Answer:
(a) 65.04%
(b) 16.91%
Solution:
As per the question:
At inlet:
Pressure of the compressor, P = 140 kPa
Temperature, T =
= 263 K
Isentropic work, W = 700 kPa
At outlet:
Pressure, P' = 700 kPa
Temperature, T' =
= 333 K
Now, from the steam table;
At the inlet , at a P = 700 kPa, T =
:
h = 243.40 kJ/kg, s = 0.9606 kJ/kg.K
At outlet, at P = 140 kPa, T =
:
h' = 296.69 kJ/kg, s' = 1.0182 kJ/kg.K
Also in isentropic process, s =
and
at 700kPa
(a) Isentropic efficiency of the compressor, ![\eta_{s} = \frac{Work\ done\ in\ isentropic\ process}{Actual\ work\ done}](https://tex.z-dn.net/?f=%5Ceta_%7Bs%7D%20%3D%20%5Cfrac%7BWork%5C%20done%5C%20in%5C%20isentropic%5C%20process%7D%7BActual%5C%20work%5C%20done%7D)
![\eta_{s} = \frac{h'_{s} - h}{h' - h} = frac{278.06 - 243.40}{296.69 - 243.40} = 0.6504 = 65.04%](https://tex.z-dn.net/?f=%5Ceta_%7Bs%7D%20%3D%20%5Cfrac%7Bh%27_%7Bs%7D%20-%20h%7D%7Bh%27%20-%20h%7D%20%3D%20frac%7B278.06%20-%20243.40%7D%7B296.69%20-%20243.40%7D%20%3D%200.6504%20%3D%2065.04%25)
(b) The temperature of the environment,
= 273 + 27 = 300 K
Availability at state 1, ![\Psi = h - T_{e}s = 243.40 - 300\times 0.9606 = - 44.78 kJ/kg](https://tex.z-dn.net/?f=%5CPsi%20%3D%20h%20-%20T_%7Be%7Ds%20%3D%20243.40%20-%20300%5Ctimes%200.9606%20%3D%20-%2044.78%20kJ%2Fkg)
Similarly for state 2, ![\Psi' = h' - T_{e}s' = 296.69 - 300\times 1.0182 = - 8.77 kJ/kg](https://tex.z-dn.net/?f=%5CPsi%27%20%3D%20h%27%20-%20T_%7Be%7Ds%27%20%3D%20296.69%20-%20300%5Ctimes%201.0182%20%3D%20-%208.77%20kJ%2Fkg)
Now, the efficiency of the compressor as per the second law;
= 67.57%