Explanation Express average temperature of oxygen. T avg = T 1 + T 2 2 (I) Here, initial temperature is T 1 and final temperature is T 2 . Express entropy change of oxygen. s 2 − s 1 = c ν , avg ln [ T 2 T 1 ] + R ln [ v 2 v 1 ] (II) Here, initial specific is s 1 , final specific entropy is s 2 , average specific heat at constant volume is c ν , avg , universal gas constant of oxygen is R , and initial and final specific volume is v 1 and v 2 respectively. Express change in specific internal energy. u 1 − u 2 = c ν , avg ( T 1 − T 2 ) (III) Express the reversible work input by setting the exergy destruction is equal to zero. X in − X out − X destroyed = Δ X system X in − X out − 0 = Δ X system X 2 − X 1 = Δ X system Δ X system = X 2 − X 1 W rev,in = ϕ 2 − ϕ 1 = − [ ( u 1 − u 2 ) − T 0 ( s 1 − s 2 ) + P 0 ( v 1 − v 2 ) ] (IV) Here, net exergy transfer by heat, work, and mass is X in − X out , exergy destruction is X destroyed , change in exergy is Δ X system , initial exergy is X 1 , final exergy is X 2 , initial exergy of oxygen is ϕ 1 , final exergy of oxygen is ϕ 2 , initial and final specific internal energy is u 1 and u 2 respectively, surrounding temperature is T 0 , initial and final specific entropy is s 1 and s 2 respectively, reversible work input is W rev,in and surrounding pressure is P 0 . Express the increase in the exergy of the oxygen. ϕ 2 − ϕ 1 = W rev,in (V) Conclusion: Refer Table A-1E, “molar mass, gas constant and critical point properties, and write the value of universal gas constant of oxygen. R = 0.06206 Btu / lbm ⋅ R Substitute 75°F for T 1 and 525°F for T 2 in Equation (I). T avg = 75°F + 525°F 2 = 300 °F Refer Table A-2E (b), “ideal gas specific heats of various common gases at various temperatures” and write the average specific heat at constant volume of oxygen corresponding to average temperature of 300°F . c ν , avg = 0.164 Btu / lbm ⋅ R Substitute 0.164 Btu / lbm ⋅ R for c ν , avg , 75°F for T 1 , 525°F for T 2 , 0.06206 Btu / lbm ⋅ R for R , 1 .5 ft 3 / lbm for v 2 and 12 ft 3 / lbm for v 1 in Equation (II). s 2 − s 1 = ( 0.164 Btu / lbm ⋅ R ) ln [ 525°F 75°F ] + ( 0

8th Edition

ISBN: 9780073398174

Author: Yunus A. Cengel Dr., Michael A. Boles

Publisher: McGraw-Hill Education

See similar textbooks

Videos

Question

Chapter 8.8, Problem 36P

To determine

The reversible work input and the increase in the exergy of the oxygen.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 8.8, Problem 35P

Chapter 8.8, Problem 37P

Students have asked these similar questions

Solve, use engineering economic tables

please show all work

Chapter 8 Solutions

Thermodynamics: An Engineering Approach

Ch. 8.8 - What final state will maximize the work output of...Ch. 8.8 - Is the exergy of a system different in different...Ch. 8.8 - How does useful work differ from actual work? For...Ch. 8.8 - Prob. 4P Ch. 8.8 - Consider two geothermal wells whose energy...Ch. 8.8 - Consider two systems that are at the same pressure...Ch. 8.8 - Prob. 7P Ch. 8.8 - Does a power plant that has a higher thermal...Ch. 8.8 - Prob. 9P Ch. 8.8 - 8–10C Can a process for which the reversible work...

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.

The reversible work input and the increase in the exergy of the oxygen.

Solution Summary: The author explains the reversible work input and the increase in the exergy of the oxygen.

Videos

The reversible work input and the increase in the exergy of the oxygen.

Want to see the full answer?

Chapter 8 Solutions