Explanation Express the complete combustion of the octane with theoretical air [ C 8 H 18 + α ( O 2 + 3.76 N 2 ) ] → ( aCO 2 + b H 2 O + c N 2 ) (I) Here, octane gas is C 8 H 18 , oxygen is O 2 , nitrogen is N 2 , carbon dioxide is CO 2 , water is H 2 O . Apply the conservation of mass principle to the carbon, hydrogen, oxygen, and nitrogen, respectively, results in four equations among the four unknowns: Perform carbon balance: a = 8 Perform hydrogen balance: 2 b = 18 b = 9 Perform oxygen balance: 2 α = 25 α = 12.5 Perform nitrogen balance: α ( 3.76 ) = c 12.5 ( 3.76 ) = c c = 47 Substitute 8 for a , 9 for b , 47 for c , and 12.5 for α in Equation (I). [ C 8 H 18 + 12.5 ( O 2 + 3.76 N 2 ) ] → ( 8CO 2 + 9 H 2 O + 47 N 2 ) Similarly express the complete combustion of the propane gas with 300% theoretical air [ C 3 H 8 + 37.5 ( O 2 + 3.76 N 2 ) ] → ( 8CO 2 + 9 H 2 O + 25 O 2 + 141 N 2 ) (II) Express the energy rate balance at steady state reduces in Equation (II). 0 = Q CV − w CV + [ n ˙ C 8 H 18 h ¯ C 8 H 18 ] 1 + [ n ˙ O 2 h ¯ O 2 + n ˙ N 2 h ¯ N 2 ] 2 − [ n ˙ CO 2 h ¯ CO 2 + n ˙ H 2 O h ¯ H 2 O + n ˙ O 2 h ¯ O 2 + n ˙ N 2 h ¯ N 2 + m ˙ 3 3 2 2 ] 3 m ˙ 3 3 2 2 = [ n ˙ C 8 H 18 h ¯ C 8 H 18 ] 1 + [ n ˙ O 2 h ¯ O 2 + n ˙ N 2 h ¯ N 2 ] 2 − [ n ˙ CO 2 h ¯ CO 2 + n ˙ H 2 O h ¯ H 2 O + n ˙ O 2 h ¯ O 2 + n ˙ N 2 h ¯ N 2 ] 3 (III) Here, the molar flow rate is n ˙ and the mass flow rate of the exiting product is m ˙ 3 . m ˙ 3 = ( m ˙ CO 2 + m ˙ H 2 O + m ˙ O 2 + m ˙ N 2 ) = ( M CO 2 n ˙ CO 2 + M H 2 O n ˙ H 2 O + M O 2 n ˙ O 2 + M N 2 n ˙ N 2 ) (IV) Dividing throughout by the molar flow rate of the fuel and solving for 3 2 2 , in Equation (I) to (III) give: 3 2 2 = ( h ¯ C 8 H 18 ) + [ 37.5 h ¯ O 2 + 141 h ¯ N 2 ] 2 − [ 8 h ¯ CO 2 + 9 h ¯ H 2 O + 25 h ¯ O 2 + 141 h ¯ N 2 ] ( 8 M CO 2 + 9 M H 2 O + 25 M O 2 + 141 M N 2 ) (V) Here, h ¯ = h ¯ f ° + Δ h ¯ , h ¯ f ° = 0 for O 2 and N 2 . Therefore, 3 2 2 = [ ( h ¯ f ° ) C 8 H 18 + 8 ( h ¯ f ° + h ¯ 990 K − h ¯ 298 K ) CO 2 − 9 ( h ¯ f ° + h ¯ 990 K − h ¯ 298 K ) H 2 O − 25 ( h ¯ 990 K − h ¯ 298 K ) O 2 − 141 ( h ¯ 990 K − h ¯ 298 K ) N 2 ] ( 8 M CO 2 + 9 M H 2 O + 25 M O 2 + 141 M N 2 ) (VI) Conclusion: Refer Table A-1, “atomic or molecular weights”, and write the molar masses of given species

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 13.8, Problem 54P

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Fundamentals of Engineering Thermodynamics, Binder Ready Version

Ch. 13.8 - Prob. 1E Ch. 13.8 - Prob. 2E Ch. 13.8 - Prob. 3E Ch. 13.8 - Prob. 4E Ch. 13.8 - Prob. 5E Ch. 13.8 - Prob. 6E Ch. 13.8 - Prob. 7E Ch. 13.8 - Prob. 8E Ch. 13.8 - Prob. 9E Ch. 13.8 - Prob. 10E

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