Section Incidence Tables

The following equations represent the vapour-liquid equilibrium for a two component mixture, see notes (section 1.2.2) .
P*1 - P*1(T) = 0
P y1 - x1 P*1 = 0
P*2 - P*2(T) = 0
P y2 - x2 P*2 = 0
x1 + x2 - 1 = 0
y1+ y2 - 1 = 0
For the purposes of setting up the incidence table, these equations may be written using the vapour mole fractions (y), instead of the partial pressures (p) in the following manner, where: pi = Pyi
P*1 - P*1(T) = 0 (1)
p1 - x1 P*1 = 0 (2)
P*2 - P*2(T) = 0 (3)
p2 - x2 P*2 = 0 (4)
x1 + x2 - 1 = 0 (5)
p1+ p2 - P = 0 (6)
As will be seen later, we are interested in the structure of sets of equations like this, in particular in which unknown variables occur in which equations. Suppose that in a given situation we know the temperature T so that the the vapour pressure functions in equations (1) and (3) may be evaluated. Suppose also that the mol fraction of one of the species is given in the liquid phase, e.g. x1.

The remaining unknowns in the equations are then:

P , P*1 , p1 , P*2 , x2 , p2

We can draw up an incidence or occurrence table for the set of equations as shown below, with a row for each equation and a column for each unknown, where an `x' indicates that the specified variable occurs in the equation and a blank that it does not.

Variable: P P*1 p1 P*2 x2 p2
Equation (1) x
Equation (2) x x
Equation (3) x
Equation (4) x x x
Equation (5) x
Equation (6) x x x

We might want to write this in `mathematical' notation we would replace the x's by ones, leaving the blanks, which are conventionally taken to imply zeros. The matrix would then look like this:

We explain the use of this type of matrix or table for ordering equations in section To summarise briefly, the matrix shows that equations (1), (3) and (5) contain only a single unknown each, and so can be solved for each of these, i.e. respectively for P*1 , P*2 and x2. It can then be seen that equation (2) now contains only one remaining unknown, p1 as P*1 is no longer unknown, and so it can be solved. In fact all the remaining unknowns can now be determined by solving equation (4) and then (6) each for a single unknown.

Use of the matrix has shown how a problem which might have been thought to involve the simultaneous solution of 6 equations in six unknowns actually involves solving only one equation at a time.

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