem if, due to the different density of the air/fuel mixtures, a syngas engine generating a certain amount of power has to be bigger (i.e. have a higher displacement) than a methane engine of the same power size. 115PDF Image | SMALL-SCALE BIOMASS POWER GENERATION
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