Adiabatic flash · energy closes the problem
Evaporation cools.
Energy decides.
Specify outlet pressure and a compressed-liquid feed temperature. Solve for outlet temperature, compositions and vapor amount with no heat transfer.
Outlet equilibrium
Teal: equilibrium outlet enthalpy. Amber: specified feed enthalpy. The intersection satisfies Q = 0. At pure or azeotropic saturation, an enthalpy interval at one temperature can fix the vapor fraction.
Follow the energy calculation Outer T iterations and inner TP equilibrium
Bracket T over 300–390 K. At every trial temperature, solve the TP flash and verify global binary tangent support. Compare outlet enthalpy with the feed, then update the temperature bracket. A state outside the property range is rejected rather than extrapolated.
Caloric model and thermodynamic consistency Explicit synthetic assumptions
Pure liquid reference enthalpies are zero at 350 K. Cp₁ = 100 and Cp₂ = 120 J mol⁻¹ K⁻¹, identical in liquid and vapor. Vaporization enthalpies are 30,000 and 35,000 J mol⁻¹, the same constants used in the Clausius vapor-pressure laws of Lab 01.
hᵢV = hᵢL + Δhvap,ᵢ
hE = −RT² ∂(gE/RT)/∂T = 0
Dimensionless Margules A or NRTL τ and α are held constant. Thus gE/RT has no explicit temperature dependence and the consistent excess enthalpy is zero, even when activity coefficients differ from one. This is a teaching model, not experimental calorimetry. It must not be applied to a temperature-dependent fit without its excess-enthalpy contribution.
Equal liquid/vapor Cp keeps Δhvap constant with T. Neglected effects: Poynting correction, pressure contribution to liquid enthalpy, heat loss, shaft work, kinetic and potential energy. Nonconvex liquid parameters are rejected; this lab does not solve energy-balanced VLLE.
LearnChemE: material and energy balances for an adiabatic flash. Calculation source · Regenerate figure.