Compressing Air by a Compressor
Air at 100 kPa and 280 K is compressed steadily to 600 kPa and 400 K. The mass flow rate of the air is 0.02 kg/s, and a heat loss of 16 kJ/kg occurs during the process. Assuming the changes in kinetic and potential energies are negligible, determine the necessary power input to the compressor.
Air is compressed steadily by a compressor to a specified temperature and pressure. The power input to the compressor is to be determined.
Assumptions 1 This is a steady-flow process since there is no change with time at any point and thus \Delta m_{ CV }=0 \text { and } \Delta E_{ CV }=0 . 2 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical-point values. 3 The kinetic and potential energy changes are zero, \Delta ke =\Delta pe =0 .
Analysis We take the compressor as the system (Fig. 5–27). This is a control volume since mass crosses the system boundary during the process. We observe that there is only one inlet and one exit and thus \dot{m}_{1}=\dot{m}_{2}=\dot{m} . Also, heat is lost from the system and work is supplied to the system.
Under stated assumptions and observations, the energy balance for this steady-flow system can be expressed in the rate form as
\underbrace{\dot{E}_{\text {in }}-\dot{E}_{\text {out }}}_{\begin{array}{l}\text { Rate of net energy transfer } \\\text { by heat, work, and mass }\end{array}}=\underbrace{d E_{\text {system }} /d t^{\nearrow ^{0(steady)}}}_{\begin{array}{c}\text { Rate of change in internal, kinetic, } \\\text { potential, etc., energies }\end{array}}=0
\begin{aligned}\dot{E}_{\text {in }} &=\dot{E}_{\text {out }} \\\dot{W}_{\text {in }}+\dot{m} h_{1} &=\dot{Q}_{\text {out }}+\dot{m} h_{2} \quad(\text { since } \Delta ke =\Delta pe \cong 0) \\\dot{W}_{\text {in }} &=\dot{m} q_{\text {out }}+\dot{m}\left(h_{2}-h_{1}\right)\end{aligned}
The enthalpy of an ideal gas depends on temperature only, and the enthalpies of the air at the specified temperatures are determined from the air table (Table A–17) to be
\begin{aligned}&h_{1}=h_{@ 280 K }=280.13 kJ / kg \\&h_{2}=h_{@ 400 K }=400.98 kJ / kg\end{aligned}
Substituting, the power input to the compressor is determined to be
\begin{aligned}\dot{W}_{\text {in }} &=(0.02 kg / s )(16 kJ / kg )+(0.02 kg / s )(400.98-280.13) kJ / kg \\&=2.74 kW\end{aligned}
Discussion Note that the mechanical energy input to the compressor manifests itself as a rise in enthalpy of air and heat loss from the compressor.
