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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 ?2 is used up. 70% of the C burns to ??2, the rest to CO; the molal ratio of CO to ?2 in the exhaust gas is 1:2. Assume that the Sulfur in the fuel burns to ??2 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

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 ?2 is used up. 70% of the C burns to ??2, the rest to CO; the molal ratio of CO to ?2 in the exhaust gas is 1:2. Assume that the Sulfur in the fuel burns to ??2 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

Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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need asap can someone please explain how does this solution happen. 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 ?2 is used up. 70% of the C burns to ??2, the rest to CO; the molal ratio of CO to ?2 in the exhaust gas is 1:2. Assume that the Sulfur in the fuel burns to ??2 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

Required: S + O2 SO2 0.09375 0.09375 0.09375 a. Orsat analysis of the exhaust gas b. % of the theoretical air which is supplied for combustion c. Equivalence Ratio Со 1 H2 2 H2 = 2 CO Solution: H2 (unburned) = 2(2.001) = 4.002 H2O = 6.5 – 4.002 = 2.498 Basis: 100 kg Crude Petroleum 80 kg C = 6.67 mol На + H2O 12 13 kg H 2.498 2.498 2.498 13 тol 1 kg N = 0.0714 mol 13 O2 theo = 6.67 + + 0.09375 –- 0.1875 (-) 14 3 kg 0 = 9.92 moles 0.1875 тol 16 3 kg S O2 air = O2 used 0.09375 mol 32 = 7.013 N2 air = 7.013 () = 26.38 O2 CO2 0.0714 4.67 4.67 (0.70)(6.67) = 4.67 N2 total = 26.38 + 2 26.4517 0: Co C 2.001 4.002 (0.30)(6.67) = 2.001
Transcribed Image Text:Required: S + O2 SO2 0.09375 0.09375 0.09375 a. Orsat analysis of the exhaust gas b. % of the theoretical air which is supplied for combustion c. Equivalence Ratio Со 1 H2 2 H2 = 2 CO Solution: H2 (unburned) = 2(2.001) = 4.002 H2O = 6.5 – 4.002 = 2.498 Basis: 100 kg Crude Petroleum 80 kg C = 6.67 mol На + H2O 12 13 kg H 2.498 2.498 2.498 13 тol 1 kg N = 0.0714 mol 13 O2 theo = 6.67 + + 0.09375 –- 0.1875 (-) 14 3 kg 0 = 9.92 moles 0.1875 тol 16 3 kg S O2 air = O2 used 0.09375 mol 32 = 7.013 N2 air = 7.013 () = 26.38 O2 CO2 0.0714 4.67 4.67 (0.70)(6.67) = 4.67 N2 total = 26.38 + 2 26.4517 0: Co C 2.001 4.002 (0.30)(6.67) = 2.001
7.013 a.) ORSAT ANALYSIS x 100% 9.92 Gas % theo air = 70.7 % CO2 4.67 12.56% Co 2.001 5.38% H2 4.002 10.76% N fuel Nair Nfuel (theo) Nair SO2 0.09375 0.25% (actual) d) ER = N2 26.4517 71.04% 20.02265 37.21845 100.00% 33.393 20.02265 9.92 0.21 Oz air b.) % theo air x 100 ER = 1.4 0, theo
Transcribed Image Text:7.013 a.) ORSAT ANALYSIS x 100% 9.92 Gas % theo air = 70.7 % CO2 4.67 12.56% Co 2.001 5.38% H2 4.002 10.76% N fuel Nair Nfuel (theo) Nair SO2 0.09375 0.25% (actual) d) ER = N2 26.4517 71.04% 20.02265 37.21845 100.00% 33.393 20.02265 9.92 0.21 Oz air b.) % theo air x 100 ER = 1.4 0, theo
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