Explanation Write the expression to calculate the equivalence ratio ( ε ) . ε = AF theo AF (I) Here, theoretical air fuel ratio is AF theo and the actual air fuel ratio is AF . Write the theoretical reaction equation between carbon dioxide and hydrogen. CO 2 + H 2 ⇄ CO + H 2 O (II) Write the actual chemical equation for the reaction of equilibrium gaseous mixture. C 8 H 18 + AF ( O 2 + 3.76 N 2 ) → ( a CO 2 + b H 2 O + c CO + d H 2 + ( 3.76 AF ) N 2 ) (III) Write the expression to calculate the equilibrium constant for the reaction ( K ) . K = N CO v CO N H 2 O v H 2 O N CO 2 v CO 2 N H 2 v H 2 ( P N total ) v CO + v H 2 O − v CO 2 − v H 2 (IV) Here, number of moles of CO is N CO , number of moles of H 2 O is N H 2 O , number of moles of CO 2 is N CO 2 , number of moles of H 2 is N H 2 , pressure is P , stoichiometric coefficients are v CO 2 , v H 2 , v H 2 O , and v CO , and the total number of moles is N total . Conclusion: Obtain the stoichiometric coefficients of substances from the Equation (II). v H 2 = 1 v CO 2 = 1 v CO = 1 v H 2 O = 1 Substitute 12.5 ( 4.76 ) for AF theo and 1.2 for ε in Equation (I). 1.2 = 12.5 ( 4.76 ) AF AF = 10.417 ( 4.76 ) Re-write the Equation (III) for the AF C 8 H 18 + 10 .417 ( O 2 + 3.76 N 2 ) → ( a CO 2 + b H 2 O + c CO + d H 2 + 39.168 N 2 ) (V) Balance for the C from Equation (V)

Fundamentals of Engineering Thermodynamics, Binder Ready Version

8th Edition

ISBN: 9781118820445

Author: Michael J. Moran, Howard N. Shapiro, Daisie D. Boettner, Margaret B. Bailey

Publisher: WILEY

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Chapter 14.6, Problem 34P

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Fundamentals of Engineering Thermodynamics, Binder Ready Version

Ch. 14.6 - 1. Why is using the Gibbs function advantageous...Ch. 14.6 - Prob. 2E Ch. 14.6 - Prob. 3E Ch. 14.6 - Prob. 4E Ch. 14.6 - Prob. 5E Ch. 14.6 - Prob. 6E Ch. 14.6 - Prob. 7E Ch. 14.6 - Prob. 8E Ch. 14.6 - Prob. 9E Ch. 14.6 - Prob. 10E

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