Rate of flow of air, m = 25 kg/min.

Initial pressure, p_{1} = 1 bar

Final pressure, p_{2} = 2.0 bar

Initial temperature, T_{1} = T_{0} = 15 + 273 = 288 K

Final temperature, T_{2} = 100 + 273 = 373 K.

Applying energy equation to compressor

W_{actual} = h_{2} – h_{1} [as Q = 0, ∆PE = 0, ∆KE = 0]

= c_{p} (T_{2} – T_{1}) = 1.005 (373 – 288) = 85.4 kJ/kg

Total actual work done/min

= 25 × 85.4 = 2135 kJ/min

= \(\cfrac{2135}{60}\) = 35.58 kJ/s = 35.58 kW

The minimum work required is given by the increase in availability of the air stream.

= 85.4 – 288 × 0.061 = 67.8 kJ/kg

Minimum work required

\(\cfrac{25\times67.8}{60}\) = 28.25 kJ/s = 28.25 kW