Thermodynamics

Chapter 4 | 177Assumptions 1 The system is stationary and thus the kinetic and potentialenergy changes are zero, KE PE 0 and E U. 2 The direction ofheat transfer is to the system (heat gain, Q in ). A negative result for Q in indicatesthe assumed direction is wrong and thus it is a heat loss. 3 The volumeof the rigid tank is constant, and thus there is no energy transfer asboundary work. 4 The water temperature remains constant during theprocess. 5 There is no electrical, shaft, or any other kind of work involved.Analysis We take the contents of the tank, including the evacuated space, asthe system (Fig. 4–15). This is a closed system since no mass crosses thesystem boundary during the process. We observe that the water fills the entiretank when the partition is removed (possibly as a liquid–vapor mixture).(a) Initially the water in the tank exists as a compressed liquid since its pressure(200 kPa) is greater than the saturation pressure at 25°C (3.1698 kPa).Approximating the compressed liquid as a saturated liquid at the given temperature,we findv 1 v f @ 25°C 0.001003 m 3 >kg 0.001 m 3 >kg1Table A–42Then the initial volume of the water isV 1 mv 1 15 kg2 10.001 m 3 >kg2 0.005 m 3The total volume of the tank is twice this amount:V tank 12210.005 m 3 2 0.01 m 3(b) At the final state, the specific volume of the water isv 2 V 2 0.01 m3 0.002 m 3 >kgm 5 kgwhich is twice the initial value of the specific volume. This result is expectedsince the volume doubles while the amount of mass remains constant.At 25°C:v f 0.001003 m 3 >kgandv g 43.340 m 3 >kg1Table A–42Since v f v 2 v g , the water is a saturated liquid–vapor mixture at the finalstate, and thus the pressure is the saturation pressure at 25°C:P 2 P sat @ 25°C 3.1698 kPa1Table A–42System boundaryP, kPaEvacuatedspacePartition2001H 2 Om = 5 kgP 1 = 200 kPaT 1 = 25 °CQ in3.172vFIGURE 4–15Schematic and P-v diagram for Example 4–6.