Thermodynamics

Chapter 7 | 365second T ds relation by setting ds 0:Thus,This result could also be obtained from the energy balance relation for anisentropic steady-flow process. Next we determine the enthalpies:State 1:State 2:Thus,Tds dh vdP1Eq. 7-242fvdP dhds 01isentropic process222w rev,in vdP dh h 2 h 11w rev,in 13194.5 2675.02 kJ>kg 519.5 kJ/kgDiscussion Note that compressing steam in the vapor form would requireover 500 times more work than compressing it in the liquid form between thesame pressure limits.1P 1 100 kPaf h 1 2675.0 kJ>kg1sat. vapor2 s 1 7.3589 kJ>kg #1Table A–52KP 2 1 MPas 2 s 1fh 2 3194.5 kJ>kg1Table A–62Proof that Steady-Flow Devices Deliver theMost and Consume the Least Work whenthe Process Is ReversibleWe have shown in Chap. 6 that cyclic devices (heat engines, refrigerators, andheat pumps) deliver the most work and consume the least when reversibleprocesses are used. Now we demonstrate that this is also the case for individualdevices such as turbines and compressors in steady operation.Consider two steady-flow devices, one reversible and the other irreversible,operating between the same inlet and exit states. Again taking heattransfer to the system and work done by the system to be positive quantities,the energy balance for each of these devices can be expressed in the differentialform asActual:dq act dw act dh dke dpeReversible:dq rev dw rev dh dke dpeThe right-hand sides of these two equations are identical since both devicesare operating between the same end states. Thus,orHowever,dq act dw act dq rev dw revdw rev dw act dq rev dq actdq rev Tds