Flash distillation (sometimes called "equilibrium distillation") is a single stage separation technique. A liquid mixture feed is pumped through a heater to raise the temperature and enthalpy of the mixture. It then flows through a valve and the pressure is reduced, causing the liquid to partially vaporize. Once the mixture enters a big enough volume (the "flash drum"), the liquid and vapor separate. Because the vapor and liquid are in such close contact up until the "flash" occurs, the product liquid and vapor phases approach equilibrium.

Simple flash separations are very common in industry, particularly petroleum refining. Even when some other method of separation is to be used, it is not uncommon to use a "pre-flash" to reduce the load on the separation itself.

Flash calculations are very common, perhaps one of the most common Ch.E. calculations. They are a key component of simulation packages like Hysis, Aspen, etc.

When designing a flash system it is important to provide enough disengaging space in the drum. Drums can also be designed as cyclone separators.

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Modeling assumptions:

- No heat losses to surroundings
- Ideal gas behavior for vapor
- Perfect mixing

Total material balance:

Component Balance (for a binary system):

Enthalpy Balance:

These modeling equations can be reduced to their steady state form by setting the time derivative to zero. In the following, let F,V, and L be mass flow rates.

We'll consider a couple different rearrangements of the steady state model. These will be useful in some of the solution methods we might try.

First, define the *fraction vaporized* of the feed as
so that the material balance can be rewritten as:

or

Later in the term, we will define a "quality" variable *q* to be
equal to the fraction liquid in the feed, so that *f + q = 1*.

Our second rearrangement will enable us to examine the relationships between the material flows and the compositions or enthalpies. First, solve the material and component balances simultaneously:

The material and enthalpy balances can be combined in the same fashion to obtain:

These equations thus define operating lines in terms of the compositions and the enthalpies, respectively.

To solve a flash distillation problem, one simultaneously solves the operating and equilibrium equations. Flash calculations can be solved directly, but usually require an iterative solution. Graphical techniques are also common. Often, the choice of technique depends on the available form of the equilibrium relationship.

**EXAMPLE:** A mixture of 50 mole % normal heptane and 50% normal
octane at 30 degrees C is continuously flash distilled at 1 standard
atmosphere so that 60 mol% of the feed is vaporized. What will be the
composition of the vapor and liquid products?

*Given:* x_{F}=0.5, f=0.6

*Find:* x, y

*Basis:* F=100 mols

Applying the mass balance yields:

The solution method really depends on what form the equilibrium data takes. If
you have an equilibrium *xy* diagram, the problem can be solved graphically by
plotting the operating line on the equilibrium diagram. The operating line is:

Sometimes, the scale of the equilibrium diagram is such that it is tricky to
locate the intercept. In that case, it is usually easier to use the slope and
some other point. The easiest other point to find it that where the 45 degree
line is crossed (y = x) and x = x_{F}, or the point (x_{F},
x_{F}). It is easy to show that this point satisfies the operating
equation (just substitute in x_{F} for both x and y).

If you then plot the operating line on the equilibrium diagram, you can read the coordinates where the two curves cross for the solution

Mathcad Example -- Flash Distillation Calculations

Hydrocarbons and water usually can be assumed to be completely immiscible for the purposes of flash calculations (one exception: high temperature systems with small amounts of water).

In this case, each liquid phase acts independently of the other and each immiscible phase (HC, W) has its own vapor pressure. The total vapor pressure is thus the sum of the vapor pressures of each phase:

Consider a stream containing water (x_{W} = 0.1) and mixed hydrocarbons
(x_{HC} = 0.9) at 80 Celsius. This will split into two immiscible
phases. The vapor pressure of the water phase will be determined by the
temperature only (so it will be about 355 mmHg).

If a second stream with x_{W} = 0.7 and x_{HC} = 0.3 at the same
temperature is considered, the vapor pressure exerted by the water phase will be
the same as well, because the water phase composition is essentially the same
(all water). Consequently, we can see that the bulk feed composition doesn't
really effect the water vapor pressure.

Dew point calculations depend on partial pressures more than vapor pressures; consequently, they don't benefit from the immiscibility. When a hydrocarbon-water mixture is cooled, a temperature will be reached where one component will begin to condense. Note however, that typically only one component will condense initially -- the water and the hydrocarbon mixture must be checked separately when determining dewpoints.

To reiterate, the relative amounts of hydrocarbon and water are unimportant in a bubble point calculation, because they depend on the vapor pressures of the immiscible phases, not on the bulk composition. Dew point calculations, however, are effected by the bulk composition.

- Smith, B.D., Design of Equilibrium Stage Processes, McGraw-Hill, 1963, pp. 105-6, 108-110.
- McCabe, W.L., J.C. Smith, P. Harriott, Unit Operations of Chemical Engineering, 5th Edition, McGraw-Hill, 1993. pp. 521- 525.
- Treybal, R.E., Mass-Transfer Operations, 3rd Edition (Reissue), McGraw-Hill, 1987. pp. 346, 348-349, 357-360, 363-365.

R.M. Price

Original: 7 January 1997

Modified: 12 January 1998, 15 January 1999; 13 February 2003

Copyright 1998, 1999, 2003 by R.M. Price -- All Rights Reserved