Solucionario Cap3 Shapiro 7 Ed

PROBLEM 3.1 H₂O + Air system boundary • two phases are present (liquid and gas). • not a pure substance because composition is different in each phase. H₂O + Air system boundary • three phases are present (solid, liquid, and gas). • not a pure substance because composition of gas phases is different than that of the solid and liquid phases. PROBLEM 3.2 two phases system boundary Vapor Oxygen Liquid Oxygen Vapor Liquid The system is a pure substance. Although the liquid is vaporized, the system remains fixed in chemical composition and is chemically homogeneous. PROBLEM 3.3 Liquid water Liquid water ice The system is a pure substance. Although the phases change, the system remains of fixed chemical composition and is chemically homogeneous. PROBLEM 3.4 air in room dish of water The system is not a pure substance during the process since the composition of the gas phase changes as water evaporates into the air. Once all of the water evaporates, the gas phase comes to equilibrium and the composition becomes homogeneous. At this point, the gas phase can be treated as a pure substance. PROBLEM 3.5 P 80 lbf/in² 312.07°F T 312.07°F P 80 lbf/in² 400°F 312.07°F T 400°F p = 80 lbf/in² T = 312.07°F two-phase liquid-vapor mixture p = 80 lbf/in² T = 400°F superheated vapor T = 400°F p = 360 lbf/in² subcooled (compressed) liquid P 70 lbf/in² 320°F T 320°F T = 320°F p = 70 lbf/in² superheated vapor P 89.6 lbf/in² 70 lbf/in² 312.07°F T 14.7 lbf/in² 0.030 lbf/in² 100°F p = 14.7 lbf/in² T = 100°F solid PROBLEM 3.6 (a) P 5 bar 151.9°C T 151.9°C p = 5 bar T = 151.9°C two-phase liquid-vapor mixture (b) P 5 bar 200°C 151.9°C T 200°C p = 5 bar T = 200°C superheated vapor (c) P 25 bar 15.54 bar 200°C T 200°C p = 2.5 MPa = 25 bar T = 200°C subcooled (compressed) liquid (d) P 6.178 bar 4.8 bar 160°C T 160°C p = 4.8 bar T = 160°C superheated vapor (e) P 100 kPa 21.76 kPa -12°C T -12°C p = 1 bar = 100 kPa T = -12°C solid PROBLEM 3.13 P T Rc = 22.09 MPa 400°C T @ = 374.15°C 20MPa 365.98°C 2MPa 212.42°C 40°C v ( a ) Table A-4 vf = 0.994 x 10 m /kg ( b ) Table A-5 vf = 0.29992 x 10 m /kg ( c ) Table A-2 - using Eq. 3.11 v'(Tp) v '( T ) = 1.0078 x 10 m'/kg PROBLEM 3.14 p 5 bar 1 bar 5 bar ( Table 4 ) 120°C P 1 . 985 bar ( Table A-2 ) Table A-2 ' @ 120 ° C : vf = 1.0603 m/kg vg = 0 . 7911 m'/kg Note : Since ( b ) is in two - phase liquid - vapor state , vf is not fixed by T = 120 ° C and p : 1.985 bar . 1 bar p V 1.985 bar T 120°C 120°C 151.9 ° 2 C