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Find the stoichiometric A/F ratio for the combustion of ethyl (C4H9OH) in a petrol engine. Calculate the A/F ratios for 0.9 & 1.3 equivalence ratios Determine the wet and dry analyses by volume of the exhaust gas for each equivalence ratio. Note we need solution steps like the picture

Find the stoichiometric A/F ratio for the combustion of ethyl (C4H9OH) in a petrol engine. Calculate the A/F ratios for 0.9 & 1.3 equivalence ratios Determine the wet and dry analyses by volume of the exhaust gas for each equivalence ratio. Note we need solution steps like the picture

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Find the stoichiometric A/F ratio for the combustion of ethyl (C4H9OH) in a petrol engine. Calculate the A/F ratios for 0.9 & 1.3 equivalence ratios Determine the wet and dry analyses by volume of the exhaust gas for each equivalence ratio. Note we need solution steps like the picture
LTE. Answer 1 of 1 Combustion ign of ethyl alchol is: C₂H5OH + 30₂ + 3x29 №₂ 21 200₂ + 3H₂O + 3× 79 №₂ 21 One mole of fuel has (2x12 + 16x 8) = 46kg 00 mass of mass of als required for Complete burning of one mole of fuel is: (3 x 32 + 3x12x20) = 412 kg. Stoichiometric 4 equivalence ratio (P) = (A/F/Sto 0-9 = 8.91 Volum for 4=0.9; 1 = 0.9 A F (A/F) actual = 9.95 xatto = 4₁² = 46 the of males of Products total no. a (A/P) achat Alf) actual (₂450H + 1-1 (30₂ + 3x29 N₂) Page No. ratto = 3x (1 + 3-762) = 14-3 times 0.334 x 100 = 17.87 stoichish. 9:17 als supplied is much Sufflined for Complete Combusttany thens Combustion equation becomes : wet fence analysis it 2 17.87 = Done x 100 = 11.2%. (0₂; = 896/2 3 x 100 = 16.8 40 4₂0 17-87 1.851 0₂ 12.54 x 100 = 70-2Y. N₂₁₂ 17-87 وله →200₂ + 3H₂0 +0.11×30₂- + 1-11 x 3 x 79 №₂ 21 = 2+3 +0.33 +12.54. 17-87
Transcribed Image Text:LTE. Answer 1 of 1 Combustion ign of ethyl alchol is: C₂H5OH + 30₂ + 3x29 №₂ 21 200₂ + 3H₂O + 3× 79 №₂ 21 One mole of fuel has (2x12 + 16x 8) = 46kg 00 mass of mass of als required for Complete burning of one mole of fuel is: (3 x 32 + 3x12x20) = 412 kg. Stoichiometric 4 equivalence ratio (P) = (A/F/Sto 0-9 = 8.91 Volum for 4=0.9; 1 = 0.9 A F (A/F) actual = 9.95 xatto = 4₁² = 46 the of males of Products total no. a (A/P) achat Alf) actual (₂450H + 1-1 (30₂ + 3x29 N₂) Page No. ratto = 3x (1 + 3-762) = 14-3 times 0.334 x 100 = 17.87 stoichish. 9:17 als supplied is much Sufflined for Complete Combusttany thens Combustion equation becomes : wet fence analysis it 2 17.87 = Done x 100 = 11.2%. (0₂; = 896/2 3 x 100 = 16.8 40 4₂0 17-87 1.851 0₂ 12.54 x 100 = 70-2Y. N₂₁₂ 17-87 وله →200₂ + 3H₂0 +0.11×30₂- + 1-11 x 3 x 79 №₂ 21 = 2+3 +0.33 +12.54. 17-87
LTE. Answer 1 of 1 Done total dry mahs = 2 + 0.33 + 12-54 = 14.87 Here dry coalysts 2 *100 = 13.45%. (02 14.87 10.33 x 100 = 2-224₁ 0₂ 14.87 12.54 x 100 = 84-49 №₂ 14.87 for d = 1.2 Actul A/F gatio = 8.96 1-2 this meat = 0.834 of the stoichiomety! 1.2 is supplied. The end is combustion Cand to assume that all the hydrogen is buorld to M₂0, Shnee H₂ atoms hous a greater affinity for oxygen then Carbon atond. The Carbon win the fuel will bun to (o and (0₂: air be complete لصالب 7:47:1 C₂H₂ +0.834 (30₂ + 3x 3.762 N₂)→ a Co₂ + bco +34₂0+ € 0.834 x3 x3.762₂ C balonce ©2=a+b 0 balance 1 + 2 x 0·834 X 3=2<+5+3 sulury above equations are get a = 1.004 b= 0.996 fance what Analyst G 1·004 x 100 = 6.97% (0₂ 14.41 The products are: textmolds 1.004 mois (O₂ +0.996 mos (0 + 3 meld H₂ + 9.41 mals N₂ total males of products. x 100 = 20.8%. 1₂0 9:17 = 1·004 +0.996. +3+9.41 14.41 3 14.41 9.41 x 100 = 65.3% N₂₂ 14.41 total dry 02-996x100=6.917. CO 14.41 mely = 1004 +0.996. +9.41
Transcribed Image Text:LTE. Answer 1 of 1 Done total dry mahs = 2 + 0.33 + 12-54 = 14.87 Here dry coalysts 2 *100 = 13.45%. (02 14.87 10.33 x 100 = 2-224₁ 0₂ 14.87 12.54 x 100 = 84-49 №₂ 14.87 for d = 1.2 Actul A/F gatio = 8.96 1-2 this meat = 0.834 of the stoichiomety! 1.2 is supplied. The end is combustion Cand to assume that all the hydrogen is buorld to M₂0, Shnee H₂ atoms hous a greater affinity for oxygen then Carbon atond. The Carbon win the fuel will bun to (o and (0₂: air be complete لصالب 7:47:1 C₂H₂ +0.834 (30₂ + 3x 3.762 N₂)→ a Co₂ + bco +34₂0+ € 0.834 x3 x3.762₂ C balonce ©2=a+b 0 balance 1 + 2 x 0·834 X 3=2<+5+3 sulury above equations are get a = 1.004 b= 0.996 fance what Analyst G 1·004 x 100 = 6.97% (0₂ 14.41 The products are: textmolds 1.004 mois (O₂ +0.996 mos (0 + 3 meld H₂ + 9.41 mals N₂ total males of products. x 100 = 20.8%. 1₂0 9:17 = 1·004 +0.996. +3+9.41 14.41 3 14.41 9.41 x 100 = 65.3% N₂₂ 14.41 total dry 02-996x100=6.917. CO 14.41 mely = 1004 +0.996. +9.41
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