Question 8.1: A three-stage compressor is required to compress air from 14...

It is shown in Section 8.3.4 that the work done is a minimum when the intermediate pressures P_{i1} and P_{i2} are related to the initial and final pressures P_{1} and P_{2} by:

P_{i1}/P_{1} = P_{i2}/P_{i1} =P_{2}/P_{i2}    (equation 8.45)

P_{1}= 140 kN/m² and P_{2} = 4000 kN/m².

∴      P_{2}/P_{1} = 28.57

∴      P_{i2}/P_{i1}=P_{2}/P_{i2}=\sqrt[3]{28.57} = 3.057 ,

P_{i1} = \underline{\underline{428 \ kN/m^2}},

and:    P_{i2} = \underline{\underline{1308 \ kN/m^2}}

The specific volume of the air at the inlet is:

v_{1}= (22.4/29)(283/273) (101.3/140) = 0.579 m³/kg

Hence, for 1 kg of air, the minimum work of compression in a compressor of n
stages is:

W =nP_{1}v_{1}(\frac{\gamma }{\gamma -1} )\left[(\frac{P_{2}}{P_{1}} )^{(\gamma-1)/n\gamma }-1 \right]        (equation 8.46)

Thus: W = (3 × 140,000 × 0.579)(1.4/0.4)[(28.57)^{0.4/3×1.4}-1] = \underline{\underline{319,170 \ J/kg}}

The isothermal work of compression is:

W_{iso}= P_{1}v_{1}(\ln P_{2}/P_{1})            (equation 8.36)

= (140,000 × 0.579 ln 28.57) = 271,740 J/kg

The isothermal efficiency = (100 ×271,740)/319,170 = \underline{\underline{85 \ \%}}

Compression cycles are shown in Figs 8a and 8b. The former indicates the effect of various values of n in PV^{n} D constant and it is seen that the work done is the area under the temperature–entropy curve. Figure 8b illustrates the three-stage compressor of this problem. The final temperature T_{2}, found from T_{2}/T_{1} = (P_{2}/P_{1}){(\gamma-1)/\gamma } is 390 K. The dotted lines illustrate the effect of imperfect interstage cooling.