Question

3. a) is the exergy of a system different in different environments? (2 marks) b) What is the second law efficiency? (2 marks) c) Consider a process during which no entropy is generated. Does the exergy destruction for thi process have to be zero? (1 mark) d) A piston cylinder device initially contains 2 L of air at 100 kPa and 25 °C. Air is now compresse to a final state of 600 kPa and 150 °C. The useful work input is 1.2 kJ. Assuming the surroundings a at 100 kPa and 25 °C, determine 0) The exergy of the air at the initial and final states 1) The minimum work that must be supplied to accomplish this compression process, and i) The second law efficiency of this process (C = 1.009kJ/kg.K Cave = 0.722kJ/kg.K) (15 marks)

Answer 1

as which The energy of a system is defined the manuum useful work that can be produced in the process through system is brought in equilibrium with the surroundings. Therefore, energy of system is different in different environments. the a determine the work that can a e. b) The recond law officiency of a syetem helps us manimum useful system produce It is defined as the ratio of useful actual work done to reversible work done by the systemer say it is the ratio of work done we can by a system in actual condition to work done by system under reversible conditions. the

(6) energy is (2 coun عار system or is no no (d) given u defined as marimum useful work that be obtained when system is brought in equilibrium with When surroundings. energy is lost entropy is generated in the syetem surroundings of entropy destruction generated there will be energy for the process. V z 2L P = 100kPa, T = 25°c T 298K 2 bookla, Tz 150°c 423K Por lookla, To= 298k, Wuseful = 1 (subscript & stands for initial state, state and 2 stands for final conditions) bunding keergy given by following relation Xz2 en (uz - Uo) + Po (V2 - vo) - To (sz-so)] here specific internal energy V2 2 2 1.2kJ o stands for surround ic for state 2) us

specific volume specific entropy S2 mas air of ñ - between or '1' entropy energy difference state 2' and o! (since initially the state ings same) Using of system and surround are the properties table kJ/kg and Mo ? 212.64 Uz 2 2. 302.88 kB) 1 kg The specific volume at initial state (0.287) (298) m RTo Po 2 V a 0.85526 m m²/kg mily (0.281) ( 423) m?ley The specific volume at final state 널 그 RT2 P2 600 2 0. 20234 m 3 kg ar

The mass be 4 relation by gas 2 PY PV a MRT kg m: 0.00233 of air initially taken can obtained Pov RTO RT, (100) (2x103) (0.287) (298) The entropy difference between 2 states be calculated by using De la Ta – R en Pa hlen Pa T, (1009) (de 423 P 293) - (0.287) en (con (1.009) (0.35) © 287)(1.79) AS ? – 0.212 - 0.212 kJ/kg k i xz> m [ue-40) + Po (uz - Vo) - To as] X = (0.00233) [(302.88-21264) + (100)(0.20234 - 0.85526) - (298) (-0.212)]

2 X = to. 205 kJ since the system has state equilibrium with surroundings at state '1' :. x=0 (10) work supplied to complete difference energies 2 states X.- X, Minimum the process in between Wmin 0.205 kJ

(71) and law efficiency عوا * = Wuseful و o 203 ( 6. 17, 12