Page 165 - Separation process principles 2

130 Chapter 4 Single Equilibrium Stages and Flash Calculations

SOLUTION where the division by 1,000 is to make the terms of the
order of 1. If the computed value of f (Tv] is not zero, the
The operating pressure corresponds to a dew-point condition for entire procedure is repeated for two or more guesses of Tv.
the vapor-distillate composition. The composition of the reilux cor-
responds to the liquid in equilibrium with the vapor distillate at its A plot of f (Tv] versus Tv is interpolated to determine
dew point. The method of false position [8] can be used to perform the correct value of Tv. The procedure is tedious because it
the iterative calculations by rewriting (4-13) in the form involves inner-loop iteration on \I, and outer-loop iteration
on Tv.
Outer-loop iteration on Tv is very successful when Eq. (3)
of Table 4.4 is not sensitive to the guess of Tv. This is the
The recursion relationship for the method of false position is based case for wide-boiling mixtures. For close-boiling mixtures
on the assumption that f {P) is linear in P such that (e.g., isomers), the algorithm may fail because of extreme
sensitivity to the value of Tv. In this case, it is preferable to
do the outer-loop iteration on \I, and solve Eq. (3) of Table 4.4
for Tv in the inner loop, using a guessed value for \I, to initi-
This assumption is reasonable because, at low pressures, K-values ate the process:
in (2) are almost inversely proportional to pressure. Two values
of P are required to initialize this formula. Choose 100 psia and
190 psia. At 100 psia, reading the K-values from the solid lines in
Figure 4.12,
Then, Eqs. (5) and (6) of Table 4.4 are solved for x and y,
0.90 0.10
+
-
f {P) = - - 1.0 = -0.40 respectively. Equation (4-17) is then solved directly for q,
2.0 0.68
since
Subsequent iterations give
k pck), psia K~ KB f IP'k'l
1 100 2.0 0.68 -0.40
2 190 1.15 0.42 +0.02 from which
3 186 1.18 0.425 -0.0020

Iterations are terminated when 1 P(~+~) - P(~+') l/ P(~+') < 0.005.
An operating pressure of 186 psia (1,282 Ha) at the partial con-
denser outlet is indicated. The composition of the liquid reflux is If \Zr from (4-20) is not equal to the value of q guessed to
obtained from xi = zi/Ki with the result
solve (4-18), the new value of II, is used to repeat the outer
Equilibrium Mole Fraction loop starting with (4- 18).
Multicomponent, isothermal-flash, bubble-point, dew-
Component Vapor Distillate Liquid Reflux
point, and adiabatic-flash calculations can be very tedious
Propylene 0.90 0.76 because of their iterative nature. They are unsuitable for
0.10
0.24
1-Butene - - manual calculations for nonideal vapor and liquid mixtures
1 .oo 1 .oo because of the complexity of the expressions for the thermo-
dynamic properties, K, hv, and hL. However, robust algo-
rithms for making such calculations are incorporated into
Adiabatic Flash
widely used steady-state simulation computer programs
When the pressure of a liquid stream of known composition, such as ASPEN PLUS, CHEMCAD, HYSYS, and PROLI.
flow rate, and temperature (or enthalpy) is reduced adiabati-
cally across a valve as in Figure 4.10a, an adiabatic-fash
calculation is made to determine the resulting temperature,
compositions, and flow rates of the equilibrium liquid and
The equilibrium liquid from the flash drum at 120°F and 485 psia
vapor streams for a specified pressure downstream of the
in Example 2.6 is fed to a distillation tower to remove the remain-
valve. For an adiabatic flash, the isothermal-flash calculation
ing hydrogen and methane. A tower for this purpose is often re-
procedure can be applied in the following iterative manner. ferred to as a stabilizer. Pressure at the feed plate of the stabilizer is
A guess is made of the flash temperature, Tv. Then \Zr, V, x, 165 psia (1,138 Ha). Calculate the percent vaporization of the feed
y, and L are determined, as for an isothermal flash, from if the pressure is decreased adiabatically from 485 to 165 psia by
steps 3 through 7 in Table 4.4. The guessed value of Tv valve and pipeline pressure drop.
(equal to TL) is next checked by an energy balance obtained
by combining Eqs. (7) and (8) of Table 4.4 with Q = 0 to give SOLUTION

This problem is most conveniently solved by using a steady-state
simulation program. If the CHEMCAD program is used with