Question

An air compressor takes air in at the state of the surroundings 100 kPa, 300 K. The air exits at 400 kPa, 200°C at the rate of 2 kg/s. Determine the minimum compressor work input. 

Answer 1

$s_0 - s_2 = S_T^0 -S_T^2-Rln(\frac{P_0}{P_2})$

$s_0 - s_2 = 6.86926-7.3303-0.287 ln(\frac{100}{400}) =-0.06317 \frac{kJ}{kg K}$ ...from Steam table

$\psi_1 = 0$ at ambient condition

$\psi_2 = h_2-h_0 + T_0(S_0-S_2)$

$\psi_2 = 475.79-300.473 + 300(-0.06317) = 156.365$ $\frac{kJ}{kg }$

$\dot{W}^{REV} = \dot{m}(\psi_2-\psi_1) = 312.73 = \dot{W}_c$