Answer:
(a) The reversible work is 207 kJ/kg
(b) The irreversibility rate is -38.39 kJ/kg
Explanation:
State1 : p1 = 100kpa, T1= 25+273 =298k
From air table, h1 =298.18 kJ/kg, s10= 1.69528 kJ/kgK
State 2a:p2=2mpa,t2=540k (actual condition 2a)
h2a= 544.35 kJ/kg,s2a0=2.29906
actual work input to the compressor =wout=h1-h2+Qin
=298.18-544.35+(-150)kJ/kg(- sign indicate heat loss)
=(-246.17)kJ/kg(-ve sign indicates the work is given into the system
a) Reversible work= Win actual - any irreversiblities present
=246.17 + irreversibilty
b) irreversibility = T0(Entopy generation Sgen) for air, Sgen
=s20-s10-Rln(p2/p1), T0=250C
=(25+273)(s2a0-s10-Rlnp2/p1+Qout/Tsurr)
= 298x[(2.29906-1.69528-0.287kJ/kgK xln(2000kpa/100) + 150 /298]
= -38.39 kJ/kg
a)Reversible work = Win actual -any irreversiblities present
=246.17 + irreversibilty
=246.17+-38.39
=207 kJ/kg
