Thermodynamics

Chapter 5 | 237Solution Air is compressed steadily by a compressor to a specified temperatureand pressure. The power input to the compressor is to be determined.Assumptions 1 This is a steady-flow process since there is no change withtime at any point and thus m CV 0 and E CV 0. 2 Air is an ideal gassince it is at a high temperature and low pressure relative to its critical-pointvalues. 3 The kinetic and potential energy changes are zero, ke pe 0.Analysis We take the compressor as the system (Fig. 5–27). This is a controlvolume since mass crosses the system boundary during the process. Weobserve that there is only one inlet and one exit and thus ṁ 1 ṁ 2 ṁ. Also,heat is lost from the system and work is supplied to the system.Under stated assumptions and observations, the energy balance for thissteady-flow system can be expressed in the rate form asE # in E # out dE system >dt 0¡ 0 (steady)q out = 16 kJ/kg˙AIRm = 0.02 kg/sP 1 = 100 kPaT 1 = 280 KP 2 = 600 kPaT 2 = 400 KW˙in = ?FIGURE 5–27Schematic for Example 5–6.⎫⎪⎪⎪⎬⎪⎪⎪⎭⎫⎪⎪⎬⎪⎪⎭Rate of net energy transferby heat, work, and massRate of change in internal, kinetic,potential, etc., energiesE # in E # outW # in m # h 1 Q # out m # h 2 1since ¢ke ¢pe 02W # in m # q out m # 1h 2 h 1 2The enthalpy of an ideal gas depends on temperature only, and theenthalpies of the air at the specified temperatures are determined from theair table (Table A–17) to beh 1 h @ 280 K 280.13 kJ/kgh 2 h @ 400 K 400.98 kJ/kgSubstituting, the power input to the compressor is determined to beW # in 10.02 kg/s2 116 kJ/kg2 10.02 kg/s21400.98 280.132 kJ/kgDiscussion Note that the mechanical energy input to the compressor manifestsitself as a rise in enthalpy of air and heat loss from the compressor.EXAMPLE 5–7 2.74 kWPower Generation by a Steam TurbineThe power output of an adiabatic steam turbine is 5 MW, and the inlet andthe exit conditions of the steam are as indicated in Fig. 5–28.(a) Compare the magnitudes of h, ke, and pe.(b) Determine the work done per unit mass of the steam flowing throughthe turbine.(c) Calculate the mass flow rate of the steam.P 1 = 2 MPaT 1 = 400°CV 1 = 50 m/sz 1 = 10 mSTEAMTURBINEW out = 5 MWSolution The inlet and exit conditions of a steam turbine and its poweroutput are given. The changes in kinetic energy, potential energy, andenthalpy of steam, as well as the work done per unit mass and the mass flowrate of steam are to be determined.Assumptions 1 This is a steady-flow process since there is no change withtime at any point and thus m CV 0 and E CV 0. 2 The system is adiabaticand thus there is no heat transfer.P 2 = 15 kPax 2 = 90%V 2 = 180 m/sz 2 = 6 mFIGURE 5–28Schematic for Example 5–7.