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Introduction to Food Processing

Mass & Energy Balance

Contents

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The Food Industry

The food industry in its broadest sense may be divided into three broad sectors

There is some overlap between these three, not to mention industries which act as intermediaries between the three sectors such as wholesalers, distributors and storage companies.

Food processing is the province of the middle of the three sectors and is involves transforming raw materials from the farm and elsewhere into food products which we buy in the shops. The amount and nature of the processing varies widely, depending on the nature of the product and can involve little more than cleaning and sorting. This latter is true in the case of fresh fruit and vegetables. Other products, on the other hand may involve a considerable amount of processing.

The function of the food industry, then is to convert raw foodstuffs into saleable products. In doing so, the food manufacturing company intends to make a profit and that is its primary purpose, and it is important not to lose sight of this. On the other hand, if a food company is to stay in business, it is in its interests to produce food products which are nutritious, and appetising and safe to eat. The functions of food processing may be summarised as follows

To produce foods which

There are a wide variety of food processing operations involving a range of physical and chemical processes. They may be divided into four broad categories, however.

This module will concern itself with the first three. Packaging and storage will be dealt with elsewhere.


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Representing Processes.

A food processing plant will usually comprise a series of processing operations through which the food is passed in sequence. The individual operations are often referred to as "unit operations". Thus a process comprises a sequence of unit operations with inputs to and outputs from each unit operation.

The simplest form of representation of a process is by means of a block diagram. In a block diagram, each unit operation or processing unit is represented by a rectangle and the flows into and out of the unit operations are represented by single lines with arrows to show direction of flow. A block diagram only usually represents the major unit operations (or processing units) and process flows. A block diagram for a single process unit is shown below (Fig 1.1).


Fig 1.1 Block Diagram

In Fig 1.1, the process has two inputs and one output. More complex processes may be represented by a series of blocks (Fig 1.2)


Fig 1.2

In Fig 1.2, each flow into or out of a process is given a label. In this case, S1 . . . S5. This means that each stream can be uniquely identified.


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Mass Balances.

Mass and energy balance is concerned with the total quantities of materials that are required by a process and is an essential first step in the design or analysis of any process.

It is based on the principle of conservation of mass, ie.

Matter cannot be created or destroyed. It can only be converted from one form to another

From this statement comes a very important corollary;

That any matter or energy entering a system must be equal to the sum of the matter leaving and the matter remaining in the system

This can be expressed in the form of a word equation;

INPUT - OUTPUT = ACCUMULATION (1)
If the conditions in a system remain constant over time, the system is said to be in steady state. In this case there is no accumulation and equation (1) reduces to

INPUT = OUTPUT (2)
Equation (2) represents the steady state form of the mass or energy balance equation and is the form in which we shall mostly be using them.


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Systems

A system is a convenient way of representing a process. Material flows into the process, is transformed in some way and ultimately leaves the process. The system approach is concerned purely with the inputs and outputs and not with the details of how the transformation occurs.


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Determining Mass Balances.

In principle, mass balances are straightforward. You simply add up the materials entering the system and the materials leaving the system. If the system is in steady state, these two totals should be the same. Any unknown streams may then be found by difference.

One way to do this is to represent the mass balances by equations relating input and output. Each processing operation will have a mass balance equation associated with it. In addition, there will be an equation relating to the inputs and outputs for the whole process. Looking at fig. 1.2, the following mass balance equations may be written. 

Stage 1: S1 + S2 = S3
Stage 2: S3 = S4 + S5
Overall: S1 + S2 = S4 + S5
Now examine the block diagram below and write mass balance equations for each stage and an overall mass balance.

Stage 1:  
Stage 2:  
Stage 3:  
Overall:  


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Component balances

In mixing and separation processes the balances discussed can be applied to each substance, not only to the total input and output.

A tank is filled with 2 m3 of water and 100 kg of sugar is added to the tank and the mixture stirred until the sugar is dissolved.

  1. Draw a block diagram of the process.
  2. What is the rate of mass input of water?
  3. What is the total rate of mass output of water and sugar?
  4. What is the rate of mass output of solution?
  5. What is the concentration of the sugar solution?

Density of water = 1000 kg m-3

i.e. we can say

MASS OF SUGAR INPUT = MASS OF SUGAR OUTPUT

even though the input sugar is pure and the output sugar is dissolved.

Similarly

MASS OF WATER INPUT = MASS OF WATER OUTPUT

To be specific, in the problem as defined

sugar input = 100 kg/h
sugar output = 100 kg/h

Similarly,

water input = 2000 kg/h
water output = 2000 kg/h

(This cannot be expressed on a volume basis, only a mass basis - hence mass balance.)

If there are n components in the system there are n component mass balances at each stage. Since the sum of these gives the overall mass balance (ie total inputs and outputs) this overall balance is not an independent equation. Hence there are n independent equations for each envelope. In practice it is usually most convenient to write down the overall balance plus the balances for (n-1) components.


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Presenting Mass Balances

In industry, mass balances are usually presented in one of two ways;

It pays to get used to these methods of presentation at an early stage.

Example

Two sugar solutions are to be blended. The first is flowing to the blender at a rate of 2000 kg/hr and has a concentration of 5% wt sugar and the second is flowing at a rate of 500 kg/hr and has a concentration of 10% sugar. Assuming steady state, carry out mass balances for;

Write your results on the block diagram below and and fill in the table. They have been started for you

Material

INPUT

OUTPUT

S1

S2

S3

Sugar

5%

 

10%

     
Water            
Total  

2000

 

500

   


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Mass Fraction

The mass fraction relates to the composition of a stream. It is defined as

The percentage composition of a component is simply the mass fraction expressed as a percentage.

Thus in the example above, the mass fraction of sugar in stream S1 is 0.05 and that of sugar in stream S2 is 0.1.

An important feature of the mass fraction is that it allows us to write mass balance equations for individual components. For example, if the mass fraction of sugar in S1 is x1, that of sugar in S2 is x2 and that of sugar in S3 is x3, then we can write the following mass balance for sugar:

S1.x1+S2.x2=S3.x3

This means that not only can we write mass balance equations for the total balance but we can write mass balance equations for each component.


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The Pearson square

The Pearson square is a method which simplifies mass balances when you know the percentages or mass fractions of the three streams. It involves writing the compositions you have to start with in the top and bottom left of a square and the composition you want to achieve in the centre of the square. You then subtract diagonally across the square, writing the answers in the top and bottom right of the square. These two figures are the ratios of the two streams needed to give the desired composition.

If x1, x2 & x3 are the mass fractions of component x in S1, S2 & S3 respectively, then the flows of S1 and S2 required to give mass fraction x3 are in the ratio |x3 x2|:|x3 x1|


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Basis

In problem 3.3 you were given the hourly flows which you used as the basis for your mass balance calculations. You could equally well have based your calculations on the sugar flow per second or per day and, although the numbers would have been different, the result would still have been a valid mass balance. In many cases, the flows for various parts of the process may well be given in different units and will therefore have to be reduced to a common basis if there is to be a valid mass balance. This process of choosing the starting point for your calculations is called CHOOSING THE BASIS and is an essential starting point in any mass balance calculation. It is usually convenient to start with either the mass flow over one unit of time, the mass required for one batch or simply 100 mass units of feed as your basis but this is not always the case.


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Procedure for carrying out Mass Balances

  1. Draw a block diagram of the process, however simple it seems.
  2. Select a basis and write it down
  3. Draw up a mass balance table and enter all the information you are given into the table.
  4. Write mass balance equations as necessary
  5. Indentify and calculate the unknown values, entering them into the table as you do.


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Mass Balance Worked Example

A pouring cream containing is produced by blending double cream containing 60% fat and full milk containing 5% fat. If there is demand for 50 t h-1 of pouring cream, what must be the flows of milk and double cream? Produce a complete mass balance table for the system.

Solution

First step with a mass balance problem is to produce a block diagram. Note that each stream is given a unique identifier.

 

The next step is to define a basis. For a basis, you may either choose one hundred mass units or you can use a quantity given in the question. In this case we will use a figure given in the question. Once a basis has been defined, a mass balance table may be drawn and a start made filling it in.

Basis: 50 tonne pouring cream.

kg S1 S2 S3
  % wt mass % wt mass % wt mass
Fat 60   5  

20

10

Aqueous 40   95   80 40
Total          

50

At this stage, that is all that can be filled in on the table. To complete the rest of the table it is necessary to write mass balance equations. These can be written for the Fat, Serum and Total quantities. To write a mass balance equation, we take advantage of the fact that the input=output rule applies for each component as well as to the total.. In this case, it is possible to write three mass balance equations.

Fat: 0.6S1 + 0.05S2 = 10   (1)
Serum: 0.4S1 + 0.95S2 = 40   (2)
Total: S1 + S2 = 50   (3)

It is possible to solve these equations using methods for simultaneous equations. However, it is much easier to use the Pearson square.

To do this, write the fat content of S1 & S2 (the givens) in the top and bottom left of the square and the fat content of S3 (the desired output) in the centre and subtract diagonally. (see overleaf)

Thus 60 20 = 40 and 20 5 = 15 and the flows of S1 and S2 should be in the ratio 15:40. To find the actual flows of S1 and S2 apply these ratios to the given flow of S3

  

Note that S1 + S2 = 50 which gives a check. Alternatively, calculate S1 as shown, then subtract the result from 50 to give S2. Either way, you have a method of checking your arithmetic.

These figures can now be entered into the mass balance table and the table completed.

kg S1 S2 S3
  % wt mass % wt mass % wt mass
Fat 60 8.2 5 1.8 20 10
Aqueous 40 5.4 95 34.6 80 40
Total   13.6   36.4   50

Note that the masses should add up across and down. This provides a useful check. The percentages will total 100 down but not across. This is because the percentages refer only to their own stream. This is a matter that often seems to lead to confusion.

Hence we will need 13.6 tonne/day of double cream and 36.4 tonne per day of full milk to produce the pouring cream.

 


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Energy Balances

All processes require energy, and in some cases also release energy. Thus, in addition to a Mass Balance it is also necessary to carry out an energy balance on a process. Energy may be transferred into and out of a process in one of two ways.

The energy possessed by materials entering and leaving the process and the work done in moving those materials is represented by the Enthalpy of the materials. Thus in the simple process below (Fig 1.3), the energy inputs and outputs are represented by


Fig 1.3 Energy Balance

The net energy flow into or out of the process is given by

Q = ΣHout - ΣHin = ΔH  (3)

If there is no change of state, then the enthalpy change, ΔH is determined from

Mass x heat capacity x change in temperature

If there is a change of state (eg. vapourisation or freezing) then ΔH is found from

Mass x Latent Heat.

These equations may then be written as.

Q = ΔH = m Cp (Tout - Tin)  (4)

Q = m λ   (5)

A common situation is when heat is transferred from one part of a process to another. In this case, the conservation law implies that the heat lost from the hot fluid must be equal to that gained by the cold fluid. Resolving such problems then involves calculating heat gains and losses using equations 4 and 5 as needed.


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Energy Balance Worked Example

250 l h-1 Orange juice is to be concentrated by heating under vacuum from 20oC to 60oC then evaporating 20% of the feed using condensing steam at 105oC. The concentrated juice, then cooled to 10oC using 400 l h-1 chilled water at 5oC. Calculate the steam and required and the cooling water exit temperature.

Data

Latent heat of steam at 105oC 2358 kJ kg-1
Latent heat of water vapour at 60oC 2242 kJ kg-1
Heat capacity of orange juice 3.8 kJ kg-1 K-1
Heat capacity of water 4.18 kJ kg-1 K-1
Density of orange juice 1030 kg m-3 (=1.03 g l-1)
Density of water 1000 kg m-3 (=1.00 g l-1)

Solution

As with the mass balance problem, the first step is to draw a diagram

 

Mass flow of feed = vol flow x density = 250 x 1.03  = 257.5 kg h-1
Heat Gained by Feed, Q = m Cp (Tout - Tin) = = 257.5 x 3.8 x 40 = 39140 kJ h-1
Heat to evaporate water, Q = m x LHt = = 257.5 x 0.2 x 2358 = 115463 kJ h-1
Total heat input to the evaporator =   = 154603 kJ h-1
Steam required = heat input/LHt of steam = 154603/2242 = 65.6 kg h-1
     
Mass flow of Concentrated juice = = 257.5 x 0.8 = 206 kg h-1
Mass flow of Cooling water = vol flow x density = 400 x 1.00 = 400 kg h-1
Heat lost by cooled juice, Q = m Cp (Tout - Tin = 206 x 3.8 x 50 = 39140 kJ h-1
Temperature change in water = Q/m Cp = 39140/(400 x 4.18) = 23.4 oC
Water outlet temperature = = 23.4 + 5 = 28.4 oC


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Problems

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