Explanation Express the closed system energy reduces. Q = Δ U = U P − U R (I) Evaluate the change in internal energy the final temperature is required using the ideal gas equation. p V = n R ¯ T V = n R ¯ T p Here, the volume is constant. Therefore, n R R ¯ T R p R = n P R ¯ T P p P T P = T R ( n R n P ) ( ρ P ρ R ) (II) 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 ) (III) 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 (III). [ C 8 H 18 + 12.5 ( O 2 + 3.76 N 2 ) ] → ( 8CO 2 + 9 H 2 O + 47 N 2 ) The number of lbmol in the reactant. n R = 1 + 12.5 ( 4.76 ) = 60.5 lbmol The number of lbmol in the product. n P = 8 + 9 + 47 = 64 lbmol Substitute 60.5 lbmol for n R and 64 lbmol for n P in Equation (III). T P = ( 537 R ) ( 60.5 lbmol 64 lbmol ) ( 3.98 atm 1 atm ) = 2020 R Express the internal energy of the energy balance reduces of reactant. U R = [ u ¯ C 8 H 18 + 12.5 u ¯ O 2 + 47 u ¯ N 2 ] R Express the internal energy of the energy balance reduces of product. U P = [ 8 u ¯ CO 2 + 9 u ¯ H 2 O + 47 u ¯ N 2 ] P Where u ¯ = h ¯ − R ¯ T and h ¯ = h ¯ f ° + Δ h ¯ . For reactant, U R = [ ( h ¯ f ° + Δ h ¯ − R ¯ T ) C 8 H 18 + 12.5 ( h ¯ f ° + Δ h ¯ − R ¯ T ) O 2 + 47 ( h ¯ f ° + Δ h ¯ − R ¯ T ) N 2 ] R = [ ( h ¯ f ° − R ¯ T ) C 8 H 18 + 12.5 ( − R ¯ T ) O 2 + 47 ( − R ¯ T ) N 2 ] (IV) For product, U P = [ 8 ( h ¯ f ° + Δ h ¯ − R ¯ T ) CO 2 + 9 ( h ¯ f ° + Δ h ¯ − R ¯ T ) H 2 O + 47 ( h ¯ f ° + Δ h ¯ − R ¯ T ) N 2 ] P = [ 8 ( h ¯ f ° + Δ h ¯ − R ¯ T ) CO 2 + 9 ( h ¯ f ° + Δ h ¯ − R ¯ T ) H 2 O + 47 ( Δ h ¯ − R ¯ T ) N 2 ] P (V) For the energy balance reduces Q = [ [ 8 ( h ¯ f ° + Δ h ¯ ) CO 2 + 9 ( h ¯ f ° + Δ h ¯ ) H 2 O + 47 ( Δ h ¯ ) N 2 − 64 R ¯ T P ] − [ ( h ¯ f ° ) C 8 H 18 − 60

Fundamentals of Engineering Thermodynamics

8th Edition

ISBN: 9781118832318

Author: MORAN

Publisher: WILEY

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Fundamentals of Engineering Thermodynamics

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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