Flash Distillation


Definition & Purpose:

Flash Distillation

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.


Modeling Equations

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Modeling assumptions:

  1. No heat losses to surroundings
  2. Ideal gas behavior for vapor
  3. Perfect mixing
F, V, and L are the volumetric flows of the feed, vapor product, and liquid product, VT the total volume of the tank.

Total material balance:

'deriv(V;T*?rho?;T,t)=F-V-L
In many cases, the total mass in the vapor phase is much less than that in the liquid phase. One can then approximate the total mass of the system by the liquid phase mass.

Component Balance (for a binary system):

'deriv(V*?rho?*x,t)=F*?rho?;F*x;F-V*?rho?;v*y-L*?rho?;l*x

Enthalpy Balance:

'deriv(V*?rho?*h,t)=F*?rho?;F*h;f-V*?rho?;V*H-L*?rho?;l*h

Steady State Model

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.

F=V+L
F=V*y+L*x
F*h;F+Q=V*H+L*h

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 f=V/F so that the material balance can be rewritten as:

1=f+'quot(L,F)
and consequently 'quot(L,F)=(1-f) . The component balance can then be written as:
x;F=f*y+(1-f)*x
or
y=_'quot(1-f,f)*x+'quot(x;F,f)
which are McCabe et al.'s Equations 21.1 and 21.2. This is an "operating line" equation -- plotted on a graph whose coordinates are vapor composition vs. liquid composition, it yields a straight line. To plot the line, you only need to know the fraction vaporized (to get the slope) and the feed composition (to get the intercept).

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:

(V+L)*x;F=V*y+L*x
V*(x;F-y)=L*(x-x;F)
to get
y = _'quot(L,V)*x + (1+'quot(L,V))*x;F
This is a line when y is plotted against x. This formula can be rearranged to get:
'quot(L,V) = 'quot(x;F - y, x - x;F)

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

H = _'quot(L,V)*h +('quot(L,V)+1)*(h;F+'quot(Q,F))
'quot(L,V)='quot(H-(h:F+'quot(Q,F)),(h;F+'quot(Q,F))-h)

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


Calculation Techniques

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: xF=0.5, f=0.6
Find: x, y
Basis: F=100 mols

Applying the mass balance yields:

V=f*F=(0.6)*(100)=60
L=F-V=100-60=40

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:

y = _'quot(1-0.6, 0.6)*x + 'quot(0.5,0.6) = _'quot(2,3)*x + 'quot(5,6)
and since both slope and intercept are known, the line can be plotted.

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 = xF, or the point (xF, xF). It is easy to show that this point satisfies the operating equation (just substitute in xF 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

Flash Distillation Example
and estimate the solution to be x=0.39, y=0.58.

If an analytical expression is available for the equilibrium curve, one simply needs to set the two equations equal to each other. Say for instance that the system has a constant relative volatility of 2.16. Then the solution occurs when y = yeq, or
_'quot(1-f,f)*x +'quot(x;F, f) = 'quot(?alpha?*x,1+x*(?alpha?-1))
_'quot(0.4,0.6)*x +'quot(0.5,0.6) = 'quot(2.16*x, 1+1.16*x)
Which can be solved to get x=0.386. This can then be substituted back into either original equation to get y=0.576.

Mathcad Example -- Flash Distillation Calculations

Water-Hydrocarbon Systems

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:

pvap;(total)(T) = pvap;(HC)(T)+pvap;(W)(T)
If the total vapor pressure is less than the system pressure, there will not be an appreciable vapor phase unless a noncondensible gas is present to make up the difference.

Consider a stream containing water (xW = 0.1) and mixed hydrocarbons (xHC = 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 xW = 0.7 and xHC = 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.

Now, say that you are considering a case where the hydrocarbon mixture is ternary and equimolar. The vapor pressure exerted by the hydrocarbon phase will be
pvap;(HC)(T) = pvap;(total)(T)-pvap;(W)(T)
where the hydrocarbon vapor pressure is a function of the temperature and the composition of the hydrocarbon phase. This means you can do bubblepoint calculations for the hydrocarbon phase as if there is no water present as long as you use the total hydrocarbon pressure instead of the system pressure in the calculations. Also note that the calculation will be the same for both cases (xHC = .9 and xHC = .3) considered above -- becaus the composition of the hydrocarbon phase is the same in both despite the different amounts of hydrocarbon present.

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.

References:

  1. Smith, B.D., Design of Equilibrium Stage Processes, McGraw-Hill, 1963, pp. 105-6, 108-110.
  2. McCabe, W.L., J.C. Smith, P. Harriott, Unit Operations of Chemical Engineering, 5th Edition, McGraw-Hill, 1993. pp. 521- 525.
  3. 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