Discussion The positive sign indicates that work is done by the system (work output).Discussion The negative sign indicates that work is done on the system (work input).vP(kPa)5001 24-6 The work done during the isothermal process shown in the figure is to be determined.Assumptions The process is quasi-equilibrium. Analysis From the ideal gas equation,P=RTvFor an isothermal process, v1=v2P2P1=(0 .2 m3/kg)600 kPa200 kPa=0 . 6 m3/kgSubstituting ideal gas equation and this result into the boundary work integral producesWb ,out=∫12P dv=mRT∫12dvv¿mP1v1lnv2v1=(3 kg )(200 kPa )(0. 6 m3)ln0 . 2 m30 . 6 m3(1 kJ1 kPa⋅m3)¿−395 . 5 kJThe negative sign shows that the work is done on the system.4-11 Refrigerant-134a in a cylinder is heated at constant pressure until its temperature rises to a specified value. The boundary work done during this process is to be determined.Assumptions The process is quasi-equilibrium. Properties Noting that the pressure remains constant during this process, the specific volumes at the initial and the final states are (Table A-11 through A-13)P1=5 00 kPaSat. liquid¿}¿¿ v1=vf @ 500 kPa=0 . 0008059 m3/kgP2=5 00 kPaT2=70° C¿¿ ¿}¿¿ v2=0 . 052427 m3/kg ¿ ¿Analysis The boundary work is determined from its definition to bem=V1v1=0 . 05 m30 .0008059 m3/kg=62. 04 kgandWb , out=∫12P dV =P(V2−V1)=mP ( v2−v1)¿(62 .04 kg )(500 kPa )( 0. 052427−0 .0008059 )m3/kg(1 kJ1 kPa⋅m3)¿1600 kJDiscussion The positive sign indicates that work is done by the system (work output).PV(m3)P = aV--2 0.10.312H2OV = const.WevT124-21 CO2 gas in a cylinder is compressed until the volume drops to a specified value. The pressure changes during the process with volume as P=aV−2. The boundary work done during this process is to be determined.Assumptions The process is quasi-equilibrium.Analysis The boundary work done during this process is determined fromWb , out=∫12P dV =∫12(aV2)dV =−a(1V2−1V1)¿−(8 kPa⋅m6)(10 . 1 m3−10 . 3 m3)(1 kJ1 kPa⋅m3)¿−53 . 3 kJDiscussion The negative sign indicates that work is done on the system (work input).4-32 An insulated rigid tank is initially filled with a saturated liquid-vapor mixture of water. An electric heater in the tank is turned on, and the entire liquid in the tank is vaporized. The length of time the heater was kept on is to be determined, and the process is to be shown on a P-v diagram.Assumptions 1 The tank is stationary and thus the kinetic and potential energy changes are zero. 2 The device is well-insulated and thus heat transfer is negligible. 3 The energy stored in the resistance wires, and the heat transferred to the tank itself is negligible.Analysis We take the contents of the tank as the system. This is a closed system since no mass enters or leaves. Noting that the volume of the system is constant and thus there is no boundary work, the energy balance for this stationary closed system can be expressed as Net energy transferby heat, work, and massChange in internal, kinetic, potential, etc. energiesEin−Eoutunderbracealignl ¿⏟¿=ΔEsystemunderbracealignl ¿⏟¿¿We , in=ΔU =m(u2−u1) (since Q=KE=PE=0 ) ¿VIΔt=m(u2−u1) ¿¿The properties of water are (Tables A-4 through A-6) P1=150 kPax1=0 . 25¿}¿¿vf=0 . 001053, vg=1 .1594 m3/kguf=466 . 97 , ufg=2052 .3 kJ/kgv2=v1=0 .29065 m3/kgsat . vapor¿ ¿ v1=vf+x1vfg=0. 001053+[0 . 25×(1 . 1594−0. 001053)]=0.29065 m3/kg ¿u1=uf+x1ufg=466 .97 +(0 .25×2052. 3)=980 .03 kJ/kg ¿ ¿¿ ¿}¿¿ u2=ug @ 0 . 29065 m3/kg=2569 .7 kJ/kg ¿¿Substituting,(110 V )(8 A ) Δt=(2 kg )(2569. 7−980 . 03)kJ/kg(1000 VA1 kJ/s)Δt=33613 s=60 . 2 minWater10 L100°C x = 0.123QTv214-35 Water contained in a rigid vessel is heated. The heat transfer is to be determined.Assumptions 1 The system is stationary and thus the kinetic and potential energy changes are zero. 2 There are nowork interactions involved 3 The thermal energy stored in the vessel itself is negligible. Analysis We take water as the system. This is a closed system since no mass entersor leaves. The energy balance for this stationary closed system can be expressed as Net energy transferby heat, work, and massChange in internal, kinetic, potential, etc . energiesEin−Eoutunderbracealignl ¿⏟¿=ΔEsystemunderbracealignl ¿⏟¿¿Qin=ΔU =m (u2−u1) (since KE=PE=0 )¿¿The properties at the initial and final states are (Table A-4) T1=100 °Cx1=0 .123¿}¿¿v1=vf+xvfg=0 . 001043+(0 . 123)( 1 . 6720−0 . 001043)=0 .2066 m3/kgu1=uf+xufg=419 . 06+(0 .123)(2087 .0)=675 .76 kJ/kgT2=15 0 ° Cv2=v1=0.2066 m3/kg ¿¿ ¿}¿¿x2=v2−vfvfg=0.2066−0 . 0010910 .39248−0.001091=0 . 5250u2=uf+x2ufg¿631.66+(0.5250 )(1927 . 4)=1643 .5 kJ/kg¿¿The mass in the system ism=V1v1=0. 100 m30 .2066 m3/kg= 0 . 04841 kgSubstituting,Qin=m(u2−u1)=(0.04841 kg )(1643. 5−675. 76 ) kJ/kg=46 . 9 kJH2OP = const.WpwWevP124-38 A cylinder is initially filled with saturated liquid water at a specified pressure. The water is heated electrically as it is stirred by a paddle-wheel at constant pressure. The voltage of the current source is to be determined, and the process is to be shown on a P-v diagram.Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2 The cylinder is well-insulated and thus heat transfer is negligible. 3 The thermal energy stored in the cylinder itself is negligible. 4The compression or expansion process is quasi-equilibrium. Analysis We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as Net energy transferby heat, work, and massChange in internal, kinetic, potential, etc . energiesEin−Eoutunderbracealignl ¿⏟¿=ΔEsystemunderbracealignl ¿⏟¿¿We,in+Wpw,in−Wb,out=ΔU (since Q =KE=PE=0) ¿We,in+Wpw,in=m(h2−h1) ¿(VIΔt )+Wpw,in=m(h2−h1) ¿ ¿since U + Wb = H during a constant pressure quasi-equilibrium process. The properties of water are (Tables A-4 through A-6) P1=175 kPasat . liquid¿}¿¿h1=hf @ 175 kPa=487 . 01 kJ/kgv1=vf @ 175 kPa=0 .001057 m3/kgP2=175 kPax2=0 .5¿¿ ¿}¿¿ h2=hf+x2hfg=487 .01+(0. 5×2213 . 1)=1593 . 6 kJ/kg ¿m=V1v1=0. 005 m30 . 001057 m3/kg=4 . 731 kg ¿ ¿Substituting,VIΔt +( 400 kJ )=(4 . 731 kg )(1593 .6−487 . 01)kJ/kgVIΔt=4835 kJV =4835 kJ(8 A )( 45×60 s)(1000 VA1 kJ/s)=223 .9 V4-57 The total internal