Explanation Express the theoretical combustion equation of benzene. C 6 H 6 ( g ) + a th ( O 2 + 3.76 N 2 ) → 6 CO 2 + 3 H 2 O + 3.76 a th N 2 (I) Here, benzene gas is C 6 H 6 , stoichiometric coefficient of air is a th , oxygen is O 2 , nitrogen is N 2 , carbon dioxide is CO 2 and water is H 2 O . Express the stoichiometric coefficient of air by O 2 balancing. a th = 6 + 1.5 = 7.5 Substitute 7.5 for a th in Equation (I). C 6 H 6 ( g ) + 7.5 ( O 2 + 3.76 N 2 ) → 6 CO 2 + 3 H 2 O + 3.76 × 7.5 N 2 C 6 H 6 ( g ) + 7.5 ( O 2 + 3.76 N 2 ) → 6 CO 2 + 3 H 2 O + 28.2 N 2 (II) Rewrite the Equation (II) in terms of actual combustion equation with 60% excess air. C 6 H 6 + 12 ( O 2 + 3.76 N 2 ) → [ 5.52 CO 2 + 0.48 CO + 3 H 2 O + 4.74 O 2 + 45.12 N 2 ] (III) Here, carbon monoxide is CO . Express the heat transfer from the combustion chamber. − Q out = ∑ N P ( h ¯ f o + h ¯ − h ¯ o − R a T ) P − ∑ N R ( h ¯ f o + h ¯ − h ¯ o − R a T ) R = ∑ N P ( h ¯ f o + h ¯ 2100 R − h ¯ 537 R − R a T ) P − ∑ N R ( h ¯ f o − R a T ) R (IV) Here, number of moles of products is N P , number of moles of reactants is N R , enthalpy of formation is h ¯ f , enthalpy is h ¯ , universal gas constant of air is R and temperature of products and reactants are T P and T R respectively. Conclusion: As the given temperature is in Rankin, perform the unit conversion of Rankin to Fahrenheit to obtain the values of enthalpies. T R = 77 °F = ( 77 + 460 ) R = 537 R Perform the unit conversion of temperature at final state from Rankin to Fahrenheit. T P = 2100 R = ( 2100 − 460 ) R = 1640 ° F Refer Equation (III), and write the number of moles of products. N P , CO 2 = 5.52 kmol N P , CO = 0.48 kmol N P , H 2 O = 3 kmol N P , O 2 = 4

CONNECT FOR THERMODYNAMICS: AN ENGINEERI

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ISBN: 9781260048636

Author: CENGEL

Publisher: MCG

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CONNECT FOR THERMODYNAMICS: AN ENGINEERI

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