Question

Problem 2 Find the missing Properties in the following table: P[kPa] 300 500 T[C] 20 31.86 V[m\/kg) X0.4196 14xe.69. 0.20 SE 11302 X 0.018YZ 200 1400 BL. Pressure from table - Vf + x (wg - Vf) V 0.001404 + 0.9 (0.02267 -0.001404) V-0.018417 cm #3) From table # 2) Temp. from table Bol.2 @ SoOkla = 151.86°C V = Vp + x (vg - Vf). 0.20 = 0.001093 + x (.37489-0.001043) x= 0.53349 1 Ys - Y Y Y (- x.) yov M = 0.63386 +3721-62388 (20 - 150) 100 - 150 V=0.419614 0.60182 -4,00 1073) 111614001073 + x=0.692093

Answer 1

Selution : PCK Pa) T( C] 0.00100|70 20 300 0•53224 151. 831 0.2 500 0.14303 1400 3. 0.8 0.01760 88 8581.9 300 > subcoolid uglon at P= 300 KPa tT=20°. V= 0.0010017 =151.831°C dey at P=500 Kla & Test = 0. 00l09084 mlg Vg = 0.37481 V = 0.2 we Kuow 1 and 0.2 = 0:00109084 fxx (0.37481- O• 0 o]09084) sding on x = 0.53224 buporheat degion at P=140okPa & T= 200c-> x -14 sAas- U =0.14303 so P = 8587.9 KPa sty at Tat = 300°C = 0.00140423 mi kg, Vg= O.021660 m³lkg V +x V4g > V =0•00140423 +o:80.02166-0 o04042) so