# P1 100 kpa a compressing a liquid p1 100 kpa b

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Unformatted text preview: now how v varies with P to perform the integration in Eq. 7–53. This relation, in general, is not readily available. But for an isentropic process, it is easily obtained from the second T ds relation by setting ds 0: T P2 = 1 MPa P2 = 1 MPa 2 1 MPa PUMP (b) COMPRESSOR 2 (a) FIGURE 7–43 Schematic and T-s diagram for Example 7–12. P1 = 100 kPa (a) Compressing a liquid P1 = 100 kPa (b) Compressing a vapor 100 kPa 1 1 s cen54261_ch07.qxd 11/18/03 9:57 AM Page 307 307 CHAPTER 7 dh υ dP (Eq. 6–24) 0 (isentropic process) T ds ds υ dP dh Thus, 2 wrev, in 1 2 υ dP dh h2 1 h1 This result could also be obtained from the energy balance relation for an isentropic steady-flow process. Next we determine the enthalpies: State 1: State 2: P2 s2 h1 s1 P1 100 kPa (sat. vapor) 1 MPa s1 2675.5 kJ/kg 7.3594 kJ/kg · K h2 (Table A–5) (Table A–6) 3195.5 kJ/kg Thus, wrev, in (3195.5 2675.5) kJ/kg 520 kJ/kg Discussion Note that compressing steam in the vapor form would require over 500 times more work than compressing it in the liquid f...
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