Engineering Equation — Solver Ees Cengel Thermo Iso

v2 = v1 "Final pressure given" P2 = 500 [kPa] T2 = temperature(Fluid$, P=P2, v=v2) u2 = intEnergy(Fluid$, P=P2, v=v2)

| Cengel Table | EES function | |--------------|---------------| | Saturated water T | v_f = volume(Water, T=T_sat, x=0) | | Saturated water P | h_g = enthalpy(Water, P=P_sat, x=1) | | Superheated | v = volume(R134a, T=T, P=P) | | Compressed liquid approx | h(T,P) ≈ h_f@T in EES: h = enthalpy(Fluid$, T=T, P=P) (EES corrects) | Engineering Equation Solver EES Cengel Thermo Iso

x = (v - v_f)/(v_g - v_f) "Or directly:" x = quality(Fluid$, P=P, h=h_mix) | Mistake | Correction | |---------|-------------| | Forgetting units | Use [kPa] , [C] , [kJ/kg] in comments or EES unit system | | Using P*v = R*T for steam | Use v = volume(Steam, P=P, T=T) | | Isentropic but wrong fluid | s2 = s1 only if reversible & adiabatic | | Confusing W_b sign | EES doesn’t enforce sign convention; write Q - W = ΔU | | Not initializing variables | EES solves iteratively; provide guesses if needed: T2 = 300 | 6. Example Problem: Cengel 7-41 (Isentropic Compression) Problem: Air at 100 kPa, 300 K is compressed isentropically to 1 MPa. Find final temp and work. v2 = v1 "Final pressure given" P2 =

P1 = 3 [MPa] T1 = 400 [C] P2 = 50 [kPa] Fluid$ = 'Steam' s1 = entropy(Fluid$, P=P1, T=T1) h1 = enthalpy(Fluid$, P=P1, T=T1) P1 = 3 [MPa] T1 = 400 [C]

"1st law for ideal gas isothermal: Δu=0" Q_in = W_b Most powerful in EES – just set ( s_2 = s_1 ) and EES finds the rest.