Introducing Food and process engineering
INTRODUCTION
This post is designed to give food
technologists an understanding of the engineering principles involved in the
processing of food products. They may not have to design process equipment in
detail but they should understand how the equipment operates. With an understanding
of the basic principles of process engineering, they will be able to develop
new food processes and modify existing ones. Food technologists must also be
able to make the food process clearly understood by design engineers and by the
suppliers of the equipment used.
Only a thorough understanding of the basic
sciences applied in the food industry - chemistry, biology and engineering -
can prepare the student for working in the complex food industry of today. This
post discusses the basic engineering principles and shows how they are
important in, and applicable to, every food industry and every food process.
For the food process engineering student, this
post will serve as a useful introduction to more specialized studies.
METHOD OF STUDYING FOOD PROCESS
ENGINEERING
As an introduction to food process
engineering, this post describes the scientific principles on which food
processing is based and gives some examples of the application of these
principles in several food industries. After understanding some of the basic
theory, students should study more detailed information about the individual
industries and apply the basic principles to their processes.
For example, after studying heat transfer
in this post, the student could seek information on heat transfer in the
canning and freezing industries.
To supplement the relatively few books on
food-process engineering, other sources of information are used, for
example:
• Specialist descriptions of particular food industries.
These in general are written from a
descriptive point of view and deal only briefly with engineering.
• Textbooks in chemical and biological process engineering.
These are studies of processing operations
but they seldom have any direct reference to food processing. However, the
basic unit operations apply equally to all process industries, including the
food industry.
• Engineering handbooks.
These contain considerable data including
some information on the properties of food materials.
• Periodicals.
In these can often be
found the most up-to-date information on specialized equipment and processes,
and increased basic knowledge of the unit operations A representative list of
food processing and engineering textbooks is in the bibliography at the end of
the book.
BASIC PRINCIPLES OF FOOD PROCESS
ENGINEERING
The study of process engineering is an
attempt to combine all forms of physical processing into a small number of
basic operations, which are called unit operations. Food processes may seem
bewildering in their diversity, but careful analysis will show that these
complicated and differing processes can be broken down into a small number of
unit operations. For example, consider heating of which innumerable instances
occur in every food industry. There are many reasons for heating and cooling -
for example, the baking of bread, the freezing of meat, the tempering of oils.
But in process engineering, the prime
considerations are firstly, the extent of the heating or cooling that is
required and secondly, the conditions under which this must be accomplished.
Thus, this physical process qualifies to
be called a unit operation. It is called 'heat transfer'.
The essential concept is therefore to
divide physical food processes into basic unit operations, each of which stands
alone and depends on coherent physical principles. For example, heat transfer
is a unit operation and the fundamental physical principle underlying it is
that heat energy will be transferred spontaneously from hotter to colder
bodies.
Because of the dependence of the unit
operation on a physical principle, or a small group of associated principles,
quantitative relationships in the form of mathematical equations can be built
to describe them. The equations can be used to follow what is happening in the
process, and to control and modify the process if required.
Important unit operations in the food
industry are fluid flow, heat transfer, drying, evaporation, contact
equilibrium processes (which include distillation, extraction, gas absorption,
crystallization, and membrane processes), mechanical separations (which include
filtration, centrifugation, sedimentation and sieving), size reduction and
mixing.
These unit operations, and in particular
the basic principles on which they depend, are the subject of this post, rather
than the equipment used or the materials being processed.
Two very important laws, which all unit
operations obey, are the laws of conservation of mass and energy.
Conservation of Mass and Energy
The law of conservation of mass states
that mass can neither be created nor destroyed. Thus in a processing plant, the
total mass of material entering the plant must equal the total mass of material
leaving the plant, less any accumulation left in the plant. If there is no
accumulation, then the simple rule holds that "what goes in must come
out". Similarly all material entering a unit operation must in due course
leave.
For example, when milk is being fed into a
centrifuge to separate it into skim milk and cream, under the law of
conservation of mass the total number of kilograms of material (milk) entering
the centrifuge per minute must equal the total number of kilograms of material
(skim milk and cream) that leave the centrifuge per minute.
Similarly, the law of conservation of mass
applies to each component in the entering materials. For example, considering
the butter fat in the milk entering the centrifuge, the weight of butter fat
entering the centrifuge per minute must be equal to the weight of butter fat
leaving the centrifuge per minute. A similar relationship will hold for the
other components, proteins, milk sugars and so on.
The law of conservation of energy states
that energy can neither be created nor destroyed. The total energy in the
materials entering the processing plant plus the energy added in the plant must
equal the total energy leaving the plant.
This is a more complex concept than the
conservation of mass, as energy can take various forms such as kinetic energy,
potential energy, heat energy, chemical energy, electrical energy and so
on.
During processing, some of these forms of
energy can be converted from one to another. Mechanical energy in a fluid can
be converted through friction into heat energy. Chemical energy in food is
converted by the human body into mechanical energy.
Note that it is the sum total of all these forms of energy that is
conserved.
For example, consider the pasteurizing
process for milk, in which milk is pumped through a heat exchanger and is first
heated and then cooled. The energy can be considered either over the whole
plant or only as it affects the milk. For total plant energy, the balance must
include: the conversion in the pump of electrical energy to kinetic and heat
energy, the kinetic and potential energies of the milk entering and leaving the
plant and the various kinds of energy in the heating and cooling sections, as
well as the exiting heat, kinetic and potential energies.
To the food technologist, the energies
affecting the product are the most important. In the case of the pasteurizer,
the energy affecting the product is the heat energy in the milk. Heat energy is
added to the milk by the pump and by the hot water passing through the heat
exchanger. Cooling water then removes part of the heat energy and some of the
heat energy is also lost to the surroundings.
The heat energy leaving in the milk must
equal the heat energy in the milk entering the pasteurizer plus or minus any
heat added or taken away in the plant.
Heat energy leaving in milk = initial heat
energy
+ heat energy added by pump
-
heat energy taken out in cooling section
-
heat energy lost to surroundings.
The law of conservation of energy can also
apply to part of a process. For example, considering the heating section of the
heat exchanger in the pasteurizer, the heat lost by the hot water must be equal
to the sum of the heat gained by the milk and the heat lost from the heat
exchanger to its surroundings.
From these laws of conservation of mass
and energy, a balance sheet for materials and for energy can be drawn up at all
times for a unit operation. These are called material balances and energy
balances.
Overall View of an Engineering
Process
Using a material balance and an energy
balance, a food engineering process can be viewed overall or as a series of
units. Each unit is a unit operation. The unit operation can be represented by
a box as shown in Fig. 1.1.
Figure 1.1 Unit operation
Into the box go the raw materials and
energy, out of the box come the desired products, byproducts, wastes and
energy. The equipment within the box will enable the required changes to be
made with as little waste of materials and energy as possible. In other words,
the desired products are required to be maximized and the undesired by-products
and wastes minimized. Control over the process is exercised by regulating the
flow of energy, or of materials, or of both.
DIMENSIONS AND UNITS
All engineering deals with definite and
measured quantities, and so depends on the making of measurements. We must be
clear and precise in making these measurements.
To make a measurement is to compare the
unknown with the known, for example, weighing a material compares it with a
standard weight of one kilogram. The result of the comparison is expressed in
terms of multiples of the known quantity, that is, as so many kilograms.
Thus, the record of a measurement consists
of three parts: the dimension of the quantity, the unit which represents a
known or standard quantity and a number which is the ratio of the measured
quantity to the standard quantity.
For example, if a rod is 1.18 m long, this
measurement can be analysed into a dimension, length; a standard unit, the
metre; and a number 1.18 that is the ratio of the length of the rod to the
standard length, 1 m.
To say that our rod is 1.18 m long is a
commonplace statement and yet because measurement is the basis of all
engineering, the statement deserves some closer attention. There are three
aspects of our statement to consider: dimensions, units of measurement and the
number itself.
Dimensions
It has been found from experience that
everyday engineering quantities can all be expressed in terms of a relatively
small number of dimensions. These dimensions are length, mass, time and
temperature. For convenience in engineering calculations, force is added as
another dimension.
Force can be expressed in terms of the
other dimensions, but it simplifies many engineering calculations to use force
as a dimension (remember that weight is a force, being mass times the
acceleration due to gravity).
Dimensions are represented as symbols by:
length [L], mass [M], time [t], temperature [T] and force [F].
Note that these are enclosed in square
brackets: this is the conventional way of expressing dimensions.
All engineering quantities used in this
post can be expressed in terms of these fundamental dimensions. All symbols for
units and dimensions are gathered in Appendix 1.
For example:
|
Length
= [L] area =
[L] 2 volume
= [L] 3 |
||
|
Velocity =
length travelled per unit time |
= |
[L] [t]
|
|
Acceleration = rate of change of velocity |
= |
[L] x 1
= [L] [t]
[t] [t]2 |
|
Pressure =
force per unit area |
= |
[F]
[L]2 |
|
Density =
mass per unit volume |
= |
[M]/ |
|
Energy = force times length |
=
|
[F] x[L] |
|
Power
= energy per unit time |
= |
[F] x [L]/[t] |
As more complex quantities are needed,
these can be analysed in terms of the fundamental dimensions. For example in
heat transfer, the heat-transfer coefficient, h, is defined as the quantity of heat energy transferred through
unit area, in unit time and with unit temperature difference:
h = [F] x [L] = [F]
[L]-1 [t]-1 [T]-1
[L]2[t] [T]
Units
Dimensions are measured in terms of units.
For example, the dimension of length is measured in terms of length units: the
micrometre, millimetre, metre, kilometre, etc.
So that the measurements can always be
compared, the units have been defined in terms of physical quantities. For
example:
• the metre (m) is defined in terms of the wavelength of light;
• the standard kilogram (kg) is the mass of a standard lump of platinum-iridium;
• the second (s) is the time taken for light of a given wavelength to
vibrate a given number of times;
• the degree Celsius (°C) is a one-hundredth part of the temperature
interval between the freezing point and the boiling point of water at standard
pressure;
• the unit of force, the newton (N), is that force which will give an
acceleration of 1 m sec-2 to
a mass of 1kg;
• the energy unit, the newton metre is called the joule (J), and
• the power unit, 1 J s-1, is called the watt (W).
More complex units arise from equations in
which several of these fundamental units are combined to define some new relationship.
For example, volume has the dimensions [L]3 and so the units are m3.
Density, mass per unit volume, similarly has the dimensions [M]/[L]3,
and the units kg/m3. A table of such relationships is given in
Appendix 1. When dealing with quantities which cannot conveniently be measured
in m, kg, s, multiples of these units are used. For example, kilometres, tonnes
and hours are useful for large quantities of metres, kilograms and seconds
respectively. In general, multiples of 103 are preferred such as millimetres
(m x 10-3) rather than centimetres (m x 10-2). Time is an
exception: its multiples are not decimalized and so although we have micro (10-6)
and milli (10-3) seconds, at the other end of the scale we still
have minutes (min), hours (h), days (d), etc.
Care must be taken to use appropriate multiplying factors when
working with these units. The common secondary units then use the prefixes
micro (µ, 10-6), milli (m,10-3), kilo (k, 103)
and mega (M, 106).
Dimensional Consistency
All physical equations must be
dimensionally consistent. This means that both sides of the equation must
reduce to the same dimensions. For example, if on one side of the equation, the
dimensions are [M] [L ]/[T]2, the other side of the equation must
also be [M] [L]/[T]2 with the same dimensions to the same powers.
Dimensions can be handled algebraically and therefore they can be divided,
multiplied, or cancelled. By remembering that an equation must be dimensionally
consistent, the dimensions of otherwise unknown quantities can sometimes be
calculated.
EXAMPLE 1.1. Dimensions of velocity
In the equation of motion of a particle
travelling at a uniform velocity for a time t, the distance travelled is given
by L = vt. Verify the dimensions of
velocity.
Knowing that length has dimensions [L] and
time has dimensions [t] we have the dimensional equation:
[v] = [L]/[t] the
dimensions of velocity must be [L][t]-1
The test of dimensional homogeneity is sometimes useful as an aid to
memory. If an equation is written down and on checking is not dimensionally
homogeneous, then something has been forgotten.
Unit
Consistency and Unit Conversion
Unit consistency implies that the units
employed for the dimensions should be chosen from a consistent group, for
example in this post we are using the SI (Systeme Internationale de Unites)
system of units. This has been internationally accepted as being desirable and
necessary for the standardization of physical measurements and although many
countries have adopted it, in the USA feet and pounds are very widely used. The
other commonly used system is the fps (foot pound second) system and a table of
conversion factors is given in Appendix 2.
Very often, quantities are specified or
measured in mixed units. For example, if a liquid has been flowing at 1.3 l
/min for 18.5 h, all the times have to be put into one only of minutes, hours
or seconds before we can calculate the total quantity that has passed.
Similarly where tabulated data are only available in non-standard units,
conversion tables such as those in Appendix 2 have to be used to convert the
units.
EXAMPLE 1.2. Conversion of grams to pounds Convert 10 grams into pounds.
From Appendix 2,
1lb = 0.4536kg and 1000g = 1kg so;
(1lb/ 0.4536kg) = 1 and (1kg/1000g)
= 1
therefore 10g =10g x (1lb/0.4536kg) x (1kg/1000g) = 2.2 x 10-2 lb
10 g = 2.2 x 10-2 lb
The quantity in brackets in the above
example is called a conversion factor. Notice that within the bracket, and
before cancelling, the numerator and the denominator are equal. In equations,
units can be cancelled in the same way as numbers. Note also that although
(1lb/0.4536kg) and (0.4536kg/1lb) are both = 1, the appropriate
numerator/denominator must be used for the unwanted units to cancel in the
conversion.
EXAMPLE 1.3. Velocity of flow of milk in a
pipe.
Milk is flowing through a full pipe whose
diameter is known to be 1.8 cm. The only measure available is a tank calibrated
in cubic feet, and it is found that it takes 1 h to fill 12.4 ft3.
What is the velocity of flow of the liquid in the pipe in SI units?
Velocity is [L]/[t] and the units in
the SI system for velocity are therefore m s-1: v =
L/t where v is the velocity.
Now V =
AL where V is the volume of a
length of pipe L of cross-sectional
area A i.e. L = V/A.
Therefore v = V/At
Checking this dimensionally
[L][t]-1 = [L]3[L]-2[t]-1 =
[L][t]-1 which is correct.
Since the required velocity is in m s-1,
volume must be in m3, time in s and area in m2.
From the volume measurement
V/t = 12.4ft3 h-1
From Appendix 2, 1 ft3 = 0.0283 m3 so 1 =
(0.0283 m3 /1 ft3 )
1 h = 60 x 60 s
so (1 h/3600 s) = 1
Therefore V/t = 12.4 ft3/h x (0.0283 m3/1 ft3) x (1 h/3600 s) =9.75 x 10-5 m3 s-1.
Also the area of the pipe A = µD2/4
= π(0.018)2 /4 m2 = 2.54x 10-4 m2
v = V/t x
1/A
= 9.75 x 10-5/2.54 x 10-4
= 0.38 m s-1
EXAMPLE 1.4. Viscosity (µ) conversion from fps to SI units The viscosity of water at 60°F is given as 7.8 x 10-4 lb ft-1 s-1. Calculate this viscosity in N s m-2.
From Appendix 2, 0.4536 kg = 1 lb
0.3048 m = 1 ft.
Therefore 7.8 x 10-4 lb ft-1
s-1 = 7.8 x 10-4 lb ft-1 s-1 x 0.4536 kg
x 1
ft
1 lb 0.3048m
= 1.16 x 10-3 kg m-1
s-1
Remembering that one Newton is the force
that accelerates unit mass at 1ms-2
1 N = 1 kg m s-2
therefore 1 N s m-2 =
1 kg m-1 s-1
Required viscosity = 1.16 x 10-3
N s m-2.
EXAMPLE 1.5. Thermal conductivity of aluminium: conversion from fps
to SI units The thermal conductivity of aluminium is given as 120 Btu ft-1
h-1 °F-1. Calculate this thermal conductivity in J m-1
s-1 °C-1.
From Appendix 2,
1 Btu = 1055 J
0.3048 m = 1 ft
°F = (5/9) °C.
Therefore 120 Btu ft-1 h-1
°F-1
=
120 Btu ft-1 h-1 °F-1 x 1055 J
x 1
ft x 1h x
1°F
1 Btu
0.3048m 3600s (5/9)°C
= 208 J m-1 s-1 °C-1
Alternatively a conversion factor 1Btu ft-1
h-1 °F-1can be calculated:
1Btu ft-1h-1°F-1
= 1Btu ft-1 h-1 °F-1
x 1055
J x 1 ft x
1h x 1°F
1 Btu 0.3048 m 3600s
(5/9)°C
= 1.73 J m-1 s-1 °C-1
Therefore 120 Btu ft-1 h-1
°F-1
= 120 x 1.73J m-1 s-1
°C-1
= 208 J m-1 s-1
°C-1
Because engineering measurements are often
made in convenient or conventional units, this question of consistency in
equations is very important. Before making calculations always check that the
units are the right ones and if not use the necessary conversion factors. The
method given above, which can be applied even in very complicated cases, is a
safe one if applied systematically.
A loose mode of expression that has
arisen, which is sometimes confusing, follows from the use of the word per, or
its equivalent the solidus, /. A common example is to give acceleration due to
gravity as 9.81 metres per second per second. From this the units of g would seem to be m/s/s, that is m s s-1
which is incorrect. A better way to write these units would be g =
9.81 m/s2 which is clearly the same as 9.81 m s-2.
Precision in writing down the units of
measurement is a great help in solving problems.
Dimensionless Ratios
It is often easier to visualize quantities
if they are expressed in ratio form and ratios have the great advantage of
being dimensionless. If a car is said to be going at twice the speed limit,
this is a dimensionless ratio, which quickly draws attention to the speed of
the car. These dimensionless ratios are often used in process engineering,
comparing the unknown with some well-known material or factor.
For example, specific gravity is a simple way to express the relative masses or
weights of equal volumes of various materials. The specific gravity is defined
as the ratio of the weight of a volume of the substance to the weight of an
equal volume of water.
SG = weight of a volume of the substance/
weight of an equal volume of water
Dimensionally, SG=([F]/[L] 3)/([F]/[L] 3) = 1
If the density of water, that is the mass
of unit volume of water, is known, then if the specific gravity of some
substance is determined, its density can be calculated from the following
relationship:
where (rho) is the density of the substance, SG is the specific gravity
of the substance and
w is the density of water.
Perhaps the most important attribute of a
dimensionless ratio, such as specific gravity, is that it gives an immediate
sense of proportion. This sense of proportion is very important to food
technologists as they are constantly making approximate mental calculations for
which they must be able to maintain correct proportions. For example, if the
specific gravity of a solid is known to be greater than 1 then that solid will
sink in water. The fact that the specific gravity of iron is 7.88 makes the
quantity more easily visualized than the equivalent statement that the density
of iron is 7880 kg m-3.
Another advantage of a dimensionless ratio
is that it does not depend upon the units of measurement used, provided the units
are consistent for each dimension.
Dimensionless ratios are employed
frequently in the study of fluid flow and heat flow. They may sometimes appear
to be more complicated than specific gravity, but they are in the same way
expressing ratios of the unknown to the known material or fact. These
dimensionless ratios are then called dimensionless numbers and are often called
after a prominent person who was associated with them, for example Reynolds
number, Prandtl number, and Nusselt number; these will be explained in the
appropriate section.
When evaluating dimensionless ratios, all
units must be kept consistent. For this purpose, conversion factors must be
used where necessary.
Precision of Measurement
Every measurement necessarily carries a
degree of precision, and it is a great advantage if the statement of the result
of the measurement shows this precision. The statement of quantity should
either itself imply the tolerance, or else the tolerances should be explicitly
specified.
For example, a quoted weight of 10.1 kg
should mean that the weight lies between 10.05 and 10.149 kg.
Where there is doubt it is better to
express the limits explicitly as 10.1 ± 0.05 kg.
The temptation to refine measurements by
the use of arithmetic must be resisted.
For example, if the surface of a rectangular tank is measured as 4.18 m
x 2.22 m and its depth estimated at 3 m, it is obviously unjustified to
calculate its volume as 27.8388 m3 which is what arithmetic or an
electronic calculator will give. A more reasonable answer would be 28 m3.
Multiplication of quantities in fact multiplies errors also.
In process engineering, the degree of
precision of statements and calculations should always be borne in mind. Every
set of data has its least precise member and no amount of mathematics can
improve on it. Only better measurement can do this.
A large proportion of practical
measurements are accurate only to about 1 part in 100. In some cases factors
may well be no more accurate than 1 in 10, and in every calculation proper
consideration must be given to the accuracy of the measurements. Electronic
calculators and computers may work to eight figures or so, but all figures
after the first few may be physically meaningless. For much of process
engineering three significant figures are all that are justifiable.
SUMMARY
1.
Food processes can be analysed
in terms of unit operations.
2.
In all processes, mass and
energy are conserved.
3.
Material and energy balances
can be written for every process.
4.
All physical quantities used in
this post can be expressed in terms of five fundamental dimensions [M] [L] [t]
[F] [T].
5.
Equations must be dimensionally
homogeneous.
6.
Equations should be consistent
in their units.
7.
Dimensions and units can be
treated algebraically in equations.
8.
Dimensionless ratios are often
a very graphic way of expressing physical relationships.
9.
Calculations are based on
measurement, and the precision of the calculation is no better than the
precision of the measurements.
PROBLEMS
1. Show that the following heat transfer equation is consistent in its units:
q = UA∆T
where q
is the heat flow rate (J s-1), U
is the overall heat transfer coefficient (J m-2s-1°C-1), A
is the area (m2) and ∆T is
the temperature difference (°C).
2.
The specific heat of apples is
given as 0.86 Btu lb-1 °F-1. Calculate this in J kg-1°C-1.
(3600 J kg-1 °C-1 =
3.6 kJ kg-1 °C-1)
3.
If the viscosity of olive oil
is given as 5.6 x 10-2 lbft-1s-1, calculate
the viscosity in SI units.
(83 x 10-3
kgm-1s-1 = 83 x 10-3
Nsm-2)
4.
The Reynolds number for a fluid
in a pipe is
where D is the diameter of
the pipe, v is the velocity of the
fluid, is the density of the fluid and µ is the viscosity of the fluid.
Using the five fundamental dimensions [M], [L], [T], [F] and [t] show that this
is a dimensionless ratio.
5.
Determine the protein content
of the following mixture, clearly showing the accuracy:
|
|
% Protein |
Weight in mixture. |
|
Maize starch |
0.3 |
100 kg |
|
Wheat flour |
12.0 |
22.5 kg |
|
Skim milk powder |
30.0 |
4.31 kg |
(3.4%)
6.
In determining the rate of heating
of a tank of 20% sugar syrup, the temperature at the beginning was 20oC
and it took 30min to heat to 80oC. The volume of the sugar syrup was
50 ft3 and its density 66.9 lbft-3. The specific heat of
sugar syrup is 0.9 Btu lb-1 °F -1.
(a) Convert the specific heat to kJ kg-1 °C-1
(b) Determine the average rate of heating, that is the heat energy
transferred in unit time, in SI units (kJs-1)
((a) 3.7 kJ kg-1 °C-1
(b) 187 kJs-1)
7.
The gas equation is PV = nRT.
If P
the pressure is 2.0 atm, V the volume
of the gas is 6 m3, R the
gas constant is 0.08206 m3atm mole-1 K-1 and T is 300 degrees Kelvin, what are the
units of n and what is its numerical
value?
(0.49 moles)
8.
The gas law constant R is given as 0.08206 m3 atm
mole-1 K-1. Find its value in:
(a) ft3mm Hg
lb-mole-1 K-1, (b) m3 Pa mole-1 K-1, (c)
Joules g-mole-1 K-1.
Assume 1atm. = 760mm Hg = 1.013x105
Nm-2. Remember 1 joule = 1Nm and in this post, mole is kg mole.
((a) 999 ft3mm Hg lb-mole-1
K-1 (b) 8313 m3 Pa mole-1 K-1(c)
8.313 J g-mole-1 K-1)
9.
The equation determining the
liquid pressure in a tank is z = Pρg where z is the depth, P is the pressure, ρ is the density and g is the acceleration due to gravity. Show that
the two sides of the equation are dimensionally the same.
10. The Grashof number (Gr) arises in the study of natural convection
heat flow. If the number is given as:
D3ρ2βg∆T /
verify the dimensions of β the coefficient of expansion of the fluid. The symbols are all
defined in Appendix 1.
( [T]-1)
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