Question 6.14: Problem: An adiabatic compressor with an isentropic efficien...
Problem: An adiabatic compressor with an isentropic efficiency of 80% takes air flowing at a rate of 0.5 kg / s from 90 kPa and 22 °C to 800 kPa. Use the air table to find the exit temperature of the air and the power input required to drive the compressor.
Find: Exit temperature T_2 of air and power required \dot{W} .
Known: Mass flow rate \dot{m} = 0.5 kg / s, inlet pressure P_1 = 90 kPa, inlet temperature T_1 = 22 °C = 295.15 K, outlet pressure P_2 = 800 kPa, isentropic efficiency η_c = 80%.
Governing equations:
Power of compressor \dot{W} _c=\dot{m} (h_i-h_e)
Isentropic compressor efficiency \eta _c=\frac{h_{2s}-h_1}{h_2-h_1}
Pressure ratio (isentropic) P_{r2}=P_{r1}\left\lgroup\frac{P_2}{P_1} \right\rgroup
Properties: Air at 295 K (approximation) has specific enthalpy h_1 = 295.17 kJ / kgK (Appendix 7), relative pressure P_{r1} = 1.3068 (Appendix 7).
Find the final relative pressure of the process,
P_{r2}=P_{r1}\left\lgroup\frac{P_2}{P_1} \right\rgroup =1.3068 \left\lgroup\frac{800 \ kPa}{90 \ kPa} \right\rgroup =11.6160.
Interpolating from the air tables (Appendix 7), the corresponding enthalpy value is h_{2s} = 552.083 kJ / kg.
The actual exit enthalpy is
h_2=\frac{h_{2s} – h_1}{\eta _c} + h_1 =\frac{552.083 \ kJ/kg – 295.17 \ kJ/kg}{0.8} +295.17 \ kJ/kg=616.311 \ kJ/kg.
Interpolating again from the air tables, the corresponding exit temperature is T_2 = 608.840 K.
The work required is
\dot{W} _c=\dot{m} (h_2 – h_1)=0.5 \ kg/s (616.311 \ kJ/kg – 295.17 \ kJ/kg)=160.571 \ kW.
Answer: The exit temperature is 609 K and the power required is 160.6 kW.