2
Material Balance
Material balance refers to the method of accounting the mass of raw materials entering a process and
the finished product and by-product streams exiting it. During food processing, the mass entering the
system may undergo mixing, heating, drying, evaporation, fermentation, or any other unit operation. In a
food industry, material comprises the ingredients, the final product, and the stock of ingredients or final
product stored in the warehouse. The word balance means reconciliation, which involves matching the
quantity of raw material received with the quantity of material produced. Thus, the concept of material
balance is the physical interpretation and application of the law of conservation of mass which states that
matter is neither created nor destroyed.
The purpose of conducting material or mass balance exercise in a food industry is manifold. The
foremost of all is the traceability of a lot of product, ingredient, or packaging material. Traceability is
critical in the current scenario to avoid risks in the supply chain and achieve compliance towards food
safety. According to ISO 9001:2015, traceability is defined as the ability to identify and trace the history,
distribution, location, and application of products, parts, materials, and services. A 100% conformity
to mass balance holds importance as it precisely traces back to all the batches of a finished product where
all of a contaminated batch of raw material was used. Apart from traceability, material balance is useful
in calculating the ingredient quantities in a recipe or formulation, the composition of the final product,
process yield, and efficiency (Farkas and Farkas, 1997). Furthermore, mass balance provides an estimate
of the levels of solid food wastes emanating from a food processing plant (Mardikar and Niranjan, 1995;
Niranjan, 1994). Therefore, obtaining a thorough understanding of the mass balance for a process would
lead to the reduction of material wastage and consequent improvement in the process yield.
The concept of material balance can be fully appreciated only after understanding the fundamental
principles of thermodynamics which is the branch of science that deals with the properties of matter.
Terminologies in thermodynamics which are of relevance to this chapter are defined as follows:
2.1 Terminologies and Definitions
1. System: It (Figure 2.1) comprises of any matter identified for investigation in the universe.
2. Surroundings: Remainder of the universe which encloses the system is known as the surrounding (Figure 2.1).
3. System boundary: The boundary which separates the system and surroundings. The boundary
may be real or imaginary. The outer jacket of a heating pan and the walls of a drying chamber
are the examples of a real boundary. An imaginary boundary is not based on a physical control
surface (Figure 2.1).
4. Closed system: A closed system (Figure 2.2) is one where no mass moves across the ­boundaries;
however, energy can be transferred across the boundary.
5. Open system: It (Figure 2.3) is one where mass and energy can move across the boundaries (in
or out) at some point during the process.
6. Unit operation: It can be defined as the subunit or stage of a process, which stands out with a
specific function and based on a logical–physical principle.
7. Process: It is defined as a set of actions which occur in a specific sequence to attain a defined
end. A process begins with the raw material and ends with the products and by-products. In
43
44
Essentials and Applications of Food Engineering
FIGURE 2.1 Schematic representation of a thermodynamic system, surroundings, and boundary.
FIGURE 2.2 Schematic representation of a closed system.
FIGURE 2.3 Schematic representation of an open system.
other words, a process is an engineering system which transforms the raw materials into desired
products, through a series of logically related unit operations with specified functions (Evranuz
and Kılıç-Akyılmaz, 2012).
8. Steady state: If all the properties of a system, such as temperature, pressure, concentration,
volume, and mass, do not vary with time at any point of the system, the system or process is
said to be at a steady state. Thus, if any variable of a steady-state system is monitored continuously, it can be noted that its value will remain constant with time. However, the properties may
vary with position within the system.
9. Unsteady state: If all the process variables corresponding to each position within the system
vary with time, the process or system is said to be at unsteady state.
10. State of equilibrium: A system is said to have attained its state of equilibrium when its properties do not change with time as the opposing forces acting on it are exactly counterbalanced. At
equilibrium, there is no net change in the system or the universe, as the driving force for change
is absent. Thus, during a processing operation, the state of equilibrium is avoided by disturbing
the system so that there is a continuous transformation of raw material to the desired product.
45
Material Balance
FIGURE 2.4 Schematic representation of a batch process.
FIGURE 2.5 Schematic representation of a continuous process.
11. Batch process: In a batch mode operation, all the materials are added to the system at the
beginning of the process; the system is then closed and the products are removed only when the
process is complete (Figure 2.4). A batch process operates in a closed system. Thus, there is no
exchange of mass between the system and the surroundings.
For example: Small-scale baking of bread, biscuits, and cookies; freeze-drying or lyophilization; and dry blending of ingredients in a flavored beverage manufacturing unit
12. Continuous process: The material flows continuously in and out of the system, throughout the
process (Figure 2.5). If the rates of mass input and output are equal, continuous processes can
be operated indefinitely. Many of the food processing operations are continuous.
For example: A continuous centrifuge separating whole milk into skim milk and cream in a
dairy industry and spray drying for the production of instant food powders (e.g., instant coffee,
milk powder).
13. Semi-batch or fed-batch process: This is a combination of batch and continuous processes. A
semi-batch process allows either input or output of mass into the system during the processing
time but not both.
For example: Coating, granulation, and fermentation (media is added to increase the yield
of production)
2.2 Fundamentals of Material Balance
According to the law of conservation of mass, the equation of material balance can be written as follows:
Mass of input material + mass of material generated =
mass of outflow material + mass of accumulated material + mass of material consumed +
mass of material lost
(2.1)
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Essentials and Applications of Food Engineering
The inflow stream is constituted by the raw material(s) or ingredients and the outflow stream comprises of the processed product(s), waste, losses, and by-products. Accumulation in the system refers to
decrease or increase in mass or moles with respect to time. Generation and consumption of mass are not
commonly encountered in food processing, except with the involvement of a chemical reaction which
results in the formation or consumption of chemical species. Thus, with respect to any food processing
operation, Eq. (2.1) can be written as
Mass of raw materials = mass of products + mass of wastes + mass of
by-products + mass of losses + mass of accumulated
(2.2)
product
2.3 Classification of Material Balance Equations
Depending on the rate of accumulation, material balance equations can be classified into steady state
and unsteady state.
2.3.1 Steady-State Material Balance
As discussed previously, under steady-state conditions, the flow rate and composition of the mass/material remain constant with time. Therefore, the mass of accumulation and losses is zero. Thus, Eq. (2.2) is
further reduced to result in Eq. (2.3):
Mass of raw materials = mass of products + mass of wastes + mass of by-products
(2.3)
Batch and semi-batch processes cannot operate under steady-state conditions as the mass of the
system increase or decrease with time, despite the total mass remains constant. However, it is feasible to carry out a continuous process at steady state (Figure 2.6) or near steady-state conditions
(Doran, 1995).
2.3.2 Unsteady-State Material Balance
In this case, mass or material accumulates in the system and the concentration or quantity of material
varies with the time. Thus, the material balance equation retains the term related to the accumulation of
component (as in Eq. (2.2)). In this case, the accumulation term is expressed as a differential term involving the rate of change of the system property with time. Then, the material balance equation is integrated
with respect to time to obtain an equation for the value of the dependent variable.
FIGURE 2.6 Schematic representation of a continuous and steady-state process.
Material Balance
47
2.4 Methodology for Conducting a Material Balance Exercise
The steps involved in carrying out a material balance exercise are depicted in Figure 2.7.
2.4.1 Data Collection
The primary step in formulating a material balance equation is to understand the process. Thus, the first
step involves the collation of all known data on the type of unit operation, mass, and composition of
inflow and outflow streams of the process as understood from the problem statement.
2.4.2 Construction of Block Diagram
The next step is to identify and set the system boundaries. This can be accomplished by constructing a
block diagram (Figure 2.8) to appreciate the various physical principles involved in a process. A block
diagram is the simplest form of a process flow sheet. It schematically describes the different steps of the
process along with the input and output streams entering into and emerging from these steps, respectively. Blocks or rectangles are used in the block diagrams to represent the different stages of the process,
with all the known data included in the illustration (Figure 2.8). In the block diagram, the unit operation
(stage) involved in the process is written within the block or rectangle and the concerned inflow/outflow
material to and from the stage is indicated by arrows. At this stage, it is important to select the unit
system to be used further in the material balance exercise and express all the quantities involved using
the chosen units in a consistent manner. Quantities along with the units should also be indicated in the
block diagram.
2.4.3 Selection of Basis and Tie Materials
The next step in solving the material balance problem is the selection of an appropriate basis for calculation and identification of the tie material. The basis is defined as the reference for a specific calculation.
It is a number, relative to which all the other values are calculated. The basis is especially useful when
the initial quantity is unknown and the expected outcome is in terms of percentage or mass fraction. In
the case of batch processes, the mass of incoming raw material is usually chosen as the basis. Generally,
numbers that are convenient to handle during the calculation are preferred as the basis. Usually, with
the mass of solids or liquids, 1 kg or 100 kg is used as the basis. Sugar, fat, protein, and salt are the other
components involved in material balance problems pertaining to food processing operations.
FIGURE 2.7 Steps in conducting a material balance exercise.
FIGURE 2.8 Block diagram for material balance calculation.
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Essentials and Applications of Food Engineering
For continuous operations, the mass flow rate of raw material or product (kg/h) or a fixed time frame
of operation is normally used as the basis of calculation (1 h or 1 day). Here, the material balance will
be arrived based on the input to the system or output of the system during this period. In multistage
processes, a block diagram for the process is drawn. System boundaries are moved along the parts of the
system to determine the unknown quantities in the material balance. Here, the basis is defined at each
stage of the process. It is important to mention the chosen basis while beginning to solve a material balance problem.
Tie material is a component which is used to relate the quantity of one process stream to that of
the other. It is the component which does not change from one stream to another. Understanding the
tie materials in a process helps in simplifying the problem and determining unknown quantities. For
example, in drying, which involves removal of water, the mass of solids remains constant and hence can
be considered as the tie material. In certain cases, such as the mixing process the tie material need not
be identified.
2.4.4 Setting Up the Equations of Material Balance
To set the simultaneous equations (equations involving two or more unknowns that should have the same
values in each equation), the material balance is devised on the entire system, known as the overall mass
balance. The material balance devised with respect to individual components is termed as the component
mass balance. At this stage, the assumptions, if any, have to be stated. The commonly used assumption is
that the system is at steady state, and consequently, the mass flow rates and compositions of the streams
do not change with time and the accumulation term is zero.
2.4.4.1 Overall Mass Balance
The overall mass balance considers the totality of the mass of process streams that enter and exit the
system (Figure 2.9).
In Figure 2.9, r1 and r 2 are the inflow streams and p1 and p2 are the outflow streams. If these streams
are entering and exiting the system at the mass flow rates of m r1, m r2, m p1, and m p2 (e.g., in the units of
kg/s) respectively, according to the law of conservation of mass
m r1 + m r 2 = m p1 + m p 2
(2.4)
In other words, the sum of mass flow rates of the inflow streams is equal to the sum of mass flow rates
of the outflow streams. Equation 2.4 represents the overall material balance around the system depicted
in Figure 2.9.
2.4.4.2 Component Mass Balance
To establish the component mass balance, the law of conservation of mass is considered for each of the
individual components of a stream. For example, if the streams r1, r 2, p1, and p2 indicated in Figure 2.9
contain a common component A of concentration x, then the component material balance on the system
with respect to A is written as
xr1m r1 + xr 2 m r 2 = x p1m p1 + x p 2 m p 2
FIGURE 2.9 Schematic of a process with two inflow and two outflow streams.
(2.5)
49
Material Balance
If n numbers of components are present in the system, then (n − 1) numbers of component balance equations can be formulated in addition to the overall mass balance equation. To determine the mass and
composition of the inflow and outflow streams, these equations must be solved simultaneously.
The concentration of an individual component in a stream is conventionally expressed in terms of
either mass fraction or mole fraction. Mass fraction of a given component is defined as its mass expressed
as a fraction of the total mass of the stream containing that component. If a stream contains a mixture
of the components A and B of mass m A and mB, respectively, then the mass fraction of component A is
given by
xA =
mA
m A + mB
(2.6)
And, the mass fraction of component B is given by
xB =
mB
m A + mB
(2.7)
Similarly, mole fraction can be defined as the number of moles of the component expressed as a fraction
of the total number of moles in the stream. If the stream contains n A moles of A and nB moles of B, mole
fractions of A and B are calculated according to Eqs. (2.8) and (2.9), respectively.
xA =
nA
n A + nB
(2.8)
xB =
nA
n A + nB
(2.9)
Conventionally, the notation x is used to denote the mass or mole fraction of a component in liquid state,
and the notation y is used for the gaseous state.
It is always desirable to write one component balance equation with respect to the tie material, as it
simplifies the solution and facilitates the determination of the unknowns (Toledo, 2007).
2.4.4.3 Recycle and Bypass
A recycle stream can be defined as a portion of the product stream which is separated and then returned
to mix again with the fresh feed and reenter the system. For instance, during drying, hot air exiting the
dryer is separated, recycled, and allowed to mix with the fresh air entering the dryer. The mixing is carried out using steam in a heat exchanger until the air attains a defined humidity and temperature.
The composition of a recycle stream may be similar or different from that of the product stream. The
separation of the recycle stream (R) from the gross product stream (G) is achieved by distillation, filtration, or extraction. The input of the inlet feed (T) to the process is made up by mixing the fresh feed (F)
and recycle stream (R) as shown in Figure 2.10.
In processes that involve recycling, material balance calculations can be made around the entire process (or) the mixing unit (or) the process (or) the separation unit. The recycle ratio or reflux ratio is
commonly used in the material balance calculations of processes that involve recycling of streams. It is
given by
Recycle ratio =
Amount of recycle
R
=
Amount of net product P
(2.10)
The quantity of net product (P) depends only on the quantity of the fresh feed. The recycle stream circulates inside the process at a constant flow rate and composition under steady-state conditions. The quality
50
Essentials and Applications of Food Engineering
FIGURE 2.10 Block diagram for a process involving recycle of stream.
of the recycle stream depends on the process conditions such as conversion, and it is fixed according to
the economic considerations.
Recycling is of significance with respect to the economics of the process. It also increases the rate of
conversion of a reactant by enhancing its chances to react and convert to the product. Other reasons for
recycling include the following:
•
•
•
•
Recovery of valuable materials
Conservation of energy by recycling high-temperature carrier fluids
Improvement of temperature control over a process
To decrease the inlet concentration of a given component by diluting it to a defined level
A bypass stream is a fraction of the fresh feed that skips one or more stages of the process and fed
forward into a later stage of the process (Figure 2.11). By varying the proportion of the bypassed
feed, the composition and properties of the product can be altered. Bypass is useful for reducing the
extent of conversion of input materials or for providing improved control over the temperature of the
streams.
Example 2.1
The drying rate of a continuous dryer is 12 kg/min of a food product containing 20% moisture
(on a wet basis) to result in a product containing 10% moisture. As the handling capacity of the
dryer is limited to a moisture content of less than 15%, a part of the dry product is recycled and
mixed with the fresh feed. Calculate the recycle ratio.
Solution
Let W, R, and P be the mass flow rates (kg/min) of evaporated water, recycle, and product,
respectively.
From Figure 2.12, an overall material balance over the entire process around boundary 1 is
given by
FIGURE 2.11 Block diagram for a process involving bypass stream.
51
Material Balance
FIGURE 2.12 Material balance around the dryer.
F =W +P
12 = W + P
The component material balance with respect to water is
0.20 × 12 = W + 0.1 P
Therefore, 2.4 = W + 0.1 (10 − W)
Thus, the evaporation rate is W = 1.55 kg/min.
The recycle stream mixes with the fresh feed to give a combined feed rate to the dryer.
Nevertheless, the overall material balance over the whole process does not give any information
about the recycle within the boundary 1, and therefore, a second overall material balance around
boundary 2 is required. Thus,
12 + R = G
(0.20 × 12) + (0.1 × R) = (0.15)G
2.4 + 0.1 R = 0.15(10 + R)
R = 18 kg/ min
Thus, the recycle ratio is equal to 18/12 or 1.5.
2.4.5 Solving the Equations of Material Balance
Solution to the material balance problem is obtained by solving the overall mass balance and component
balance equations simultaneously. The equations should be written in an order such that the calculations
are simplified. For example, the equation with only one unknown is written first, as it can be solved at
once, thereby eliminating an unknown from the subsequent calculations.
2.4.6 Material Balance for a Drying Process
The following section presents a stepwise procedure to establish the material balance for a drying process with an example.
Example 2.2
Determine the weight reduction that would result when potatoes are dried from 85% to 5%
moisture.
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Essentials and Applications of Food Engineering
Solution
Unit operation: Drying
System: Dryer
Type of process: Batch
Boundary: Real boundary formed by the walls of the drying chamber
Raw material: Potatoes having moisture content of 85%
Product: Dried potatoes having moisture content of 5%
Waste stream: Moisture removed from the potatoes during drying.
In this example, moisture and solid content of potatoes are the two components.
The decrease in moisture content will result in a corresponding increase in the solid
concentration.
• The solid content is considered as the tie material.
•
•
•
•
•
•
•
•
In the block diagram shown in Figure 2.13,
R = Mass of potatoes fed to the dryer (kg)
P = Mass of dried potato chips (kg)
W = Mass of moisture removed during drying (kg)
MC = Moisture content (%)
In the given example, the mass of potatoes (in kg) entering the dryer is assumed and considered
as the basis for calculation
Basis: 100 kg of potato inflow into the dryer.
∴R =100 kg
Overall mass balance equation
R=W +P
(2.11)
i. Component balance with respect to moisture content
(0.85) R = (1)W + (0.05) P
(2.12)
ii. Component balance with respect to solid content (tie material)
(0.15) R = (0)W + (0.95) P
(2.13)
(or)
(0.15) R = 0.95( P)
(2.14)
∴ P = 0.16( R)
(2.15)
Thus, substituting Eq. (2.15) in Eq. (2.12),
(0.85) R = W + (0.05)(0.16)( R)
FIGURE 2.13
Block diagram to explain the process flow during drying of potato chips.
(2.16)
53
Material Balance
Since, R = 100,
W = 84.2 kg
From Eq. (2.11),
P = 100 − 84.2 = 15.8 kg
∴ P = 15.8 kg
Hence, weight reduction on drying of potato is 100 − 15.8 = 84.2 kg.
2.4.7 Material Balance for a Mixing Process
Explained in the following example are the steps involved in estimating the amounts of different ingredients to be mixed to prepare a food product.
Example 2.3
A 100 kg of a frankfurter formulation is to be prepared from lean beef, pork fat, and water. The
composition of ingredients and the required composition of the final product are as follows:
• Lean beef—15% fat, 65% water, 20% protein
• Pork fat—85% fat, 10% water, 5% protein
• Frankfurter—20% fat, 15% protein, 65% water
Calculate the amount of lean beef, pork fat, and water required to prepare the frankfurter
formulation.
Solution
Step 1: Data collection
From the preceding problem statement, the following information can be inferred:
Unit operation: Mixing
Type of process: Batch
System: Mixer
Boundary: Real boundary formed by the walls of the mixer
Raw materials: Lean beef, pork fat, and water (of known composition as given in the
problem statement)
• Product: 100 kg of frankfurter (of known composition as given in the problem
statement)
•
•
•
•
•
Step 2: Construction of block diagram (See Figure 2.14)
Step 3: Selection of basis
Basis: 100 kg of frankfurter with composition 20% fat, 15% protein, 65% water.
Step 4: Setting up the equations of material balance
FIGURE 2.14 Block diagram for the mixing process.
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Essentials and Applications of Food Engineering
i. Overall mass balance
In the block diagram shown in Figure 2.14,
P = amount of frankfurter produced (kg)
X = amount of lean beef used to prepare the frankfurter (kg)
Y = amount of pork fat used to prepare the frankfurter (kg)
W = amount of water used to prepare the frankfurter (kg)
f = fat content (%)
w = water content (%)
p = protein content (%)
Therefore, the overall mass balance can be written as
X +Y +W = P
(2.17)
Since, P = 100 kg (given)
∴ X + Y + W = 100
(2.18)
ii. Component mass balance
P = 100 kg (given data)
I. Protein: 0.20( X ) + 0.05(Y ) + W (0) = 0.15(100)
0.20( X ) + 0.05(Y ) = 15
(2.19)
II. Fat: 0.15( X ) + 0.85(Y ) + W (0) = 0.20(100)
0.15( X ) + 0.85(Y ) = 20
(2.20)
III. Water: 0.65( X ) + 0.10(Y ) + W (1) = 0.65(100)
0.65( X ) + 0.10(Y ) + W = 65
(2.21)
Step 5: Solving the equations of material balance
First, solving the simultaneous Eqs. (2.19) and (2.20) and changing the sign of Eq. (2.20), in order
to eliminate one of the unknown variables, we get
(0.20 × 17)( X ) + (0.05 × 17)(Y ) = (15 × 17)
(−) (0.15 × 1)( X ) (−) (0.85 × 1)(Y ) = (−)(20 × 1)
3.4 X +0.85Y = 255
−0.15X –0.85Y = −20
3.25X = 235
∴ X = 72.31 kg
Substituting the value of X = 72.31 kg in Eqs. (2.19) or (2.20), the value of Y can be obtained.
Accordingly,
Y = 10.76 kg
Substituting the values of X = 72.31 kg and Y = 10.76 kg in the Eq. (2.18), the value of W can be
obtained. Accordingly,
W = (100) – (72.31 + 10.76)
∴ W = 16.93 kg
55
Material Balance
Hence, for the preparation of 100 kg frankfurter with 20% fat, 15% protein, and 65% water,
72.31 kg of lean beef 10.76 kg of pork fat and 16.93 kg of water are required.
This problem can also be solved using the determinant method. The matrices for the protein
and fat balance equations are
 0.15

 0.2
X=
Y=
20
15
0.15
0.2
0.15
0.2
0.15
0.2
0.85
0.05
0.85
0.05
20
15
0.85
0.05
0.85   X   20 
=


0.05   Y   15 
=
(20)(0.05) − (15)(0.85)
= 72.31
(0.15)(0.05) − (0.20)(0.85)
=
(0.15)(15) − (0.2)(20)
= 10.76
(0.15)(0.05) − (0.20)(0.85)
From the overall mass balance: W = 100 − 72.31 − 10.76 = 16.93 kg
2.4.8 Material Balance for an Evaporation Process
The following discussion is on setting up a material balance around a single-effect evaporator which
concentrates a dilute liquid food product. The application of material balance, in this case, is to calculate
the amount of water evaporated and the product collected per unit time of the process.
Example 2.4
A single-effect evaporator concentrates 500 kg/h of juice from 10% to 40%. Calculate the amount
of water evaporated and the product collected per hour.
Solution
Step 1: Data collection
•
•
•
•
•
•
•
Unit operation: Evaporation
Type of process: Continuous
System: Single-effect evaporator
Boundary: Real boundary formed by the outer jacket of the evaporator chamber
Raw material: 500 kg juice of concentration 10%
Product: Juice of concentration 40%
Waste stream: Water removed from juice during evaporation
Step 2: Construction of block diagram (Figure 2.15)
FIGURE 2.15 Block diagram for the evaporation process.
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Essentials and Applications of Food Engineering
Step 3: Selection of basis and tie materials
Basis: 1 h of operation
In the block diagram,
R = Amount of dilute orange juice fed to the evaporator (kg)
P = Amount of orange juice concentrate (kg)
W = Amount of water removed from the orange juice during evaporation (kg)
TS = Total solid content (%)
Tie material: During evaporation, water is removed from the juice to concentrate it. The mass
of solids remain constant and hence can be considered as the tie material.
Step 4: Setting up the equations of material balance
i. Overall mass balance
R = P+W
(2.22)
500 = P + W
(2.23)
Since, R = 500 kg (given)
ii. Component mass balance
In this system, water and solids are the two components of the material involved
(inflow of dilute juice and outflow of concentrated juice). Thus, writing the component
mass balance equations with respect to water and solid content as follows:
I. Solids
(0.10) R = (0.40) P + (0)W
(2.24)
(0.10)(500) = (0.40) P
50 = ( 0.40 ) P
(2.25)
II. Water
(1 − 0.10) R = (1 − 0.40) P + (1)W
(0.90) R = (0.60) P + W
(2.26)
(0.90)(500) = (0.60) P + W
450 = (0.60) P + W
(2.27)
Step 5: Solving the equations of material balance
From Eq. (2.25),
P = 50/0.40 = 125
∴P = 125 kg
Substituting the value of P = 125 kg in Eq. (2.25),
450 = (0.60)(125) + W
∴W = 375 kg
Amount of water evaporated per unit time: 375 kg
Amount of product collected per unit time: 125 kg
In many instances, two or more unit operations may be combined to result in a final product.
The principle and steps for the calculation remain the same. The block diagram should represent
the different stages. Material balance for such cases is explained with the following examples:
57
Material Balance
Example 2.5
Onion slices containing 85% moisture is dehydrated in the dryer using hot air. The dry product
has 5% moisture content. Find the quantity of fresh onion required to produce 1 ton dry product
per day. Preparation losses like peeling and trimming will be 10%.
Solution
Step 1: Data collection
•
•
•
•
•
•
•
•
Unit operations: (1) Slicing (peeling and trimming), (2) drying
Type of process: Continuous
System: (1) Slicer, (2) dryer
Boundary: Real boundary formed by the walls of the slicer and drying chamber
Raw material: To the dryer: onion slices containing 85% moisture content
Product: Dry onion containing 5% moisture content at the rate of 1 ton/day
Losses: From slicer: 10% preparation losses (peels and trimmings)
Waste stream: From dryer: moisture removed from the onion slices during drying
Step 2: Construction of block diagram
In the block diagram shown in Figure 2.16,
R = Amount of fresh onion fed to the slicer (kg)
L = Amount of onion peel waste (kg)
X = Amount of onion after slicing (kg)
P = Amount of dried onion slices (kg)
W = Amount of moisture removed during drying (kg)
M = Moisture content (%)
Step 3: Selection of basis and tie materials
Basis: 1 day
Tie material: Solid content of onion slices
Step 4: Setting up the equations of material balance
i. Overall mass balance
Overall mass balance around the slicer: R = X + L
(2.28)
R = X + (0.10) R
∴ X = 0.9( R)
(2.29)
Overall mass balance around the dryer: X = P + W
(2.30)
Since P = 1 ton (or) 1000 kg,
X = 1000 + W
FIGURE 2.16 Block diagram for the onion dehydration process.
(2.31)
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Essentials and Applications of Food Engineering
ii. Component mass balance
In this system, moisture and solid content of onion slices are the two components. The
component mass balance equations are considered around the dryer as the mass of
abovementioned components are changed only during the drying operation.
I. Moisture balance
(0.85) X = (0.05) P + (1)W
(2.32)
(0.15) X = (0.95) P + (0)W
(2.33)
II. Solid balance
(0.15) X = (0.95)(1000)
(0.15) X = 950
(2.34)
Step 5: Solving the equations of material balance
From Eq. (2.34), X = 950/0.15
∴ X = 6333.33 kg
From Eq. (2.29), R = X/0.9 = 6333.33/0.9 = 7037.04 kg
∴ R = 7037.04 kg
From Eq. (2.28),
L = 7037.04 − 6333.33 = 703.71 kg
∴ L = 703.71 kg
From Eq. (2.31), W = 6333.33 − 1000
∴ W = 5333.33 kg
Thus, 7037.04 kg of fresh onion was taken for the processing. In the first stage of slicing, 703.71 kg
was separated as waste in the form of peels and trimmings. Subsequently, 6333.33 kg of onion
slices were dried to yield 1 ton of dried product per day, during which 5333.33 kg of moisture
was removed.
Example 2.6
For the production of marmalade, fruit pulp is mixed with sugar and pectin and the mixture is
boiled to about 65% solid concentration. Find the amount of fruit pulp, sugar, and pectin that
must be used for the production of 100 kg marmalade, if the solid content of fruit pulp is 10%, the
ratio of sugar to fruit pulp in the recipe is 56:44, and the ratio of sugar to pectin is 100.
Solution
Step 1: Data collection
•
•
•
•
•
•
•
•
•
Unit operations: (1) Mixing, (2) evaporation
Type of process: Batch
System: (1) Mixer, (2) evaporator
Boundary: Real boundary formed by the walls of mixer and evaporator
Raw material: (1) To mixer: fruit pulp with 10% solids, sugar, and pectin
(2) To evaporator: mixture of fruit pulp, sugar, and pectin
Product: (1) From mixer: mixture of fruit pulp, sugar, and pectin
(2) From evaporator: marmalade of 65% solid concentration
Waste stream: From evaporator: moisture removed during the boiling and concentration of the mixture of fruit pulp, sugar, and pectin
Ratio of sugar to fruit pulp: 56:44
Ratio of sugar to pectin: 100
Step 2: Construction of block diagram
59
Material Balance
FIGURE 2.17 Block diagram for the manufacturing of marmalade.
Step 3: Selection of basis and tie materials
Basis: 100 kg of marmalade
In the block diagram shown in Figure 2.17,
F = Amount of fruit pulp (kg)
P = Amount of pectin (kg)
S = Amount of sugar (kg)
W = Amount of water removed during marmalade-making (kg)
M = Amount of marmalade produced (kg) = 100 kg
TS = Total solid content (%)
Tie material: Solid content of the (fruit pulp + sugar + pectin) mixture
Step 4: Setting up the equations of material balance
i. Overall mass balance
F + S + P = M +W
(2.35)
ii. Component mass balance
The solid balance equations are considered around the evaporator as the mass of solids is
altered only during the evaporation.
Component mass balance around the evaporator
Solid balance
(0.10) F + (1) S + (1)( P) = (0.65) M + (0)W
(2.36)
0.10 F + S + P = 65
(2.37)
Step 5: Solving the equations of material balance
It is given that the ratio of sugar to fruit pulp in the recipe is 56:44.
F 44
44
Therefore, =
and F =
S
S 56
56
44
Substituting F =
S and P = 0.01S in Eq. (2.37),
56

 44  
(0.10)   S  + S + 0.01S = 65
56 

S = 59.71 kg
And, F = 46.92 kg (calculated from the known ratio of fruit pulp to sugar)
Since, the ratio of sugar to pectin is 100
∴ P = 0.597 kg
Therefore, for the production of 100 kg of marmalade with 65% solid concentration, 59.71 kg
sugar, 46.92 kg of fruit pulp, and 0.597 kg of pectin are required.
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Essentials and Applications of Food Engineering
2.5 Material Balance for Food Standardization
Standardization can be defined as a process which is done to ensure that a consistent level of product
quality is maintained throughout the period of its availability. The most common example of food
standardization is that of milk practised in the dairy industry. The concentration of fat and protein in
milk vary with season and other factors. Hence, standardization of milk is essential to maintain a constant fat content and consistency, irrespective of the period of supply. According to the Food and Drug
Administration (CFR, 2017), standardized milk is the lacteal secretion obtained by the complete milking
of one or more healthy cows, that in the final package form for beverage use shall contain not less than
8.25% of milk solids not fat and not less than 3.25% milk fat. Milk may be adjusted by separating part
of the milk fat therefrom, or by adding thereto cream, concentrated milk, dry whole milk, skim milk,
concentrated skim milk, or nonfat dry milk. Thus, the process of milk standardization can be classified
into two types, as that involving the partial removal of cream from milk to achieve fat reduction or the
mixing of skimmed milk and whole milk to achieve the desired fat content. Thus, material balance plays
a major role in the standardization of milk.
The most simple and commonly adopted material balance approach in the food standardization process is the Pearson square. This can also be considered as an easy method to solve simultaneous equations limited to two variables.
The steps involved in using the Pearson square concept are listed as follows:
1. Draw a square.
2. At the center of the square, write the value of the desired percentage of fat, fx (or any other
component) to be present in the standardized milk (or any other food to be standardized).
3. At the upper left-hand corner of the square, write the higher constituent content, f w (% fat concentration of the most concentrated fat source used; e.g., whole milk or cream).
4. At the bottom left-hand corner of the square, place the lower constituent content, fs (% fat concentration of the lesser concentrated fat source used; e.g., skim milk).
5. Subtract diagonally the higher value from the smaller value between those at the center and left
side corners as shown in Figure 2.18.
6. Write the obtained values on the diagonally opposite right-hand corners of the square.
7. More specifically, write the “parts milk” on the top right-hand side and write “parts skim” on
the bottom right-hand side of the Pearson square.
8. The numbers, thus, written on the right corners of the Pearson square are the solutions of material balance equations with two variables. Thus, these values provide the proportions of the
ingredients to be mixed to obtain the desired standardized food product.
FIGURE 2.18 The concept of Pearson square for mass balance.
Material Balance
61
It is to be noted that the sum of values on the right-hand side corners of the square equals the difference
between the values at the top left and bottom left-hand side corners (Figure 2.18).
The application of Pearson square for mass balance calculation is explained with the following
example:
Example 2.7
A dairy industry has whole milk and skim milk with 6.5% and 0.04% fat content, respectively.
Calculate the amounts of whole milk and skim milk needed to produce 1000 kg of standardized
milk with 4.5% fat.
Solution
I. Data collection
• Desired fat percentage in the standardized milk: 4.5%
• Higher constituent content, f w = 6.5%
• Lower constituent content, fs = 0.04%
• Amount of standardized milk to be produced = 1000 kg
II. Construction of Pearson square
Therefore, from the Pearson square (Figure 2.19), it could be inferred that the
Proportion of whole milk = 4.46/6.46
Amount of whole milk required = (4.46/6.46) × 1000 = 690.4 kg
Proportion of skimmed milk = 2/6.46
Amount of skimmed milk required = (2/6.46) × 1000 = 309.6 kg
Thus, 690.4 kg of whole milk and 309.6 kg of skim milk are required to produce
1000 kg of standardized milk with 4.5% fat content.
Pearson square is not confined to milk standardization but can also be used to calculate the
amounts of any two components to be mixed together to result in a desired final concentration.
Examples could be that of
i.
ii.
iii.
iv.
Mixing of fruit juice and sugar syrup to prepare a fruit squash
Mixing of fruit pulp and sugar in the jam preparation
Mixing rations for animal feeding
In the meat industry to produce meat products such as sausages to a particular fat
content
v. Blending of wines and other alcoholic beverages to provide products of a specified
alcohol concentration.
However, the Pearson square concept is limited to the blending of only two components. When
more than two components are involved, more complex mass balance equations have to be used
(Mullan, 2006).
FIGURE 2.19 Pearson square for milk standardization.
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Essentials and Applications of Food Engineering
2.6 Application of Material Balance in Food Product Traceability
Including a material balance check exercise on at least one key ingredient or packaging material of a
product is essential to test the efficacy of a traceability system in a food industry. Here, the purpose of
mass balance exercise is to check whether the manufacturer can account for any input material that is
involved in the production of a lot or batch of the product. The input material can be a food ingredient,
semi-processed product, or packaging material. Mass balance and traceability are extremely crucial
in the reported event of a critical food safety (contamination) or regulatory (mislabeling) issue, which
demands a complete recall of a particular lot of the product from the market. The application of mass
balance in a traceability exercise is explained by the following example. Following are the steps involved
in conducting the traceability exercise.
1. To conduct the reconciliation exercise and obtain details on the quantity of the product that has
been sold, retained in the inventory, and disposed of for reasons if any.
2. The collection of abovementioned details gives the total quantity of unaccounted material.
3. Determine the percentage effectiveness of the traceability procedure using the following
formula:
% Effectiveness of traceability procedure =
B+C + D
× 100
A
Example 2.8
A complete reconciliation/mass balance exercise is to be conducted for the ingredient traceability in a seasoning mix product. The list of details to be collected for the exercise is given in
Table 2.1.
Details of reconciliation for the product containing a raw ingredient of the recalled product
are listed in Table 2.2.
TABLE 2.1
General Details of the Lot
Name of the Raw Material (Ingredient)
Name of the recalled product in which the ingredient was used
Production date
Reason for recall
Lot number of raw material
Lot number of product
Dispatch invoice number
Date of delivery
Name of the customer
Whey Powder
Seasoning mix
02/08/2017 (2nd August 2017)
Report of allergy due to incidental and
undeclared dairy ingredient
WP080117
SM080217
SM-2023
05/08/2017 (5th August 2017)
XYZ
TABLE 2.2
Details of Mass Balance
Total Amount of Product Produced (A)
10000 Packs
Amount still on inventory (B)
Amount delivered to customers (C)
Incidental disposal (D)
Total
3000 packs
5900 packs
50 packs
8950 packs
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Material Balance
where,
•
•
•
•
A: Total amount of product produced
B: Amount still on inventory
C: Amount delivered to customers
D: Incidental wastage (product dropped on ground and so on)
Thus, total accounted: 8950 packs; total unaccounted: 1050 packs.
% Effectiveness of traceability = (3000 + 5900 + 50) × 100
10000
= 89.5%