Thermodynamics

Chapter 17 | 827Disregarding potential energy changes and heat transfer, the compressorwork per unit mass of air is determined from Eq. 17–8:w in c p 1T 02 T 01 2 11.005 kJ>kg # K21519.5 K 286.8 K2 233.9 kJ/kgThus the work supplied to the compressor is 233.9 kJ/kg.Discussion Notice that using stagnation properties automatically accountsfor any changes in the kinetic energy of a fluid stream.17–2 ■ SPEED OF SOUND AND MACH NUMBERAn important parameter in the study of compressible flow is the speed ofsound (or the sonic speed), which is the speed at which an infinitesimallysmall pressure wave travels through a medium. The pressure wave may becaused by a small disturbance, which creates a slight rise in local pressure.To obtain a relation for the speed of sound in a medium, consider a pipethat is filled with a fluid at rest, as shown in Fig. 17–7. A piston fitted in thepipe is now moved to the right with a constant incremental velocity dV, creatinga sonic wave. The wave front moves to the right through the fluid atthe speed of sound c and separates the moving fluid adjacent to the pistonfrom the fluid still at rest. The fluid to the left of the wave front experiencesan incremental change in its thermodynamic properties, while the fluid onthe right of the wave front maintains its original thermodynamic properties,as shown in Fig. 17–7.To simplify the analysis, consider a control volume that encloses the wavefront and moves with it, as shown in Fig. 17–8. To an observer travelingwith the wave front, the fluid to the right will appear to be moving towardthe wave front with a speed of c and the fluid to the left to be moving awayfrom the wave front with a speed of c dV. Of course, the observer willthink the control volume that encloses the wave front (and herself or himself)is stationary, and the observer will be witnessing a steady-flow process.The mass balance for this single-stream, steady-flow process can beexpressed asorrAc 1r dr2A 1c dV2By canceling the cross-sectional (or flow) area A and neglecting the higherorderterms, this equation reduces toNo heat or work crosses the boundaries of the control volume during thissteady-flow process, and the potential energy change, if any, can beneglected. Then the steady-flow energy balance e in e out becomesh c22m # right m # leftc dr r dV 0 h dh 1c dV222(a)V0PPistondVh + dhP + dP chPr + dr rdVP + dPMovingwave frontStationaryfluidFIGURE 17–7Propagation of a small pressure wavealong a duct.Control volumetraveling withthe wave fronth + dh c – dV c hP + dPPr + dr rFIGURE 17–8Control volume moving with the smallpressure wave along a duct.Pxx