Explanation Write the chemical reaction for the combustion of Butane ( C 4 H 10 ) with 200% theoretical amount of air. { ( 0 .65CH 4 +0 .08H 2 + 0 .18N 2 + 0 .03O 2 + 0 .06CO 2 ) + a th ( O 2 + 3.76 N 2 ) } → x CO 2 + y H 2 O + z N 2 (I) Here, stoichiometric coefficient for air is a th , and B , D , E , and F are the stoichiometric coefficients. Write the expression to calculate the partial pressure of the moisture in the air ( P v , in ) . P v , in = ϕ air P sat @ 25 ° C (II) Here, humidity of the air is ϕ air , and saturation pressure at the temperature of 25 ° C is P sat @ 25 ° C . Write the expression to calculate the number of moles of moisture in the air ( N v , in ) . N v , in = ( P v , in P total ) N total (III) Here, partial pressure of the products is P total , and number of moles of the products is N total . Write the expression to calculate the mass of the mixture of gases ( m total ) . m total = N A M A + N B M B + ....... + N Z M Z (IV) Here, number of moles of the compound A in the product is N A , the molecular weight of the compound A is M A , number of moles of compound B is N B , molecular weight of compound B is M B Write the expression to calculate the air fuel ratio ( AF ) . AF = m air m fuel (V) Here, mass of the air is m air and the mass of the fuel is m fuel . Conclusion: Balance for the Carbon from Combustion reaction Equation (I). 0.65 + 0.06 = x x = 0.71 Balance for the hydrogen from Combustion reaction Equation (I). 0.65 × 4 + 0.08 × 2 = 2 y y = 1.38 Balance for the oxygen from Combustion reaction Equation (I). 0.03 + 0.06 + a th = x + y 2 Substitute 0.71 for x and 1.38 for y . 0.03 + 0.06 + a th = 0.71 + 1.38 2 a th = 1.31 Balance for the Nitrogen from Combustion reaction Equation (I). 0.18 + 3.76 a th = z Substitute 1.31 for a th . 0.18 + 3.76 ( 1.31 ) = z z = 5.106 Rewrite the complete balanced chemical reaction for combustion as follows: { ( 0 .65CH 4 +0 .08H 2 + 0 .18N 2 + 0 .03O 2 +0 .06CO 2 ) + ( 1.31 ) [ O 2 + 3.76 N 2 ] } → { ( 0.71 ) CO 2 + ( 1.38 ) H 2 O + ( 5.106 ) N 2 } { ( 0 .65CH 4 +0 .08H 2 + 0 .18N 2 + 0 .03O 2 +0 .06CO 2 ) + 1.31 [ O 2 + 3.76 N 2 ] } → { 0.71 CO 2 + 1.38 H 2 O + 5.106 N 2 } From the Table A-4 of “Saturated water–Temperature table”, select the saturation pressure at the temperature of 25 ° C as, P sat @ 25 ° C = 3.1698 kPa Substitute 0.70 for ϕ air , and 3.1698 kPa for P sat @ 25 ° C in Equation (II). P v , in = ( 0.70 ) ( 3.1698 kPa ) = 2.219 kPa Substitute 2.219 kPa for P v , in , 101.325 kPa for P total , and ( 6.24 + N v , in ) for N total in Equation (III). N v , in = ( 2.219 kPa 101.325 kPa ) ( 6.24 ) = 0

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ISBN: 9780100257054

Author: CENGEL

Publisher: YUZU

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Chapter 15.7, Problem 27P

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The theoretical amount of air fuel ratio.

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EBK THERMODYNAMICS: AN ENGINEERING APPR

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