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Engineering Equation Solver Ees - Cengel Thermo Iso

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)

"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"

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

"1st law" Q_in - W_b = m*(u2 - u1) Rule: ( v_1 = v_2 ), ( W_b = 0 ), ( Q = \Delta U ). Engineering Equation Solver EES Cengel Thermo Iso

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).

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)

"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) P1 = 300 [kPa] T1 = 60 [C] m = 0

P1 = 200 [kPa] P2 = P1 T1 = 25 [C] m = 1 [kg] Fluid$ = 'R134a' v1 = volume(Fluid$, P=P1, T=T1) u1 = intEnergy(Fluid$, P=P1, T=T1) h1 = enthalpy(Fluid$, P=P1, T=T1)

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

"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$, P=P2, T=T2) T=T1) u1 = intEnergy(Fluid$

| 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) |

"Steady-flow compressor work" w_comp_in = h2 - h1 "kJ/kg"