distillation models for simulation

Distillation Models for Simulation

B. Tech. Petrochemical EngineeringMulticomponent Distillation

University College of Engineering (A) KakinadaDept. of Petroleum Engineering & Petrochemical Engineering

Jawaharlal Nehru Technological University Kakinada

Prof. K. V. RaoProgramme DirectorPetroleum Courses

JNTUK

Distillation Models for Simulation

Why should you know process modeling and simulation of distillation columns?

In recent years the use of commercial process simulation packages for distillation column design has increased dramatically.

Process simulation packages have made performing design calculations for distillation column significantly easier.

To arrive at a viable design, the design engineer should have basic understanding of column modeling and simulation procedures.

What do you expect from this presentation? Modeling and simulation based on nesting of

equations of Thiele –Geddes method.

Modeling and simulation based on the method of Holland.

THIELE-GEDDES METHOD

Figure1: Configuration of a conventional distillation column with liquid and vapor streams including product streams for the programs

T. G. method is a simulation method to predict Product distribution for a given multi-component Distillation column.

1. Specify the following variables

i. Number of components in the liquid mixture for separation, nc

ii. Feed rate, F: composition of each component, xi,F , saturated liquid feed Q=1

iii. Product recovery, D/F or B/F;iv. Number of stages in rectifying section, nr and in

stripping section ns. Feed plate location, nt= ns+nr+2

v. Reflux ratio, R. The reflux is assumed to be saturated liquid.

vi. Liquid and vapor flow rates in the rectifying and stripping section are assumed (RL and RV in rectifying section, RL=D*R; RV=L+D; SL & SV in SL=RL+Q*F ; SV= RV-(1-Q) *F )

vii. Type of condenser :totalviii. Top pressure of the column and allowable

pressure drop per plate, ix. ‘K’ values as function of T for all the

components at the given pressure

2. Assume a linear temperature and pressure profile of the column.

i. Calculate bubble point and dew point temperature for the assumed compositions of the bottom and top product. The temperatures may be taken as the initial bottom temperature (Reboiler temperature or Nth plate temperature) and top temperatures. Alternatively, if the key components are significant (Heavy key in the bottom product and light key component in the top product),the boiling temperatures of these components at the given pressure may be taken as the initial bottom (Tbott) and top temperature(Ttop).

ii. Calculate T T= (Tbot-Ttop) /(np-1)

iii. Calculate each plate temperature in the column, Tj = Tj-1+T (j=2,np)

iv. Calculate the pressure of each plate for given condenser pressure, top plate pressure and P

Pj = Pj-1 +P (j=2,np)

3. Calculate the values of equilibrium constant, Ki,j for each component on each plate.(If Ki is function of both temperature and pressure, then correlation for each component may be used)

Ki,j = exp[ Ai – Bi / (Tj + Cj) ] /Pj (j=1,np ; i=1,nc)

4. Initialize the ratios of the compositions yi,1/xi,D and yi,N/xi,B

(i) yi,1 / xi,D = 1 ( i=1,nc) (ii) yi,N+1/ xi,B = Ki,N+1 ( i=1,nc)

5. Calculate the values of the ratios of the components for each plate in the rectifying section.

(i) xi,j /xi,D = (yi,j /xi,D) / Ki,j ( j=1,nr ; I=1, nc)(ii) yi,j+1 / xi,D = (RL* (xi,j / xi,D) + D) / RV (j=1,nr ; i=1to nc)

6. Calculate the ratios of compositions, xi,j / xi,B and yi,j / xi,B for each component on each plate in the stripping section.(i) xi,(np+1-j) / xi,B = (SV* yi,(np+1-j+1) / xi,B + B) /SL (j=

1,ns+1; i=1,nc) (ii) yi,(np+1-j) /xi,B = Ki,(np+1-j) * xi,(np+1-j) / xi,B (j= 1,ns+1 ; i=1,nc)

7. Calculation of the ratio compositions, (xi,B /xi,D) xi,B/xi,D = (yi,NF / xi,D ) /( yi,NF / xi,B) (i=1,nc)

8. Calculate the top product composition xi,D from the overall component balance of the column and then xi,B.

xi,D = F.xi,F / (D+B (xi,B / xi,D)), x i,B = (xi,B / xi,D) xi,D

9. Normalize the product compositions, xi,D and xi,B xi,D,cal=xi,D / xi,D

xi,B,cal = xi,B / xi,B

10. Obtain the liquid compositions of all the components on each plate .

(i) Rectifying section xi,j = (xi,j /xi,D) xi,d,cal (j=1,nr ; i=1,nc) (ii) Stripping section xi,np-j = (xi, np-j /xi,B) xi,B,cal (j=1,ns+1; i=1,nc)

11. Normalization of the liquid compositions on each plate in the column

xi,j / xi,j (j=1,np; i=1,nc)

12. Correct the temperature profile using bubble method.i. Ki,j = exp[ Ai – Bi / (Tj + Cj) ] /Pj (j=1,np ;

i=1,nc)ii. Calculate the yi,j =Ki,j* xi,j (j=1,np; i=1,nc)iii. Calculate sum of yij (j=1,np; i=1,nc)iv. F = sumy-1v. If Absolute value of F is less than a error

tolerance of 10-4 or 10-6, T[j] is corrected by using Newton’s method.

vi. T[j] = T[j] – F/DF. Where DF is derivative of F in Temperature.

vii. Then the calculation is repeated from (i) to (vi) until the condition in (v) is satisfied.

13. Change the iteration number.iter =iter+1;

14. Repeat the calculations from the step3. No convergence procedure used except for 20 or more iterations. After convergence, plate compositions and temperature are obtained along with the values of xi,D and xi,B .

Should it be necessary to study the effect of variables like reflux ratio, feed plate location, feed composition on product profile, one variable at a time, may be changed and the results may be obtained. Thus, several simulations will lead to decide upon the best possible separation.

Holland’s Method Tridiagonal Matrix for material balance in terms

of component vapor flow rates and absorption factors.

Thomas Algorithm to solve the tridiagonal

matrix.

KB method for correcting temperature profile.

“Ѳ” method of convergence.

The ‘’ Method of Convergence To achieve fast convergence of the problems

dealing with the simulation of distillation columns, the ‘’ method combined with Kb method was proposed by Holland .

In the iterative procedure, the improved sets of liquid mole fractions on each plate in a column required for the calculation of a new temperature profile are obtained using the corrected product compositions (top and bottom).

These product compositions should satisfy two conditions. i) the overall component material balance and ii. the criterion of the sum of the top product compositions, which is equal to unity.

Thus for all the components,

Fxi,F = D(xi,D)cor + B(xi,B)cor xi,D =1

The corrected values of (xi,D/xi,B)cor and the

calculated values of (xi,D/xi,B) cal are related by a multiplier ‘’

( xi,B / xi,D )cor = ( xi,B / xi,D)cal

The value of becomes unity when the convergence is reached.

The KB method

The new temperature profile for the next iteration is calculated on the basis of the corrected liquid mole fractions and the temperature profile of the previous iteration by using the KB method.

This method is considered as a modified bubble point method. For any plate, it is applied as

Where, i,j = Ki,j / Kj,b , the relative volatility of component i at the temperature of plate j , and Kj,b is the K value of base component ( in the calculations , heavy key component is considered as the base component).

As the value of Kj,b is function of temperature, the new temperature of each plate can be calculated using K equation rewritten for T.

Tridiagonal Matrix Equation

A set of variables-the component-flow rates in the vapor and liquid phases-are introduced, namely,

Also, the flow rates of component i in the distillate and bottoms are represented by

and the flow rates of component i in the vapor and liquid parts of the feed by

jijji yVv jijji xLl and

Dii xDd Bii xBb and

FiFFi yVv FiFFi xLl and

The equilibrium relationship may be restated in an equivalent form in terms of the component-flow rates

and as follows. Then, expression may be restated in the form

jijiji xKy

jiv jil jijiji xKy

Where the absorption factor and the stripping are defined as follows

jiA jiS

j

jiji

j

ji

Ll

KVv

jijiji lSv jijiji vAl and

==jiAjiS1

jji

j

VKL

An equivalent set of component-material balances is obtained by enclosing each stage (j=1, 2, ….., N, N+1) by a component-material balance.

Corresponding set of material balance for each component i are as follows

010 iii vdl

02110 iiii vlvl

)2,.......,4,3,2(,0,1,1 fjvlvl ijjijiij

Fifiififif vvlvl ,1,1,2

Continued in next slide

Fiiffifiif lvlvl ,1,1

),....,2,1(,0,1,1 Nffjvlvl ijjijiij

01, iiNiN bvl

The ’s may be eliminated by use of the equilibrium relationship. For the case of a total condenser, and

have the same composition, and thus

ji l

il0 id

ii dDL

l )( 00

For a partial condenser, and hence may be restated as follows

Dii xy 0 iii xKy 000

ii

Di xLLDK

Dx 000

0 )(

or

iii dAl 00

DKL

Ai

i0

00 Where

Continued in next slide

The expression given above may be used to represent both a partial condenser and a total condenser, provided that A0i is set equal to for a total condenser.

Also, the form of AN+1i differs slightly from that

for Aji because of the double representation of the reboiler by the subscripts N+1 and B.

Thus, the equilibrium relationship may be restated in the form

DL0

iNiNiN xKy 111

BiNiN

iNN BxBVK

yV )( 1111

iNiNi vAb 11

111

NiNiN VK

BA

or

Where

When the ‘s and are eliminated from the above MB equations, the following result is obtained

jil ib

0)1( 10 iii vdA

0)1( 2110 iiiii vvAdA

)2,......,3,2(,0)1( ,1,1,1 fjvvAvA ijjijiijij

Fifiifififif vvvAvA ,1,1,2,2 )1(

Fiiffifiifif lvvAvA ,1,1,1 )1(

),......,2,1(0)1( ,1,1,1 NffjvvAvA ijjijiijij

0)1( 11,, iNiNiNiN vAvA

(23)

This set of equations may be stated in the matrix form

iii fvA Where

Continued in next slide

Tinnifiifiiii vvvvvvdv ]......[ 1121

TFiFii lvf ]00...00...00[

)1( jiji A

The procedure proposed by Boston and Sullivan for Thomas algorithm is used here in the following form

Continued in next slide

Again after the and have been computed, the values of flow rates, are computed by use of the following equations for all components on each plate

After the recurrence formulas have been applied for each component i and the set of component vapor rates have been found, the corresponding set of liquid rates

are then calculated. These sets of calculated flow rates are used in

conjunction with the θ method of the convergence and the Kb method in the determination of an improved set of temperatures.

sf ' sg '121 ,,......,, iiiNiN vvvv

Thiele-Geddes Method

Example 1: Separation of Phenol-Cresol mixture

Feed rate, F: 100 moles/hr; Feed composition, xiF: Phenol 0.35; o-cresol 0.15; m-cresol 0.30; xylenol 0.20Feed enthalpy, Q: 1; Distillate, D: 33.1;Reflux, R: 10.0; Theoretical plates in rectifying section, NR: 16Theoretical plates in stripping section, NS: 8; Pressure drop per plate = 4mmHgPressure in the condenser = 150mmHg; Total Condenser and Partial Reboiler

INPUT DATA FILE FOR EXAMPLE 1

PHENOL-CRESOL-SEPARATION100 1 33.1 10.0 4 16 8 410 426 150 4 0 1 1 10.35 16.4279 3490.89 -98.590.15 15.9148 3305.37 -108.00.30 17.2878 4274.42 -74.090.20 16.2424 3724.58 -102.4

Output Data FilePHENOL-CRESOL-SEPARATIONnumber of components= 4Iterations=21TOP AND BOTTOM PRODUCT COMPOSITIONS

-----------------------------------------------XD XB

-----------------------------------------------0.955389 0.0502730.044574 0.2021960.000037 0.4485110.000000 0.299020

---------------------------------------------

------------------------------------------------------------------plate liquid compositons no T X1 X2 X3 X4------------------------------------------------------------------ 1 404.68 0.943438 0.056491 0.000071 0.000000 2 405.47 0.929879 0.069991 0.000130 0.000000 3 406.26 0.914587 0.085182 0.000232 0.000000 4 407.04 0.897452 0.102142 0.000405 0.000001 5 407.83 0.878386 0.120914 0.000698 0.000002 6 408.62 0.857319 0.141483 0.001193 0.000005 7 409.41 0.834203 0.163766 0.002019 0.000013 8 410.21 0.809000 0.187581 0.003387 0.000032 9 411.02 0.781665 0.212623 0.005630 0.000082 10 411.85 0.752106 0.238422 0.009266 0.000207 11 412.71 0.720131 0.264270 0.015084 0.000515 12 413.62 0.685366 0.289131 0.024239 0.001264 13 414.59 0.647136 0.311472 0.038341 0.003051 14 415.69 0.604331 0.329041 0.059424 0.007204 15 416.99 0.555308 0.338599 0.089576 0.016516 16 418.61 0.498040 0.335812 0.129817 0.036331 17 420.71 0.430971 0.315923 0.177779 0.075327 18 421.69 0.394371 0.339072 0.189805 0.076751 19 422.76 0.354032 0.360260 0.206830 0.078878 20 423.95 0.310334 0.376802 0.230557 0.082307 21 425.30 0.263991 0.385213 0.262571 0.088224 22 426.86 0.216162 0.381477 0.303462 0.098898 23 428.70 0.168544 0.361785 0.351308 0.118363 24 430.88 0.123365 0.323877 0.399796 0.152962 25 433.45 0.083167 0.268800 0.437203 0.210830 26 436.43 0.050273 0.202196 0.448511 0.299020-------------------------------------------------------------------

EXAMPLE 2

Separation of Toluene-Ethyl benzeneFeed rate, F: 100 moles/hrFeed composition, xiF: Benzene 0.022; Toluene 0.074; Ethyl Benzene 0.434; Styrene 0.470Feed enthalpy, Q: 1;Distillate, D: 5.58;Reflux, R: 4.36;Theoretical plates in rectifying section, NR: 5Theoretical plates in stripping section, NS: 3;Pressure drop per plate = 3mmHgPressure in the condenser = 160mmHg; Total Condenser and Partial Reboiler

INPUT DATA FILE FOR EXAMPLE 2

SEPERATION-OF-TOLUENE_ETHYL-BENZENE100 1 5.58 4.36 4 5 3 330 360 160 3 0 1 3 10.022 15.9008 2788.51 -52.360.074 16.0137 3096.52 -53.670.434 16.0195 3279.47 -59.950.470 16.0193 3328.57 -63.72

Output Data File

SEPERATION-OF-TOLUENE_ETHYL-BENZENEnumber of components= 4Iterations=21TOP AND BOTTOM PRODUCT COMPOSITIONS

----------------------------------------------------XD XB

----------------------------------------------------0.364715 0.0017470.485727 0.0496690.126900 0.4521480.022658 0.496435

----------------------------------------------------

---------------------------------------------------------------plate liquid compositons no T X1 X2 X3 X4--------------------------------------------------------------- 1 336.98 0.130411 0.481448 0.310103 0.078039 2 346.73 0.045858 0.338564 0.459399 0.156179 3 352.98 0.023200 0.209843 0.527835 0.239123 4 356.78 0.017361 0.134006 0.531047 0.317586 5 359.14 0.015566 0.096276 0.498194 0.389964 6 360.76 0.014831 0.078579 0.449805 0.456785 7 361.89 0.010152 0.077442 0.453027 0.459379 8 362.98 0.006537 0.073897 0.456444 0.463123 9 364.14 0.003793 0.065885 0.458979 0.471344 10 365.66 0.001747 0.049669 0.452148 0.496435-----------------------------------------------------------------------------------------------------------------------------

Holland’s Method

Output Data File for Example 1

PHENOL-CRESOL-SEPARATION

Number of Components=4

ITERATIONS=7

THETA=0.999997

TOP AND BOTTOM PRODUCT COMPOSITIONS

----------------------------------------------

XD XB

--------------------------------------------

0.955610 0.050363

0.044354 0.202270

0.000036 0.448413

0.000000 0.298954

----------------------------------------------

------------------------------------------------------------------plate liquid compositons no T X1 X2 X3 X4------------------------------------------------------------------ 1 403.96 0.943704 0.056227 0.000070 0.000000 2 404.76 0.930191 0.069681 0.000127 0.000000 3 405.56 0.914948 0.084825 0.000227 0.000000 4 406.36 0.897865 0.101738 0.000397 0.000001 5 407.16 0.878850 0.120463 0.000686 0.000002 6 407.96 0.857834 0.140989 0.001173 0.000005 7 408.77 0.834768 0.163233 0.001987 0.000012 8 409.58 0.809614 0.187016 0.003338 0.000032 9 410.40 0.782325 0.212038 0.005556 0.000081 10 411.24 0.752809 0.237830 0.009158 0.000204 11 412.11 0.720874 0.263690 0.014927 0.000508 12 413.02 0.686145 0.288584 0.024021 0.001250 13 414.00 0.647945 0.310985 0.038049 0.003022 14 415.11 0.605156 0.328641 0.059052 0.007151 15 416.41 0.556122 0.338306 0.089142 0.016430 16 418.04 0.498790 0.335628 0.129365 0.036218 17 420.15 0.431574 0.315809 0.177380 0.075237 18 421.13 0.394997 0.339009 0.189341 0.076653 19 422.21 0.354670 0.360271 0.206291 0.078768 20 423.40 0.310965 0.376910 0.229946 0.082179 21 424.76 0.264590 0.385430 0.261910 0.088070 22 426.33 0.216697 0.381790 0.302800 0.098713 23 428.17 0.168982 0.362149 0.350721 0.118148 24 430.35 0.123684 0.324217 0.399362 0.152737 25 432.93 0.083361 0.269033 0.436958 0.210648 26 435.92 0.050363 0.202270 0.448413 0.298954-------------------------------------------------------------------

Output Data File for Example-2

SEPERATION-OF-TOLUENE_ETHYL-BENZENE

Number of Components=4

ITERATIONS=12

THETA=0.999995

TOP AND BOTTOM PRODUCT COMPOSITIONS

-----------------------------

XD XB

----------------------------

0.365042 0.001727

0.486792 0.049605

0.125836 0.452212

0.022330 0.496456

-----------------------------

--------------------------------------------------------------plate liquid compositons no T X1 X2 X3 X4-------------------------------------------------------------- 1 336.44 0.130484 0.483426 0.308783 0.077307 2 346.19 0.045840 0.340179 0.458695 0.155286 3 352.46 0.023165 0.210704 0.527798 0.238333 4 356.27 0.017322 0.134361 0.531299 0.317018 5 358.64 0.015526 0.096394 0.498429 0.389652 6 360.28 0.014790 0.078602 0.449871 0.456737 7 361.42 0.010100 0.077454 0.453105 0.459340 8 362.51 0.006488 0.073888 0.456532 0.463091 9 363.69 0.003757 0.065848 0.459070 0.471325 10 365.20 0.001727 0.049605 0.452212 0.496456---------------------------------------------------------------

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Continued in next slide

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Continued in next slide

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Continued in next slide

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Thank You