Chemical Engineering Calculations 2 Combustion of Liquids 1. Crude petroleum oil is generally considered to be formed from animal and vegetable debris accumulating in sea basins or estuaries and decomposed by anaerobic bacteria resulting in a black viscous product. A typical elemental analysis shows 80% C, 13% H, 1% N, 3% O, and 3% S. During a certain combustion, air supplied is less than the theoretical so that all of the 𝑂2 is used up. 70% of the C burns to 𝐶𝑂2, the rest to CO; the molal ratio of CO to 𝐻2 in the exhaust gas is 1:2. Assume that the Sulfur in the fuel burns to 𝑆𝑂2 and the Nitrogen combines with the nitrogen from air. Calculate: a) Orsat analysis of the exhaust gas b) % of the theoretical air which is supplied for combustion c) Equivalence Ratio air Given: exhaust gas fuel 80% C 13% H 1% N 3% O 70 % C → 𝐶𝑂2 30% C → CO 𝐶𝑂 𝐻2 = 1 2 3% S Required: C a. Orsat analysis of the exhaust gas b. % of the theoretical air which is supplied for combustion c. Equivalence Ratio 4.67 C Solution: + + 2.001 S Basis: 100 kg Crude Petroleum 80 𝑘𝑔 𝐶 = 6.67 𝑚𝑜𝑙 12 13 𝑘𝑔 𝐻 = 13 𝑚𝑜𝑙 1 1 𝑘𝑔 𝑁 = 0.0714 𝑚𝑜𝑙 14 3 𝑘𝑔 𝑂 = 0.1875 𝑚𝑜𝑙 16 3 𝑘𝑔 𝑆 = 0.09375 𝑚𝑜𝑙 32 + 0.09375 O2 CO2 4.67 (0.70)(6.67) = 4.67 1 2 O2 CO 4.002 (0.30)(6.67) = 2.001 O2 SO2 0.09375 0.09375 𝐶𝑜 1 = 𝐻2 2 H2 = 2 CO H2 (unburned) = 2(2.001) = 4.002 H2O = 6. ...
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