Explanation Write the stoichiometric reaction of octane with excess of 400% theoretical air. C 3 H 18 + 4 ( 12.5 ) ( O 2 + 3.76 N 2 ) → x CO 2 + y H 2 O + w O 2 + z N 2 (I) Write the chemical exergy of each element using Gibbs energy equation. e C 3 H 18 − c h = [ g ¯ Re actants − g ¯ Re actants ] ( T 0 , p 0 ) + [ e Reactants − c h − e Reactants − c h ] (II) Here, Gibbs free energy of each element is g ¯ , temperature of the reaction is T 0 , pressure of the reaction is p 0 and exergy of each element is e − c h . Conclusion: Balance carbon on both sides of the chemical equation. 8 = x Balance hydrogen on both sides of the chemical equation. 18 = 2 y y = 9 Balance oxygen on both sides of the chemical equation. 4 ( 12.5 ) × 2 = 2 x + y + 2 w 4 ( 12.5 ) × 2 = 2 ( 8 ) + 9 + 2 w w = 37.5 Balance nitrogen on both sides of the chemical equation. 4 ( 12.5 ) × 2 × 3.76 = 2 z z = 188 Write the balanced combustion equation. C 3 H 18 ( l ) + 4 × 1.25 ( O 2 + 3.76 N 2 ) → 8CO 2 + 9 H 2 O + 37.5 O 2 + 188 N 2 Rewrite Equation (II) using the stoichiometric equation. e C 8 H 18 − c h = [ g ¯ Re actants − g ¯ Re actants ] ( T 0 , p 0 ) + [ e Compounds − c h − e Elements − c h ] e C 8 H 18 − c h = [ { [ g ¯ C 8 H 18 + ( 4 × 12.5 ) g ¯ O 2 + ( 4 × 12.5 × 3.76 ) g ¯ N 2 ] − [ 8 g CO 2 + 9 g H 2 O + 37.5 g O 2 + 188 g N 2 ] } ( T 0 , p 0 ) + ( 8 e CO 2 − c h + 9 e H 2 O − c h + 37.5 e O 2 − c h − ( 4 × 12.5 ) e O 2 − c h − 188 e N 2 − c h ) ] e C 8 H 18 − c h = [ ( g ¯ C 8 H 18 + 50 g ¯ O 2 + 188 g ¯ N 2 − 8 g CO 2 − 9 g H 2 O − 37.5 g O 2 − 188 g N 2 ) ( T 0 , p 0 ) + ( 8 e CO 2 − c h + 9 e H 2 O − c h + 37

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ISBN: 9781119391630

Author: MORAN

Publisher: WILEY

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Chapter 13.8, Problem 28CU

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The rate of exergy destruction within the reactor.

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Chapter 13.8, Problem 27CU

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FUND OF ENGINEERING THERMO W/WILEY PLU

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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