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

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Chapter 8.8, Problem 36P

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