Cálculos Termodinâmicos de Propriedades

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PROBLEM 6.5 
 
//(a) 
pa = 0.20 //bar 
sa = 4.3703 // kJ/kg K 
xa = x_sP("Water/Steam", sa, pa) 
ha = hsat_Px("Water/Steam", pa, xa) 
 
 
//(b) 
pb = 10 //bar 
ub = 3124.4 // kJ/kg 
ub = u_PT("Water/Steam", pb, Tb) 
sb = s_PT("Water/Steam", pb, Tb) 
 
//(c) 
Tc = -28 // oC 
xc = 0.8 
pc = Psat_T("R134A", Tc) 
sc = ssat_Px("R134A", pc, xc) 
 
//(d) 
Td = 20 // oC 
sd = 5.0849 // kJ/kg K 
pd = Psat_T("Ammonia", Td) 
sd = ssat_Px("Ammonia", pd, xd) 
ud = usat_Px("Ammonia", pd, xd) 
 
Results 
IT: h = 1431 kJ/kg 
Problem 6.3: h = 1430.55 kJ/kg 
 
Results 
IT: s = 7.762 kJ/kg∙K 
Problem 6.3: s = 7.7622 kJ/kg∙K 
 
Results 
IT: s = 0.7649 kJ/kg∙K 
Problem 6.3: s = 0.7649 kJ/kg∙K 
 
Results 
IT: u = 1332 kJ/kg 
Problem 6.3: u = 1331.94 kJ/kg