Assignment title: Information

MODULE TITLE : MASS AND ENERGY BALANCE TOPIC TITLE : EXAMPLES OF MASS BALANCE LESSON 1 : PHYSICAL CHANGES MAEB - 2 - 1 © Teesside University 2011 Published by Teesside University Open Learning (Engineering) School of Science & Engineering Teesside University Tees Valley, UK TS1 3BA +44 (0)1642 342740 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording or otherwise without the prior permission of the Copyright owner. This book is sold subject to the condition that it shall not, by way of trade or otherwise, be lent, re-sold, hired out or otherwise circulated without the publisher's prior consent in any form of binding or cover other than that in which it is published and without a similar condition including this condition being imposed on the subsequent purchaser. 1 ________________________________________________________________________________________ INTRODUCTION ________________________________________________________________________________________ In this lesson, and the next, we will look at solving examples of mass balances involving both non-reacting (physical changes only) and reacting (chemical changes) systems. The principles used to solve the problems can be applied to most processes as we will be simply using blocks to indicate processes, thus the method is independent of the process actually occurring. The problems will all be steady state problems where the conditions within the process do not change and will involve both overall and component mass balances. We will start off with simple balances and move onto more complex ones in the course of the lesson. Some of the work will reinforce the work already done in earlier lessons. ________________________________________________________________________________________ YOUR AIMS ________________________________________________________________________________________ At the end of this lesson, you should be able to solve mass balance problems involving: • only physical changes • recycle streams • by-pass streams. Teesside University Open Learning © Teesside University 2011 2 ________________________________________________________________________________________ STUDY ADVICE ________________________________________________________________________________________ The examples will involve both simple mathematics and the use of simultaneous equations and you should be familiar with solving these types of equations. You will also need a calculator to solve the problems in this lesson. Teesside University Open Learning © Teesside University 2011 3 ________________________________________________________________________________________ SIMPLE EXAMPLES ________________________________________________________________________________________ Example 1 150 kg min–1 of a mixture containing 55% w/w methanol, 45% water is to be separated in a distillation column to give a top product containing 100% methanol and a bottom product containing 100% water. Calculate the flow rate of top (T) and bottom (B) product in kg s–1. Solution Let us represent this on a block diagram. The answer is fairly simple and you may have already solved it but we will go through the principles. T kg s–1 100% methanol 150 kg min–1 55% w/w methanol 45% w/w water Distillation B kg s–1 100% water The first step is ensure the units are consistent. The feed rate is kg min–1 and the products flow rates are required in kg s–1 so we need to convert the feed to 150 150 60 kg min kg s feed 2 5 – – 1 1 = = . Teesside University Open Learning © Teesside University 2011 4 Let us first look at an overall mass balance. Overall mass balance i.e. feed in products out = 2 5. ........ = + T B ................................... 1( ) This does not help much as there are 2 unknowns, T and B, so let us do a mass balance over methanol. Balance over methanol methanol in feed methanol in products = ed is 55% w/w methanol 55% of 2.5 kg s 1 = = – 1. – 375 kg s So meth 1 anol in kg s 1 = 1 375 . – All the methanol leaves as the top product so methanol out = T Therefore kg s 1 T = 1 375 . – From equation 1: 2 5 = + . = + B = = T B 1 375 B . 2 5 1 375 .–. kg s 1 1 125 . – Teesside University Open Learning © Teesside University 2011 5 We can check this by doing a mass balance over the water: feed is 45% w/w water 45% of 2.5 kg s 1 = – = 1 125 s . –1 kg As all the water leaves as 100% bottom product then B = 1.125 kg s–1 (the same as calculated earlier). IN (kg s–1 Material ) OUT (kg s–1) Total (kg s–1) Methanol 1.375 1.125 2.5 TOP 1.375 0 1.375 BOTTOM 0 1.125 1.125 1.375 1.125 2.5 Let us slightly complicate matters! The process is not performing well and on measuring the bottom product flow and concentration we find 1.15 kg s–1 of bottom product is leaving and it contains 5% w/w methanol. What is the new top product rate and concentration? Doing an overall mass balance feed in products out = 2.5 =+=+ T = T BT . – .5 1 35 = . – kg s 1 25 11 1 15 . Let us do a balance over the methanol (we could have done it on the water and we will do this later as a check) Teesside University Open Learning © Teesside University 2011 6 methanol in methanol in top (unknown?) meth = + anol in bottom ........... 2( ) methanol in: bottom o 1 = = ×= feed 55%of 2.5 0.55 2.5 1.375 kg s 5 f %1.15 0.05 1.15 0.0575 kg s 1 =×= – = – Therefore, substituting in equation (2) 1 375 . = + methanol in top 0.0575 methanol in top = 1 3175 . – kg s 1 This is in a total of 1.35 kg s–1 (T) calculated earlier. concentration 1.3175 = × 1.35 97 6.% anol (2.4% w/w water) w/w meth = 100 We can check this by doing a balance on the water. water in water in bottom water in top 2.5 0. × 45 1.15 0.95 1.35 0.024 1.125 1.0925 0.0324 = + =×+× =+= 1.1249 The difference is due to rounding errors. Teesside University Open Learning © Teesside University 2011 7 IN (kg s–1 Material ) OUT (kg s–1) Total (kg s–1) Methanol 1.375 1.125 2.5 TOP 1.3175 0.0324 1.35 BOTTOM 0.0575 1.0925 1.15 1.375 1.1249 2.5 Let us do another example: Example 2 A distillation column is required to separate a mixture containing 46% w/w X and 54% w/w Y to produce a top product containing 98 % w/w X and a bottom product of 95% w/w Y. Calculate the flow of top (T) and bottom (B) product. Solution The first thing is draw a block diagram with the given information. T 98% w/w X 2% w/w Y Feed (F) 46% w/w X 54% w/w Y Distillation B 5% w/w X 95% w/w Y Overall mass balance FT B = + Teesside University Open Learning © Teesside University 2011 8 Mass balance over X 46% w/w of of 5% of i.e. 0 46 F TB = + 98 . FT B = + 0 98 0 05 . . % Mass balance over Y 54% w/w of of 5% of i.e. 0 54 FT B = + 2 9 . FT B = + 0 02 0 95 . . % We are not given any of the values of F, T and B so how do we start? Although we have 3 equations and three unknowns, we cannot solve these as simultaneous equations as they are not all independent; adding the second and third together will give the first equation. So we are going to assume a feed F of 100 kg. This amount is irrelevant as we shall see later, but we have chosen 100 as it is a nice round number – it is actually easy to work with percentages when dealing with 100! 100 100 TB T B =+ = 0 46 100 0 98 0 05 ×= + . .. or ± T B but T = 100 – B, so substitute for T × = ( ) + 0 46 100 0 98 100 0 05 . . –. 0 93 52 46 98 0 98 0 05 = + B = . –. . B B B B B = = kg 0 93 52 55 91 . . Teesside University Open Learning © Teesside University 2011 9 (At this point if we had not used 100 as feed but left it as F, we would have 0 52 B = , i.e. 55.91% of F. ) 0 93 . F 100 T B = 100 55 91 44 09 T = = – – . . kg As we started with 100 kg of feed the % of the feed which becomes top product T =× = × = F 100 44 09 100 100 44 09 . . % and % of the feed which becomes bottom product B =× = × = F 100 55 91 100 100 55 91 . . % As we now have a % of feed converted, we can calculate the actual amounts of T and B for any value of F, hence the reason we said earlier that the value of F was irrelevant. Can you check this answer by doing a balance over Y using a feed of 50 kg? ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ________________________________________________________________________________________ Teesside University Open Learning © Teesside University 2011 10 50 50 0 54 50 0 02 0 95 ×= + . .. or – TB T B =+ = T B Substitute 50 – B for T × = ( ) + 0 54 50 0 02 50 0 95 . .– . 27 1 0 02 0 95 0 9 3 26 B . B B B B B T B T = + –. . = 26 = = . . kg 27 96 0 93 50 = – 50 27 96 22 04 – . . kg = = As we started with 50 kg of feed the % of the feed which becomes top product 100 27 96 T =× = × = F 100 55 92 . . % 50 And % of the feed which becomes bottom product 100 22 04 B =× = × = F 100 44 08 . . % 50 The slight difference in % between balances over X and Y is due to rounding errors. Teesside University Open Learning © Teesside University 2011 11 ________________________________________________________________________________________ MORE COMPLEX EXAMPLES USING RECYCLE STREAMS ________________________________________________________________________________________ Example 3 An extraction process is used to separate sugar from sugar beet by mixing with hot water to dissolve the sugar and then filtration to remove the insoluble beet. 200 kg of sugar beet containing 12% w/w sugar, 88% w/w insoluble residue is to be processed using 100 kg of water. The beet residue after filtration contains 95% insoluble residue, water and un-dissolved and dissolved sugar. The sugar solution is 18% w/w sugar in solution. Calculate the amount of: (a) beet residue after filtration (b) solution formed (c) un-dissolved sugar in the residue. Solution As always the best starting point is to draw a block diagram. Can you do this in the space below? ________________________________________________________________________________________ Teesside University Open Learning © Teesside University 2011 12 You may have drawn this as two units, one block which mixes the water and the beet and then another for the filter used to separate the mixture. This is not wrong and may be useful in some processes, however, we are going to combine both into one mix and filter block as shown below: 200 kg sugar beet 12% w/w sugar Beet (95% w/w insoluble residue) Mix and filter 100 kg water Solution (18% w/w sugar) (a) Let the unknown amount of beet on the filter be R and the unknown amount of solution be S, Doing an overall mass balance: mass in mass out + = += R S 300 ..... ......... ............ 3( ) 200 100 kg .......... = Doing a mass balance over the sugar sugar in sugar out in 12% of 200 kg 0.12 200 out 18% of unknown % of So unknown % of 24 0 18 = + . S R ..................... 4( ) = = = ×= = + 24 kg S R We cannot solve this as there are three unknowns R, S and % of sugar in residue. Teesside University Open Learning © Teesside University 2011 13 Doing a mass balance over insoluble beet residue insoluble residue in insoluble residue out = esidue in of 200 0.88 200 176 kg = = ×= 88% r The insoluble residue only appears in the beet on the filter and forms 95% w/w of R. So residue out of 95 0 95 % . = = R R out in 0 95 176 R . ∴== kg R 176 = kg = 185 26 . . 0 95 By experience and looking at the question we should have done the balance over insoluble residue first as there is only one stream in and one stream out that contains the residue which helps make the balance easier. However, it is never a waste doing balances as we can use these as checks or to solve other parts of the problem later on. (b) From equation 3 R S + = kg 300 300 185 26 114 74 S R ∴= = = 300 – –. . lution produced kg = 114 74 . so Teesside University Open Learning © Teesside University 2011 14 (c) From our sugar balance (equation 4), 18% of S is sugar = 0.18 × 114.74 = 20.65 kg Thus, amount of water in solution 114.74 = – 20.65 = 94.09 kg From equation 4 unknown % of kg unknown % 24 0 18 = + 24 20 65 = + R = = 24 20 65 3 35 –. . == = 3.35 3.35 S R unknown % of . unknown % of . R R 185 26 1 81 . . % The rest of the beet on the filter must be water as no other materials can be present i.e. % water 100 95 (% insolubles) 1.81 = – –(% sugar) = 3.19% Mass of water in beet on filter 3.19% of = R 0 0319 185 26 5 91 . .. × = kg = Teesside University Open Learning © Teesside University 2011 15 This water will contain 18% w/w sugar, the same as the solution formed 5 91 . kg of water 82% and dissolved sugar is 1 = 8% 5.91 18 82 = kg × = 1 30 . Thus the amount of un-dissolved sugar = total sugar in beet on filter dissolved su – gar = = 3.35 1.30 2.05 kg – Summarising these in a table: Material IN (kg) Insoluble beet 100 176 300 in solution 94.09 0 20.65 24 114.74 OUT (kg) in beet left on filter 5.91 176 3.35 (1.3 dissolved, 2.05 un-dissolved) 185.26 Total out 100 176 24 300 The figures balance. Teesside University Open Learning © Teesside University 2011 16 Try to solve the next problem for yourself before checking our solution. Example 4 An extraction process uses a solvent (T) to remove a solute A from a mixture of 15% w/w A in S. After separation, the original mixture – known as raffinate – is required to have only 2% w/w A and the solution produced – known as extract – is required to be 20% w/w A in T. The feed to be separated flows into the process at 500 kg min–1. Calculate the amount of: (a) raffinate leaving the process (b) extract leaving the process (c) solvent T required. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ________________________________________________________________________________________ Teesside University Open Learning © Teesside University 2011 17 Solution First the diagram! Solvent T X kg min–1 Feed (F) 500 kg min–1 15% w/w A 85% w/w S Extraction Raffinate R kg min–1 2% w/w A 98% w/w S Extract E kg min–1 20% w/w A 80% w/w T Overall balance FXER +=+ 500 +=+ XER (a) We now need to look at the situation to see if we can easily calculate one of the components. We can see that S is only present in the feed in and only leaves in the raffinate and T enters as 100% purity and leaves only in extract. Either of these would be a good starting point but we do not have figures for T but do for S in. So let us do a balance over S. Teesside University Open Learning © Teesside University 2011 18 ∴ = and S in 85% of feed 0.85 500 425 kg min S out 9 0 98 425 R . kg min R 1 = = 425 1 = = ×= 8% of = – R R 0 98 = . 433 67 . . – 0 98 The amount of raffinate (R) = 433.67 kg min–1. (b) From this balance, we can now calculate the amount of A in the raffinate, since we have found that 425 kg min–1 were S and total flow is 433.67 kg min–1. The difference is A = 8.67 kg min–1. (Check 2% of R = 0.02 × 433.67 = 8.67 kg min–1) We can now do a balance over component A. Balance over A in 15% of feed 0.15 500 75 kg min out A in e 1 = = ×= xtract (20% of ) A in raffinate = (8.67 kg min from above) E . . – E + –1 75 0 2 8 67 ∴=+ 0 2 E 75 8 67 66 33 –. . – = kg min . 1 = (this is amount of A in extract) kg min E 1 = = 66 33 331 65 . . . – 0 2 Thus the amount of extract = 331.65 kg min–1. Teesside University Open Learning © Teesside University 2011 19 (c) We now have values for R, E and feed rates so returning to our overall balance FXER +=+ 500 331 65 433 67 765 32 X += + = X ... . –00 265 32 = . – kg min 1 765 32 5 = Thus the amount of solvent required is 265.32 kg min–1 As a check let us do a balance over T T out is 80% of X = = solvent T in 265.32 kg min extract 0.8 331.65 265.32 kg min 1 =× = – –1 So this balances. We can summarise the results in a table. Material IN (kg) 75 425 265.32 765.32 in extract 66.33 0 265.32 331.65 OUT (kg) in raffinate 8.67 425 0 433.67 Total out 75 425 265.32 765.32 Teesside University Open Learning © Teesside University 2011 20 Example 5 In the process described above, the extract is now passed to an evaporator which separates the solvent T from the solute A. The process is not 100% efficient and 2% w/w T is left in the solid A. The purified solvent is recycled back to the process. Calculate the amount of recycled solvent and fresh solvent make-up required. Solution Modifying the diagram to account for these changes gives us Solvent make-up Solvent 265.32 kg Feed (F) 500 kg min–1 15% w/w A 85% w/w S Extraction Raffinate 433.67 kg min–1 2% w/w A 98% w/w S Recycled solvent 98% w/w A 2% w/w T Extract 331.95 kg min–1 20% w/w A 80% w/w T Evaporation Nothing else has changed except the addition of the evaporation system and solvent make-up so we can just concentrate on that. Let the amount of solvent removed in the evaporator be x kg min–1 and the amount of product removed = y kg min–1. Teesside University Open Learning © Teesside University 2011 21 265.32 kg min Solvent make-up –1 x kg min–1 100% T Extract 331.65 kg min–1 66.33 kg min–1A 265.32 kg min–1T y kg min–1 Evaporation 98% w/w A 2% w/w T Let us do all possible mass balances over the evaporator as the first step, to see what they reveal. Overall mass balance extract feed solvent evaporated product rem = + oved 331.65 = + x y Mass balance over A amount in extract amount in product = 66.33 = × y 0 98 . Mass balance over T amount in extract amount removed for solven = t recycle amount in product 265.32 = + x 0.02y + Teesside University Open Learning © Teesside University 2011 22 Looking at these, the easiest one to solve is the balance over A, 0 98 66 33 . . y = y = 67 68 . – kg min 1 which gives us: From the overall balance 331 65 67 68 = + . . x kg min 1 = = x 331 65 67 68 263 97 .–. . – Thus the amount of solvent recycled = 263.97 kg min–1. From earlier, the solvent required to be fed into the extractor = 265.32 kg min–1 Thus solvent make-up = 265.32 – 263.97 = 1.35 kg min–1 How can we check this? ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ________________________________________________________________________________________ Teesside University Open Learning © Teesside University 2011 23 As a check do the balance over the evaporator for T 265.32 = 263.97 + 0.02 × 67.68 = 263.97 + 1.35 = 265.32 Material IN (kg) OUT (kg) Total out 75 425 1.35 501.35 In raffinate 8.67 425 0 433.67 66.33 1.35 67.68 In product 0 75 425 1.35 501.35 Teesside University Open Learning © Teesside University 2011 24 ________________________________________________________________________________________ BY-PASS STREAMS ________________________________________________________________________________________ A final example, this time involving a by-pass stream. Example 6 In an industrial process a solid mixture is made up of two solids of different particle size. The original mixture consists of 83% w/w A (which has a large particle size) and 17% B which has a smaller particle size. A and B can be separated by using a sieve which retains most of the large particles but allows all the smaller particles to pass through. The composition of the material passing through the sieve is 5% w/w A, 95% w/w B. The final product required is a 50% w/w mixture of A and B. Determine the amount and composition of the material that is not used in the process and the minimum amount of material required to be sieved. Solution We could actual sieve the whole material and then remix the calculated amounts of the individual streams to give us the required product. This would, however, involve a larger sieve and/or a longer time for the sieving operation, both of which involve additional costs. We could sieve only part of the material and allow some to bypass the sieve, mixing the streams at the end in the right proportion to give us the required product. Since the original material has a much higher % of A than that required in the final product, A will need to be removed in the system. If possible, removing this as 100% A would be ideal and the sieving allows this (only the large particles of A remain on the sieve, B completely passes through). Therefore it will be the material passing through the sieve that will be mixed with the feed to give us the required product. Teesside University Open Learning © Teesside University 2011 25 Let us try to make some sense of this by drawing a diagram. Let x = the amount of material bypassing the sieve, F = the amount of feed, W = amount of waste material, y = amount of material passing through the sieve and P = the amount of final product. Feed (F) 83% w/w A 17% w/w B F - x Sieve W 100% w/w A x y 5% w/w A 95% w/w B Product P 50% w/w A 50% w/w B Overall mass balance feed product waste i.e. in out = ( ) F PW = ( ) + ( ) FPW = + Overall balance over A 83 100 50 i.e. of of of F WP %%% = + 0 83 = W P + 0 5. F . Overall balance over B 17 50 i.e. of of % % F P = 0 17 0 5 F P = . . Teesside University Open Learning © Teesside University 2011 26 From the last balance 0 17 0 5 . . F P = we can calculate that P F = 0 34 . This means that the final product is only 34% of the feed. From the overall balance this means W = 0.66F, i.e. the waste amount is 66% of the amount of feed and is pure A. To determine the amount of by-pass we need to look at just the sieve. Overall balance on the sieve feed into sieve = F x – Material leaves as waste (W) and amount passing the sieve (y) W F 0 66 = . 0 66 = + – . Fx Fy . – .................. .................................... 5( ) 0 34 Fxy = Balance over B for the sieve (B chosen as only one stream in and out) feed to sieve 17% of = ( ) = ( ) out 95% = Fx Fx – .– 0 17 of y y 0 95 = . ( ) = 0 17 0 95 Fx y .– . Teesside University Open Learning © Teesside University 2011 27 Substitute for y from equation (5) 0 17 0 95 0 34 . – ..– 0 17 0 17 0 323 0 Fx F . –. . – ( ) = ( ) Fx Fx = . 95 x 0 78 0 153 x F . . x F = 0 196 = . This means the amount by-passing the sieve (x) is 19.6% of F. This has been solved using algebraic symbols rather than 'real' numbers. Another way to solve this is to assume a value for the feed. As in previous examples, we are going to let the feed be 100 kg. Doing the overall balances: FPW = + overall balance over B feed in is 17% of F = × 0 17 100 . B is only present as 50% of the product = 0.5P 0 17 100 0 5 × = . . so kg P P 0 17 100 = × = . 0 5 . 34 Teesside University Open Learning © Teesside University 2011 28 From overall balance 100 34 = + W kg 66 = W For a feed of 100 kg, 66 kg is waste and it is 100% A. Therefore 66% of the feed is wasted as pure A. To determine the amount of feed that by-passes the sieve (x) we need to look at just the sieve. Doing a balance over the sieve and having calculated above that W = 66 kg, overall balance feed in waste amount passing through sieve 1 00 – = + 66 = + x y 34 = + x y 34 y x = – Balance over A ( ) = + 0 83 100 66 0 05 .– . 83 0 83 66 0 05 –. . x y = + x y 17 0 0 = 5 0 83 y x + . . Teesside University Open Learning © Teesside University 2011 29 Substitute for y = 34 – x from overall balance 17 0 05 34 0 83 17 1 7 0 05 0 83 15 3 0 = ( ) + .–. = + .–. . 78 = . . x = kg 19 62 . x x x x x i.e. 19.62 kg out of a feed of 100 kg by-passes the sieve which is 19.62% of the feed or a minimum of 80.38% must be sieved. How can we check this? ................................................................................................................................................... ................................................................................................................................................... ________________________________________________________________________________________ Let us check this by finding y = 34 – x and substituting this into the balance over B, i.e. y = 34 – 19.62 = 14.38 kg of 100 19.62 of ( ) = 17 95 %– % 0 17 80 38 0 95 y 14 38 . ( ) = .. . 13 66 13 66 . . × = A balance! Teesside University Open Learning © Teesside University 2011 30 Calculate the amounts of material in each stream. Your answers should be the same as in the table that follows. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ________________________________________________________________________________________ Summarising all the results in a table Material Feed IN 83 17 100 Feed split Final OUT To sieve 66.72 13.66 80.38 By pass 16.28 3.34 19.62 Product from sieve 0.72 13.66 14.38 Final product 17 17 34 Waste 66 0 66 Feed in Required product We have completed our look at mass balances involving physical changes. In the next lesson we will look at chemical changes and see how these differ. For now, test your understanding by attempting the Self-Assessment Questions which follow. Teesside University Open Learning © Teesside University 2011 31 ________________________________________________________________________________________ ________________________________________________________________________________________ ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... Teesside University Open Learning © Teesside University 2011 32 ________________________________________________________________________________________ SELF-ASSESSMENT QUESTIONS ________________________________________________________________________________________ All these questions are at the steady state. 1. A distillation column is required to separate 75 kg s–1 of a mixture of 35% w/w benzene and 65% w/w toluene. The maximum amount of benzene in the bottom product is 2% w/w and the top product must be at least 99% w/w benzene. Calculate the rate of production of the top and bottom products. 2. An evaporator is used to concentrate the amount of soluble solids in a feed from 3.5% w/w to a product containing 55% w/w solids by evaporating off the water solvent. Calculate the amount of water evaporated and the rate of production of the product. 3. In an extraction process, hexane is used to remove soya oil from soya beans containing 13% w/w oil. The maximum strength solution formed at the operating conditions of the extractor is 20% w/w and all the soluble oil is to be removed. The hexane/oil mixture is then separated from the soya bean residue by filtration but 5% w/w of the filter cake is hexane/oil solution which remains with the residue. The hexane/oil mix is then separated in a distillation column which produces 100% w/w hexane as top product and 99.8% w/w oil as the bottom product. For every 100 kg of feed, calculate: (a) the amount of hexane required at the extractor (b) the amount of oil removed as product (c) the amount of hexane make-up required. Teesside University Open Learning © Teesside University 2011 33 4. An air conditioning system is required to produce air containing 0.2% w/w moisture from a supply of 1.5% w/w moisture. The conditioning unit produces air with a moisture content of 0.1% w/w. In order to save cost and wear on the conditioner, some of the input air by-passes the conditioner and mixes with the conditioned air to give the required product. Calculate the maximum % of the input air that may by-pass the conditioner to give the required product. Teesside University Open Learning © Teesside University 2011 34 ________________________________________________________________________________________ ANSWERS TO SELF-ASSESSMENT QUESTIONS ________________________________________________________________________________________ T kg s–1 99% w/w benzene 1% w/w toluene 75 kg s–1 35% w/w benzene 65% w/w toluene D I S T I L L A T I O N B kg s–1 2% w/w benzene 98% w/w toluene Let the amount of tops = T kg s–1 and bottoms = B kg s–1. Doing balances over the process (in = out) Overall 75 = + T B Over benzene 0 35 75 0 99 0 02 ×= + . .. 26 25 0 99 0 02 .. . T B T B = + Teesside University Open Learning © Teesside University 2011 35 From the overall T = 75 – B Substitute in the benzene balance 26 25 0 99 75 0 02 .. – . 26 25 74 25 0 99 0 0 . . –. . 0 97 48 B . B = ( ) + = + B B B 2 B = = kg s 1 49 48 . – From the overall balance T = = 75 49 48 25 52 –. . – kg s 1 Check using balance over toluene 0 65 75 0 01 25 52 0 98 49 48 ×= × + × . . .. . 48 75 0 26 48 .. . = + 49 48 75 = . A balance! Teesside University Open Learning © Teesside University 2011 36 100% w/w water 3.5% w/w soluble solids 96.5% w/w water E V A P O R A T I O N 55% w/w soluble solids 45% w/w water Not given a feed rate so assume 100 kg. Let x = water evaporated in kg and y = product rate in kg. Overall Over water From overall balance, Substitute in water balance 100 = + x y 0 965 100 1 0 0 45 . .. ×= + x y y x = 100 – 96 5 0 45 100 = + ( ) x x . .– = + x x 45 0 45 – . 51 5 0 55 = . . 64 93 x . = 100 93 64 6 36 y = = kg x kg –. . Teesside University Open Learning © Teesside University 2011 37 Check by balance over soluble solids 3.5% of solids in feed 55% solids in produc = t 0 035 100 0 55 × = . . y kg the same as a y bove. 6 36 = . If you did the balance over solids rather than water first you would have got your answer with less maths. 3. Let the amount of feed = 100 kg, the amount of hexane = H kg, the product = P kg, the extract = E kg, the raffinate = R kg and the hexane make-up = X kg. Put these onto a block diagram. 13% w/w oil 87% w/w bean residue H Raffinate R 5% w/w hexane/oil mix 95% w/w bean residue 100% hexane X, hexane make-up Extractor/filter Extract E 80% w/w hexane 20% w/w oil D I S T I L L A T I O N P 0.2% w/w hexane 99.8% w/w oil Teesside University Open Learning © Teesside University 2011 38 As the first part of the question asks about the extractor, the first thing to do is to look at balances over the extractor/filter. 100 kg 13% w/w oil 87% w/w bean residue Raffinate R H 100% hexane Extractor/filter 5% w/w hexane/oil mix 95% w/w bean residue Extract E 80% w/w hexane 20% w/w oil (a) Overall mass balance over the extractor/filter 100 + =+ HER As the residue all appears in the raffinate we can do a simple balance over the residue. 87% of 100 kg feed 95% of = R 87 0 95 R R = . 87 . = 91 58 . kg = 0 95 Teesside University Open Learning © Teesside University 2011 39 Mass balance over the oil. Here is where you may have gone 'wrong'. Before we can do this we need to look carefully at the raffinate's composition. It states 5% hexane/oil mix. The mix is 20% w/w oil, 80% w/w hexane, the same as the extract composition. So of the 5% w/w of the hexane/oil mix, 1% w/w is oil and 4% w/w is hexane. Now doing the balance: 13 1 % % of 100 kg 20% of of = + 0 13 100 0 20 × = . . 13 0 20 0 92 E R E 0 01 91 58 + × . . E = + . . = ( ) = 13 0 92 – . E 0 20 . 0 4. kg 6 Mass balance over hexane H ER = + 80 4 % % 0 8 60 4 0 04 91 58 51 98 =× + × = . .. . . of of kg We can check these answers in the overall balance. 100 51 98 60 4 91 58 100 HER + =+ + =+ . .. 151 98 151 98 . . = Thus the amount of hexane required to be fed to the extractor is 51.98 kg. Teesside University Open Learning © Teesside University 2011 40 (b) The oil removed as product in the extract from the extractor = 20% of the extract and all this is recovered in the distillation column. So oil as product of 20 0 2 60 4 = =× 12 08 = % .. E kg . (c) All of this oil is recovered in the distillation column as 99.8% of the product So P = = 12 08 . . kg 0 998 12 1 . For the last part we can either do an overall balance over the whole process or do an overall balance over hexane. For the overall balance over the whole process feed in hexane make up raffinate out produc + =+ t out 100 100 91 58 12 1 XRP +=+ X += + 91 58 12 1 X = + = 3 68 kg . . . . .– 100 Teesside University Open Learning © Teesside University 2011 41 Check by doing an overall balance on the hexane: hexane in as make-up hexane out in raffinat = e hexane out in product + X R = + 0 2. % P 4%of of 0 04 91 58 0 002 12 1 =× + × . .. . 3 66 0 02 3 68 . . . kg =+= Material IN (kg) OUT (kg) Total out 13 87 Bean residue Hexane 3.68 103.68 In raffinate 0.92 87 3.66 91.58 12.08 0.02 In product 0 12.1 13 87 3.68 103.68 4. No amount is given so assume 100 kg to be treated. Let the amount by-passing the conditioner = x kg, the product rate be P kg and water out = W kg. First the diagram! 100 kg input air 1.5% w/w water 100 - x Conditioned air 0.1% w/w water P, product air 0.2% w/w water Conditioner W, water out x Teesside University Open Learning © Teesside University 2011 42 Doing balances over the whole process: overall 100 = + product air out water out P W = + Overall balance over water 15 02 .% .% of input of product = + W Overall balance over air 98 5. % of input (100 kg) 99.8% of product = From this last equation 98 5 0 998 . . P kg P = 98 70 = . From overall balance 100 98 70 W kg = + . 1 30 = . W Doing balances over the conditioner: overall 100 ( ) = + x W – ( ) = + 100 1 30 x – . conditioned air out conditioned air out conditioned air out 98.7 = – x Teesside University Open Learning © Teesside University 2011 43 Over water 1 5 100 0 1 .% – .% of of conditioned air o ( ) x W= + ut Substitute for conditioned air from overall balance 0 015 100 1 30 0 001 98 7 . – . . .– ( ) = + ( ) 1 5 0 015 1 .–. 0 101 0 014 x x 30 0 099 0 001 x = . . – . + x x = . . 0 101 . x .214 kg = = 0 014 . 7 Thus the maximum air flow by-passing the conditioner = 7.214 kg out of every 100 kg = 7% in real terms. Check by doing a balance over air 98 5 99 9 .% – .% of 100 of conditioned air ( ) x = = of 98 7 99 9 × ( ) = × 14 0 985 100 7 214 0 999 98 7 7 2 . –. . .–. ( ) x . % ( ) . – 91 40 91 40 . . = We can also check by doing balances over the section where mixing of conditioned air and by-pass air occurs. Overall x x + ( ) 98 7 98 7 .– . = Teesside University Open Learning © Teesside University 2011 44 Over water 0 015 7 21 0 1 5 0 1 98 7 0 2 .% .% . – .% × + 001 98 7 7 21 0 002 98 7 . .. of of x xP + ( ) = ( ) = × .–. . . 0 1996 0 1974 . . = Difference due to the small numbers involved and the rounding errors. Teesside University Open Learning © Teesside University 2011 45 ________________________________________________________________________________________ SUMMARY ________________________________________________________________________________________ In this lesson we have worked through several different examples of the use of mass balances where no reaction occurs in the process and all materials in the system remain the same. Balances here are relatively straightforward and the biggest problem is usually deciding upon which balance you should do first. In reality this does not matter as all mass balances must balance eventually. It is useful to remember that where materials enter and leave in only one stream these are the best starting points for balances. It is important that you carry out checks on the balance but be aware that rounding errors may mean the balances are not perfect. You will need to use your own judgement as to how accurate you require the answers (normally we would expect less than 0.5% difference). The basis for a mass balance is also an important consideration. Check the information given and choose a suitable basis for the calculation if none is given. Usually choosing values such as 100 kg of feed or 100 kg of product, help as 100 is a number that enables easy conversion to and from percentages. Teesside University Open Learning © Teesside University 2011