2-46
Since this is a rigid tank, it is a constant volume process. The constant volume can be found from v = V/m = 2.5 m3/5 kg = 0.5 m3/kg. When liquid is completely vaporized, it is saturated vapor so v = vg. If you look on the saturated table, the vg that corresponds to 0.5 m3/kg occurs at about 140 °C.

2-50
a) Initially there are two phases so the T has to be Tsat at 800 kPa = 170.43 °C.
b) Add the masses of each phase to get total mass. The mass of each phase is found by dividing the volume of each phase by its specific volume. So mf = Vf/vf = 0.1 m3/0.001115 m3/kg = 89.69 kg. The mg = 0.9m3/0.2404 m3/kg = 3.74 kg. The total mass is 93.43 kg.
c) At the final state, water is superheated vapor so V = mv. Look up the specific volume in the superheated tables at 800 kPa and 350 °C. So V=(93.43 kg)(0.3544 m3/kg) = 33.1 m3.

2-54
This problem is just a matter of looking things up in the tables. Remember that the saturated liquid approximation for a compressed liquid involves the saturated liquid value of that property at the given temperature. Listed below are the two different comparisons.

Property        Sat. liquid @ 100 °C    Comp. liq @ 15 Mpa & 100 °C     Percent error
v (m3/kg)       0.001044        0.0010361       0.76
u(kJ/kg)        418.94  414.74  1.01
h (kJ/kg)       419.04  430.28  2.61

2-56
b) At its final state, the device contains saturated mixture so the T = Tsat at 1 MPa = 179.91 °C.

c) To find the change in volume, you need the specific volumes at the initial and final states since DV = m(v2 - v1). Look in superheated tables at 1.0 MPa and 300 °C for the initial specific volume. For the final specific volume, the quality is 0.5, since half is liquid and half vapor. So v2 = vf + x2vfg = 0.001127 + 0.5*(0.19444-0.001127) = 0.0978 m3/kg. The total mass is 0.8kg, so putting numbers in gives a volume change of -0.128 m3. You would expect the volume to decrease since the water is condensing.

2-59
The maximum vapor pressure in the air is the saturation pressure at the given temperature. The Psat of water at 30 °C is 4.246 which is less than 5.2 kPa. Therefore the claim is false.

2-104
To find the volume of the tank, you need to find the final volume (which is the volume of the tank). So V = mv2. To find specific volume, look it up in superheated tables. To find mass, go back to the initial state where you are given the volume, and the phase description, saturated liquid at 0.8 MPa. So m = V1/v1 = 0.01 m3/0.0008454 m3/kg = 11.83 kg. At P = 200 kPa and T = 25°C (the final conditions), v = 0.11625 m3/kg. So Vtank = (11.83 kg)(0.11625 m3/kg) = 1.375 m3.