Question 3.12: Estimate the power required to compress 1000 m^3/h air from ...

Take the inlet pressure, P_{1} , as 1 atmosphere = 101.33 kN/ m^{2}, absolute.
Outlet pressure, P_{2} ,= 700 + 101.33 = 801.33 kN/ m^{2} , absolute.
For equal work in each stage the intermediate pressure, P_{i} ,

= \sqrt{\left(1.0133 \times 10^{5} \times 8.0133 \times 10^{5} \right) }=\underline{\underline{2.8495 \times 10^{5}} } N/ m^{2}

For air, take ratio of the specific heats, Υ, to be 1.4.
For equal work in each stage the total work will be twice that in the first stage.
Take the inlet temperature to be 20 °C, At that temperature the specific volume is
given by  v_{1}=\frac{29}{22.4}\times \frac{293}{273}= 1.39 m^{3} /kg

Work done, -W = 2\times 1.0133 \times 10^{5} \times 1.39 \times \frac{1.4}{1.4-1} \left[\left(\frac{2.8495}{1.0133} \right) ^{\left(1.4-1\right)/1.4 } -1\right]

= 338,844 J/kg = 339 kJ/kg

From Figure 3.7, for a compression ratio of 2.85 the efficiency is approximately 84%. So work required
=339/0.84 = \underline{\underline{404} } kJ/kg

Mass flow-rate = \frac{1000}{1.39\times 3600 }=0.2 kg/s

Power required = 404 × 0.2 = \underline{\underline{80 kW} }