Is the surrogate isothermal flash for an adiabatic flash unique?

Question

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1. What the question is really asking

You have two inlet streams, S₁ and S₂, that are mixed adiabatically (no heat is added or removed) at a common pressure P.
The mixture leaves the mixer as a single stream S′ that will later “flash’’ into a vapor stream V and a liquid stream L while staying at the same pressure P.

Because the mixer is adiabatic, the only way to determine the final thermodynamic state of S′ is to satisfy both

the Rachford–Rice equation (RRE) – which tells you, for any guessed temperature T, what vapor fraction β(T) would be in equilibrium at that T and P, and
the overall energy balance – which forces the enthalpy of the equilibrium split (liquid + vapor) to equal the total enthalpy that entered the mixer.

The usual engineering practice is therefore an iterative “isothermal‑flash surrogate’’:

Guess a temperature T.
Solve the RRE at (P, T) → obtain β(T).
Compute the enthalpy of the split streams (using β(T) and the K‑values).
Compare that enthalpy with the known inlet enthalpy; adjust T and repeat.

The core question is:

Does this procedure always converge to a single (unique) solution, or could there be two different temperatures (and thus two different equilibrium splits) that both satisfy the energy balance?

In other words: Is the surrogate isothermal flash for an adiabatic flash unique?

2. Full step‑by‑step analysis

Below we work under the standard assumptions stated in the problem:

Ideal‑gas vapor phase and ideal‑solution liquid phase.
Constant pressure P throughout the flash.
Known overall mole fractions zᵢ (obtained by mixing the two inlet streams).
Known overall enthalpy H_in (sum of the enthalpies of the two inlet streams).
Known temperature‑dependent K‑values Kᵢ(T) (e.g. from Wilson, NRTL, etc.).

2.1 Write the two governing equations

Rachford–Rice (mass‑balance) equation – for any temperature T

[ f(T,\beta)=\sum_{i=1}^{N}\frac{z_i\bigl(K_i(T)-1\bigr)}{1+\bigl(K_i(T)-1\bigr)\beta}=0 \tag{1} ]

For a fixed T, (1) is a monotonic function of β that has a single root in the interval ([0,1]) (provided at least one component is more volatile than the other). Hence for every T we obtain a unique vapor fraction β(T).
Energy‑balance equation – the enthalpy of the equilibrium split must equal the inlet enthalpy

[ H_{\text{calc}}(T) \equiv \beta(T) H_V(T) + \bigl[1-\beta(T)\bigr] H_L(T) = H_{\text{in}} \tag{2} ]

where

[ H_V(T)=\sum_i y_i(T) \, \bar h_i^{\,V}(T) ,\qquad H_L(T)=\sum_i x_i(T) \, \bar h_i^{\,L}(T) ]

and the equilibrium compositions are

[ y_i(T)=\frac{K_i(T)z_i}{1+\bigl(K_i(T)-1\bigr)\beta(T)},\qquad x_i(T)=\frac{z_i}{1+\bigl(K_i(T)-1\bigr)\beta(T)} . ]

The molar enthalpies (\bar h_i^{\,V},\; \bar h_i^{\,L}) are smooth, monotonic functions of T for ideal phases (they are linear in T if constant‑Cp is assumed).

2.2 Reduce the problem to a single scalar function

Define a residual function that measures the mismatch between the calculated and the known enthalpy:

[ \Phi(T) \equiv H_{\text{calc}}(T) - H_{\text{in}} . \tag{3} ]

Finding the adiabatic flash state is therefore equivalent to solving

[ \boxed{\Phi(T)=0} ]

with T the only unknown.

All the other quantities (β, xᵢ, yᵢ, H_V, H_L) are functions of T obtained from the steps above.

2.3 Prove that (\Phi(T)) is monotonic

We need to show that Φ(T) is strictly monotonic (either always increasing or always decreasing) over the physically relevant temperature range ([T_{\min},T_{\max}]). If that holds, the equation (\Phi(T)=0) can have at most one root → uniqueness.

2.3.1 Derivative of β(T)

From the implicit function theorem applied to (1):

[ \frac{d\beta}{dT}= -\frac{\partial f/\partial T}{\partial f/\partial \beta}. ]

The denominator (\partial f/\partial \beta) is negative because each term in (1) has the form

[ \frac{z_i (K_i-1)}{\bigl[1+(K_i-1)\beta\bigr]^2} ]

which is positive; the sum multiplied by a minus sign gives a negative denominator.

The numerator (\partial f/\partial T) contains (\partial K_i/\partial T). For most real components, the K‑value decreases with temperature (the more volatile component becomes relatively less volatile as temperature rises). Hence (\partial K_i/\partial T <0) for the lighter components and (>0) for the heavier ones, but the net sum is negative for a typical binary or multicomponent mixture that actually flashes. Consequently

[ \frac{d\beta}{dT}>0 . ]

Interpretation: As the guessed temperature rises, the equilibrium vapor fraction β also rises.

2.3.2 Derivative of the enthalpy term

Write (2) as

[ H_{\text{calc}}(T)=\beta(T) \sum_i y_i(T) \bar h_i^{V}(T) + \bigl[1-\beta(T)\bigr] \sum_i x_i(T) \bar h_i^{L}(T). ]

Both (\bar h_i^{V}(T)) and (\bar h_i^{L}(T)) are increasing functions of T (positive heat capacities). The compositions (x_i, y_i) are also smooth functions of T (through K‑values and β). Differentiating (H_{\text{calc}}(T)) gives three positive contributions:

Direct temperature rise of the pure‑component enthalpies → positive.
Increase of β with T → more vapor (usually higher enthalpy) → positive.
Shift of compositions toward the more volatile species (because K‑values move) → also raises the mixture enthalpy (vapor is richer in low‑boiling, high‑enthalpy species).

Hence

[ \frac{d H_{\text{calc}}}{dT} \;>\; 0 \qquad\text{for all } T\text{ in the flashing region}. ]

Since (H_{\text{in}}) is a constant,

[ \frac{d\Phi}{dT}= \frac{d H_{\text{calc}}}{dT}>0 . ]

Thus Φ(T) is strictly increasing.

2.4 Existence of a root

Because the mixture is adiabatically mixed, the inlet enthalpy H_in must lie between the enthalpy of the all‑liquid state (β = 0) and the all‑vapor state (β = 1) evaluated at the same pressure. Define

[ \Phi_{\text{liq}} \equiv H_{\text{calc}}(T_{\text{liq}})-H_{\text{in}},\qquad \Phi_{\text{vap}} \equiv H_{\text{calc}}(T_{\text{vap}})-H_{\text{in}}, ]

where (T_{\text{liq}}) is the temperature at which the RRE gives β = 0 (the bubble‑point temperature for the overall composition) and (T_{\text{vap}}) the temperature at which β = 1 (the dew‑point temperature).

At T = T_liq we have β = 0 ⇒ (H_{\text{calc}} = H_L). Since the inlet mixture contains some vapor‑forming energy, (H_{\text{in}} > H_L) and therefore (\Phi_{\text{liq}} < 0).
At T = T_vap we have β = 1 ⇒ (H_{\text{calc}} = H_V). Because the inlet mixture cannot have more enthalpy than a fully vaporized stream at the same pressure, (H_{\text{in}} < H_V) and (\Phi_{\text{vap}} > 0).

Consequently

[ \Phi(T_{\text{liq}}) < 0 < \Phi(T_{\text{vap}}). ]

A continuous, strictly increasing function that changes sign must cross zero exactly once (Intermediate Value Theorem).

Hence there exists a unique temperature T* that satisfies Φ(T*) = 0, and the corresponding β* and equilibrium compositions are unique as well.

2.5 Summary of the logical chain

Step	What we prove	Consequence
1.	For any fixed T, the RRE (1) has a single root β(T) in [0,1]	Vapor fraction is a well‑defined function of T
2.	β(T) is monotonic increasing with T (dβ/dT > 0)	Higher temperature → more vapor
3.	The mixture enthalpy (H_{\text{calc}}(T)) is strictly increasing with T	The residual Φ(T) is monotonic
4.	Φ(T) is negative at the bubble‑point temperature and positive at the dew‑point temperature	Φ(T) must cross zero
5.	Because Φ(T) is continuous and monotonic, the crossing occurs once	Unique solution for the adiabatic flash

3. Final answer

Yes.
Under the usual assumptions of ideal liquid/vapor phases, known K‑values, and a single pressure, the surrogate isothermal‑flash iteration converges to a single, unique thermodynamic state for the adiabatic flash. The uniqueness follows from:

the Rachford–Rice equation giving a unique vapor fraction for each temperature,
the monotonic increase of the calculated equilibrium enthalpy with temperature, and
the fact that the inlet enthalpy always lies between the enthalpy of the all‑liquid and all‑vapor limits.

Therefore the pair ((T^*,\beta^*)) that satisfies both the RRE and the overall energy balance is unique; no second temperature can produce the same total enthalpy while also meeting phase‑equilibrium constraints.

4. Common mistakes when tackling this type of problem

Mistake	Why it is wrong	How to avoid it
Assuming β is independent of T.	β is defined by the RRE; it changes with temperature because K‑values change.	Always solve the RRE for β after picking a temperature; treat β(T) as a function.
Treating the enthalpy balance as linear in T.	The enthalpy of a mixture involves both temperature‑dependent pure‑component enthalpies and composition changes (via K‑values).	Keep the full expression (2) and differentiate it; note the three positive contributions to dH/dT.
Neglecting the sign of ∂f/∂β.	A wrong sign leads to an incorrect conclusion about monotonicity of β(T).	Write out ∂f/∂β explicitly; it is negative because each term is a positive fraction divided by a squared denominator.
Checking only one temperature limit (e.g., only bubble point).	You need both limits to guarantee that Φ changes sign.	Evaluate Φ at both the bubble‑point (β = 0) and the dew‑point (β = 1) temperatures.
Assuming ideal‑gas heat capacities are zero.

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