Question

A mass of 2.4 kg of water contained in a piston-cylinder assembly expand from an initial state where T1 = 574°C. P1 = 700 kPa to a final state where T2 = 209°C, p2 = 300 kPa, with no significant effects of kinetic and potential energy. It is claimed that the water undergoes an adiabatic process between these states, while developing work. Evaluate entropy production in Joules/k.

Please show all your work neatly and use the tables from Fundamentals of Engineering Thermodynamics 8th edition by Michael J Moran, Howard N. Shapiro

Answer 1

Here, we formulate our answer with the help of second law of thermodynamics as follows:-

Given :- det Tomass of water =2.4kg ams • Initial condition Estate 1) Pia Pressure at state is 700 kPa 0.mpa Toe Temperature at state l 3574°C o final condition (state 20:- P22 pressure at state 2 = 300 kPa TQ7 Temperature at state 2 3209°C Assumptions! -y Adiabatic system, il Heat transfer = Q =0 jiy no significance eppect of kinetic and potential energy ie system is in equillibrium with initial and final state Priy gystem is closed system Now, entropy balance for closed system is given by, [ Entropy change] sentropy transfer? I entropy within the across system . It l production | System I boundry | Lin system ne 182-51 = (do) +6 - Lo-where, a S, and so are the specific entropy alt state i and state e respectively o de p's entropy transfer o is an entropy produce

- Now , according to second Law on thermodynamic, antropy production must be zero or positive ine boo2 ive process will occur only in con bao these condition, Now, from the table we calculate entropy at state land 2 as follows!. For state i !-(9100 kpa and 574°c) we used interpolation to find entropy at statel esi) as, i at 700 kpa: • for 600°c = 95.801956 KIIKgk. and for sooc o go 7.9299 kJ legk. Golve per 100°C -> AS,= 39-569-0.2657. . Ir por 100°C-0A3,-0.2657 then for 26°c o AS220.06908.2. si 0523 0.069082 • Therefore for 574 c and aoo kPa entropy is, Si= 8,1956 -0.069082 Isi 28.126518 KJ 1 kg kl for state 2%-( 300 kPa and 200°c) we used interpolation to find entropy at state 2 (82) as, at 300 KPI :- o for 240°c o sa a 904774 kJ l kyk per 200°c osb 27.3115 kJ/kg k

o lie for 40°C -0 AS1 = 39-563 0.1659 - If for 40°c Đ03120.1659 then for 31°C 3 05220.12857 . There more for 209*c and 300 kpq entropy is j 522 entropy at - 052 - 240°C = 7.4774 - 0112857 (52 = 2.3488 kJ 1kg k] - Now put se 713488 kJ / kg kogo. = 8.126 518 kJ lage I and Q=0 in en ☺ OD 82-812 8 de ) + 6 7.3488-212651e e segregato -0.7976 = 6 ,'6 = -0.7976. KII kg k. 6- mis x C-0.9976) kJ/k. 22-4 * (-0.7776) x1000 Ilk ABS-1866.4572 g/k] AoS - in 163-1866. 4572 Tol Lo Hence, process will not occur and our claim is not correct ii) Entropy production is B=-1866.4572 J/K !

Given :- det Tomass of water =2.4kg ams • Initial condition Estate 1) Pia Pressure at state is 700 kPa 0.mpa Toe Temperature at state l 3574°C o final condition (state 20:- P22 pressure at state 2 = 300 kPa TQ7 Temperature at state 2 3209°C Assumptions! -y Adiabatic system, il Heat transfer = Q =0 jiy no significance eppect of kinetic and potential energy ie system is in equillibrium with initial and final state Priy gystem is closed system Now, entropy balance for closed system is given by, [ Entropy change] sentropy transfer? I entropy within the across system . It l production | System I boundry | Lin system ne 182-51 = (do) +6 - Lo-where, a S, and so are the specific entropy alt state i and state e respectively o de p's entropy transfer o is an entropy produce
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