1. The problem statement, all variables and given/known data A rigid vessel of volume 2.5 m3 contains steam at a pressure of 6 bar and a dryness fraction of 0.75. If the contents of the vessel are heated, determine: i) the pressure at which the steam becomes dry saturated, ii) the temperature of the steam when the pressure reaches 10.0 bar, iii) the heat transfer to the steam in attaining state b). 2. Relevant equations Just notation Vg= specific volume of saturated vapour 3. The attempt at a solution So i'm just really confused about this question Firstly its rigid so i know that the volume is a constant. So surely when the steam becomes dry saturated M*Vg = 2.5m^3 To find mass i did 0.75 x Vg at 6 bar, i found this in steam tables to be 0.3156 Vg = 0.2367 so mass of wet vapour = 2.5/0.2367 = 10.56 Kg, but mass of the whole contents is 10.56/0.75 = 14.08 kg When the steam becomes dry saturated M*vg = 2.5 so 2.5/14.08 = Vg Vg= 0.1775, vg@11 bar = 0.1774 so i made the conclusion that it must be at 11 bar. I then took a glance at my teachers solution he put up today, he has done something a lot different. He stated that Vg=0.75x0.3156 as did i. ∴Vg = 0.2367 He then stated that this Vg lies between 8 and 9 bar on the steam tables So he did this: @9 bar vg = 0.2149 @8 bar vg= 0.2403 P=0.2403-0.2367/0.2403-0.2149 = 0.14 So P = 8.14 Bar. I really don't follow what he's done here, i don't see any equation i've been given in this form.