Engineering Equation Solver Ees Cengel Thermo Iso May 2026

"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)

"Isentropic turbine work" W_s = h1 - h2s "kJ/kg"

"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. 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.

"Given final state: superheat to T2" T2 = 80 [C] v2 = volume(Fluid$, P=P2, T=T2) u2 = intEnergy(Fluid$, P=P2, T=T2) h2 = enthalpy(Fluid$, P=P2, T=T2) "Isothermal boundary work for ideal gas" W_b =

P1 = 300 [kPa] T1 = 60 [C] m = 0.5 [kg] Fluid$ = 'Water' v1 = volume(Fluid$, P=P1, T=T1) u1 = intEnergy(Fluid$, P=P1, T=T1) s1 = entropy(Fluid$, P=P1, T=T1)

"1st law" Q_in - W_b = m*(u2 - u1) Rule: ( v_1 = v_2 ), ( W_b = 0 ), ( Q = \Delta U ). Find final temp and work

"Closed system boundary work" W_b = m * P1 * (v2 - v1) "kPa*m^3 = kJ"