Engineering Equation Solver Ees Cengel Thermo Iso May 2026

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.

"Given" P1 = 100 [kPa] T1 = 300 [K] P2 = 1000 [kPa] Fluid$ = 'Air' "EES treats as ideal gas with var cp" s1 = entropy(Fluid$, P=P1, T=T1) "Isentropic" s2 = s1 T2 = temperature(Fluid$, P=P2, s=s2) h1 = enthalpy(Fluid$, T=T1) h2 = enthalpy(Fluid$, T=T2) Engineering Equation Solver EES Cengel Thermo Iso

"Isothermal boundary work for ideal gas" W_b = m R T ln(v2/v1) "Negative if compressed" "Alternatively:" W_b = m R T ln(P1/P2) x = (v - v_f)/(v_g - v_f) "Or

"Steady-flow compressor work" w_comp_in = h2 - h1 "kJ/kg" "Given" P1 = 100 [kPa] T1 = 300

Q_in = m*(u2 - u1) "W=0" Cengel use: Often for ideal gas (( Pv = RT )) or phase change (but constant T in two-phase region implies constant P).

"Isentropic expansion" s2 = s1 h2s = enthalpy(Fluid$, P=P2, s=s2) T2s = temperature(Fluid$, P=P2, s=s2) x2s = quality(Fluid$, P=P2, s=s2) "If in two-phase"

| 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) |