Air enters a compressor at 100 kPa and 25 ⁰C. It is compressed to 2 MPa and exits the compressor at 540 K. The compressor is at a higher temperature than the surroundings, therefore 140 kJ is lost to the surroundings (as heat) per kg of air flowing through the compressor. You can neglect the kinetic and potential energy. (a) What is the reversible work? (answer in kJ/kg, as a negative number) (b) What is the irreversibility rate? (answer in kJ/kg)

(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