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

See similar textbooks

Concept explainers

Properties of Pure Substances

A substance with fixed chemical composition throughout is called a pure substance like water, air, and nitrogen. A pure substance does not have to be a single element or compound. A mixture of two or more phases of a pure substance is also a pure substan…

Videos

Question

Chapter 14.6, Problem 41P

To determine

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

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 14.6, Problem 40P

Chapter 14.6, Problem 42P

Students have asked these similar questions

For some viscoelastic polymers that are subjected to stress relaxation tests, the stress decays with time according to a(t) = a(0) exp(-4) (15.10) where σ(t) and o(0) represent the time-dependent and initial (i.e., time = 0) stresses, respectively, and t and T denote elapsed time and the relaxation time, respectively; T is a time-independent constant characteristic of the material. A specimen of a viscoelastic polymer whose stress relaxation obeys Equation 15.10 was suddenly pulled in tension to a measured strain of 0.5; the stress necessary to maintain this constant strain was measured as a function of time. Determine E (10) for this material if the initial stress level was 3.5 MPa (500 psi), which dropped to 0.5 MPa (70 psi) after 30 s.

For the flows in Examples 11.1 and 11.2, calculate the magnitudes of the Δ V2 / 2 terms omitted in B.E., and compare these with the magnitude of the ℱ terms.

Calculate ℛP.M. in Example 11.2.

Chapter 14 Solutions

Fundamentals of Engineering Thermodynamics

Ch. 14.6 - 1. Why is using the Gibbs function advantageous...Ch. 14.6 - Prob. 2E Ch. 14.6 - Prob. 3E Ch. 14.6 - Prob. 4E Ch. 14.6 - Prob. 5E Ch. 14.6 - Prob. 6E Ch. 14.6 - Prob. 7E Ch. 14.6 - Prob. 8E Ch. 14.6 - Prob. 9E Ch. 14.6 - Prob. 10E

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.