Explanation Write the complete combustion equation for methane with theoretical amount of air. CH 4 + 2 ( O 2 + 3.76 N 2 ) → CO 2 + 2 H 2 O + 7.52 N 2 (I) Write the chemical balanced equation for the reaction between carbon dioxide and the hydrogen. CO 2 + H 2 ⇄ CO + H 2 O (II) Write the expression to calculate the equilibrium constant ( K ) for the reaction in Equation (II). K = N CO v CO N H 2 O v H 2 O N CO 2 v CO 2 N H 2 v H 2 ( P N total ) v CO + v H 2 O − v CO 2 − v H 2 (III) Here, number of moles of CO is N CO , number of moles of H 2 O is N H 2 O , number of moles of CO 2 is N CO 2 , number of moles of H 2 is N H 2 , pressure is P , total number of moles of mixture is N total , stoichiometric coefficients are v CO , v H 2 O , v CO 2 , and v H 2 . Write the steady state energy balance equation for the reaction of Gaseous methane in the reactor. E ˙ in − E ˙ out = Δ E ˙ system ︸ = 0 E ˙ in = E ˙ out (IV) Here, the net energy rate input is E ˙ in , net energy rate output is E ˙ out , and the change in net energy of the system is Δ E ˙ system . Conclusion: Obtain the stoichiometric coefficients from the Equation (II). v CO = 1 v H 2 O = 1 v CO 2 = 1 v H 2 = 1 Re-write the Equation (I) for the combustion of methane with 80% theoretical amount of air. CH 4 + 0.8 × 2 ( O 2 + 3.76 N 2 ) → [ x CO 2 + ( 1 − x ) CO+ ( 2.2 − x ) H 2 O + ( x − 0.2 ) H 2 + 6.016 N 2 ] (V) Calculate the total number of moles: n = x + ( 1 − x ) + ( 2.2 − x ) + ( x − 0.2 ) + ( 6.016 ) = 9.016 From the Table A-27 of “logarithms to the base 10 of the Equilibrium constant K ”, obtain the equilibrium constant for Equation (II) at 1700 K as log 10 K = 0.530 K = 3.3884 Substitute 3.3884 for K , ( 1 − x ) for N CO , ( 2.2 − x ) for N H 2 O , x for N CO 2 , ( x − 0.2 ) for N H 2 , 1 for v CO , 1 for v H 2 O , 1 for v CO 2 , and 1 for v H 2 in Equation (III). 2.5322 = ( x − 1 ) ( 2.2 − x ) x ( x − 0.2 ) ( P N total ) 1 + 1 − 1 − 1 x = 0.5674 Substitute 0.5674 for x , and re-write the Equation (V). CH 4 + 0.8 × 2 ( O 2 + 3.76 N 2 ) → [ 0.5674 CO 2 + 0.4326 CO+ 1.6326 H 2 O + 0.3674 H 2 + 6.016 N 2 ] (VI) Thus, The equilibrium composition of the mixture is ( 0.5674 CO 2 , 0.4326 CO, 1.6326 H 2 O , 0.3674 H 2 , 6.016 N 2 ) . Re-write the energy balance Equation (IV) for the burning of gaseous methane from Equation (VII). 0 = { Q ˙ c v n ˙ CH 4 − W ˙ c v n ˙ CH 4 ︸ = 0 + ( h ¯ f o ) CH 4 + 1.6 ( h ¯ f o + h ¯ − h ¯ o ) O 2 + 6.016 ( h ¯ f o + h ¯ − h ¯ o ) N 2 − [ 0.5674 ( h ¯ f o + h ¯ − h ¯ o ) CO 2 + 0.4326 ( h ¯ f o + h ¯ − h ¯ o ) CO + 1.6326 ( h ¯ f o + h ¯ − h ¯ o ) H 2 O + 0.3674 ( h ¯ f o + h ¯ − h ¯ o ) H 2 + 6

Fundamentals of Engineering Thermodynamics

8th Edition

ISBN: 9781118412930

Author: Michael J. Moran, Howard N. Shapiro, Daisie D. Boettner, Margaret B. Bailey

Publisher: WILEY

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Chapter 14.6, Problem 41P

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The amount of heat transfer between the reactor and the surroundings for the methane gas

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Chapter 14.6, Problem 40P

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