Engineering Equation Solver Ees Cengel Thermo Iso

For a nozzle, the stagnation enthalpy remains constant. EES uses the h0 function.

m = 5 [kg] T1 = 20 [C] P1 = 300 [kPa] s1 = entropy(R134a, T=T1, P=P1) T2 = -10 [C] s2 = s1 "Isentropic condition" P2 = pressure(R134a, T=T2, s=s2) "EES iterates internally" Engineering Equation Solver EES Cengel Thermo Iso

"State 2s - Isentropic outlet (The 'Iso' part)" s2s = s1 "Isentropic condition" h2s = enthalpy(Steam, P=P2, s=s2s) T2s = temperature(Steam, P=P2, s=s2s) "Often superheated or wet" For a nozzle, the stagnation enthalpy remains constant

P2 = 182.4 kPa . Done.

For example, to find the enthalpy of steam at a specific pressure and temperature, one merely writes: For a nozzle

However, the most critical feature for a thermodynamics student is the .