## Answers

Solution a) water at [u= 1800 kolky us internal energy 250°c is We need to find this point subcooled region, or wet region Super heoted region. From steom table Ογ At energy 250°C - T Uf. internal at saturoted liquid line Ug> Internal energy at saturated Vapour line Uf = 1080.7 leg lkg ug = 2601.8 kolleg The given energy is in internal between of Uf and wet region U= 1800 bo kolleg ug this meon it is u. Usta Cug-us) ac ( 260118-1080-7 1800 & 1080.7 et >2-0-4728

Pressure P= Psot at 250°C P= 3976.2 kPa - specific volume V = 1 + xreg- of at 250°c 20 0.001252 +04728 (0.050085 - 0.001252) 0.0243 44 m² 20 leg A fil А T=C Point A represent the water at 250°C and 1800 kg/lag V, 0.024344 U= with = 0.4728 and P= 3976.2 kPa b) R134a at 75°C and P = 1200 kPa of R134a from refrigeration table the saturation temperature is 46.29°C Herc the temperature is 75°C higher thon saturation temperature

point is ic the state Super heated P= 12oolc Pa T= 75°C state In superheotid X = 1 dryness fraction Weo need to T= 75°C and find 2o and u at from superheated 1200lepa table of R134 a. 277.24 + ( 286.75 - 277.21) x (75-70) 10 interpolotion of u voluc between T: 70° and sou to get the u volue of 75°C lu= 281.98 281.98 killeg V- 0.019502 + (0.020529-0.019502) *(75-70) 80-70 interpolotion between 70°c and 80°C to volue at 75°C get V- 0.0200155 melkg

P TA B The at 75°C Point and U- 281.96 leg/leg B represent the R134a 1200 lepa with x=1 and V= 0.0200155 m 55 milleg

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