Question

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Ext-2.2 (35 points) An insulated, rigid tank is divided into two compartments by a frictionless, thermally conducting piston. One compartment initially contains 0.6 m' of saturated water vapor at 200 °C and the other compartment contains 0.8 m' of superheated water vapor at 210 bar, 780 °C. The piston is released and equilibrium is attained. The mass of piston is small enough to be neglected in this problem. (a) Draw a schematic diagram of the tank and identify the system or systems for the solution of the following parts of the problem. (4 points) (6) Write down the general equations for mass balance, energy balance, and entropy balance for the system or systems and modify them as applied to this problem (6 points) Determine the final pressure, in bar, and the final temperature, in "C. (17 point) (a) Evaluate the amount of entropy produced, in kJ/K. (8) Note: You will save substantial amount of time if you use the IT program. If IT program is used you must include a copy of the program code and the output data.

Answer 1

Soun (a) Jinendate given date, V = 0.6m3 V2 = 0.8m3 T2 200c P2 = 210 bar P = 15:54B48 T22 780°C (from & team table) given that chamber is rigid & Insulated so ota 20 8 & wao (6) - From first Land of the modynamics. da-dutaw do ao hizo - duzo - Ufa Vi & uf- ui 20 032 U tu . con represents the energy Balance Mass Balance za mitm -2 eon. ③ – from con Uz - v, tu Mzev. T3= m. c. I, & m..Tz M₂ T3 = Matit M₂ TL - Pro MRT 7 m2 PV RT ? ? ? 1 3 3 x T3 = PV LXT, + P2 V2 x 12 - P3432 PN, the le RTI » Ps - Pix+ Pu Vz- PINT P₂ V2

from ean m, + m₂ 2 M3 & Tea P33 To put vame of Ps from ean We get T32 PN+P V2 we know final that from ean ③ Pressure 22 PNtP2 2 put the names of . lz 2 Pi,V, & P then 15.sux 0.67 210x0.8 1-4 126.66 Bar Auswer P₂² and final Tempo T. Ton PVTBN 15.54x0.6+ 210808 15.54 x 0.6 200 t + a 210x0.8 780 [T3 = 616.8°C Auswer 676.8°C tuganer.

(d). Lem 473 Entsory ean for the system is given by as apen (T) + Ren (VF) troon steam Tables we put the values of CP8R → AS - 1996* en 1053 ) + 0.46 ) DS= 1.597 + 0.132 as 1.729 KI IGR - mo 15.54x10sx 0.6 RT, 2 0.468103 x 472 4.28 mg M22 210x105x 0.8 0.46x103 10532 346 8 kg. & final moss (Ma) = marm, & 34.68+ 4.28 38.96 kg. - DS- 38.96 8 1.729. as a 67,36 KJ