Chemical Engineering Technology IV: Unit operations MODULE C Only study guide for CEM4M3CE±E/1/2006±2008 PDF

Preview

Summary

Chemical Engineering Technology IV: Unit operations MODULE C Only study guide for CEM4M3CE±E/1/2006±2008 Complied byH.G.J. Potgieter Moderated by: Dr. M Smit UNIVERSITY OF SOUTH AFRICA PRETORIA # 2005 University of South Africa All rights reserved Printed and published by the University of South Afr...

Description

Chemical Engineering Technology IV: Unit operations

MODULE C

Only study guide for CEM4M3CE±E/1/2006±2008

H.G.J. Potgieter Moderated by: Dr. M Smit

Complied by

UNIVERSITY OF SOUTH AFRICA PRETORIA

# 2005 University of South Africa All rights reserved Printed and published by the University of South Africa Muckleneuk, Pretoria CEM4M3C±E/1/2006±2008 97732354 3B2

In accordance with the Copyright Act 98 of 1978 no part of this material may be reproduced, republished, redistributed, transmitted, screened or used in any form without prior written permission form UNISA. Where materials have been used from other sources permission must be obtained directly for the original source.

A4 6 pica Style

(iii)

LCP409-R/2/2006-2008

Contents

Chapter

Page

1

DISTILLATION

00

2

MULTICOMPONENT DISTILLATION

00

3

RIGORIOUS DISTILLATION DESIGN METHOD

00

4

EVAPORATION

00

5

ADSORPTION

00

6

CRYSTALLISATION

00

8

MULTICOMPONENT ABSORPTION/STRIPPING

00

REFERENCES

00

SUPPLYMENTARY MATERIAL

00

1

CEM4M3-C/1

CHAPTER 1

Distillation CONTENTS 1.1 1.1.1 1.1.2 1.1.3 1.1.4 1.1.5 12 1.2.1 1.3 1.3.1

INTRODUCTION Objectives McCabe ± Thiele Method Minimum Reflux Ratio, Rm Number of stages at total reflux Batch Distillation MULTIPLE FEED AND SIDE STREAMS Objectives PONCHON ± SAVARIT METHOD FOR TRAY TOWERS(2) Objective

00 00 00 00 00 00 00 00 00 00

1.1 INTRODUCTION 1.1.1 Objectives Brief revisions of the McCabe ± Thiele method and batch distillation are given in this section Refer to the following sketch (1) of a distillation column that operates with a total condenser and a reboiler that vapourises a part of the liquid that leaves the bottom stage. When a partial condenser is used the top product would be mixture of vapour and liquid. In the sketch the more volatile component is referred to as the light key (LK) and the less volatile component as the heavey key (HK). In multi component systems the LK is the most volatile component in the bottom product (bottoms) and the HK the least volatile component in the top product (distillate).

1.1.2 McCabe ± Thiele Method This method is based on the assumption of constant ± molar ± overflow (equi ± molar overflow). The liquid and vapour molar flow rates in the top part of the column (the rectification section) do not change from stage to stage. This is also the case for the bottom part (the stripping section) but the flow rates can be different from that in the rectification section. It is assumed that equilibrium is attained in each stage and such a stage is called an equilibrium stage. The vapour that leaves the partial reboiler is also assumed to be in equilibrium with the liquid that leaves it. The reboiler is thus considered to be a n equilibrium stage. The vapour leaving the reboiler is called the boilup. The following specifications are required to use this method successfully:

2

Figure 1.1: J.D. Seader and E.J. Henley The total feed rate, F. The mol fraction of a component (normally the light one) of the feed, ZF. The phase condition of the feed at the column pressure. Vapour ± liquid equilibrium data. The mol fraction of the light component of the distillate, XD. The mol faction of the light component of the bottoms, XB. The reflux ratio, R or a factor times Rm (minimum reflux ratio). The type of condenser (partial or total) and the type of reboiler (normally partial). The relationship between vapours and liquids at equilibrium is frequently expressed by: y ˆ Kx . Where y = mol fraction of light component in the vapour x = mol fraction of light component in the liquid Various methods are available for estimating the K ± values. The relative volatility, a, indicates the ease or difficulty with which components can be separated.

3 a1;2 ˆ

K1 K2

CEM4M3-C/1 where 1 refers to light key and 2 to the heavey key

The closer a is to 1 the more difficult the separation. It can be assumed that Raoult's law applies when the components form ideal solutions and ideal gas law applies in the vapour phase ± thus: Ps Ps Ps K1 ˆ 1 and K2 ˆ P2 and a1;2 ˆ Ps1 where P1s and P2s are the vapour pres2 P sures. It can be shown that: y1 ˆ

a1;2 x1 1 ‡ x1 …a1;2 ÿ 1†

(1)

The amounts of distillate, D and bottoms, B are found by doing a molar balance. The reflux ratio, R = Ln/S determines the liquid flow rate, Ln which remains constant in the rectification section. The vapour flow rate is given by Ln + D which also remains constant in the rectification section. It can be shown that the top operating line is given by: yn ˆ

Ln D R xD xn‡1 ‡ xn‡1 ‡ xD or yn ˆ Vn R‡1 Vn R‡1

(2)

 xD  Equation (2) is a straight line passing through (xD ; xD ) and 0; R‡1 The bottom operating line is given by: ym ˆ

Lm B xm‡1 ÿ xB VM Vm

(3)

Lm . Equation (3) is also a straight line passing through (xB ; xB ) with a slope of V m It is worth remembering that: the compositions of the vapour and liquid leaving a stage is obtained from the equilibrium curve and that the composition of the vapour entering a stage in terms of the liquid leaving stage is given by the operating line. the physical condition of the feed determines the flow rate of the liquid flowing from the feed tray to the stripping section. If the feed is for instance a liquid at its boiling point Lm ˆ Ln ‡ F. The quantity, q is defined as heat to vapourise 1 mol of feed. molar latent heat of the feed It can be shown that the equation of the q ± line is given by: yq ˆ

zf q xq ÿ qÿ1 qÿ1

Equation (4) passes through (xf ; xf ) with a slope of When the feed is (a) a cold liquor q > 1 slope is positive (b) liquor at boiling point q = 1 slope is vertical

q : qÿ1

(4)

4 (c) partly vapour 0 < q < 1 (d) saturated vapour q = 0 (e) superheated vapour q < 0

slope is negative slope is horizontal slope is positive

Procedure 1. 2. 3. 4. 5. 6.

7. 8. 9. 10.

Plot equilibrium curve Draw 458 line Draw top ooperating line Draw q ± line Draw bottom operating line by connecting (xB ; xB ) with the intersection of the top operatiang line and the q ± line. Draw a horizontal line from (xD ; xD ) to the equilibrium curve ± drop a vertical line from this intersection to the top operating line. This completes the determination of the first stage. Repeat this procedure until a vertical line from the equilibrium curve has to be dropped to the bottom operating line (past (xf ; xf ). The above procedure is carried out till a vertical line from the equilibrium curve passes (xB ; xB ). Count the number of theoretical stages. Number of theoretical stages/efficiency = number of actual stages. Number of actual stages ± 1 = number of actual trays.

Example 1 200 k mol/h of a mixture containing 55 mol% benzene and 45 mol% toluene is fed to a continuous distillation column. The feed is at its boiling point. a = 3,09 and the reflux ration is 1,6. The distillate must contain 95 mol% benzene and the bottoms 5 mol% benzene. Determine: (a) the number of theoretical stages (b) the feed stage (c) the number of actual trays if the overall efficiency is 60%. F = 200 = D + B

(1)

200 6 0,55 = 110 = 0,95 D + 0,05 B

(2)

Substitute (1) in (2) y ˆ

B = 88,9

D = 111,1

3; 09x 1 ‡ 2; 09x x

y

0

0

0,2

0,435

0,4

0,67

0,6

0,82

0,8

0,92

1,0

1,0

Top operating line through (0,95; 0,95) and (0; 0,95/2,6) = (0; 0,365).

5

CEM4M3-C/1

q ± line is vertical through (0,55; 0,55). Bottom operating line throught (o,55; 0,05) and intersection of q ± line and the top operating line. The construction is shown below. (a) 9 theoretical stages are required (b) the theoretical feed stage is number 4 from the top (c) 9/0,6 ± 1 = 14 actual trays

1.1.3 Minimum Reflux Ration, Rm The minimum reflux ration is determined by drawing a top operating line from (xD ; xD ) to the intersection of the q ± line and the equilibrium curve. A separation requires an infinite number of stages at Rm and can thus not be used in practice. Rm is, however, used as a starting point and an actual R would be Rm multiplied by a factor that is much bigger than one. In the above example the line passes through (0,95; 0,95) and (0,55; 0,78). R The slope of the top operating line is given by: . R‡1 The slope of this line is

Rm 0; 95 ÿ 0; 78 ˆ 0; 425 thus Rm ˆ 0; 74. ˆ 0; 95 ÿ 0; 55 Rm ‡ 1

6

Problems 1. Repeat the above example but with a feed that is 60% vapour. Answer: 9+ theoretical stages; 5 theoretical feed stage. 2. A mixture that contains 40 mass % benzene and 60 mass % ethyl- benzene must be separated into a distillate containing 95 mol % benzene and a bottoms containing 5 mol % benzene. The feed is a liquid at 308C. The bubble point of the feed is 1048C its latent heat of vapourisation is 36300 KJ/kmol and its specific heat is 160 kJ/kmol K. Determine Rm and the number of actual trays if the efficiency of the trays is 55% and R = 1,5 Rm. a = 6,8 Answer: Rm = 0,318; number of trays = 12

1.1.4 Number of stages at total reflux It is implied in this case that no products are withdrawn and is thus of no practical value. It can, however, be used as a starting point in distillation calculations. The two operating lines merge with the 458 line and stages are stepped off from xD to xB .

1.1.5 Batch distillation This is an unsteady state distillation process that is frequently used for small scale operations. This type of column consists of a boiler (also called the still) on top of which a distillation column is installed. A whole batch is charged to the boiler. The vapour is condensed and part of the condensate is returned as reflux. As the distillation process proceeds the composition in the boiler changes continuously. This results in the decrease of the lighter component in the distillate. In order to maintain a constant distillate composition the reflux ration can be adjusted continuously or the column can be operated initially with a higher concentration of the light component at a given reflux ratio. This ratio is kept constant which will result in a lower concentration of the light component. The distillation is stopped when the required distillate composition is given by the average. A batch distillation column is only fitted with a rectification section. The operating line of a rectification section thus applies. Only the constant reflux method will be considered here. Consider the boiler to be initially charged with S1 mols of liqid with a mol fraction x1 of the light component. The composition of the distillate is xD with R1 the reflux ratio. The distillation is stopped when S2 mols remain with mol fraction x2 , It is necessary to increase the reflux ratio to R2 in order to maintain the distillate composition at xD if the number of trays remains the same. A total mol balance gives: S1 ÿ S2 ˆ D A light component balance gives: S1 xs1 ÿ S2 xs2 ˆ D xD hx ÿ x i s1 s2 From these equations it follows that D ˆ S1 xD ÿ xs2 xD = A(say). The intercept of the operating line on the Y ± axis is R‡1

(5)

7

CEM4M3-C/1

xD ÿ1 (6) A Equations (5) and (6) allows one to determine the final reflux ratio that is required to obtain a given final concentration in the boiler and the quantity of distillate. Z x s2 S1 dxs ˆ . It can also be shown that: In S2 xD ÿ xs x Thus R ˆ

s1

Example 2(1) A batch disttillation column with three theoretical stages (the boiler being the first) is charged with 100 kmol of a mixture of containing 32 mol & n ± hezane and the rest n ± octane. average xD ˆ 0; 6 and constant R = 1,0. Determine the amount of distillate if the final mol fraction of n ± hexane in the boiler is 0,1. a ˆ 3; 7. The equilibrium and 458 line are plotted. A trial ± and ± error procedure is then required. Assume that the first xD = 0,85 A = 0,425 connect (0,85; 0,85) with this A ± value. Step off three stages starting from xs2 = 0,32. The third stage ends with (0,85; 0,85). The diagram and the table below illustrate the rest of the procedure.

8 xs

xD

A

0,32

0,85

0,425

1,9

0,16

0,6

0,3

2,3

0,1

0,5

0,25

2,5

0,05

0,3

0,15

4

1 xD ÿ xs

1 is shown in the follwing graph. The area is determined xD ÿ xs between xs = 0,32 and 0,1 and found to be 0,532.

The plot of xs versus

S1 In = 0,532 thus S2 = 58,7 and D = 41,3 S2 DxD ˆ S1 xs1 ÿ S2 xs2 ˆ 32ÿ 5,87 thus xD ˆ 0; 63

9

CEM4M3-C/1

1.2 MUTIPLE FEED AND SIDE STREAMS 1.2.1 Objectives The methods that are required to solve these types of problems are presented here. The McCabe ± Thiele method can also be used for multiple feeds and/or side streams. A multiple feed system is shown below.

The equation of the operating line in the rectification section of a seciton of a column with one feed is given by: Ln D yn ˆ xn‡1 ‡ xD Vn Vn The slope of this equation is

Ln L ˆ . V Vn

This remains the slope of the operating line above F1 if there are no side streams. The slope of the operating line between F1 and F2 (no F) is given by L'/V' and it intersects the top operating line at the intersection of the q ± line and the top operating line. The operating line of the stripping section is again drawn from (xB ; xB ) and the intersection of the second q ± line and the second top operating line. Refer to the sketches(1) below.

10

Figure 1.2: Page 389, Seader In the figure on the left F1 is a saturated vapour at its dew point while F2 is a liquid at its dubble point. The third feed stream and side stream are not present. V' = V ± F1 or V = V' + F1 as F1 is in the vapour phase.  = L' + F2 = L + F2 L The figure on the right shows a side stream Ls that is withdrawn as a satureated liquid between the top of the column and the feed. Side streams may be withdrawn from the rectification and stripping sections as saturated vapours or saturated liquids. L' = L ± Ls and V' = V The equations of the operating lines will now be drived and the constructions will be illustated. Refer to the sketches(1) below.

11

Figure 1.3: Page 390, Seader

Figure 1.4: Page 390, Seader

CEM4M3-C/1

12 A material balance over section 1 gives: Vnÿ1 Ynÿ1 ˆ Ln xn ‡ DxD

(7)

A material balance over section 2 gives: Vsÿ1 Ysÿ1 ˆ L0sÿ1 xsÿ1 ‡ Ls xs ‡ DxD

(8)

These equations can be simplified to: y ˆ

L0 Ls xs ‡ DxD L D x ‡ and y ˆ x ‡ xD V V V V

(9)

By equating the two equations of (9) the intersection of the two operating lines are found to be at x ˆ xs L0 Ls xs ‡ DxD The intersection of y ˆ x and y ˆ x ‡ occurs at V V Ls xs ‡ DxD (10) x ˆ Ls ‡ D

Example 3(2) A mixture of H2O and ethyl alcohol (EtOH) contianing 0,16 mol fraction EtOH is continuously distilled in a tray fractionation column to give a product containing 0,77 mol faction EtOH and a wast containing 0,02 mol fraction EtOH. It is proposed to withdraw 25% of the EtOH in the feed as a liquid side stream with a mol fraction of 0,5 EtOH. Determine the number of theoretical trays required and the tray from which the side stream should be withdrawn if the feed is a liquid at its bubble point and the reflux ratio is 2. x

0,019

,072

,097

,124

,166

,234

,261

,327

,396

,508

,52

,57

,676

,747

,894

y

0,17

,389

,437

,47

,509

,544

,558

,583

,612

,656

,66

,68

,738

,781

,894

Basis 100 kmol feed EtOH in feed = 1006 0,16 = 16 kmol 25% of EtOH in side stream = 4 kmol ; Water in side stream = 4 kmol Thus Ls = 8 kmol and xs = 0,5 Top operating line: coordinates: (0,77; 0,88) and (0; 0,77/3) = (0; 0,257) Overall balance: F = D + L + B = 100 = D + 8 + B Thus

92 = D + B

(1)

EtOH Balance 16 = 0,77 D + 4 + 0,02 B

(2)

13 With (1) and (2) it is found that

CEM4M3-C/1

B = 78,45 kmol D = 13,55 kmol

L = 2 D = 2 6 13,55 = 27,1 kmol V= D + L + 13,55 + 27,1 = 40,65 kmol = V' L' = L ± Ls = 27,1 ± 8 = 18,1 kmol Second Operating Line: yˆsˆ

Ls xs ‡ DxD 8  0; 5 ‡ 13; 55  0; 77 ˆ 0; 67 ˆ 8 ‡ 13; 55 Ls ‡ D

Second coordinate where xs = 0,5 intersects the top operating line Bottom Operating Line: Coordinates (0,02; 0,02) and where x ˆ zF = 0,16 intersects the second operating ling. From the construction below one finds that the number of theoretical trays are 9 and the side steam is withdrawn from the 6th theoretical tray from the top.

14

Problems 3.(1) Two feed streams containing water and acetic acid are fed to a continuous distillation column. Feed 1 enters as a liquid at its bubble point relatively high up in the column and contains 75 kmol/h water (W) and 25 kmol/h acetic acid (A). The second feed, F2 enters lower down, is 50% vapour and contains 50 kmol/h W and 50 kmol/h A. The column is operated at a reflux ratio of 3,0 Rm. The distillate contains 95 mol % W while the bottom product contains 95 mol % A. Determine the number of theoretical trays and the feed trays. x

,0055

,053

,125

,206

,297

,51

,649

,803

,9594

y

,0112

,133

,24

,338

,437

,63

,751

,866

,972

Answer: 16; 9; 13 4. Repeat example 2 by using the relevant equations. That is the equilibrium and operating line equations. Answer: D = 38,3; S2 = 61,7; xD = 0,67 5.(1) Determine the number of theoretical stages and the locations of the feed and side stream when 100 kmol/h of a mixture of A and B is distilled at atmospheric pressure. The mol fraction of A in the feed is 0,26. The distillate contains 95 mol % A and the bottoms 95 mol % B. The side stream is withdrawn as a liquid from the rectification section at a rate of 10 kmol/h that contains 40 mol % A. Relative volatility is 2,23 and reflux ratio is 5. Answer: 14; 6; 9±10

1.3 PONCHON ± SAVARIT METHOD FOR TRAY TOWERS(2) 1.3.1 Objective This method can be used to design distillation columns for binary, non ± ideal systems where equimolar overflow is invalid. The McCabe ± Thiele method assumes constant molar overflow which implies that the molar latent heats are constant and that heat of mixing is negligible. For non-ideal systems where the above assumptions are not valid, energy as well as material balances and phase equilibrium relationships have to be utilized to do a proper design. This can be a very tedious process (except that rig...