Monte Carlo Marching Distillation: Simultaneous solution of the MESH equations with no derivatives,...

8/20/2019 Monte Carlo Marching Distillation: Simultaneous solution of the MESH equations with no derivatives, no initial valu…

1/56

Monte Carlo Marching Distillation: Simultaneous solution

of the MESH equations with no derivatives, no initial values,

and no doubt about completion.

Jacob H. LashoverFebruary 29, 2016 Baton Rouge, Louisiana

Abstract: The mathematical solution technique named Monte Carlo Marching

(MCM) by the author (Lashover, 2012) was previously used to solve flashvaporization and dew point problems for multicomponent, non-ideal solutions of

non-electrolytes. It was developed for chemical engineering calculations to avoidtaking the complicated calculus derivatives required for Newton-Rhapson solutions

and to guarantee convergence. This work extends use of the MCM method to thedesign of distillation towers which require significantly more model equations and

variables to describe a separation cascade or collection of contacting stages. Themodel equations which include descriptor variables such as liquid and vapor flow

rates, compositions, and equilibrium temperatures are called MESH equations andcan be solved to accomplish a given separation or to reduce the required amount of

the mass- or energy- separating agent. (Seader, 1998) The equations are put into a

linear programming form called constraint equations which are designed to be

convex. (Conley, 1984) Similar to “Maximum Likelihood” methods, the constraintequations are added to form a convex objective function which can thus be

minimized. (Banbura, 2010) Using the concept of arbitrary hyper-rectangles

(Faloutsos, 1903), the variables are placed in hyper-rectangles which are sized by the boundary values of the variables: 0 to 1 for mole fractions of

compositions, low boiler to high boiler boiling points for bubble points, etc.The values of the variables are simultaneously improved from their lower boundary values toward their higher boundary values by randomly analyzing linear

regions of the hyper-rectangles. (Eckhardt, 1987) A focus factor, FF, is used in

the calculations to reduce the rectangle widths by a power. When the power is

two, the range is bisected. Essentially, the widths of the hyper-rectangles arereduced as the objective function is minimized by better values of the variables

until the rectangles close around the final solution. Successful solutions are

demonstrated for an ideal binary column presented by Cutlip and Shacham (Cutlip,2008) , and a multicomponent, non-ideal column using the temperature and liquid

composition dependent NRTL activity coefficient model. (Renon, 1968)

Multicomponent results for methanol, 2-propanol, and water are compared to aChemSep solution. (Kooijman, 2006) Complete computer code is included.


2/56

Key Words: Algorithm, analytical solution, constraints, mathematical method,

optimization, mathematical modeling, QB64 computer language, calculus

derivatives, convergence, non-linear equations, convex objective function,

distillation, chemical engineering, linear programming, Monte Carlo, NRTL,

and MESH equations.

INTRODUCTION: “Newton’s method is one of the most versatile and robustmethods for solving systems of nonlinear equations. It is, in fact, the method of

choice in ChemSep. Newton’s method works extremely well if the initial estimateof the solution vector is ”near” the actual solution. Outside this ”near” region, the

method may converge slowly, oscillate, or diverge. The question to which there isno simple answer is, ‘What exactly is it near?’ Having said this it must also be

stated that Newton’s method is a much more reliable procedure than repeatedsubstitution. An important issue not addressed here is how to tell when the

problem has been solved; that is, when convergence has been obtained. A properdiscussion of this topic is beyond the scope of this work. “ (Kooijman, 2006)

The above statements in “The ChemSep Book” provide the reasons for thiswork. In addition to these problems, some designers have difficulty in performing

the calculus derivatives of the activity coefficient models required by Newton’smethod. The Monte Carlo Marching method requires no derivatives and no initial

estimates of the solution vector. In fact, the initial values of the variables may be

zeroes. It has always converged, and the value of the objective function clearly

signals when the problem has been solved by diminishing to very near zero andremaining constant at the solution. The binary column solution required 0.71

seconds while the more difficult ternary, non-ideal problem required 45 seconds asa finer solution grid was required. Results tables are included and show a

maximum 0.005 difference with Cutlip and Kooijman’s compositions, less than 0.3

degrees C on temperatures, and less than 1% on the flows. Overall mass and

energy balances close within 10-5. The computer used was a 2009 Dell Inspiron

546 Desktop with 6 GB of RAM, and an AMD Phenom™ 9650 Quad-Core

Processor operating at 2.30 GHz. VBA compatible QB64 programming

language was used. These calculation times can obviously be significantlyreduced by the newer computer processors.

Summarizing in layman’s language, if you’ve ever struggled to obtain

convergence with a Newton-Raphson method, “You’re going to love this!”


3/56

Development of the Monte Carlo Marching model

The Monte Carlo Marching (MCM) method presented here provides robust

convergence performance without derivatives and is insensitive to initial estimatesof the variables. Application of MCM was demonstrated in an earlier paper

(Lashover, 2012) on several non-linear models including Isaac Newton’s cubic

polynomial, y^3 – 2y -5 = 0, (Cajori, 1911) which Newton used to first

demonstrate his Newton procedure.

“Marching”, “Binary-chop”, “Interval-halving” or “Bi-section” is a simple

convergence procedure which guarantees a solution to “bounded” variables. The

MCM method is an extension of this procedure based on work by Stanislaus Ulam

(Ulam, 1991) and William Conley, both mathematicians.1 The author has

extended Ulam and Conley’s work into the field of chemical engineering. Two

examples are presented.

Example No. 1

Consider the solution for probability, Pr, when the odds are known. These

odds are calculated as (1 – Pr) / Pr. This non-linear equation is not solvable using

algebra. The following simple QB64 program using MCM easily solves for Pr.

The probability, Pr, varies from 0 to 1. When solving for O, odds, using

“Marching”, the lower limit, PL is set to zero and the upper limit, PU, is set equal

to 1.0. The initial value of Pr is calculated as Pr = (PL + PU)/ 2 = 0.50. If the first

iteration produces a negative difference, DEL, between the actual value of O, odds,

and the calculated odds, OO, PL is set equal to Pr. PU is left at 1.0. If the

difference, DEL, is positive, PU is set equal to Pr, and PL is left at 0. The next

1 Ulam is credited with invention of the Monte Carlo computation method and co-inventor with Edward Teller, atheoretical physicist, of the H-Bomb. The author was motivated early in his career by meeting Edward Teller in

1963 where they both presented papers and sat on a panel discussion at an ACS meeting in Gatlinburg, Tennessee.

William Conley, mathematics professor at the University of Wisconsin at Green Bay, first melded linear

programming and the Monte Carlo method for typical operations research problems. Conley assisted the author,

one of the original SABER (Society for American Baseball Research) statisticians, with writing a baseball game

simulator which he used to optimize the batting order for Tulane University’s Division I college baseball team. The

program was successfully used by Tulane in the 1980s.


4/56

value of Pr to be tried is again calculated as Pr = (PL + PU)/2 and the process is

repeated. So, if Pr is too low, the lower limit, PL, is raised to Pr as no solution

exists below Pr. Similarly, if Pr is too high, values above PU can be eliminated.

When the “correct” probability is calculated there will be no difference (within

tolerance) between the actual and calculated odds. The error or difference, DEL,eventually becomes very small, and below the acceptable tolerance, TOL. This

method always converges and eliminates one-half of the range of feasible solutions

after each iteration. This can be represented algebraically by the function f(x) = R /

(Xn) where R = the range of solutions, X = 2, and n = 1, 2, 3….n, the number of

the iteration. The limit of this function as n approaches infinity is 0 where the

entire feasible range of solutions has been examined. It does not produce an exact

solution like algebra or calculus can, however it will be shown that this

convergence procedure permits use of constraints which are not easily includedwhen using algebra or calculus.

The QB64 statements demonstrating this method follow:

REM SOLVE NONLINEAR ODDS/PROBABILITY EQUATION

INPUT " ODDS= ", O

LP = 0.0: UP = 1.0: TOL = 0.003

2 PR = (LP + UP) / 2.0

OO = (1 - PR) / PR

DEL = O - OO

IF ABS(DEL)


5/56

ODDS FOR WHICH PROBABILITY IS NEEDED= 2

LP UP PR OO O DEL LP UP CYCLE

0 1 0.5 1 2 +1 0 0.5 1

0 0.5 0.25 3 2 -1 0.25 0.5 2

0.25 0.5 0.375 1.667 2 +0.333 0.25 0.375 3

LP UP PR OO O DEL LP UP CYCLE

0.25 0.3750 0.3125 2.2 2 -0.200 0.3125 0.375 4

0.3125 0.3750 0.3438 1.909 2 +0.091 0.3125 0.3438 5

0.3125 0.3438 0.3281 2.048 2 -0.048 0.3281 0.3438 6

0.3281 0.3438 0.336 1.976 2 +0.024 0.3281 0.3360 7

0.3281 0.3360 0.332 2.012 2 -0.012 0.3320 0.3360 8

0.3320 0.3360 0.334 1.994 2 +0.006 0.3320 0.3340 9

0.3320 0.3340 0.333 2.000 2 0.000 10

P= .333Example No. 2 See Lashover, J. H. (2012). Monte Carlo Marching. Baton Rouge, Louisiana:"Academia.edu" for a more detailed explanation of this example.)

Consider the chemical engineering calculation of the fraction of vapor

flashed, V, from a liquid stream whose composition varies from 0 to 100 % or

from 0 to 1.0 as a fraction. The composition of the liquid stream (feed) is known

as are its temperature and pressure, however the compositions of both the vapor

and liquid streams formed are unknown. To complicate matters further, the

composition of the vapor stream formed is related to the composition of the liquidstream formed by a complex, non-linear relationship, i.e. the composition of the

vapor stream is dependent on the yet unknown liquid stream. The mass of the feed

must also equal the total mass of the resulting liquid and vapor streams

(conservation of mass). There have been many solutions presented, including the


6/56

author's using MCM, but others fail frequently, especially when V is close to 0 or

1.

When solving for V using “Marching”, the lower limit, VL is set to zero and

the upper limit, VU, is set equal to 1.0. The liquid and vapor concentrationsformed are also bounded between 0 and 1 when expressed as mole fractions and

the total molar concentration of each stream must also sum to unity. An overall

mass balance must be satisfied and calculation of the total pressure must equal the

pressure specified. When one mole of feed is flashed and the total pressure is

equal to one atmosphere, all variables can have a lower limit of 0 and an upper

limit of 1. Since the vapor composition is dependent on the liquid composition, the

liquid compositions become the unknown independent variables. With three

components, there are then four unknown independent variables when V is

included. Actually, there are only two unknown compositions as the third equals

1.0 minus the sum of the other two, however, all three are considered unknown to

avoid losing the constraint requiring the sum of the compositions to be unity.

In MCM, the simple marching technique shown previously is applied to the

multiple variables. These unknown variables are essentially improved from their

lower boundary values toward their higher boundary values by analyzing linear

regions which are inverse multiples of powers of two. This procedure emulates

the guaranteed conversion procedure of “Marching” and satisfies the mathematicaltheories of hyper-rectangles. (Faloutsos, 1903) Remember, after the first iteration

of marching, the region left to study has decreased by ½ or 1/ (2^1). After the

second iteration, the region left to study has decreased to ¼ or 1/ (2^2), and so

forth. The feasible region of solutions for V, after being reduced to ½ by searching

either 0.5 to 1.0 or 0 to 0.5 is next reduced to ¼ by searching either 0.75 to 1.0 or 0

to 0.25. By normalizing all independent variables to 0 to 1 or 0 to 100, no one

variable can overly influence the objective function. As shown in the distillation

examples which follow, this normalization has not always been found necessary.

Finally, the objective function is the sum of the convex constraint equations

which are crafted as the absolute differences of the constraint values from the

desired constraint values at each iteration, i.e., for the V calculation, G1 = ABS(Σ

X – 1.0) would be one constraint, G2 = ABS(ΣY – 1.0) would be the second, G3

= ABS(∑P – 1.0) would be the third, and G4 = ABS(∑Z – V*∑Y – (1 – V)*∑X).


7/56


8/56


9/56

Solving for V Using MCM

Computer Output

Iteration 1

V= .283724138736725

OLD OPT= 1D+30 G= 85.29139956077

NEW OPT= 85.29139956077

SUMX= .8772386085987092

SUMY= .9174543451798476

SUMPT= 92.96106152534807

SUMZX= .9911135121617614

G1= 8.363938474651931 G2= 12.2761391401291 G3= 8.254565482015238 G4=

56.3967564639

7374 G= 85.29139956077

X(1)= .2280518072843552 X(2)= .3864084398746491 X(3)= .262778361439705

Y(1)= .3562745191479074 Y(2)= .3508095455125981 Y(3)= .2103702805193421

LL(1)= 1 LL(2)= 1 LL(3)= 1 LL(4)= 1

UL(1)= 100 UL(2)= 100 UL(3)= 100 UL(4)= 100

R(1)= 49 R(2)= 49 R(3)= 49 R(4)= 49

ZX(1)= .4973556831422566 ZX(2)= .146236918257682 ZX(3)= .3475209107618227

PP(1)= 36.09951565266172 PP(2)= 35.54577719906401 PP(3)= 21.31576867362234

COUNTI= 1 COUNTJ= 1


10/56

Final Iteration

V= .9451196705718721

OLD OPT= .108056010212107 G= 8.869474744795088D-02

NEW OPT= 8.869474744795088D-02

SUMX= .9991877469223368

SUMY= .9999986734315671

SUMPT= 101.3248655854536

SUMZX= .990000458304824

G1= 1.344145464514668D-04 G2= .0812253077663172 G3= 1.326568432915565D-04 G4= 7.20236829189065D-03 G= 8.869474744795088D-02

X(1)= .254672596779671 X(2)= 6.184798915739828D-02 X(3)= .6826671609852676

Y(1)= .514203322045091 Y(2)= .1550985014029116 Y(3)= .3306968499835646

LL(1)= 25.43944087766766 LL(2)= 6.15487926417003 LL(3)= 68.2508029011053 LL(4)=

94.5072167416216

UL(1)= 100 UL(2)= 100 UL(3)= 100 UL(4)= 100

R(1)= .048828125 R(2)= .048828125 R(3)= .048828125 R(4)= .048828125

ZX(1)= .495000398096542 ZX(2)= .1485001911742847 ZX(3)= .3464998690339974

PP(1)= 52.10165160621884 PP(2)= 15.71535565465002 PP(3)= 33.50785832458468

COUNTI= 5505 COUNTJ= 12


11/56

Table 2. Calculation of V, fraction of vapor flashed, for a 50, 15, 35 mole % mixtureof MeOH, IPA, & H2O, respectively, at 78 C and 101.325 kPa. using

MCM.

LL(1), LL(2), AND LL(3) are lower limits of liq. phase mole%, LL(4) is lower limit of V.

R()'s are ranges of variables to be searched for feasible solutions.

OPT is last optimum value of objective function.G is new optimum found during present iteration.

X's and Y's are liquid and vapor compositions in mole fraction, respectively.

∑X's and ∑Y's are sums of liquid and vapor compositions.

∑kPa and ∑moles are the sums of the vapor pressures and total moles fed.

Total feed is 100 moles while total pressure is 101.325 kPa.

Iteration I 1 2 32 1034 1408 1638 1680 6224

Iteration J 1 1 1 3 3 4 4 13

LL(1) 1 1 1 22.02 22.18 25.61 20.93 25.45

LL(2) 1 1 3.804 1 1 4.79 1.1775 6.175LL(3) 1 1 1 34.15 33.43 44.16 56.58 68.25

LL(4) 1 1 18.08 24.33 1 66.47 76.95 94.5

R(1) 49 71.81 68.7 25 25 12.5 1.25E+01 2.44E-02

R(2) 49 87.64 96.2 20.61 24.07 12.5 1.25E+01 2.44E-02

R(3) 49 75.28 81.72 25 25 12.5 1.25E+01 2.44E-02

R(4) 49 77.37 12.5 25 25 12.5 1.25E+01 2.44E-02

OPT 1.00E+30 85.29 77.93 25.37 15.58 10.87 6.62E+00 8.87E-02

G 85.29 77.93 29.86 19.04 11.82 10.29 6.33E+00 8.75E-02

V 28.37 68.08 37.54 46.69 67.27 78.16 8.86E+01 9.45E+01

X(1) 0.2281 0.197 0.3865 0.3327 0.326 0.3074 2.64E-01 2.55E-01X(2) 0.3864 0.538 0.1309 0.148 0.1111 0.08579 8.26E-02 6.18E-02

X(3) 0.2628 0.3272 0.3112 0.4206 0.4865 0.5548 6.34E-01 6.83E-01

ƩX"S 0.8773 1.0622 0.8285 0.9012 0.9236 0.9421 9.81E-01 9.99E-01

Y(1) 0.3563 0.2928 0.6558 0.5656 0.5769 0.5687 5.06E-01 5.14E-01

Y(2) 0.3508 0.4568 0.1432 0.1867 0.1644 0.1499 1.76E-01 1.55E-01

Y(3) 0.2104 0.304 0.1926 0.2521 0.2712 0.2887 3.19E-01 3.31E-01

ƩY'S 0.9175 1.0536 0.9916 1.0044 1.0126 1.0073 1.00E+00 1.00E+00

ƩkPa 92.96 106.76 100.47 101.77 102.6 102.07 1.01E+02 1.01E+02

Ʃmoles 99.91 98.94 99.11 99.05 99.01 99 9.90E+01 1.00E+02

Redlich-Kister model for activity coefficients; Raoult's Law for vapor phase


12/56

V X1 X2 X3 Y1 Y2 Y3

Derivative 94.34 0.2579 0.06269 0.6794 0.5145 0.1552 0.3302

MCM 94.56 0.2545 0.06175 0.6831 0.5141 0.1551 0.3308

Δ -0.22 0.0034 0.00094 -0.0037 0.0004 0.0001 -0.0006

Table 3. Comparison of MCM results with results using derivatives


13/56

Figure 2. MCM Logic Diagram


14/56

Figure 3. MCM Solution of Flash Vaporization


15/56

Figure 4. MCM Constraint Values vs. Iteration Number


16/56

USING MCM TO SOLVE DISTILLATION

The distillation column to be specified in the two demonstration problems is

shown below. For MCM, the stages are numbered from top to bottom but with

the condenser being Stage# 0 and the bottom stage being Stage# 3 for both problems. The condenser is assumed to condense all of the vapor that enters it thus

being defined as a “total condenser.” Since the condenser is not an equilibrium

stage, no further enrichment takes place there, the composition of the Distillate, D,

and the Reflux stream, L(0), will be the same as that of the vapor, V(1) leaving

Stage# 1. The feed will enter Stage# 2 which will be numbered Stage# 4 for

purposes of determining its temperature and enthalpy. The Feed, F, the Distillate,

D, and the Bottoms stream, BTMS, are all liquids at their boiling points, therefore

the Heat added to the Reboiler, QR , will be equal to the heat removed by the

Condenser, QC. Thus QR – QC = 0 and the Feed enthalpy, QF, will equal the total

heat removed from the column by the Distillate, QD and the BTMS, QBTMS. Thus

QF - QD - QBTMS = 0 = DELQ. The column will be considered to be at constant

atmospheric pressure with no heat loss and an efficiency of 100 %.

The stage numbers will be represented by S% (% designating an integer)

and the components by I% thus the component liquid compositions will be X(S%,

I%) and their vapor compositions will be Y(S%, I%). The enthalpy of the liquid

streams will be labelled hL (S%) while that of the vapor will named HV(S%).Activity coefficients will be named GAM(S%, I%), while pure component vapor

pressures will be called VP(S%, I%), equilibrium constants = VP(S%, I%) / P will

be named K(S%, I%), and temperatures will be T(S%). Vapor and liquid molar

flows will be V(S%) and L(S%), respectively. Reflux ratio will be RR and Boil-up

ratio will be BR.

The Cutlip problem was selected because it had been solved and would

provide confirmation of the MCM method. Likewise, a multi-component problem

was designed that could be corroborated by comparison with a ChemSep solution.The Cutlip problem was done in English units for direct comparison to the Cutlip

problem while Metric units were used for the ternary non-ideal problem.


17/56

Figure 5. Three Equilibrium Stage Tower


18/56

The Algorithm:

As noted in the examples already shown, the first step in the MCM

method is to define the unknown variables to be solved. Per the Gibbs

phase rule, the following must be specified for each column:

(The numbers in parentheses are the number of degrees of freedom, i.e.,

the number of stages = 3, but it is considered only one degree of

freedom.)

1) The number of stages. (1) There are 3 stages but one is the

reboiler leaving technically 2.

2)

The location of all feeds. (1)

3) The location of all side streams. (0)

4)

For each feed stream, the three component flows (3), and the

pressure and vapor fraction. (2) Remember that there are

two component flows for the binary column.

5) The pressure in each stage. (2) The reboiler and condenser are

counted separately.

6) The heat duty of each stage except condensers and reboilers.

(2) Set at zero—no heat loss.

7) The pressure in the condenser. (1)

8) The pressure in the reboiler. (1)

9)

Specify one variable at the top (condenser). (1) Distillate

flow is given here.

10) Specify one variable at the bottom (reboiler). (1) Heat duty

is given here.


19/56

Therefore, the total number of degrees of freedom is 14 for the

binary column and 15 for the ternary column. For a total

condenser, the liquid phase composition of the condenser, and

hence the reflux and distillate, are equal to the composition ofthe vapor leaving the first stage. The bottoms flow, L(3) leaving

the reboiler is calculated from an overall material balance as

BTMS = F – D, and once the bottoms compositions are known,

the distillate compositions can be calculated from overall

component balances as the feed composition is given. The liquid

and vapor flows from each stage are also related by material

balance thus only one of the liquid and vapor flow set need be

specified. Units of the flows are g-moles/hour. Compositions

are mole fraction, temperatures are in Centigrade, and pressure

is mm Hg. Enthalpies are in calories/g-mole.

Finally, the following variables are selected as the unknowns to

be simultaneously solved:

Vapor flows: V(1), V(2), and V(3) There is no vapor flow,

V(0), from the condenser.

Liquid compositions: For each stage there will be 3 for the

ternary system. Just 2 for the binary column, and in both cases

since they must sum to 1.0 when expressed as mole fractions,

there is one less composition needed after the first one or two

are selected.

X(0, 1), X(0, 2), X(0, 3) Stage# 0Calculated by material balance.2

2 F * Z(1) = XD(1) * D + XB(1) * BTMS

XD(1) = (F * Z(1) - XB(1) * BTMS) / D


20/56

X(1, 1), X(1 ,2), X(1, 3) Stage# 1

X(2, 1), X(2, 2), X(2, 3) Stage# 2

X(3, 1), X(3, 2), X(3, 3) Stage# 3

Liquid flows: L(0), L(1), L(2), and L(3) As noted above L(3)

= BTMS leaving only 3. These are calculated from material

balances once the vapor flows are selected. This leaves 9

unknowns for the binary and 12 for the ternary system when the

bottoms compositions are selected first.

Temperatures: T(0), T(1), T(2), T(3) can be calculated as

the bubble point temperature of each stage once the liquid

compositions are selected. The feed temperature can easily be

calculated from it’s known composition and pressure. It is

designated as T(4) here and along with the given composition is

used to calculate the feed enthalpy, hL. There is no vapor phase

in the feed. T(0) can be calculated as the bubble point of the

vapor compositions from V(1) which are the same as the liquid

compositions in the condenser. MCM is also used to calculatethe bubble points so no temperature guesses or derivatives are

required.

Bounds on these variables are 0 to 1.0 for composition

and low boiler temperature to high boiler temperatures for

boiling points. Since V(1) = L(0) + D and L(0) = RR x D,

selecting a high reflux rate like 20 for the ternary leads to a high

for L(0) of 20 x 30 = 600 moles/hr. or 600 + 30 = 630 for

V(S%) max. The minimum V(1) must be at least 30+ in order

for there to be liquid for reflux. We can then proceed to begin

selecting values for the unknowns and calculating constraints

for the objective function to be minimized.


21/56

CONSTRAINT EQUATIONS

REM STAGE 0 (CONDENSER)G0 = ABS(V(1) - D - L!(0))

G01 = ABS(V(1) * Y(1, 1) - L!(0) * X(0, 1) - D * X(0, 1))

G02 = ABS(V(1) * Y(1, 2) - L!(0) * X(0, 2) - D * X(0, 2))

G03 = ABS(V(1) * Y(1, 3) - l!(0) * X(0, 3) - D * X(0, 3))

H0 = ABS(V(1) * SUMHV(1) - L!(0) * SUMHL(0) - D * SUMHL(0) - QC)

H0 = H0 / 1000

REM STAGE 1

G1 =ABS(L!(0) + V(2) - L!(1) - V(1)) 'OVERALL MOLE BALANCE

G11=ABS(-V(1) * Y(1,1)-L!(1) * X(1,1)+V(2) * Y(2,1)+L!(0) * X(0,1))

G12=ABS(-V(1) * Y(1,2)-L!(1) * X(1,2)+V(2) * Y(2,2)+L!(0) * X(0,2))

G13=ABS(-V(1) * Y(1,3)-L!(1) * X(1,3)+V(2) * Y(2,3)+L!(0) * X(0,3))SUMY(1) = Y(1, 1) + Y(1, 2) + Y(1, 3)

SUMY(1) = ABS(1. - SUMY(1))

H1=ABS(-V(1)*SUMHV(1)+V(2)*SUMHV(2)-L!(1)*SUMHL(1)+L!(0)*SUMHL(0))

H1 = H1 / 1000

REM STAGE 2

G2 = ABS(L!(1) + F + V(3) - L!(2) - V(2))

G21=ABS(L!(1)*X(1,1)-V(2)*Y(2,1)-L!(2)*X(2,1)+V(3)*Y(3,1)+ZF(1)*F)



SUMY(2) = Y(2, 1) + Y(2, 2) + Y(2,3)


H2=ABS(-V(2)*SUMHV(2)+V(3)*SUMHV(3)+SUMHL(4)*F+L!(1)*SUMHL(1)-

L!(2)*SUMHL(2))

H2 = H2 / 1000

REM STAGE 3

G3 = ABS(L!(2) - L!(3) - V(3))

G31 = ABS(L!(2) * X(2, 1) - V(3) * Y(3, 1) - BTMS * X(3, 1))

G32 = ABS(L!(2) * X(2, 2) - V(3) * Y(3, 2) - BTMS * X(3, 2))

G33 = ABS(L!(2) * X(2, 3) - V(3) * Y(3, 3) - BTMS * X(3, 3))

SUMY(3) = Y(3, 1) + Y(3, 2) + Y(3, 3)


H3 = ABS(-V(3) * SUMHV(3) + QR + L!(2) * SUMHL(2) - L!(3) * SUMHL(3))

H3 = H3 / 1000

GOSUB ENERGY

100 G = G01 + G02 + G03 + G11 + G12 + G13 + G21 + G22 + G23 + G31 +

G32 + G33

G = G + H0 + H1 + H2 + H3 + SUMY(0) + SUMY(1) + SUMY(2) + SUMY(3) +

DELQ


22/56

DISCUSSION

Note that the constraint equations are of the convex form

where the right hand side of the equation is subtracted from theleft hand side to result in 0 when all of the variables are correct.

The absolute value of these differences is used and these values

are added to form a convex objective function which can be

minimized. These constrain equations are MESH equations in

convex form.

There are three component material balances and one

energy balance equation for each stage, a total of 12 equationsfor the ternary system, 9 equations for the binary system

matching the number of unknowns. Note that the energy

differences (errors) have been scaled by dividing each of them

by 1000 to make the values more closely match the other errors.

The equations for G0, G1, G2, and G3 are the overall material

balances for each stage, but are redundant as they have already

been used to relate the vapor and liquid flows. An overallenergy balance relating the energy input by the feed and reboiler

to that removed by the distillate, bottoms, and condenser is also

included as are the Sum of the Y’s. Note that the Y’s are used in

the mass balances incorporating all of the unknown variables

into these equations. Experienced distillation simulator users

know that with non-linear equations there are multiple solutions

possible so it is extremely important that the overall mass andenergy balances close. The overall component mass balances

have been forced to close by calculating the distillate

compositions from the bottoms’ compositions using the overall

mass balance, F = D + B, broken into individual component


23/56

balances. The overall energy balance is also redundant to the

stage energy balances but was included for insurance.

The vapor compositions for each stage, except for the

feed and condenser which create no vapor, are calculated duringthe bubble point calculations. The sum of these Y’s must also

be equal to 1.0 and are included in the objective function

calculated during each iteration for the subscript, I. When an

infeasible value such as 0 or (-) for any of the unknown

variables is encountered, the algorithm returns to its start and

tries another solution with I incremented by 1. The subscript ,

J, remains the same until all I’s are tried, then is incremented by1 to tighten the hyper-rectangles around the best solution so far.

J is used as a power of the focus factor, FF, which has been

selected as 2.0 for the n-butane-n-pentane problem, and selected

as 1.3 for the ternary problem to provide a finer grid for

searching. Selection of the proper focus factor is learned from

experience with various problems. The author has solved all

problems so far with FF = 1.3 or FF = 2.0. Computer code for each problem is included in the

Appendices and contains comments to guide the reader.

Intermediate calculation output is also provided so that the

reader can follow the paths of the objective function values as

they are reduced to a minimum and remain constant signifying

the completion of the solution. The PRINT statements can be

bypassed by placing an apostrophe in front of the statement.The intermediate output provided by the PRINT statements is

essential until code debugging and certainty of the correct

MESH equations are completed.


24/56

RESULTS:

I. n-butane & n-pentane from Cutlip & Shacham


25/56

Convergence of n-butane & n-pentane solution

WELCOME TO RIGDIS14.BAS

THE TIME IS 18:05:13 AND THE DATE IS 02-23-2016

COMPONENTS ARE USED IN ORDER OF BOILING POINTS, LOWEST TO HIGHEST

BP'S IN F ARE 31.129 AND 96.87 F, RESPECTIVELY

IST SUBSCRIPT IS STAGE NO., 2ND SUBSCRIPT IS COMPONENT NO.

NEW OPT= 13.56237 I= 2 J= 1

NEW OPT= 8.60752 I= 117 J= 1

NEW OPT= 7.099331 I= 272 J= 1

NEW OPT= 6.90924 I= 8 J= 2

NEW OPT= 6.548571 I= 32 J= 2

NEW OPT= 6.342526 I= 163 J= 2NEW OPT= 2.496014 I= 221 J= 2

NEW OPT= 1.376465 I= 467 J= 3

NEW OPT= 1.261207 I= 294 J= 4

NEW OPT= 1.129066 I= 215 J= 5

NEW OPT= 1.100345 I= 228 J= 5

NEW OPT= .9700624 I= 462 J= 5

NEW OPT= .53333 I= 562 J= 5

NEW OPT= .5306512 I= 106 J= 6

NEW OPT= .4136833 I= 115 J= 6

NEW OPT= .3883325 I= 100 J= 7

NEW OPT= .3694802 I= 126 J= 7

NEW OPT= .3689947 I= 158 J= 7NEW OPT= .3177557 I= 164 J= 7

NEW OPT= .3021875 I= 227 J= 8

NEW OPT= .2983455 I= 484 J= 8

NEW OPT= .2756652 I= 523 J= 8

NEW OPT= .2595588 I= 694 J= 8

NEW OPT= .2451617 I= 717 J= 8

NEW OPT= .240488 I= 7 J= 9

NEW OPT= .2306435 I= 18 J= 9

NEW OPT= .2234408 I= 21 J= 9

NEW OPT= .2156426 I= 68 J= 9

NEW OPT= .2144745 I= 77 J= 9

NEW OPT= .2067866 I= 83 J= 9

NEW OPT= .2043912 I= 87 J= 9

NEW OPT= .2017308 I= 226 J= 9

NEW OPT= .2000239 I= 310 J= 9

NEW OPT= .1970954 I= 311 J= 9

NEW OPT= .1866049 I= 351 J= 9

NEW OPT= .1760544 I= 383 J= 9

NEW OPT= .170598 I= 427 J= 9

NEW OPT= .168051 I= 598 J= 9


26/56

NEW OPT= .1592648 I= 717 J= 9

NEW OPT= .1579064 I= 777 J= 9

NEW OPT= .1510153 I= 884 J= 9

NEW OPT= .1504792 I= 13 J= 10

NEW OPT= .1491413 I= 19 J= 10


NEW OPT= .1417301 I= 76 J= 10

NEW OPT= .1386222 I= 97 J= 10

NEW OPT= .1362681 I= 124 J= 10

NEW OPT= .1357231 I= 186 J= 10

NEW OPT= .1347106 I= 215 J= 10

NEW OPT= .1327947 I= 234 J= 10

NEW OPT= .1302254 I= 281 J= 10

NEW OPT= .12947 I= 395 J= 10

NEW OPT= .1279244 I= 398 J= 10

NEW OPT= .126572 I= 420 J= 10


NEW OPT= .1212975 I= 445 J= 10

NEW OPT= .1199046 I= 447 J= 10

NEW OPT= .11675 I= 475 J= 10

NEW OPT= .1163187 I= 485 J= 10

NEW OPT= .1113952 I= 492 J= 10

NEW OPT= .1082548 I= 493 J= 10

NEW OPT= .1063951 I= 511 J= 10

NEW OPT= .1050479 I= 577 J= 10

NEW OPT= .1043972 I= 599 J= 10

NEW OPT= .1023048 I= 603 J= 10

NEW OPT= .102284 I= 612 J= 10

NEW OPT= .098626 I= 638 J= 10

NEW OPT= 9.856462E-02 I= 970 J= 10

NEW OPT= 9.705922E-02 I= 2 J= 11

NEW OPT= 9.659583E-02 I= 7 J= 11

NEW OPT= 9.657159E-02 I= 40 J= 11

NEW OPT= 9.640448E-02 I= 96 J= 11

NEW OPT= 9.626424E-02 I= 255 J= 11

NEW OPT= 9.609228E-02 I= 473 J= 11

NEW OPT= 9.586024E-02 I= 522 J= 11

NEW OPT= 9.585699E-02 I= 757 J= 11

NEW OPT= 9.571949E-02 I= 17 J= 12

NEW OPT= 9.571089E-02 I= 139 J= 12

NEW OPT= 9.552763E-02 I= 193 J= 12NEW OPT= 9.543725E-02 I= 628 J= 12

NEW OPT= 9.543505E-02 I= 756 J= 12

NEW OPT= 9.529868E-02 I= 816 J= 12

NEW OPT= .0952848 I= 220 J= 13

NEW OPT= 9.524148E-02 I= 340 J= 13

NEW OPT= 9.522845E-02 I= 429 J= 13

NEW OPT= 9.515925E-02 I= 124 J= 14

NEW OPT= 9.514897E-02 I= 483 J= 14


27/56

NEW OPT= 9.513015E-02 I= 602 J= 14

NEW OPT= .0951127 I= 787 J= 14

NEW OPT= 9.509851E-02 I= 96 J= 15

NEW OPT= 9.509684E-02 I= 269 J= 15

NEW OPT= 9.509414E-02 I= 348 J= 15

NEW OPT= 9.506063E-02 I= 360 J= 15

NEW OPT= 9.505975E-02 I= 498 J= 15NEW OPT= 9.505945E-02 I= 991 J= 15

ELAPSED TIME= .71484375 SECONDS

OPT= 9.505945E-02 I= 1001 J= 16

G01= 3.439188E-05 G02= 3.308803E-05 H0= 8.758545E-03

G11= 9.143346E-05 G12= 5.759504E-05 H1= .0274367

G21= 1.297146E-04 G22= 9.071827E-05 H2= 3.665925E-03

G31= 4.639508E-05 G32= 4.095277E-05 H3= 4.650269E-02

SUMY1= 1.192093E-06 SUMY2= 3.20673E-05 SUMY3= 5.245209E-06

DELQ= 8.636401E-03 QQC= 9913.636

OVERALL COUNT= 15000


28/56

II. MeOH-IPA-H2O comparison with ChemSep


29/56

Convergence of MeOH-IPA-H2O SolutionWELCOME TO DISMCM07.BAS

THE TIME IS 17:16:50 AND THE DATE IS 02-23-2016

COMPONENTS ARE USED IN ORDER OF BOILING POINTS,LOW TO HIGH

COMPONENTS ARE METHANOL, 2-PROPANOL, AND WATER

BP'S IN C ARE 64.55, 82.25, AND 100. C, RESPECTIVELY1ST SUBSCRIPT IS STAGE NO., 2ND SUBSCRIPT IS COMPONENT NO.

G= 7817.1500676441 I= 7 J= 1 NEW OPT= 7817.1500676441

G= 4690.94383561659 I= 63 J= 1 NEW OPT= 4690.94383561659

G= 4489.91825568842 I= 312 J= 1 NEW OPT= 4489.91825568842

G= 4488.71401505321 I= 1046 J= 1 NEW OPT= 4488.71401505321

G= 4120.66780269603 I= 1579 J= 1 NEW OPT= 4120.66780269603

G= 3503.64139490078 I= 2163 J= 1 NEW OPT= 3503.64139490078

G= 2809.98256802775 I= 2213 J= 1 NEW OPT= 2809.98256802775

G= 2800.4278138952 I= 1242 J= 2 NEW OPT= 2800.4278138952

G= 2618.44267113462 I= 1992 J= 2 NEW OPT= 2618.44267113462

G= 2219.54018317678 I= 56 J= 3 NEW OPT= 2219.54018317678

G= 1802.42512445764 I= 870 J= 3 NEW OPT= 1802.42512445764

G= 1740.23882593035 I= 739 J= 5 NEW OPT= 1740.23882593035

G= 1272.68609459433 I= 976 J= 5 NEW OPT= 1272.68609459433

G= 1252.6357041528 I= 126 J= 6 NEW OPT= 1252.6357041528

G= 948.059635421107 I= 1481 J= 6 NEW OPT= 948.059635421107

G= 883.574640018997 I= 2637 J= 6 NEW OPT= 883.574640018997

G= 551.308735366156 I= 1252 J= 7 NEW OPT= 551.308735366156

G= 491.337852186098 I= 1855 J= 8 NEW OPT= 491.337852186098

G= 452.39434903912 I= 2363 J= 9 NEW OPT= 452.39434903912

G= 433.495599014635 I= 1673 J= 11 NEW OPT= 433.495599014635

G= 426.428005765711 I= 346 J= 12 NEW OPT= 426.428005765711

G= 408.438109011682 I= 880 J= 12 NEW OPT= 408.438109011682

G= 406.267188700355 I= 1456 J= 12 NEW OPT= 406.267188700355G= 378.664312716753 I= 2984 J= 12 NEW OPT= 378.664312716753

G= 365.279482749328 I= 30 J= 13 NEW OPT= 365.279482749328

G= 341.764777434676 I= 43 J= 13 NEW OPT= 341.764777434676

G= 320.079789475042 I= 1136 J= 14 NEW OPT= 320.079789475042

G= 319.010077856613 I= 1194 J= 14 NEW OPT= 319.010077856613

G= 309.465691626442 I= 1340 J= 14 NEW OPT= 309.465691626442

G= 307.754102992359 I= 1459 J= 14 NEW OPT= 307.754102992359

G= 305.209950805526 I= 1534 J= 14 NEW OPT= 305.209950805526

G= 305.014917929711 I= 1654 J= 14 NEW OPT= 305.014917929711

G= 228.027517057871 I= 1674 J= 14 NEW OPT= 228.027517057871

G= 211.250747879507 I= 1714 J= 14 NEW OPT= 211.250747879507


G= 182.030781498348 I= 77 J= 15 NEW OPT= 182.030781498348

G= 177.026329358877 I= 213 J= 15 NEW OPT= 177.026329358877

G= 170.412827068251 I= 282 J= 15 NEW OPT= 170.412827068251

G= 153.022912002767 I= 443 J= 15 NEW OPT= 153.022912002767

G= 149.914258130783 I= 750 J= 15 NEW OPT= 149.914258130783

G= 123.355499115842 I= 828 J= 15 NEW OPT= 123.355499115842

G= 97.0841176441661 I= 994 J= 15 NEW OPT= 97.0841176441661


30/56

G= 92.5949724676948 I= 2494 J= 15 NEW OPT= 92.5949724676948

G= 88.626361937812 I= 2553 J= 15 NEW OPT= 88.626361937812

G= 85.4843395120787 I= 1207 J= 17 NEW OPT= 85.4843395120787

G= 76.0364313360713 I= 1426 J= 17 NEW OPT= 76.0364313360713

G= 74.752070185055 I= 1903 J= 17 NEW OPT= 74.752070185055


G= 57.9305197370146 I= 276 J= 18 NEW OPT= 57.9305197370146

G= 57.3130779750471 I= 467 J= 18 NEW OPT= 57.3130779750471

G= 50.6595563661554 I= 824 J= 18 NEW OPT= 50.6595563661554

G= 46.9185777883877 I= 1600 J= 18 NEW OPT= 46.9185777883877

G= 33.2190849109639 I= 1659 J= 18 NEW OPT= 33.2190849109639

G= 31.1479108366452 I= 10 J= 20 NEW OPT= 31.1479108366452

G= 28.935906815921 I= 916 J= 20 NEW OPT= 28.935906815921

G= 27.2964246584842 I= 292 J= 21 NEW OPT= 27.2964246584842

G= 24.9289176676958 I= 242 J= 22 NEW OPT= 24.9289176676958

G= 16.5797175097385 I= 530 J= 22 NEW OPT= 16.5797175097385


G= 11.0069629989503 I= 1988 J= 23 NEW OPT= 11.0069629989503

G= 10.9773741166885 I= 91 J= 26 NEW OPT= 10.9773741166885

G= 10.4083736231255 I= 708 J= 26 NEW OPT= 10.4083736231255

G= 9.88356578393092 I= 124 J= 27 NEW OPT= 9.88356578393092

G= 9.55697291947769 I= 513 J= 27 NEW OPT= 9.55697291947769

G= 9.38718723385838 I= 531 J= 27 NEW OPT= 9.38718723385838

G= 9.1753945773423 I= 752 J= 27 NEW OPT= 9.1753945773423

G= 7.88019362329578 I= 849 J= 27 NEW OPT= 7.88019362329578

G= 7.78279143030387 I= 984 J= 27 NEW OPT= 7.78279143030387

G= 5.78818000950896 I= 1238 J= 27 NEW OPT= 5.78818000950896

G= 5.11384103056108 I= 352 J= 28 NEW OPT= 5.11384103056108

G= 4.87377960666485 I= 43 J= 29 NEW OPT= 4.87377960666485

G= 4.2463866159891 I= 627 J= 29 NEW OPT= 4.2463866159891

G= 4.23637468512125 I= 829 J= 29 NEW OPT= 4.23637468512125

G= 3.78995470641092 I= 1069 J= 29 NEW OPT= 3.78995470641092

G= 3.4808251707264 I= 2876 J= 30 NEW OPT= 3.4808251707264

G= 3.21750748648925 I= 1572 J= 31 NEW OPT= 3.21750748648925

G= 3.20179204716282 I= 1717 J= 31 NEW OPT= 3.20179204716282

G= 2.95679615413781 I= 829 J= 33 NEW OPT= 2.95679615413781

G= 2.74296629896886 I= 1011 J= 33 NEW OPT= 2.74296629896886

G= 2.59972413597935 I= 1177 J= 33 NEW OPT= 2.59972413597935

G= 2.56564719793815 I= 1907 J= 33 NEW OPT= 2.56564719793815

G= 2.32391873128655 I= 2735 J= 33 NEW OPT= 2.32391873128655


G= 1.90124469792828 I= 2766 J= 34 NEW OPT= 1.90124469792828

G= 1.76062834546378 I= 2994 J= 34 NEW OPT= 1.76062834546378

G= 1.65499579147665 I= 1987 J= 35 NEW OPT= 1.65499579147665

G= 1.62096088133278 I= 2628 J= 36 NEW OPT= 1.62096088133278

G= 1.602354350631 I= 707 J= 37 NEW OPT= 1.602354350631

G= 1.59817529380331 I= 1329 J= 37 NEW OPT= 1.59817529380331


31/56

G= 1.53654006743516 I= 1484 J= 37 NEW OPT= 1.53654006743516

G= 1.51941892914619 I= 1995 J= 37 NEW OPT= 1.51941892914619

G= 1.46356249523146 I= 2261 J= 37 NEW OPT= 1.46356249523146

G= 1.43737712725776 I= 311 J= 38 NEW OPT= 1.43737712725776

G= 1.43263431568158 I= 720 J= 38 NEW OPT= 1.43263431568158

G= 1.38874991664128 I= 731 J= 38 NEW OPT= 1.38874991664128


G= 1.30606393629554 I= 2333 J= 39 NEW OPT= 1.30606393629554

G= 1.26377169552792 I= 269 J= 40 NEW OPT= 1.26377169552792

G= 1.25723129977059 I= 2377 J= 40 NEW OPT= 1.25723129977059

G= 1.25714159610264 I= 1003 J= 42 NEW OPT= 1.25714159610264

G= 1.24925519226137 I= 1107 J= 42 NEW OPT= 1.24925519226137

G= 1.24708315569576 I= 1124 J= 43 NEW OPT= 1.24708315569576

G= 1.2425015332705 I= 1428 J= 43 NEW OPT= 1.2425015332705

G= 1.23720890838308 I= 1190 J= 44 NEW OPT= 1.23720890838308

G= 1.23701853612049 I= 3000 J= 45 NEW OPT= 1.23701853612049

G= 1.23603633525977 I= 43 J= 46 NEW OPT= 1.23603633525977


G= 1.23136432390398 I= 141 J= 48 NEW OPT= 1.23136432390398

G= 1.22926242284027 I= 562 J= 48 NEW OPT= 1.22926242284027

G= 1.22802072811206 I= 2001 J= 50 NEW OPT= 1.22802072811206

START TIME= 62210.903 FINISH TIME= 62256.234

ELAPSED TIME= 45.3309999999983 SECONDS

FINAL RESULTS

BOTTOM,STAGE= 3 X1= .375998126966507 X2= .231066869116218 X3=

.392935003917275

STAGE= 2 X1= .507903390190903 X2= .232151477558479 X3=

.259945132250618

STAGE= 1 X1= .652298339650017 X2= .190642120616716 X3=

.157059539733267

COND., STAGE= 0 X1= .789338879214408 X2= .127507156548963 X3=

8.31540139071657E-2

V(1)= 391.207429289046

V(2)= 385.279448536624

V(3)= 381.237768763931

CONDENSER TEMP., T(0)= 67.689453125

T(1)= 70.046875

T(2)= 72.54296875

T(3)= 75.0390625

FEED TEMP., T(4)= 72.8203125


32/56

REFLUX, L(0)= 361.2086

L(1)= 355.2796

L(2)= 451.2384

BOTTOMS,L(3)= 70

REFLUX RATIO= 12.0402862548828

BOIL-UP RATIO= 5.44626349545249

LAST OBJ. CALC., MAY NOT BE MIN.VALUE.

G= 1.2411560748514 I= 3001 J= 51 OPT= 1.22802072811206


G01= 8.3375987467782E-4 G02= 1.6611736015954E-3 G03=

8.35323018942208E-4 H0= 3.44388386601645E-2

G11= 8.03936258425952E-4 G12= 1.78070623340435E-3 G13=

9.93409256316916E-4 H1= 4.78141324082159E-2G21= 4.13536897028624E-4 G22= 1.49741325416551E-3 G23=

1.90883635664135E-3 H2= 9.27941712609439E-2

G31= 3.8371328077566E-4 G32= 1.37788062235628E-3 G33=

1.7507501192665E-3 H3= 5.14147309300448E-2

SUMY0= 1.00037700167585 SUMY1= 0 SUMY2= 0

SUMY3= 0

DELQ= 5.24692882990787E-5 QQC= 3439200.53812674



33/56

REFERENCES:

Banbura, M. (2010). "Maximum Likelihood Estimation of Factor Models on Data Sets with Arbitrary,

Pattern of Missing Data". Social Science Research Network . Retrieved September 1, 2012, from

SSRN Electronic Library (Working Paper Series No. 1189/May 2010,: h.

Conley, W. C. (1984). Computer Optimization Techniques Revised Edition. Princeton, N. J.: Petrocelli

Books, Inc.

Cutlip, M. B. (2008). Problem Solving in Chemical and Biochemical Engineering with POLYMATH, EXCEL,

and MATLAB, 2nd Ed. Upper Saddle River, N. J.: Prentice Hall.

Eckhardt, R. (1987). "Stan Ulam, John von Neumann, and the Monte Carlo method". Los Alamos Science,

Special Issue (15),, 131-137.

Faloutsos, C. (1903). Analysis of the n-dimensional quadtree decomposition for arbitrary hyper-

rectangles. Retrieved September 15, 2012, from

http://drum.lib.umd.edu/bitstream/1903/678/2/CS-TR-3381.pdf .

Kooijman, H. a. and Ross Taylor (2006). Modeling Separation Processes, Version 7.04 [2100/6] LITE free

license at http://www.chemsep.com.

Lashover, J. H. (2012). "Monte Carlo Marching". Academia.edu, ResearchGate.com, and Scribd.com.

Renon, H. &. (1968). "Local Compositions in Thermodynamic Excess Functions for Liquid Mixtures", NRTL

(non-random, two liquid). AIChE Journal , 135-144.

Seader, J. D. (1998). Separation Process Principles. New York, N. Y.: John Wiley & Sons, Inc.

Ulam, S. M. (1991). Adventures of a Mathematician. Berkeley, CA: U. of California Press.


34/56

APPENDIX A COMPUTER CODE FOR N-BUTANE & N-

PENTANE PROBLEM

REM RIGDIS14.BAS (Remove apostrophe (') in front of PRINT statements to print

intermediate output.)

REM 12.8 RIGOROUS DISTILLATION CALCULATIONS FOR A SIMPLE SEPARATIONTOWER REV. 02/23/16

REM PAGE 551 OF PROBLEM SOLVING IN CHEMICAL AND BIOCHEMICAL

ENGINEERING 2ND ED.

REM BY MICHAEL B. CUTLIP AND MORDECHAI SHACHAM

REM PRENTICE HALL 1999

DECLARE SUB BUBBLE

DECLARE SUB CONSTRAINT

DECLARE SUB ENTHALPY

DECLARE SUB ENERGY

DECLARE SUB PHASE

1 CLS: ON ERROR GOTO 1000

DEFSNG A-Z

DIM BP(10), VP(10, 10), L(10)

DIM V(10), TC(10), X(10, 10), Y(10, 10)

DIM K(10, 10), HL(10, 10), HV(10, 10)

DIM Z(10), A(10), B(10), C(10)

DIM SUMHL(10), SUMHV(10)

DIM AA(10), LL(10), UL(10), AAV(10), LLV(10), ULV(10), XX(10), TF(10),

SUMY(10)

PRINT: PRINT " WELCOME TO RIGDIS14.BAS "

PRINT " THE TIME IS " + TIME$ + " AND THE DATE IS " + DATE$PRINT " COMPONENTS ARE USED IN ORDER OF BOILING POINTS, LOWEST TO

HIGHEST"

PRINT " BP'S IN F ARE 31.129 AND 96.87 F, RESPECTIVELY"

PRINT " IST SUBSCRIPT IS STAGE NO., 2ND SUBSCRIPT IS COMPONENT NO."

INPUT " NAME OF FILE TO STORE OUTPUT= ", FI$

OPEN "C:\QB64\" + FI$ + ".TXT" FOR OUTPUT AS #1

PRINT #1,: PRINT #1, " WELCOME TO RIGDIS14.BAS "

PRINT #1, " THE TIME IS " + TIME$ + " AND THE DATE IS " + DATE$

PRINT #1, " COMPONENTS ARE USED IN ORDER OF BOILING POINTS, LOWEST TO

HIGHEST"PRINT #1, " BP'S IN F ARE 31.129 AND 96.87 F, RESPECTIVELY"

PRINT #1, " IST SUBSCRIPT IS STAGE NO., 2ND SUBSCRIPT IS COMPONENT

NO."

REM SPECS

N% = 2 'NO. OF COMPONENTS

F = 1.0 'MOLES/HR

Z(1) = 0.23 'MOLES/HR


35/56

Z(2) = 1.0 - Z(1)

BTMS = 0.75 'MOLES/HR

D = F - BTMS

PT = 120 'PSI

QR = 10000 'BTU/HR

QC = 10000

BP(1) = 71.554 'AT 120 PSI = 160.8 FBP(2) = 115.67 'AT 120 PSI = 240.21 F

NT% = 3 'NUMBER OF STAGES

REM ANTOINE CONSTANTS

A(1) = 11.7324: B(1) = 2154.9: C(1) = 238.7 'TEMPERATURE IN CENTIGRADE

AND PRESSURE IN PSI

A(2) = 11.8875: B(2) = 2477.07: C(2) = 233.2

REM TEST ANTOINE CONSTANTS USING PURE COMPONENT BOILING TEMPERATURES

REM BP'S IN C AT 120 PSI ARE 71.554 AND 115.67, RESPECTIVELY

REM BP'S IN F AT 120 PSI ARE 160.8 AND 240.21, RESPECTIVELY

REM FOR SCALING PURPOSES, TEMPERATURES WILL BE CALCULATED IN C ANDCONVERTED TO F FOR ENTHALPY EQUATIONS

REM F = 1.8 * C + 32

VAP1! = EXP(A(1) - B(1) / (71.554 + C(1)))

VAP2! = EXP(A(2) - B(2) / (115.67 + C(2)))

PRINT: PRINT " VAPOR PRESSURE OF PURE COMPONENTS AT THEIR BOILING

POINTS SHOULD BE 120 PSIA"

PRINT " PRESSURE AT BP FOR N-BUTANE= "; VAP1!; " PSI"

PRINT " PRESSURE AT BP FOR N-PENTANE= "; VAP2!; " PSI"

INPUT " PRESS ENTER TO CONTINUE", YES

S% = 4: X(4, 1) = Z(1): X(4, 2) = Z(2) ' CHECK WHETHER FEED IS ONE OR

TWO PHASES

GOSUB BUBBLE

GOSUB PHASE

REM FOR THIS 3 STAGE BINARY COLUMN, THERE ARE 9 UNKNOWNS:

REM COMPOSITION OF ONE COMPONENT ON EACH STAGE = 3

REM TEMPERATURE (BP) OF EACH STAGE = 3

REM VAPOR FLOW LEAVING EACH STAGE = 3

REM ALL OTHER VARIABLES ARE DEPENDENT ON THESE INDEPENDENT VARIABLES

REM TEMPERATURE IS DETERMINED IN C WHICH WILL RANGE FROM THE LOWEST

BP TO THE HIGHEST

OPT = 1E+30: TS# = TIMER: RANDOMIZE (-2222)

COUNTI% = 0: COUNTJ% = 0: COUNT& = 0

FOR II = 1 TO 10

AA(II) = 0.: LL(II) = 0.: UL(II) = 0.: AAV(II) = 0.: LLV(II) = 0.:

ULV(II) = 0.

NEXT II

FF = 2


36/56

FOR J = 1 TO 15

ZZZ = 1000

FOR I = 1 TO ZZZ

COUNT& = COUNT& + 1

REM GET COMPOSITION VALUES FOR STAGES 3 & 2. FOR A BINARY X2

= 1.0 - X1 SO X1 DETERMINES X2REM SINCE F, D, AND B ARE KNOWN, SELECTING X(3, 1) SETS X(0,

1) BY OVERALL MATERIAL BALANCE

REM X(0, 1) ALSO SETS Y(1, 1) SINCE FOR A TOTAL CONDENSER X(0,

1) = Y(1, 1)

FOR S% = 3 TO 1 STEP -1

IF AA(S%) - 100! / FF ^ J < 0 THEN GOTO 15

GOTO 17

15 LL(S%) = 0!

GOTO 16

17 LL(S%) = AA(S%) - 100! / FF ^ J

16 IF AA(S%) + 100! / FF ^ J > 100 THEN GOTO 18GOTO 19

18 UL(S%) = 100! - LL(S%)

GOTO 20

19 UL(S%) = AA(S%) + 100! / FF ^ J - LL(S%)

20 XX(S%) = LL(S%) + RND * UL(S%)

NEXT S%

'PRINT: PRINT " LIQUID COMPOSITIONS ON STAGES 1 - 3"


X(S%, 1) = XX(S%) / 100!: X(S%, 2) = 1.0 - X(S%, 1)

'PRINT " S%= "; S%; " X1= "; X(S%, 1); " X2= "; X(S%, 2)

NEXT S%

X(4, 1) = 0.23: X(4, 2) = 0.77 'FEED STAGE GIVEN DATA

'INPUT " PRESS ENTER TO CONTINUE", YES

REM CALCULATE DISTILLATE COMPOSITION BY OVERALL MATERIAL

BALANCE

REM F * Z(1) = XD(1) * D + XB(1) * BTMS

REM XD(1) = (F * Z(1) - XB(1) * BTMS) / D

X(0, 1) = (F * Z(1) - X(3, 1) * BTMS) / D

IF X(0, 1) 3. THEN GOTO 188


37/56

GOTO 199

188 ULV(S%) = 3! - LLV(S%)

GOTO 200

199 ULV(S%) = AAV(S%) + 3! / FF ^ J - LLV(S%)

200 V(S%) = LLV(S%) + RND * ULV(S%)

'PRINT " S= "; S%; " V(S%)= "; V(S%)

NEXT S%

IF V(1)


38/56

'PRINT " DELQ= "; DELQ; " QQC= "; QQC

'PRINT " OVERALL COUNT= "; COUNT&

PRINT #1,: PRINT #1, " NEW OPT= "; OPT; " I= "; I; " J= "; J

'PRINT #1,: PRINT #1, " G= "; G; " I= "; I; " J= "; J

'PRINT #1, " G01= "; G01; " G02= "; G02; " H0= "; H0

'PRINT #1, " G11= "; G11; " G12= "; G12; " H1= "; H1'PRINT #1, " G21= "; G21; " G22= "; G22; " H2= "; H2

'PRINT #1, " G31= "; G31; " G32= "; G32; " H3= "; H3

'PRINT #1, " SUMY1= "; SUMY(1); " SUMY2= "; SUMY(2); " SUMY3=

"; SUMY(3)

'PRINT #1, " DELQ= "; DELQ; " QQC= "; QQC

'PRINT #1, " OVERALL COUNT= "; COUNT&


AA(S%) = X(S%, 1) * 100! ': PRINT " STAGE= "; S%; " X1= ";

X(S%, 1); " X2= "; X(S%, 2)

'PRINT #1, " STAGE= "; S%; " X1= "; X(S%, 1); " X2= ";

X(S%, 2)NEXT S%

FOR S% = 1 TO 3

AAV(S%) = V(S%) ': PRINT " V(S%)= "; V(S%)

'PRINT #1, " V(S%)= "; V(S%)

NEXT S%

COUNTI% = COUNTI% + 1 ': PRINT: PRINT " COUNTI= "; COUNTI%


450 NEXT I

COUNTI% = 0

COUNTJ% = COUNTJ% + 1 ': PRINT: PRINT " COUNTJ= "; COUNTJ%

NEXT J

TFF# = TIMER

PRINT: PRINT " START TIME= "; TS#; " FINISH TIME= "; TFF#

ELAPSE# = (TFF# - TS#): PRINT " ELAPSED TIME= "; ELAPSE#; " SECONDS"

PRINT #1, " ELAPSED TIME= "; ELAPSE#; " SECONDS"

PRINT " FINAL COMPOSITIONS"

PRINT: PRINT " X(0, 1)= "; X(0, 1); " X(0, 2)= "; X(0, 2)

PRINT: PRINT " X(1, 1)= "; AA(1) / 100.; " X(1, 2)= "; (1.0 - AA(1) /

100.)

PRINT: PRINT " X(2, 1)= "; AA(2) / 100.; " X(2, 2)= "; (1.0 - AA(2) /100.)

PRINT: PRINT " X(3, 1)= "; AA(3) / 100.; " X(3, 2)= "; (1.0 - AA(3) /

100.)

PRINT: PRINT " V(1)= "; AAV(1)





39/56

PRINT: PRINT " TF(0)= "; TF(0)




PRINT: PRINT " L(0)= "; L(0)PRINT: PRINT " L(1)= "; L(1)

PRINT: PRINT " L(2)= "; L(2)

PRINT: PRINT " L(3)= "; L(3)

PRINT: PRINT " REFLUX RATIO= "; RR

PRINT: PRINT " BOIL-UP RATIO= "; BR


GOSUB CONSTRAINT

PRINT: PRINT " OPT= "; OPT; " I= "; I; " J= "; J

PRINT " G01= "; G01; " G02= "; G02; " H0= "; H0PRINT " G11= "; G11; " G12= "; G12; " H1= "; H1

PRINT " G21= "; G21; " G22= "; G22; " H2= "; H2

PRINT " G31= "; G31; " G32= "; G32; " H3= "; H3

PRINT " SUMY1= "; SUMY(1); " SUMY2= "; SUMY(2); " SUMY3= "; SUMY(3)

PRINT " DELQ= "; DELQ; " QQC= "; QQC

PRINT " OVERALL COUNT= "; COUNT&

PRINT #1,: PRINT #1, " OPT= "; OPT; " I= "; I; " J= "; J

PRINT #1, " G01= "; G01; " G02= "; G02; " H0= "; H0

PRINT #1, " G11= "; G11; " G12= "; G12; " H1= "; H1

PRINT #1, " G21= "; G21; " G22= "; G22; " H2= "; H2

PRINT #1, " G31= "; G31; " G32= "; G32; " H3= "; H3

PRINT #1, " SUMY1= "; SUMY(1); " SUMY2= "; SUMY(2); " SUMY3= ";

SUMY(3)

PRINT #1, " DELQ= "; DELQ; " QQC= "; QQC

PRINT #1, " OVERALL COUNT= "; COUNT&

995 INPUT " PRESS ENTER TO CONTINUE", YES

CLOSE #1

GOTO 999

1000 PRINT "**********"; CHR$(7)

PRINT " ERROR NUMBER "; ERR; " HAS OCCURRED! ON LINE#= "; ERL

PRINT " PRESS ENTER TO ABORT OR ANY OTHER KEY TO CONTINUE: "CHOICE$ = INPUT$(1)

IF ASC(CHOICE$) = 13 THEN

ON ERROR GOTO 0

ELSE

RESUME NEXT

END IF

CLS

GOTO 1


40/56

999 END

ENTHALPY:

'PRINT #1,: PRINT #1, " CALCULATE LIQUID AND VAPOR ENTHALPIES FOR EACH

STAGE."

'PRINT: PRINT " CALCULATE LIQUID AND VAPOR ENTHALPIES FOR EACH STAGE."FOR S% = 0 TO 4: SUMHL(S%) = 0.: SUMHV(S%) = 0.: NEXT S%

FOR S% = 0 TO 4

HL(S%, 1) = (29.6 * TF(S%) + 0.04 * TF(S%) ^ 2) * X(S%, 1) ':

PRINT " STAGE= "; S%; " HL(S%, 1)= "; HL(S%, 1)

HL(S%, 2) = (38.5 * TF(S%) + 0.025 * TF(S%) ^ 2) * X(S%, 2) ':

PRINT " STAGE= "; S%; " HL(S%, 2)= "; HL(S%, 2)

SUMHL(S%) = HL(S%, 1) + HL(S%, 2) ': PRINT " STAGE#= "; S%; "

SUMHL(S%)= "; SUMHL(S%)

IF S% = 0 OR S% = 4 THEN SUMHV(S%) = 0.: GOTO 80 'NO VAPOR FROM

CONDENSER AND FEED TRAY

HV(S%, 1) = (8003 + 43.8 * TF(S%) - 0.04 * TF(S%) ^ 2) * Y(S%, 1)': PRINT " STAGE= "; S%; " HV(S%, 1)= "; HV(S%, 1)

HV(S%, 2) = (12004 + 31.7 * TF(S%) + 0.007 * TF(S%) ^ 2) * Y(S%,

2) ': PRINT " STAGE= "; S%; " HV(S%, 2)= "; HV(S%, 2)

SUMHV(S%) = HV(S%, 1) + HV(S%, 2) ': PRINT " STAGE#= "; S%; "

SUMHV(S%)= "; SUMHV(S%)

80 NEXT S%


RETURN

BUBBLE:

TL = BP(1)

TU = BP(2)

TC(S%) = (TL + TU) / 2: TF(S%) = 1.8 * TC(S%) + 32

COUNTD% = 0: LMAX% = 1000

50 REM CHECK ITERATIONS VS LMAX%

IF COUNTD% > LMAX% THEN PRINT " COUNTD >= MAX IN BUBBLE SUB"; COUNTD%:

GOTO 60 'INPUT " PRESS ENTER TO CONTINUE", YES

VP(S%, 1) = EXP(A(1) - B(1) / (TC(S%) + C(1))): VP(S%, 2) = EXP(A(2) -

B(2) / (TC(S%) + C(2)))

K(S%, 1) = VP(S%, 1) / PT: K(S%, 2) = VP(S%, 2) / PT

PTT = X(S%, 1) * VP(S%, 1) + X(S%, 2) * VP(S%, 2) ': PRINT " PTT= ";

PTT; " TC= "; TC(S%)DELP = ABS(PT - PTT)

IF DELP


41/56

111 'PRINT: PRINT " FOR STAGE NO.= "; S%; " BUBBLE TEMP.= "; TC(S%); "

ITER. CT. = "; COUNTD%

'PRINT #1,: PRINT #1, " FOR STAGE NO.= "; S%; " BUBBLE TEMP.= ";

TC(S%); " ITER. CT. = "; COUNTD%

TF(S%) = 1.8 * TC(S%) + 32 ': PRINT " BUBBLE TEMP. IN F= "; TF(S%)

Y(S%, 1) = (VP(S%, 1) * X(S%, 1)) / PT ': PRINT " Y(S%, 1)= "; Y(S%,

1)Y(S%, 2) = (VP(S%, 2) * X(S%, 2)) / PT ': PRINT " Y(S%, 2)= "; Y(S%,

2)

SUMY(S%) = Y(S%, 1) + Y(S%, 2)

'PRINT " S%= "; S%; " SUMY= "; SUMY(S%)

'PRINT #1, " SUMY= "; SUMY(S%)

60 'INPUT "PRESS ENTER TO CONTINUE", YES

RETURN

CONSTRAINT:

REM STAGE 0 (CONDENSER'G0 = ABS(V(1) - D - L(0))

G01 = ABS(V(1) * K(1, 1) * X(1, 1) - L(0) * X(0, 1) - D * X(0, 1))

G02 = ABS(V(1) * K(1, 2) * X(1, 2) - L(0) * X(0, 2) - D * X(0, 2))

H0 = ABS(V(1) * SUMHV(1) - L(0) * SUMHL(0) - D * SUMHL(0) - QC)

H0 = H0 / 1000

REM STAGE 1

'G1 = ABS(L(0) + V(2) - L(1) - V(1)) 'OVERALL MOLE BALANCE

G11 = ABS(-((V(1) - L(0)) * K(1, 1) + L(1)) * X(1, 1) + V(2) * K(2, 1)

* X(2, 1))

G12 = ABS(-((V(1) - L(0)) * K(1, 2) + L(1)) * X(1, 2) + V(2) * K(2, 2)

* X(2, 2))

SUMY(1) = Y(1, 1) + Y(1, 2)


H1 = ABS(-V(1) * SUMHV(1) + V(2) * SUMHV(2) - L(1) * SUMHL(1) + L(0) *

SUMHL(0))

H1 = H1 / 1000

REM STAGE 2

'G2 = ABS(L(1) + F + V(3) - L(2) - V(2))

G21 = ABS(L(1) * X(1, 1) - (V(2) * K(2, 1) + L(2)) * X(2, 1) + V(3) *

K(3, 1) * X(3, 1) + Z(1) * F)

G22 = ABS(L(1) * X(1, 2) - (V(2) * K(2, 2) + L(2)) * X(2, 2) + V(3) *

K(3, 2) * X(3, 2) + Z(2) * F)

SUMY(2) = Y(2, 1) + Y(2, 2)SUMY(2) = ABS(1. - SUMY(2))

H2 = ABS(-V(2) * SUMHV(2) + V(3) * SUMHV(3) + SUMHL(4) + L(1) *

SUMHL(1) - L(2) * SUMHL(2))

H2 = H2 / 1000

REM STAGE 3

'G3 = ABS(L(2) - L(3) - V(3))

G31 = ABS(L(2) * X(2, 1) - (V(3) * K(3, 1) + BTMS) * X(3, 1))


42/56

G32 = ABS(L(2) * X(2, 2) - (V(3) * K(3, 2) + BTMS) * X(3, 2))

SUMY(3) = Y(3, 1) + Y(3, 2)


H3 = ABS(-V(3) * SUMHV(3) + QR + L(2) * SUMHL(2) - L(3) * SUMHL(3))

H3 = H3 / 1000

GOSUB ENERGY 'GET DELQ

100 G = G01 + G02 + G11 + G12 + G21 + G22 + G31 + G32 + H0 + H1 + H2 +

H3 + SUMY(1) + SUMY(2) + SUMY(3) + DELQ

'PRINT: PRINT " G= "; G; " I= "; I; " J= "; J; " OPT= "; OPT; "

OVERALL COUNT= "; COUNT&

'PRINT " G01= "; G01; " G02= "; G02; " H0= "; H0

'PRINT " G11= "; G11; " G12= "; G12; " H1= "; H1

'PRINT " G21= "; G21; " G22= "; G22; " H2= "; H2

'PRINT " G31= "; G31; " G32= "; G32; " H3= "; H3

'PRINT " SUMY1= "; SUMY(1); " SUMY2= "; SUMY(2); " SUMY3= "; SUMY(3)

'PRINT " DELQ= "; DELQ'PRINT #1,: PRINT #1, " G= "; G; " I= "; I; " J= "; J; " OPT= "; OPT;

" OVERALL COUNT= "; COUNT&

'PRINT #1, " G01= "; G01; " G02= "; G02; " H0= "; H0

'PRINT #1, " G11= "; G11; " G12= "; G12; " H1= "; H1

'PRINT #1, " G21= "; G21; " G22= "; G22; " H2= "; H2

'PRINT #1, " G31= "; G31; " G32= "; G32; " H3= "; H3

'PRINT #1, " SUMY1= "; SUMY(1); " SUMY2= "; SUMY(2); " SUMY3= ";

SUMY(3)

'PRINT #1, " DELQ= "; DELQ


RETURN

PHASE: 'CHECK TO SEE IF FEED IS ONE OR TWO PHASES

PRINT: PRINT " CHECK TO SEE IF FEED IS ONE OR TWO PHASES "

PRINT #1,: PRINT #1, " CHECK TO SEE IF FEED IS ONE OR TWO PHASES "

REM CALCULATE BUBBLE PRESSURE OF FEED

PBUBL = Z(1) * VP(S%, 1) + Z(2) * VP(S%, 2)

REM CALCULATE DEW PRESSURE OF FEED

PDEW = 1 / ((Z(1) / VP(S%, 1)) + (Z(2) / VP(S%, 2)))

PRINT " BUBBLE PRESSURE OF FEED= "; PBUBL; " DEW PRESSURE OF FEED= ";

PDEW

PRINT #1, " BUBBLE PRESSURE OF FEED= "; PBUBL; " DEW PRESSURE OF FEED=

"; PDEW

REM ESTIMATE FRACTION VAPOR BY INTERPOLATIONVB = (PBUBL - PT) / (PBUBL - PDEW)

PRINT " ESTIMATE VB FOR PT= "; PT; " VB ESTIMATE= "; VB

PRINT #1, " ESTIMATE VB FOR PT= "; PT; " VB ESTIMATE= "; VB

IF VB


43/56

ENERGY:

'PRINT: PRINT " TEST OVERALL ENERGY BALANCE"

'PRINT " QF + QR = QD + QBTMS + QC "

'PRINT #1,: PRINT #1, " TEST OVERALL ENERGY BALANCE"

'PRINT #1, " QF + QR = QD + QBTMS + QC "

QD = D * SUMHL(0): QBTMS = BTMS * SUMHL(3): QF = F * SUMHL(4): QR =

10000'PRINT " QD= "; QD; " QBTMS= "; QBTMS; " QF= "; QF; " QR= "; QR

'PRINT #1, " QD= "; QD; " QBTMS= "; QBTMS; " QF= "; QF; " QR= "; QR

DELQ = ABS((QF - QBTMS - QD) / 10000)

'PRINT " DELQ= "; DELQ

'PRINT #1,: PRINT #1, " DELQ= "; DELQ

QQC = QF + QR - QD - QBTMS

'PRINT " QQC= "; QQC: PRINT #1, " QQC= "; QQC

BR = V(3) / BTMS

RR = L(0) / D

'PRINT " BR= "; BR; " RR= "; RR;: PRINT #1, " BR= "; BR; " RR= "; RR;

" QQC= "; QQC

'INPUT " PRESS ENTER TO CONTINUE", YESRETURN

APPENDIX B COMPUTER CODE FOR MEOH-IPA-H2O

PROBLEM

REM DISMCM07.BAS (Remove apostrophe (') in front of PRINT statements to print

intermediate output.)

#COMPILE EXE

#BREAK ON#DEBUG ERROR ON

#DEBUG DISPLAY ON

FUNCTION PBMAIN () AS LONG

REM RIGOROUS MULTICOMPONENT NONIDEAL DISTILLATION CALCULATIONS

REM USING MONTE CARLO MARCHING REV. 02/23/16 C

1 CLS: ON ERROR GOTO 1000

DEFDBL A-Z

DEFINT i, j, l, n

DIM BP(10), VP(10, 10), L!(10), MW(10)DIM V(10), TC(10), X(10, 10), Y(10, 10)

DIM K!(10, 10), HL(10, 10), HV(10, 10), XX(10)

DIM ZF(10), EA(10), EB(10), EC(10), PC(10), SUMY(10)

DIM SUMHL(10), SUMHV(10), W(10, 10), COMPONENT$(10)

DIM AAC(10), LL!(10), UL(10), AAV(10), LLV!(10), ULV(10)

DIM AA(10), BB(10), CC(10), AAA(10), BBB(10), CCC(10)

DIM AV(10), BV(10), CV(10), T(10), GAM(10, 10)

DIM HLF(10), ENTHFL(10)


44/56

DIM A(3, 3), tau(3, 3), G(3, 3), Alph(3, 3)

DIM Term1(3), Term2(3, 3), Sum3(3), alpha(3)

DIM lngam!(3), Aij(3), Aji(3), Ap(3), Bp(3), Cp(3)

PRINT: PRINT " WELCOME TO DISMCM07.BAS "PRINT " THE TIME IS " + TIME$ + " AND THE DATE IS " + DATE$

PRINT " COMPONENTS ARE USED IN ORDER OF BOILING POINTS, LOW TO HIGH"

PRINT " COMPONENTS ARE METHANOL, 2-PROPANOL, AND WATER"

PRINT " BP'S IN C ARE 64.47. 82.52, AND 100. C, RESPECTIVELY"

PRINT " 1ST SUBSCRIPT IS STAGE NO., 2ND SUBSCRIPT IS COMPONENT NO."

INPUT " NAME OF FILE TO STORE OUTPUT= ", FI$

OPEN "C:\QB64\" + FI$ + ".TXT" FOR OUTPUT AS #1

PRINT #1,: PRINT #1, " WELCOME TO DISMCM07.BAS "

PRINT #1, " THE TIME IS " + TIME$ + " AND THE DATE IS " + DATE$

PRINT #1, " COMPONENTS ARE USED IN ORDER OF BOILING POINTS,LOW TOHIGH"

PRINT #1, " COMPONENTS ARE METHANOL, 2-PROPANOL, AND WATER"

PRINT #1, " BP'S IN C ARE 64.55, 82.25, AND 100. C, RESPECTIVELY"

PRINT #1, " 1ST SUBSCRIPT IS STAGE NO., 2ND SUBSCRIPT IS COMPONENT

NO."

REM FOR SYSTEM MEOH/IPA/H2O UNITS ARE MOLE/HR, CAL/HR, DEG.C,

REM AND MOLE FRACTION, MM HG

COMPONENT$(1)=" METHANOL": COMPONENT$(2)= " IPA":COMPONENT$(3)= " H2O"

'PRINT: PRINT COMPONENT$(1); COMPONENT$(2); COMPONENT$(3)

REM SPECS

N% = 3 'NO. OF COMPONENTS

F = 100.0 'MOLES/HR

NF% = 2

TF = 73.26

ZF(1) = 0.50 'MOLE FRACTION

ZF(2) = 0.20

ZF(3) = 0.30

BTMS = 70. 'MOLES/HR 50

D = F - BTMS

PT = 760 'MM HG

QR = 3439427

QC = QR

BP(1) = 64.55 'AT 760 MM HGBP(2) = 82.25 'AT 760 MM HG

BP(3) = 100.00 'AT 760 MM HG

NT% = 3 'NUMBER OF STAGES

X(4, 1)= ZF(1): X(4, 2) = ZF(2): X(4, 3)= ZF(3)'FEED STAGE GIVEN DATA

REM ENERGY DATA FOR MEOH, IPA, AND H2O

MW(1) = 32.0422: MW(2) = 60.0959: MW(3) = 18.0153

REM MULTIPLY J/G BY 0.238846 TO GET CALORIES AND BY MW TO GET CAL/MOLE


45/56

AA(1)= 2.33: BB(1)= 0.00782: CC(1)= 0.0000377 'MEOH LIQ.CP J/G @T(C)

AA(2)= 2.320857: BB(2)= 0.01242953: CC(2)= 0.0 'IPA LIQ CP J/G @T(C)

AA(3)= 4.188: BB(3)= -0.000569: CC(3)= 0.00000849 'H2O LIQ CP J/G

@T(C)

AAA(1)= 1218.4: BBB(1)= -1.3849: CCC(1)= -0.006402 'HVAP MEOH J/G

@T(C)

AAA(2)= 801.13: BBB(2)= -1.767583: CCC(2)= 0.0 'HVAP IPA J/G @T(C)AAA(3)= 2481.1: BBB(3)= -1.821: CCC(3)= -0.004236 'HVAP H2O J/G @ T(C)

AV(1)= 42.63215:BV(1)= 0.07690246: CV(1)= 0.0/32.0422'MEOH VAP CP

J/G@T(C)

AV(2)= 1.40664: BV(2)= 0.00353197: CV(2)= 0.0 'IPA VAP CP J/G @T(C)

AV(3)= 1.813: BV(3)= 0.0007439: CV(3)= 0.0000001123 'H2O VAP CP J/G

@T(C)

TC(1)= 512.64: TC(2)= 508.3: TC(3)= 647.13: H= .238846# 'JOULES TO

CAL.

PC(1)= 79.9112 * 760: PC(2)= 46.9973 * 760: PC(3)= 217.666 * 760

REM LIQ. CP'S = AA + BB*TC + CC*TC^2

REM HVAP'S = AAA + BBB*TC + CCC*TC^2REM VAP. CP'S = AV + BV*TC + CV*TC^2

REM REDUCED TEMPERATURE, TR = T/TC WITH T AND TC IN KELVIN

REM ANTOINE VAPOR PRESSURE CONSTANTS--MM Hg and T, C AND LOGe

EA(1) = 18.5875: EB(1) = 3626.55: EC(1) = -34.29 'MEOH

EA(2) = 18.6929: EB(2) = 3640.2: EC(2) = -53.54 'IPA

EA(3) = 18.3036: EB(3) = 3816.44: EC(3) = -46.13 'H2O

'PRINT: PRINT " CHECK FOR PT = 760 MM HG AT BPT."

FOR I% = 1 TO N%

VP(1, I%)= EXP(EA(I%) - EB(I%) / (BP(I%) + 273.15 + EC(I%)))

'PRINT: PRINT " COMPONENT= "; COMPONENT$(I%)

'PRINT " BOILING POINT PRESSURE CHECK= "; VP(1, I%)

'PRINT " DELTA P= "; 760 - VP(1, I%)

'PRINT " PCT ERROR= "; 100 * (760 - VP(1, I%)) / VP(1, I%)

'PRINT " A= "; EA(I%); " B= "; EB(I%); " C= "; EC(I%)

NEXT I%

REM FOR THIS 3 STAGE COLUMN, THERE ARE 12 UNKNOWNS:

REM COMPOSITION OF TWO COMPONENTS ON EACH STAGE = 6

REM TEMPERATURE (BP) OF EACH STAGE = 3

REM VAPOR FLOW LEAVING EACH STAGE = 3

REM ALL OTHER VARIABLES ARE DEPENDENT ON THESE INDEPENDENT VARIABLES

REM TEMP. IS DETERMINED IN C WHICH WILL RANGE FROM THE LOW BP TO HIGH

BP

OPT = 1E+30: TS# = TIMER: RANDOMIZE (-2222)

COUNTI% = 0: COUNTJ% = 0: COUNT& = 0

FOR II = 1 TO 10

AAC(II) = 0.: LL!(II) = 0.: UL(II) = 0.: AAV(II) = 0.: LLV!(II) =

0.

ULV(II) = 0.

NEXT II


46/56

FF = 1.3

FOR J = 1 TO 50

ZZZ = 3000

FOR I = 1 TO ZZZ

COUNT& = COUNT& + 1

REM GET COMPOSITION VALUES FOR STAGES 3 & 2.

REM SINCE F, D, AND B ARE KNOWN, SELECTING X(3, 1) SETS X(0,

1),

REM X(0, 2) AND X(0, 3) BY OVERALL MATERIAL BALANCE


IF AAC(S%) - 100! / FF ^ J < 0 THEN GOTO 15

GOTO 17

15 LL!(S%) = 0!

GOTO 16

17 LL!(S%) = AAC(S%) - 100! / FF ^ J

16 IF AAC(S%) + 100! / FF ^ J > 100 THEN GOTO 18GOTO 19

18 UL(S%) = 100! - LL!(S%)

GOTO 20

19 UL(S%) = AAC(S%) + 100! / FF ^ J - LL!(S%)

20 XX(S%) = LL!(S%) + RND * UL(S%)

NEXT S%

'PRINT: PRINT " LIQUID COMPOSITIONS ON STAGES 3 - 1"

X(3, 1) = XX(6) / 100!

X(3, 2) = XX(5) / 100!: X(3, 3) = 1.0 - X(3, 1) - X(3, 2)

IF X(3, 3)


47/56

IF X(0, 2)


48/56

'PRINT : PRINT " CALC. LIQ. AND VAP. ENTHALPIES PER STAGE."

GOSUB ENTHALPY

GOSUB CONSTRAINT

IF G < OPT THEN GOTO 350

GOTO 450

350 OPT = G': PRINT: PRINT " NEW OPT= "; OPT; " I= "; I; " J=

"; J

'PRINT: PRINT " G= "; G; " I= "; I; " J= "; J; " OPT= "; OPT


'PRINT " G01= "; G01; " G02= "; G02; " G03= "; G03; " H0= ";

H0

'PRINT " G11= "; G11; " G12= "; G12; " G13= "; G13; " H1= ";

H1

'PRINT " G21= "; G21; " G22= "; G22; " G23= "; G23; " H2= ";

H2

'PRINT " G31= "; G31; " G32= "; G32; " G33= "; G33; " H3= ";H3

'PRINT " SUMY0= "; SUMY(0); " SUMY1= "; SUMY(1); " SUMY2=

";SUMY(2)

'PRINT " SUMY3= "; SUMY(3)

'PRINT " DELQ= "; DELQ ; " QQC= "; QQC

PRINT #1,: PRINT #1, " G= "; G;" I= "; I;" J= "; J;" NEW OPT=

";OPT


'PRINT #1, " G01= "; G01; " G02= "; G02; " G03= "; G03; " H0=

"; H0

'PRINT #1, " G11= "; G11; " G12= "; G12; " G13= "; G13; " H1=

"; H1

'PRINT #1, " G21= "; G21; " G22= "; G22; " G23= "; G23; " H2=

"; H2

'PRINT #1, " G31= "; G31; " G32= "; G32; " G33= "; G33; " H3=

"; H3

'PRINT #1, " SUMY0= ";SUMY(0);" SUMY1= ";SUMY(1);" SUMY2=

";SUMY(2)

'PRINT #1, " SUMY3= ";SUMY(3)

'PRINT #1, " DELQ= "; DELQ; " QQC= "; QQC



AAC(6) = X(3, 1) * 100!: AAC(5) = X(3, 2) * 100!AAC(4) = X(2, 1) * 100!: AAC(3) = X(2, 2) * 100!

AAC(2) = X(1, 1) * 100!: AAC(1) = X(1, 2) * 100!

X(3, 3) = 1.0 - X(3, 1) - X(3, 2)

X(2, 3) = 1.0 - X(2, 1) - X(2, 2)

X(1, 3) = 1.0 - X(1, 1) - X(1, 2)


49/56

'PRINT " STAGE= ";3;" X1= ";X(3,1);" X2= ";X(3,2);" X3=

";X(3,3)


";X(2,3)


";X(1,3)

'PRINT " STAGE= ";0;" X1= ";X(0,1);" X2= ";X(0,2);" X3=";X(0,3)

FOR S% = 1 TO 3

AAV(S%) = V(S%)': PRINT " V(S%)= "; V(S%)

NEXT S%

COUNTI% = COUNTI% + 1 ': PRINT: PRINT " COUNTI= "; COUNTI%


450 NEXT I

COUNTI% = 0

COUNTJ% = COUNTJ% + 1 ': PRINT: PRINT " COUNTJ= "; COUNTJ%

NEXT J

TFF# = TIMER

PRINT: PRINT " START TIME= "; TS#; " FINISH TIME= "; TFF#

PRINT #1, : PRINT #1, " START TIME= "; TS#; " FINISH TIME= ";TFF#

ELAPSE# = (TFF# - TS#): PRINT " ELAPSED TIME= "; ELAPSE#;" SEC."

PRINT #1, " ELAPSED TIME= "; ELAPSE#; " SECONDS"


PRINT #1, " FINAL RESULTS"

PRINT " FINAL RESULTS"

X(0, 1)=X(0, 1): X(0, 2)=X(0, 2): X(0, 3)= X(0, 3)

X(1, 1)=AAC(2)/100.:X(1,2)=AAC(1)/100.:X(1,3)=(1.0-X(1,1)-X(1,2))

X(2, 1)=AAC(4)/100.:X(2,2)=AAC(3)/100.:X(2,3)=(1.0-X(2,1)-X(2,2))

X(3, 1)=AAC(6)/100.:X(3,2)=AAC(5)/100.:X(3,3)=(1.0-X(3,1)-X(3,2))

PRINT " STAGE= "; 3;" X1= ";X(3,1);" X2= ";X(3,2);" X3= ";X(3,3)








PRINT: PRINT " T(0)= "; T(0)





PRINT: PRINT " L(0)= "; L!(0)



50/56



PRINT: PRINT " REFLUX RATIO= "; RR

PRINT: PRINT " BOIL-UP RATIO= "; BR


PRINT #1, " BOTTOM,STAGE= ";3;" X1= ";X(3,1);" X2= ";X(3,2);" X3=

";X(3,3)

PRINT #1, " STAGE= ";2;" X1= ";X(2,1);" X2= ";X(2,2);" X3=

";X(2,3)

PRINT #1, " STAGE= ";1;" X1= ";X(1,1);" X2= ";X(1,2);" X3=

";X(1,3)

PRINT #1, " COND., STAGE= ";0;" X1= ";X(0,1);" X2= ";X(0,2);" X3=

";X(0,3)

PRINT #1, : PRINT #1, " V(1)= "; AAV(1)

PRINT #1, : PRINT #1, " V(2)= "; AAV(2)

PRINT #1, : PRINT #1, " V(3)= "; AAV(3)

PRINT #1, : PRINT #1, " CONDENSER TEMP., T(0)= "; T(0)

PRINT #1, : PRINT #1, " T(1)= "; T(1)

PRINT #1, : PRINT #1, " T(2)= "; T(2)

PRINT #1, : PRINT #1, " T(3)= "; T(3)

PRINT #1, : PRINT #1, " FEED TEMP., T(4)= "; T(4)

PRINT #1, : PRINT #1, " REFLUX, L(0)= "; L!(0)

PRINT #1, : PRINT #1, " L(1)= "; L!(1)

PRINT #1, : PRINT #1, " L(2)= "; L!(2)

PRINT #1, : PRINT #1, " BOTTOMS,L(3)= "; L!(3)

PRINT #1, : PRINT #1, " REFLUX RATIO= "; RR

PRINT #1, : PRINT #1, " BOIL-UP RATIO="; BR

PRINT: PRINT " G= "; G; " I= "; I; " J= "; J; " OPT= "; OPT

PRINT " OVERALL COUNT= "; COUNT&

PRINT " G01= "; G01; " G02= "; G02; " G03= "; G03; " H0= "; H0

PRINT " G11= "; G11; " G12= "; G12; " G13= "; G13; " H1= "; H1

PRINT " G21= "; G21; " G22= "; G22; " G23= "; G23; " H2= "; H2

PRINT " G31= "; G31; " G32= "; G32; " G33= "; G33; " H3= "; H3

PRINT " SUMY0= "; SUMY(0); " SUMY1= "; SUMY(1); " SUMY2= "; SUMY(2)

PRINT " SUMY3= "; SUMY(3)

PRINT " DELQ= "; DELQ; " QQC= "; QQC

PRINT #1, : PRINT #1, " LAST OBJ. CALC., MAY NOT BE MIN.VALUE."PRINT #1, " G= "; G; " I= "; I; " J= "; J; " OPT= "; OPT


PRINT #1, " G01= "; G01; " G02= "; G02; " G03= "; G03; " H0= "; H0

PRINT #1, " G11= "; G11; " G12= "; G12; " G13= "; G13; " H1= "; H1

PRINT #1, " G21= "; G21; " G22= "; G22; " G23= "; G23; " H2= "; H2

PRINT #1, " G31= "; G31; " G32= "; G32; " G33= "; G33; " H3= "; H3

PRINT #1, " SUMY0= ";SUMY(0);" SUMY1= ";SUMY(1);" SUMY2= ";SUMY(2)

PRINT #1, " SUMY3= "; SUMY(3)


51/56

PRINT #1, " DELQ= "; DELQ; " QQC= "; QQC


995 INPUT " PRESS ENTER TO CONTINUE", YES

CLOSE #1

GOTO 999

1000 PRINT "**********"; CHR$(7)

PRINT " ERROR NUMBER "; ERR; " HAS OCCURRED ON LINE#= "; ERL

RESUME NEXT

999 END

BUBBLE:

TL = 64.5

TU = 100.T(S%) = (TL + TU) / 2

COUNTD% = 0: LMAX% = 1000

50 REM CHECK ITERATIONS VS LMAX%

IF COUNTD% > LMAX% THEN GOTO 60'INPUT "PRESS ENTER TO CONTINUE",YES

GOSUB NRTL

VP(S%, 1) = EXP(EA(1) - EB(1) / (T(S%) + 273.15 + EC(1)))

VP(S%, 2) = EXP(EA(2) - EB(2) / (T(S%) + 273.15 + EC(2)))

VP(S%, 3) = EXP(EA(3) - EB(3) / (T(S%) + 273.15 + EC(3)))

K!(S%, 1)=VP(S%, 1)/PT: K!(S%,2)=VP(S%,2)/PT: K!(S%,3)=VP(S%,3)/PT

PTT=GAM(S%,1)*X(S%,1)*VP(S%,1)+GAM(S%,2)*X(S%,2)*VP(S%,2)

PTT = PTT + GAM(S%,3)*X(S%,3)*VP(S%,3)

'PRINT " PTT= "; PTT; " T= "; T(S%)

DELP = ABS(PT - PTT)

IF DELP


52/56

RETURN

CONSTRAINT:

REM STAGE 0 (CONDENSER)

G0 = ABS(V(1) - D - L!(0))G01 = ABS(V(1) * Y(1, 1) - L!(0) * X(0, 1) - D * X(0, 1))

G02 = ABS(V(1) * Y(1, 2) - L!(0) * X(0, 2) - D * X(0, 2))

G03 = ABS(V(1) * Y(1, 3) - l!(0) * X(0, 3) - D * X(0, 3))

H0 = ABS(V(1) * SUMHV(1) - L!(0) * SUMHL(0) - D * SUMHL(0) - QC)

H0 = H0 / 1000

REM STAGE 1

G1 =ABS(L!(0) + V(2) - L!(1) - V(1)) 'OVERALL MOLE BALANCE

G11=ABS(-V(1) * Y(1,1)-L!(1) * X(1,1)+V(2) * Y(2,1)+L!(0) * X(0,1))

G12=ABS(-V(1) * Y(1,2)-L!(1) * X(1,2)+V(2) * Y(2,2)+L!(0) * X(0,2))

G13=ABS(-V(1) * Y(1,3)-L!(1) * X(1,3)+V(2) * Y(2,3)+L!(0) * X(0,3))

SUMY(1) = Y(1, 1) + Y(1, 2) + Y(1, 3)SUMY(1) = ABS(1. - SUMY(1))

H1=ABS(-V(1)*SUMHV(1)+V(2)*SUMHV(2)-L!(1)*SUMHL(1)+L!(0)*SUMHL(0))

H1 = H1 / 1000

REM STAGE 2

G2 = ABS(L!(1) + F + V(3) - L!(2) - V(2))




SUMY(2) = Y(2, 1) + Y(2, 2) + Y(2,3)


H2=ABS(-V(2)*SUMHV(2)+V(3)*SUMHV(3)+SUMHL(4)*F+L!(1)*SUMHL(1)-

L!(2)*SUMHL(2))

H2 = H2 / 1000

REM STAGE 3

G3 = ABS(L!(2) - L!(3) - V(3))

G31 = ABS(L!(2) * X(2, 1) - V(3) * Y(3, 1) - BTMS * X(3, 1))

G32 = ABS(L!(2) * X(2, 2) - V(3) * Y(3, 2) - BTMS * X(3, 2))

G33 = ABS(L!(2) * X(2, 3) - V(3) * Y(3, 3) - BTMS * X(3, 3))

SUMY(3) = Y(3, 1) + Y(3, 2) + Y(3, 3)


H3 = ABS(-V(3) * SUMHV(3) + QR + L!(2) * SUMHL(2) - L!(3) * SUMHL(3))

H3 = H3 / 1000

GOSUB ENERGY

100 G = G01 + G02 + G03 + G11 + G12 + G13 + G21 + G22 + G23 + G31 +

G32 + G33

G = G + H0 + H1 + H2 + H3 + SUMY(0) + SUMY(1) + SUMY(2) + SUMY(3) +

DELQ

G = G + G0 + G1 + G2 + G3


53/56

'PRINT: PRINT " G= "; G; " I= "; I; " J= "; J; " OPT= "; OPT


'PRINT " G0= "; G0; " G1= "; G1; " G2= "; G2; " G3= "; G3

'PRINT " G01= "; G01; " G02= "; G02; " G03= "; G03; " H0= "; H0

'PRINT " G11= "; G11; " G12= "; G12; " G13= "; G13; " H1= "; H1

'PRINT " G21= "; G21; " G22= "; G22; " G23= "; G23; " H2= "; H2

'PRINT " G31= "; G31; " G32= "; G32; " G33= "; G33; " H3= "; H3'PRINT " SUMY0= "; SUMY(0);" SUMY1= "; SUMY(1); " SUMY2= "; SUMY(2)

'PRINT " SUMY3= "; SUMY(3)

'PRINT " DELQ= "; DELQ; " QQC= "; QQC

'PRINT #1,: PRINT #1, " G= "; G; " I= "; I; " J= "; J; " OPT= "; OPT


'PRINT #1, " G0= "; G0; " G1= "; G1; " G2= "; G2; " G3= "; G3

'PRINT #1, " G01= "; G01; " G02= "; G02; " G03= "; G03; " H0= "; H0

'PRINT #1, " G11= "; G11; " G12= "; G12; " G13= "; G13; " H1= "; H1

'PRINT #1, " G21= "; G21; " G22= "; G22; " G23= "; G23; " H2= "; H2

'PRINT #1, " G31= "; G31; " G32= "; G32; " G33= "; G33; " H3= "; H3

'PRINT #1, " SUMY0= ";SUMY(0);" SUMY1= ";SUMY(1);" SUMY2= ";SUMY(2)

'PRINT " SUMY3= "; SUMY(3)'PRINT #1, " DELQ= "; DELQ; " QQC= "; QQC


RETURN

ENERGY:

'PRINT: PRINT " TEST OVERALL ENERGY BALANCE"

'PRINT " QF + QR = QD + QBTMS + QC "

'PRINT #1,: PRINT #1, " TEST OVERALL ENERGY BALANCE"

'PRINT #1, " QF + QR = QD + QBTMS + QC "

QD = D * SUMHL(0): QBTMS = BTMS * SUMHL(3): QF = F * SUMHL(4)

QR = 3439381

'PRINT " QD= "; QD; " QBTMS= "; QBTMS; " QF= "; QF; " QR= "; QR

'PRINT #1, " QD= ";QD; " QBTMS= ";QBTMS; " QF= ";QF; " QR= ";QR

DELQ = ABS((QF - QBTMS - QD) / QR)

'PRINT " DELQ= "; DELQ

'PRINT #1,: PRINT #1, " DELQ= "; DELQ

QQC = QF + QR - QD - QBTMS

'PRINT " QQC= "; QQC: PRINT #1, " QQC= "; QQC

BR = V(3) / BTMS

RR = L!(0) / D

'PRINT " BR= ";BR;" RR= ";RR;: PRINT #1, " BR= ";BR;" RR= ";RR

'PRINT " QQC= "; QQC


RETURN

WILSON:

'PRINT:PRINT " CALCULATE WILSON ACTIVITY COEFFS. FOR STAGE= ";S%"

VW = 18.07: VM = 40.73: VI = 76.92 'CM^3/MOL

W(1, 1) = 1: W(2, 2) = 1: W(3, 3) = 1

W(1, 2) = 1.03116: W(2, 1) = 1.08036

W(1, 3) = 0.60703: W(3, 1) = 0.92588

W(2, 3) = 0.1258: W(3, 2) = 0.7292


54/56

FOR KK% = 1 TO N%

Q1 = 0

FOR C% = 1 TO N%

Q1 = Q1 + X(S%, C%) * W(KK%, C%)

NEXT C%

Q2 = 0FOR II% = 1 TO N%

Q3 = 0

FOR C% = 1 TO N%

Q3 = Q3 + X(S%, C%) * W(II%, C%)

NEXT C%

Q2 = Q2 + (X(S%, II%) * W(II%, KK%)) / Q3

NEXT II%

GAM(S%, KK%) = EXP(1. - LOG(Q1) - Q2)

'PRINT: PRINT " S= "; S%; " KK= "; KK%

'PRINT " GAM(S%, KK)= "; GAM(S%, KK%)

NEXT KK%

'INPUT " PRESS ENTER TO CONTINUE", YESRETURN

ENTHALPY:

FOR S% = 0 TO 4

SUMHL(S%) = 0: SUMHV(S%) = 0

FOR II% = 1 TO N%

HL(S%,II%)=(AA(II%)*T(S%))+((BB(II%)/2.0)*(T(S%)^2))+((CC(II%)/3.0)*(T

(S%)^3))

HL(S%, II%) = HL(S%, II%) * H * MW(II%) 'CAL/MOLE

SUMHL(S%) = SUMHL(S%) + HL(S%, II%) * X(S%, II%)

IF S% = 0 THEN GOTO 77

IF S% = 4 THEN GOTO 77

HV(S%, II%) = AAA(II%) + BBB(II%) * T(S%)

Monte Carlo Marching Distillation: Simultaneous solution of the MESH equations with no derivatives,...

Documents

Transcript of Monte Carlo Marching Distillation: Simultaneous solution of the MESH equations with no derivatives,...