Respuesta :
Answer:
(a) 65.04%
(b) 16.91%
Solution:
As per the question:
At inlet:
Pressure of the compressor, P = 140 kPa
Temperature, T = [tex] - 10^{\circ}C[/tex] = 263 K
Isentropic work, W = 700 kPa
At outlet:
Pressure, P' = 700 kPa
Temperature, T' = [tex]60^{\circ}C[/tex] = 333 K
Now, from the steam table;
At the inlet , at a P = 700 kPa, T =[tex] 60^{\circ}C[/tex]:
h = 243.40 kJ/kg, s = 0.9606 kJ/kg.K
At outlet, at P = 140 kPa, T =[tex] - 10^{\circ}C[/tex]:
h' = 296.69 kJ/kg, s' = 1.0182 kJ/kg.K
Also in isentropic process, s = [tex]s'_{s}[/tex] and [tex]h'_{s} = 278.06 kJ/kg.K[/tex] at 700kPa
(a) Isentropic efficiency of the compressor, [tex]\eta_{s} = \frac{Work\ done\ in\ isentropic\ process}{Actual\ work\ done}[/tex]
[tex]\eta_{s} = \frac{h'_{s} - h}{h' - h} = frac{278.06 - 243.40}{296.69 - 243.40} = 0.6504 = 65.04%[/tex]
(b) The temperature of the environment, [tex]T_{e} = 27^{\circ}C[/tex] = 273 + 27 = 300 K
Availability at state 1, [tex]\Psi = h - T_{e}s = 243.40 - 300\times 0.9606 = - 44.78 kJ/kg[/tex]
Similarly for state 2, [tex]\Psi' = h' - T_{e}s' = 296.69 - 300\times 1.0182 = - 8.77 kJ/kg[/tex]
Now, the efficiency of the compressor as per the second law;
[tex]\eta' = \frac{\Psi' - \Psi}{h' - h} = \frac{- 8.77 - (- 44.78)}{296.69 - 243.40} = 0.6757[/tex] = 67.57%
