Question

Air, initially at the dead state, is compressed isothermally and with no friction inside a piston cylinder device until it reaches 1/10 of its original volume. a) What is the change of the energy of the air due to this process? (10 pts) b) What is the change of exergy of air due to this process? Briefly comment on the difference (or lack of difference) between the findings in part(a) and part (b). (15 pts) c) What is the amount of work per unit mass of air required to achieve this process? (15 pts) d) Confirm that the exergy is balanced during the process. (10 pts)

Answer 1

20(fomo) Given that, Afs, initially at the dead state, is compressed ico- thermally and with no foiction inside a piston cylinder device until it reaches yo of its oopginal volume. Initial - Po, To and i Applying isothermal Final = P2, To and 4/10 condition: No friction - P₂ (t) = Pot 1 P2 = 10P] a) ir System 'consider Kothermal process occurs 80, OV=0/->0 6 8 = U-Tostov . so so = exergy = CUAL)-TO (51-52)+7 (41-42) - Po(tr) +7 AS os = pin (77P-Ren () - isothermal) OS=-R In (10Me) --Reln (10) so, f50 = 9 tot + To (-8.Info)) @ In part @ we applied energy equation Coo) simply we can gay in isothermal poocess, so, tempesafuse charge not occurs, so, change in energy also not occurs But in (b) 'we alio fovend change in entropy as well as change in volume to find Maximum uxful wook (Pxergy). The change in exergy is simply in terms of difference of availability function (0)

1. CC) we know,' Wisothermal = lot, In (40) =Pot, In (T) Weso = -Pov ln (10) Here are sign repre- sents isothermal work. Wilso is work per whit mass (kJ/kg) (d) From Equation we can clearly see that sø is not Zero and having same value depends on Po, tep Valere and that diffesence occur because of same exergy distraction. e C