Thermodynamics

give a negative result when work is done on the system. To avoid the negativesign, Eq. 7–51 can be written for work input to steady-flow devicessuch as compressors and pumps as2w rev,in vdP ¢ke ¢pe1(7–53)The resemblance between the v dP in these relations and P dv is striking.They should not be confused with each other, however, since P dv is associatedwith reversible boundary work in closed systems (Fig. 7–41).Obviously, one needs to know v as a function of P for the given processto perform the integration. When the working fluid is incompressible, thespecific volume v remains constant during the process and can be taken outof the integration. Then Eq. 7–51 simplifies tow rev v 1P 2 P 1 2 ¢ke ¢pe1kJ>kg2(7–54)For the steady flow of a liquid through a device that involves no work interactions(such as a nozzle or a pipe section), the work term is zero, and theequation above can be expressed asv 1P 2 P 1 2 V 2 2 1 V 1 g 1z22 z 1 2 0(7–55)which is known as the Bernoulli equation in fluid mechanics. It is developedfor an internally reversible process and thus is applicable to incompressiblefluids that involve no irreversibilities such as friction or shockwaves. This equation can be modified, however, to incorporate these effects.Equation 7–52 has far-reaching implications in engineering regardingdevices that produce or consume work steadily such as turbines, compressors,and pumps. It is obvious from this equation that the reversible steadyflowwork is closely associated with the specific volume of the fluid flowingthrough the device. The larger the specific volume, the larger the reversiblework produced or consumed by the steady-flow device (Fig. 7–42). Thisconclusion is equally valid for actual steady-flow devices. Therefore, everyeffort should be made to keep the specific volume of a fluid as small as possibleduring a compression process to minimize the work input and as largeas possible during an expansion process to maximize the work output.In steam or gas power plants, the pressure rise in the pump or compressoris equal to the pressure drop in the turbine if we disregard the pressurelosses in various other components. In steam power plants, the pump handlesliquid, which has a very small specific volume, and the turbine handlesvapor, whose specific volume is many times larger. Therefore, the work outputof the turbine is much larger than the work input to the pump. This isone of the reasons for the wide-spread use of steam power plants in electricpower generation.If we were to compress the steam exiting the turbine back to the turbineinlet pressure before cooling it first in the condenser in order to “save” theheat rejected, we would have to supply all the work produced by the turbineback to the compressor. In reality, the required work input would be evengreater than the work output of the turbine because of the irreversibilitiespresent in both processes.Chapter 7 | 363= –∫ 12v dPw rev(a) Steady-flow system2= ∫1w revw revP dv(b) Closed systemw revFIGURE 7–41Reversible work relations for steadyflowand closed systems.w = –∫2v dP12w = – ∫1 v dP2w = –∫ 1v dPFIGURE 7–42The larger the specific volume, thegreater the work produced (orconsumed) by a steady-flow device.