The Author Professor Simons is a Chartered Engineer and Fellow of the Institution of Chemical Engineers (IChemE) and is a member of the Executive Committee of the Standing Conference of Chemical Engineering Professors. He is also vice-chair of the committee of the IChemE’s Particle Technology Subject Group and serves on the committee of the Royal Society of Chemistry’s Process Technology Group, which initiated the short course on which this book is based. Professor Simons’ expertise lies in particle technology, in general, and, more specifically, in the development of low carbon technologies in regards to the reduction of carbon dioxide emissions, particularly in carbon capture and storage, for which he was an expert reviewer of the Intergovernmental Panel on Climate Change’s special report on this subject.

Concepts of Chemical Engineering 4 Chemists

By Stefaan J. R. Simons

The Royal Society of Chemistry

Copyright © 2007 The Royal Society of Chemistry
All rights reserved.
ISBN: 978-0-85404-951-6

Contents

Chapter 1 Process Analysis – The Importance of Mass and Energy Balances Eric S. Fraga, 1,
Chapter 2 Introduction to Chemical Reaction Engineering George Manos, 21,
Chapter 3 Concepts of Fluid Flow Tim Elson, 55,
Chapter 4 An Introduction to Heat Transfer Tim Elson and Paola Lettieri, 96,
Chapter 5 An Introduction to Mass-Transfer Operations Eva Sorensen, 139,
Chapter 6 Scale-Up in Chemical Engineering Tim Elson, 171,
Chapter 7 An Introduction to Particle Systems Paola Lettieri, 203,
Chapter 8 An Introduction to Process Control Eva Sorensen, 249,
Chapter 9 Economic Appraisal of Large Projects Ken Sutherland, 272,
Chapter 10 Hazard Studies and Risk Assessment Robert Thornton, 309,
Subject Index, 342,

CHAPTER 1

Process Analysis – The Importance of Mass and Energy Balances

ERIC S. FRAGA

1.1 INTRODUCTION

Process engineering includes the generation, study and analysis of process designs. All processes must obey some fundamental laws of conservation. We can group these into conservation of matter and conservation of energy. Given a set of operations, if we draw a box around this set, the amount of mass going in must equal the amount going out; the same applies to the energy. Mass and energy balance operations are fundamental operations in the analysis of any process. This chapter describes some of the basic principles of mass and energy balances.

1.1.1 Nomenclature and Units of Measurement

In carrying out any analysis, it is important to ensure that all units of measurement used are consistent. For example, mass may be given in kg (kilogrammes), in lb (pounds) or in any other units. If two quantities are given in different units, one quantity must be converted to the same unit as the other quantity. Any book on chemical engineering (or physics and chemistry) will have conversion tables for standard units.

There are seven fundamental quantities that are typically used to describe chemical processes, mass, length, volume, force, pressure, energy and power, although some of these can be described in terms of others in the list. For example, volume is length raised to the power 3; power is energy per unit time, pressure is force per area or force per length squared, and so on.

Chemical engineering uses some standard notation for many of the quantities we will encounter in process analysis. These are summarised in Table 1 where T is time, M is mass and L is length.

In describing processes, the variables that describe the condition of a process fall into two categories:

(i) extensive variables, which depend on (are proportional to) the size of the system, such as mass and volume, and

(ii) intensive variables, which do not depend on the size of the system, such as temperature, pressure, density and specific volume, and mass and mole fractions of individual system components.

The number of intensive variables that can be specified independently for a system at equilibrium is known as the degrees of freedom of the system.

Finally, it is important to note that the precision of quantities is often not arbitrary. Measuring tools have limits on the precision of measurement. Such measures will have a particular number of significant figures. Calculations with measurements may not result in an increase in the number of significant figures. There are two rules to follow to determine the number of significant figures in the result of calculations:

(i) When two or more quantities are combined by multiplication and/or division, the number of significant figures in the result should equal the lowest number of significant figures of any of the multiplicands or divisors. In the following example, one multiplicand has three significant figures and the other, four. Therefore, the result must have no more than three significant figures regardless of the number of figures that are generated by the calculation:

3:57 x 4:286 = 15:30102 [??] 15:3

(ii) When two or more numbers are either added or subtracted, the positions of the last significant figure of each number relative to the decimal point should be compared. Of these positions, the one farthest to the left is the position of the last permissible significant figure of the sum or difference. It is important to make sure that all the numbers are represented with the same exponent if scientific notation is used:

1:53 x 103 – 2:56 = (1:53 x 103) – (0:00256 x 103)

= (1:53 – 0:00256) x 103

= 1:52744 x 103

= 1:53 x 103

1.2 MASS BALANCES

Chemical processes may be classified as batch, continuous or semi-batch and as either steady-state or transient. Although the procedure required for performing mass, or material, balances depends on the type of process, most of the concepts translate directly to all types.

The general rule for mass balance in a system box (a box drawn around the complete process or the part of the process of interest) is:

input + generation – output – consumption = accumulation (1)

where,

(i) input is the material entering through the system box. This will include feed and makeup streams;

(ii) generation is the material produced within the system, such as the reaction products in a reactor;

(iii) output is the material that leaves through the system boundaries. These will typically be the product streams of the process;

(iv) consumption is the material consumed within the system, such as the reactants in a reactor;

(v) accumulation is the amount of material that builds up within the system.

In a steady-state continuous process, the accumulation should always be zero, which leads to a more simple mass balance equation:

input + generation = output + consumption (2)

In the case of systems with no reaction, where mass is neither generated nor consumed, the result is even simpler:

input = output (3)

1.2.1 Process Analysis Procedure

The analysis of the mass balance of a process typically follows a number of steps:

(i) Draw and label a diagram for the process, clearly indicating the information given by the problem definition and the values that have been requested.

(ii) Choose a basis of calculation if required. If no extensive variables (e.g. amount or flow rate of a stream) have been defined, a basis of calculation is required and this must be an extensive variable.

(iii) Write down appropriate equations until zero degrees of freedom are achieved. In other words, write down enough equations so that the number of equations equals the number of unknown variables and such that all the unknown variables are referred to in the equations. Possible sources of equations include the following:

(a) Mass balances. For a system with n species, n mass balance equations may be written down. These mass balance equations may be drawn from a total mass balance and from individual species mass balances.

(b) Process specifications and conditions such as, for example, the separation achieved by a distillation unit or the conversion in a reactor.

(c) Definitions such as the relationship between density, mass and volume or the relationship between mole fraction and total mass.

(iv) Identify the order in which the equations should be solved.

(v) Solve the equations for the unknown values.

These steps are illustrated by the following example in Section 1.2.2.

1.2.2 Example 1: Mass Balance on a Continuous Distillation Process

Suppose that a 1000 kmol h-1 feed stream, consisting of 30.0% by mole n-pentane and the remainder n-hexane, is to be separated into 95.0% pure pentane and 95.0% pure hexane streams using a distillation column. Determine the flow rates of the output streams through the use of mass balances, assuming steady-state operation. We will assume three digits of significance for this example.

.2.2.1 Solution.

(i) The first step is to draw and label a flowsheet diagram indicating the process steps and all the streams. Figure 1 shows the layout of the distillation unit labelled with both the known variables and the variables we wish to determine. The system box for this example is also indicated. The streams we wish to consider are those that intersect this system box and consist of the feed stream and the two output streams. All other streams can be ignored in solving this example. The notation we have used is that F is the feed stream, D the distillate or tops product stream (which will be primarily pentane, the lighter of the two species), B the bottoms product stream (primarily hexane), p refers to pentane andh to hexane. We are interested in finding the values for [??]D and [??]B, as indicated by boxed question marks in the figure. There are three streams and two unknowns. Strictly speaking some of the mole fractions are also unknown, but we have directly incorporated the equations we will use to determine values for XF,pXD,h and XB,p into the diagram, noting that by definition the mole fractions of the components in a stream add up to one. Note that we have assumed that the pentane- rich stream is the distillate output of the unit and that the hexane- rich stream is the bottoms product. This is based on their relative volatilities; pentane has a lower boiling point than hexane.

(ii) As this is the simplest case described above (steady-state, continuous and no reaction), we can use the simplest mass balance equation:

input = output

(iii) As indicated above, we have two unknowns. Therefore, we need to generate two independent equations that will allow us to solve for these unknowns. For mass balance problems, the general rule is that we can define nc equations if there are nc components in the streams involved in the mass balance problem defined by the system box chosen. The one exception is that if the system box includes only a splitter, there is only one independent equation that can be defined. This is because the splitter does not change the compositions of the streams involved, only the amounts or flows. For this example we have two components, and so we can define two mass balance equations. The choice of equations is a total mass balance and two individual component balances:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (4)

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (5)

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (6)

(iv) We can choose any two of these three equations to solve our problem. For this example, we will choose the total mass balance, Equation (4), and the pentane mass balance, Equation (6). Together with the mole fraction definitions already labelled on the diagram, we are left with 0 degrees of freedom. The degrees of freedom are defined as the difference between the number of unknowns and the number of equations relating these unknowns. For the example, we have five unknowns – [??]D,[??]B, XF,p, XD,h and XB,p – and five equations, three equations from the mole fractions in each stream adding up to one and two equations from the mass balances (Equations (4) and (6)). Therefore, we can solve these equations simultaneously to find the values of the unknowns.

(v) Before solving the equations, it is worth planning ahead to determine which order to solve the equations in. The full set of equations are given here:

[??]F = [??]D + [??]B (7)

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (8)

(XF,p = 1 – (XF,h (9)

XD,h = 1 – XD,p (10)

XB,p = 1 – XB,h (11)

We can solve for XF,p immediately using Equation (9), as we know XF,b. Likewise, we can solve for both, XD,h with Equation (10), and XB,p, with Equation (11), given the specifications on the output streams (XD,p and XB,h. Finally, we solve the two mass balance equations, Equations (7) and (8), simultaneously.

(vi) The actual solution can now proceed:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (12)

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (13)

We now rearrange this last equation, Equation (13), to have [??]B on the left hand side alone:

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This value can now be plugged back into the first mass balance equation, which we previously rearranged with [??] on the left hand side, Equation (12):

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(vii) The example has been solved. We can now use the unused mass balance equation, in this case being the hexane component mass balance, Equation (5), to provide a check for consistency:

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So the results are at least consistent, which gives some confidence in their correctness.

For processes involving reactions, the mass balance equation, Equation (3), used in the first example is not sufficient. Equations (1) or (2) must be used. The presence of reactions means that the generation and consumption terms in these equations are non-zero. The key difference between a simple separation process and a process involving reactions is the need to define these extra terms in terms of the amounts of the components in the system. Extra equations are required to satisfy the extra degrees of freedom introduced by having these terms present.

Two concepts are often used to describe the behaviour of a process involving reactions: conversion and selectivity. Conversion is defined with respect to a particular reactant, and it describes the extent of the reaction that takes place relative to the amount that could take place. If we consider the limiting reactant, the reactant that would be consumed first, based on the stoichiometry of the reaction, the definition of conversion is straight forward:

conversion [equivalent to] amount of reactant consumed/ amount of reactant fed (14)

Selectivity is a concept that applies to processes with multiple simultaneous reactions. It is used to quantify the relative rates of the individual reactions. However, any discussion about multiple reactions and the analysis of these is beyond the scope of this chapter. Refer to the further reading material identified at the end of this chapter for more information.

1.2.3 Example 2: Mass Balance on a Process with Reaction

Suppose an initially empty tank is filled with 1000 mol of ethane and the remainder with air. A spark is used to ignite this mixture and the following combustion reaction takes place:

2C2H6 + 7O2 [right arrow] 4CO2 + 6H2O (15)

Assume that the amount of air provides twice the stoichiometric requirement of oxygen for this reaction, and that air is composed of 79% nitrogen and the remainder oxygen. Suppose that the reaction reaches a 90% conversion. What is the composition of the mixture in the tank at the end of the reaction?

1.2.3.1 Solution.

(i) As in the first example, the first step is to draw and label a diagram, as shown in Figure 2. This example is a batch problem, and so arrows indicate material flow in the sense of loading and discharging the tank. The subscripts for mass amounts are the stream index (1 for the original contents of the tank and 2 for the final contents after combustion) and the species involved. The key is E for ethanol, O for oxygen, N for nitrogen, C for carbon dioxide and W for water.

(ii) As in the first example, the boxed question marks highlight the variables that need to be determined. As there are five highlighted variables, we will require at least five equations. As there are five species involved, five mass balance equations can be defined. In this case, we should use Equation (1) but with accumulation set to zero to indicate that the tank is empty to start with and also empty at the end. This means that the mass balance equation we should use is identical in form to Equation (2):

input + generation = output + consumption

Our five mass balance equations, therefore, are as follows:

n1,E + ng,E = n2,E + nc,E (Ethane balance) n1,O + ng,O = n2,O + nc,O (Oxygen balance) n1,N + ng,N = n2,N + nc,N (Nitrogen balance) n1,C + ng,C = n2,C + nc,C (Carbon dioxide balance) n1,W + ng,W = n2,W + nc,W (water balance) (16)

where the subscript g is used to indicate the amount generated and the subscript c indicates the amount consumed. At this point, all the variables except for n1,E are unknown. We have 19 unknown variables. As we have just defined 5 equations, we have 14 degrees of freedom remaining. To solve this problem, therefore, we need to define at least 14 more equations.

(iii) We can write down new equations relating the unknown and known variables by making use of the stoichiometric coefficients given by Equation (15). It is helpful to write all of these in terms of one of the consumption or generation terms. In this case, given that the key process specification is the conversion of ethane, it helps to write the equations in terms of the amount of ethane consumed:

ng,E = 0 (No ethane is generated) ng,O = 0 (No oxygen is generated) nc,O = 7/2 nc,E ng,N = 0 (No nitrogen is generated) nc,N = 0 (No nitrogen is consumed) ng,C = 4/2nc,E nc,C = 0 (No carbon dioxide is consumed) ng,W = 6/2nc,E nc,W = 0 (No water is consumed)

This set of nine equations reduces the degrees of freedom to 5 as no new variables have been introduced.

(iv) Further equations can be defined on the basis of the specifications of the feed and the conversion of the reaction:

n1,O = 2 x 7/2n1,E (Twice as much as required) n1,N = 0.79 x n1,O/0.21 (Nitrogen is remainder of air) n1,C = 0 (No carbon dioxide in the feed) n1,W = 0 (No water in the feed) 0.90 = nc,E/n1,E (Conversion of ethane) (18)

This set of five equations also introduces no new unknown variables. The result is that we have 19 equations and 19 unknowns giving zero degrees of freedom.

(v) The system of 19 equations, comprising the three sets of equations above, Equations (16), (17) and (18), can be solved as follows. Given n1,E = 1000 mol, the initial amount of ethane, we

evaluate the equations in set Equation (18):

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Now determine the amounts generated and consumed for each species using the set of Equation (17):

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Finally, we use the mass balance equation, Equation (16), to determine the amount of each species in the output:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

These two examples have illustrated some of the key steps in solving mass balance problems. In particular, the number of mass balance equations that are available for ensuring you have zero degrees of freedom and the use of the system box to limit yourself to the streams that are of interest. These steps are also key points for the analysis of energy balances in processes, the topic of Section 1.3 in this chapter.

(Continues…)Excerpted from Concepts of Chemical Engineering 4 Chemists by Stefaan J. R. Simons. Copyright © 2007 The Royal Society of Chemistry. Excerpted by permission of The Royal Society of Chemistry.
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