mass balances session 1 of 6 - introduction to mass balances (course notes) rev 4 without answers

EG2002: Process Engineering Rev 4H by JC on 06/10/2011

Page | 1

1. Introduction

My name is John Cavanagh - your tutor for the next three weeks over which I will be introducing

the concept of material balances.

I am providing course notes and will be making recommendations for further reading so will use

my lectures simply to introduce and discuss key concepts whilst hopefully providing a different

perspective on some of the material. You will be expected to take notes during lectures as I will

not be providing copies of my lecture slides.

2. Overview of the lecture series

The course notes are divided into six sessions but will be delivered in nine lectures over the next

three weeks.

Session 1 - Introduction to material balances

Session 2 - Material balances involving change of composition

Session 3 - Degrees of freedom and problems involving multiple processes

Session 4 - Material balances involving chemical reaction

Session 5 - Material balances at the molecular level and problems involving recycles and purges

Session 6 – Solution of complex material balances

3. Recommended reading

Whilst course notes are provided students are still expected to undertake additional reading in

order to more fully understand the subject. However, as everyone learns in different ways it is

important to find a reference book that meets your own personal needs. That said, I highly

recommend that you consult the following two well established text books both of which are

cited extensively throughout these course notes.

Sinnott, R and Towler, G. eds., 2009. Chemical Engineering Design 5th ed. Elsevier Ltd.

(ISBN 978-0-7506-8551-1)

Felder, R and Rousseau, R., 2005. Elementary Principles of Chemical Processes 3rd ed. John Wiley

and sons, Inc. (ISBN 978-0-471-37587-6)

Introduction to Material Balances

(Material Balances - Session 1 of 6)


Page | 2

4. Learning Outcomes

As this is the first session it is appropriate to begin by outlining the learning outcomes for all six

sessions.

Having completed this course you should understand the importance and key concepts of

material balances. Presented with a suitable problem you should be able to draw and mark-up

simple flow diagram, select a basis of calculation, select suitable system boundary(s), analyse the

degrees of freedom, write down the material balance equations to describe the process and solve

these equations to determine any unknowns. Having done all this you should be able to present

your results professionally in the form of a stream list.

It is recognised that not everyone studying this course will have a background in chemistry but

such students should leave with understanding of how to write down and balance an equation

describing a simple chemical reaction. They should also understand and be able to make use of

the concept of a ‘mole’.

At the end of this first session you should understand what a material balance is and why they

are so important within the process industries. You should also understand and be able to write

down the general equation for conservation of mass.

5. Why are mass (and energy) balances so important to Process /Chemical Engineers?

Imagine you are CEO of a company wishing to design and build a new process.

In all likelihood you will only be considering such an investment because you hope to make a

healthy profit that will pay back the initial investment in the shortest possible time.

What do you need to know in order to make the right decision? A decision that could potentially

make or break your organisation.

Clearly you need to know what you are going to make and where you are going to make it so it

might be useful, albeit outside the scope of this course, for you to spend a few moments thinking

about what things that might influence your choice?

What else do you need to know?

The production rate of the new facility is important as it not only sets the scale of the investment

and infrastructure requirements but also determines the quantities of products you will need to

sell and raw materials you will need to purchase.

OK, so you know the ‘what’, the ‘where’ and the ‘how much’ but what else do you need to know?

You need to be sure that you can you make whatever it is safely without damaging the

environment but you also need to know what it will cost you to make as only then can you value

the investment and decide whether or not to go ahead.

Questions you may have might include…..


Page | 3

What raw materials are required? At what rate?

What by-products and/or waste streams are there?

What are the energy requirements?

What will it cost to build?

What inventories and process conditions are required and what hazards do they present.

What intermittent and continuous effluents are there to air, land and water?

All reasonable questions but how will you answer these?

If what you are proposing is very similar to something you (or someone else) have done before it

may be possible to start by making comparisons then scaling up or down for factors such as size

or complexity. However, you will rapidly need to get into more detail, initially to provide better

estimates of capital expenditure (CAPEX) and operating expenditure (OPEX) but then to actually

specify and build the equipment. Without material and energy balances it would be very difficult

if not impossible to answer many of these questions.

Material and Energy balances are the starting point and fundamental building blocks of most, if

not all, process designs. Most projects are aborted based solely on initial material balances and

the associated economic case. For the few that continue material and energy balances are the

starting point for pretty much every aspect of the design, be it process, mechanical, civil or

electrical in nature. Even once a process is operational the material and energy balance is the

basis for production monitoring systems and any troubleshooting that is required to resolve

emergent process issues.

This is something that engineers from all disciplines will need to grasp in order to function

effectively in the process industries be it oil and gas, chemicals, petrochemicals, pharmaceuticals,

food and drink, utilities or waste treatment.

6. Mass and Energy

According to Sinnott and Towler (page 52) the loss of mass associated with the production of

energy, as described by Einstein’s equation E=mC2, is only significant in nuclear reactions. In fact

whilst Einstein proposed the concept of mass-energy equivalence he also proposed that both

total mass and total energy are in fact conserved separately.

This, albeit a very interesting discussion, is only mentioned here in this context as justification for

why the conservation of mass and that of energy are always, with the possible exception of the

nuclear industry, dealt with separately. That said, in reality, it is almost always necessary to carry

out a material balance in order to carry out an energy balance as the latter is usually reliant on

data generated by the former.

In this series of lectures we deal exclusively with material balances but return to discuss energy

balances later in the course.


Page | 4

7. Material, Mass and Weight

Material is anything made of matter( a somewhat poorly defined concept), Mass is a property of

matter and weight is the force exerted by a mass on earth as a consequence of it mass and

acceleration due to gravity.

In studying this area and in industry you will come across references to material balances, mass

balances and less commonly weight balances. However, unless there is some specific reason to

believe otherwise you can generally consider these terms as equivalent and interchangeable. In

the case of mass and weight the fact that both sides of the ‘balance’ are subject to the same

force of due to gravity mean that this term can be cancelled making any disambiguation rather

pointless.

8. What do we mean by Mass Balance?

8.1 Conservation of Mass

Imagine you are inside a bubble which has permeable membrane such that material can enter

and/or leave as necessary. Now consider the fate of a molecule of oxygen which passes through

the membrane in to the bubble. There are a number of different things that might happen next.

The molecule can

pass back through the membrane and out of the bubble back to whence it came

remain within the bubble (accumulate)

be used up by your breathing perhaps (consumed)

Now consider the fate of a molecule of carbon dioxide which also enters the bubble

It can also course pass back out of the bubble or accumulate within it and whilst it cannot

realistically be consumed by the human body it could be consumed if you happened to be holding

your favourite potted plant. On the other hand it could gain a friend as carbon dioxide is made

(generated) by the same process which consumed the oxygen.

If we had knowledge of the material entering the bubble, the capability of the human body to

convert oxygen to carbon dioxide and the capacity for the bubble to accumulate gases we could

perhaps predict both the quantity and composition of the air leaving the bubble. In doing so we

would in effect be conducting a simple material balance over the bubbles membrane.

It has already been said that mass can neither be created nor destroyed (other than in a nuclear

process) and based on the above, albeit rather simplistic example, it is possible to write the

following expression to describe this conservation of mass.

Material Out = Material In + Generation – Consumption – Accumulation

(Sinnott & Towler, 2009 page 53)


Page | 5

Where, ‘Material Out’ refers to material leaving the system boundary whilst ‘Material In’ refers to

material entering the system boundary. ‘Generation’ and ‘Consumption’ refer to material

generated and consumed within the system and ‘Accumulation’ refers to an increase of material

within the system although this can also decrease (usually expressed as ‘negative accumulation’).

The above expression known as the general mass balance equation can be applied (and holds

true) across any given boundary whether considering overall flows, component flows or flows of

individual elements. It no doubt holds true even at the subatomic level although this is not

considered within this course since modelling at this level would serve no useful purpose in the

context of industrial processes.elder and Rousseau (page 86) define two types of balances;

1. Differential balances indicate what is happening at an instant of time where each term of

the balance equation has a rate such as kg/hr. - usually applied to continuous processes.

2. Integral balances indicate what happens between two instants in time where each term

of the balance equation may be stated as some given quantity per batch or per day or

even per hour provided it is recognised that the flow throughout the period being studied

may well not be uniform- usually applied to batch processes

Felder and Rousseau correctly obverse that differential balances are generally applied to

continuous processes whilst integral balances are generally applied to batch processes. However,

it is worth noting that integral balances can be (and are) easily derived from differential balances

produced for continuous processes in order to produce balances showing such things as annual

production figures. This information can be far more useful than instantaneous balances for

evaluating the financial performance of a particular unit as they take into account on stream time

(the proportion of the year the process is running) and the proportion of maximum capacity that

the process is operating at which may be limited by technical or commercial constraints.

Differential balances are often more useful for design and troubleshooting of continuous

processes as they provide information about what is happening at a particular period of time

which may well be masked by the averaging which can result from integration over time.

Having introduced the general mass balance equation it is worthwhile pausing to examine it in

more detail. This expression is the foundation on which the next five lectures are built so it is

essential that everyone has a clear understanding before we take the next step.

8.2. Material OUT = Material IN

Consider a simple pipe, just like the copper ones in figure 1 (overleaf) which most of you will have

supplying water to your bathroom at home.


Page | 6

Fig. 1: Selection of Copper Pipes

Imagine you are filling your bath and water is flowing through such a pipe. This can be

represented diagrammatically as in figure 2, with the arrow indicating the direction of flow.

Fig. 2: Cross-section of a pipe with water flowing through it

As you we see over the coming sessions we ALWAYS start by converting a written problem into a

process flow diagram. A good process diagram is clear and concise showing all relevant items of

equipment and direction of material flows. All streams should be labelled and information such

as flow rate and/or process conditions marked (although the same information could be

presented in a stream list as we will see later). This helps in clarifying the problem, structuring

your calculations and (importantly) communicating your approach and findings to the reader.

This is regarded as a critical part of any mass balance calculation for which marks will always be

awarded. Expectations will become clearer over the next few lectures as we cover some

examples.

Having drawn our diagram the next step is always to mark on the boundary(s) over which we will

be conducting our material balance. In this example it is rather obvious but it is best to develop

good habits right from the start.

So let’s draw our ‘boundary’ as shown in figure 3, marked in red where M1 and M2 represent the

mass of material entering and leaving the chosen boundary in any given time. This boundary is

analogous to the permeable skin of the bubble we discussed earlier. You should also note that

the streams have been given numbers to uniquely identify them and arrows to indicate the

direction of flow.

Bath Supply


Page | 7

Fig. 3: Cross-section of a pipe with water flowing through it showing mass balance envelope and

stream numbers

The rate water enters and leaves the boundary in fig 3 should be the same (unless we have a

leak!) or in other words;

Mass of water leaving per unit time= Mass of water entering per unit time

(Interestingly this still applies even in the case of a leak although the mass of water leaving

through both the end of the pipe and the leak would then equal the mass of water entering)

Or expressing this more generally we have

Material OUT = Material IN

This is a material balance in its simplest form although if you compare this formula with the

generalised mass balance formula described earlier there are some other terms (shown in

brackets below) which are missing.

Material Out = Material In (+ Generation – Consumption – Accumulation)

We will now go on to discuss the ‘missing’ terms in turn.

8.3. Accumulation

In this context accumulation is essentially the ‘build-up of material’

In order to investigate this concept let us return to the previous example but this time consider a

situation where the copper pipe is initially empty (and to simplify things) oriented vertically with

water entering from the bottom as in figure 4 (overleaf)

1 2

M1 M2


Page | 8

fig.4: Cross-section of a pipe with water flowing through it showing mass balance envelope with

inlet marked as stream ‘1’ and outlet marked as stream ‘2’

So does ‘material OUT = material IN’ still hold true in this example? What do you think?

In order to answer this question we should consider what happens over time. This is shown in

Figure 5 (below) which assumes that the pipe holds 20kg of water when full and water is entering

at the rate of 10kg/hr.

Time = 0hr Time = 1hr Time = 2hr Time = 3hrs

Figure 5: Mass balance after 0, 1, 2 and 3 hours where ṁn = mass flowrate of stream n

ṁ1= 10kg/hr

ṁ2= 0kg/hr

ṁ1= 10kg/hr

ṁ2= 0kg/hr

ṁ1= 10kg/hr ṁ1= 10kg/hr

ṁ2= 0kg/hr ṁ2= 10kg/hr

System

boundary for

Mass balance

purposes

2 2 2 2

1 1 1 1

2

1


Page | 9

To start with the pipe is empty and must be filled. After 1hr the pipe is half filled but nothing has

left the system boundary at this point. After 2hrs the pipe is completely full but again nothing has

actually left the system boundary. Clearly, up until the point when the pipe is full

Material OUT ≠ Material IN

But rather

Material OUT = Material IN - Accumulation

This type of situation where flows can change with time is known as an ‘un-steady state’. The

modelling of such is often described as dynamic modelling.

However from 2hrs onwards water cannot enter without displacing an equal amount of water

out the other end. Once the pipe is full it is again true to say that


The process is now said to be in ‘steady-state’

In real life situations a process is always in unsteady state to some degree especially during start-

up and turn down. However, due primarily to the difficulties associated with modelling the steady

state the vast majority of process modelling has to date been done on a steady state basis with

appropriate design margins and the assumption that appropriately specified control systems will

keep us at least close to steady state for the vast majority of the time. In reality the use of steady

state modelling has proven itself successful over the years when linked with good engineering

judgement and input from experienced plant managers and control engineers as required. Only

occasionally do we get caught out but when it happens it tends to happen ‘big-time’ causing

major operating difficulties and/or delaying plant start-up by weeks or even months whilst

appropriate modifications are made to equipment and control systems.

In most, though not all, problems encountered it will normally be safe to ignore accumulation

but you will be expected to state the assumption of ‘steady-state’ at the start of any calculations

as part of simplifying general mass balance equation.

In order to further illustrate this concept let us consider a cylindrical vessel similar to the one

shown below.


Page | 10

We can represent such a vessel, which we can assume is partially full, as shown in the diagram

below.

In steady state the level remains constant and material OUT = material IN

ṁ2= ṁ1

However, if we increase the flowrate of stream 1, whilst holding that of stream 2 constant, we

will observe the level in the drum will start to rise and we will no longer be in steady state. The

vessel will continue to fill until the excess flow spills out of the vent at the top unless or until the

flowrate of stream 1 is reduced or conversely that of stream 2 is increased to the point where the

flows again balance. Steady state is restored but it should be noted that the level in the vessel is

now higher than it was before. In order to restore the level to its original position one would have

to temporarily reduce the flowrate of stream 1 or increase that of stream 2 until the level

returned to its original position.

8.4. Generation and Consumption

Returning to the full material balance equation

Material OUT = Material IN + Generation- Consumption – Accumulation.

From an either overall or an elemental material balance perspective both Generation and

Consumption are always zero. This is because mass can neither be created nor destroyed, other

than in a nuclear reaction.

However if a chemical reaction takes place one chemical species can of course be converted to

another chemical species. When we burn methane, for example, it ‘reacts’ with oxygen to

produce carbon dioxide and water. In effect methane and oxygen are consumed whilst carbon

dioxide and water are generated. The overall mass and elemental balance is of course still

preserved but we exchange one molecule for another so if we are looking only at a balance on a

ṁ1

ṁ2

Vent

1

2

F

F


Page | 11

particular molecule we can have both generation and consumption. Only if we have no chemical

reaction, are ignoring changes in composition or are balancing down at the elemental level can

we cancel out these terms from the general mass balance equation.

As always it is essential that all assumptions are stated when carrying out calculations. Marks will

be awarded for assumptions and conversely deducted where they are either lacking or not

explicitly stated.

9. Example Problems

The following examples are rather simplistic but are intended to get you thinking along the right

lines and for you to start to develop a structured methodology for answering such questions.

Mass balance problems can get very complicated and if you want to have any chance at solving

them you should try to develop a standard approach that works and stick with it. You will note

that Sinnott and Towler tend to prefer a more ad hoc approach to solving their examples and

whilst this can be an effective way of solving simple problems it is suggested that you try to adopt

a more structured approach even when, as will often be the case to start with, the examples

seem to have an intuitive solution.

In going through the following example try not worry too much about the nomenclature as we

will standardise this at the start of the next lecture.

[PTO]


Page | 12

Example 1

Question

The flowrate into and out of a 1m3 vessel is 10kg/day and 5 kg/day respectively calculate how

long it will take to fill the vessel if it is initially empty. Density of water = 1000kg/m3

Solution

As always, start with a flowsheet.

Flowsheet

Basis: Differential, Mass kg/day

General Equation


But assume no chemical reactions (Generation and Consumption = 0) reducing the above formula

to

Material OUT = Material IN – Accumulation

Overall mass balance

ṁ2 = ṁ 1 – Accumulation

Rearranging Accumulation = ṁ1 - ṁ 2= 10-5 = 5kg/day

Time to fill vessel = 1000/5 = 200 days

Tank

Capacity = 1m3

= 1000 kg

ṁ1 = 10 kg/day ṁ2 = 5 kg/day

1 2


Page | 13

Example 2

Question

The flowrate into and out of a process is of 10 kg/hr of Reactant A 20kg/hr of Reactant B. Carry

out a mass balance to determine the amount of product assuming that the flow of both

reactants increases by 1 kg per hour for each hour of the day returning to the original rates at

the start of each day. The cycle is repeated for 5 out of 7 days and the plant shuts down for

three weeks overhaul every year.

Solution

As always, start with a flowsheet.

Flowsheet

Basis = Integral, 1 year and mass (tonne)

The calculation could be done on a differential basis at a specific rate e.g. maximum and /or

minimum rates however the answers would differ. This is explored in the problems set at the end

of these lecture notes but think about the value of doing it?

General Equation


But assume steady state (Accumulation = 0) and no chemical reactions (Generation and

Consumption = 0) reducing the above formula to


Overall mass balance

m3= m1 + m2

Next we need to calculate m1 and m2

Over each day ṁ1 increases from 10 to 34 kg /hr which is an average of (10+34)/2 = 22te/hr so

total daily flow = 22 x 24 = 528 kg.

This could have been calculated by adding up the flow over each hour or even by integration as

the flowrate increases linearly (shown below).

Process

3

1

2

Product

Reactant A

Reactant B

1

m1 m3

m2


Page | 14

Since flow increases linearly over time using formula for a straight line

ṁ1 = gradient x time + intercept

Over 24 hr flow increases from 10te/hr to 34te/hr so gradient = 24/24 = 1 with intercept = 10

ṁ1 = t+10

So total flow over 1 hour is found as follows

m1 (daily basis) =

=

= = 528kg

m1 (annual basis) fed to process per year = (52-3) x (5/7) x 528 = 18,480 kg

Similarly for m2

Over each day m2 increases from 20 to 44 kg /hr which is an average of (20+44)/2 = 32te/hr so

total daily flow = 32 x 24 = 768 kg.

m2 (annual basis) fed to process per year = (52-3) x (5/7) x 768 = 26,880 kg

So returning to the mass balance formula

m3= m1 + m2 = 28880+18480 = 47,360 tonne (annual basis)

Example 3

Question

Water flows into a 1000 tonne tank at 100 UK Gallons per minute whilst water leaves the same

tank at 240 m3 per day. If the tank is initially half full how long will it take to completely fill the

tank? (Assume 1US Gallon = 3785 cm3)

Solution

To start with draw a flowsheet labelling the streams and indicating the direction of flow.

The above flowsheet should be marked up with the givens which should be converted to the same

basis. In this case a mass basis would seem most appropriate as we have the tanks capacity in

tonnes.

Basis is ‘mass flow’ using units of tonne/hr

Tank

Capacity = 1000

tonne

1 2

Volumetric Flow =

10 UK gal/min

Volumetric Flow =

1m3/day

11


Page | 15

The next step is to convert the ‘givens’ into the selected basis and mark these up on the flowsheet.

In this question you have not been given the density of water. It is often the case that you do not

have all the available data to hand in which case you can try to find it in books etc or if necessary

make an intelligent assumption.

As no data provided assume density of water as 1000kg/m3.

Mass flow of stream 1 (ṁ1) = 100 US gal/min =100 x 60 US gal /hr = (100*60*3785)/1E6 m3/hr

= (100*60*3785*1000)/1E6 kg/hr

= (100*60*3785*1000)/(1E6*1000) tonne/hr = 23 tonne/hr (0 d.p.)

Mass flow of stream 2 (ṁ2) = 240 m3/day =240/24 m3/hr = (240*1000)/24 kg/hr

= (240*1000)/(24*1000) te/hr = 10 tonne/hr (0 d.p.)

These figures can be marked up on the flowsheet or if necessary due to the quantity of data

tabulated for ease of reference.

General Mass balance Equation is

Material OUT = Material IN + Generation - Consumption – Accumulation.

This problem is clearly NOT steady state as there is an imbalance between the material entering

and that leaving i.e.

ṁ1 = 23 tonne/hr ≠ ṁ2 = 10 tonne/hr

However, there is no obvious reaction so we can state

Assume no reaction so Generation and Consumption = 0 thus reducing the general mass balance

equation to

Material OUT = Material IN – Accumulation

So substituting in the givens 10 = 23 – Accumulation

or Accumulation = 23-10 = 13 tonne/hr

The storage tank has a capacity of 1000 tonne and is initially half full giving an initial ullage (space

left at the top of the tank) of 500 tonne.

Time to fill tank, assuming given rates remain constant = 500/13 = 38 hrs (0 d.p.)

1 2

Tank

Capacity = 1000

tonne ṁ2 = 10 tonne/hr ṁ1 = 23 tonne/hr


Page | 16

10. Assignment

10.1 Reading

It is recommended that you take a look at Sinnott and Towler Chapter 2 and/or Felder and

Rousseau Chapter 4 as early as possible within this series of lectures. You may well not

understand much beyond the start of each reference at the moment but you can always revisit

the relevant parts later and it will put you in a much stronger position to ask questions in lectures

and answer tutorial questions. I will give more specific reading in later lectures.

10.2 Problems

The following problems are quite simple and could easily be solved intuitively but as mentioned

earlier, things get a lot more complex and so it is important to develop a methodology that works

and stick with it. So in answering the following questions imagine ALL of the marks are for the

method. Remember to include a marked up flowsheet, define your basis of calculation, write

down your assumptions and solve the problem by application of the generalised mass balance

equation. Try to attempt the question before the next lecture but if not make sure you attempt it

and bring your efforts to the next tutorial for discussion and/or help as appropriate.

1. Sewerage flows into a primary treatment unit of a waste treatment facility from two sources.

The first source is a continuous flow of 10 tonne/hr whilst the second is rather intermittent

increasing linearly from 0 tonne/hr to 10 tonne/hr over a half hour period then reducing at the

same rate over the next half hour. Perform a material balance over the primary treatment unit to

determine the load on the secondary unit. Try this using both a differential and an integral

approach (hint the key to solving this one is in the selection of a suitable basis). Which do you

think is the most appropriate approach in this case and why?

2. According to Wikipedia Loch Ness has a total water capacity of 7.4 km3, as surface area of 56.4

km2 and a water catchment area of 1,775 km2. This had not been a particularly dry summer and

the Loch is only 80% full so you remain concerned water maybe leaking away into a newly

discovered system of underground caves. Your measurements indicate that the level is dropping

at a worrying 1% per week. You estimate that water enters the river from at a rate of 1 litre per

day for each m2 of catchment area and leaves via streams etc at a rate of 1000 tonne/hr. In

addition you estimate that the loss due to evaporation is 1 litre per m2 of surface area. Carry out

a mass balance in order to investigate the issue further and calculate any unexplained losses.

Additionally you have been asked to consider what impact the large community of subsurface

monsters drinking the water have on the situation? (Assume density of water = 1000 kg/m3)

mass balances session 1 of 6 - introduction to mass balances (course notes) rev 4 without answers

Documents

material balances session

concept of material

energy balances

course notes

material balance equations

chemical reaction session

purges session

process engineering