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Basis and units
total mass and composition
Types of process situations
continuous processes

The first step is to look at the three basic categories: materials in, materials out and materials stored. Then the materials in each category have to be considered whether they are to be treated as a whole, a gross mass balance, or whether various constituents should be treated separately and if so what constituents.

To take a simple example, it might be to take dry solids as opposed to total material; this really means separating the two groups of constituents, non-water and water. More complete dissection can separate out chemical types such as minerals, or chemical elements such as carbon.
The choice and the detail depend on the reasons for making the balance and on the information that is required. A major factor in industry is, of course, the value of the materials and so expensive raw materials are more likely to be considered than cheaper ones, and products than waste materials.

Basis and Units

Having decided which constituents need consideration, the basis for the calculations has to be decided. This might be some mass of raw material entering the process in a batch system, or some mass per hour in a continuous process.
It could be: some mass of a particular predominant constituent, for example mass balances in a bakery might be all related to 100 kg of flour entering; or some unchanging constituent, such as in combustion calculations with air where it is helpful to relate everything to the inert nitrogen component; or carbon added in the nutrients in a fermentation system because the essential energy relationships of the growing micro-organisms are related to the combined carbon in the feed; or the essentially inert non-oil constituents of the oilseeds in an oil-extraction process. Sometimes it is unimportant what basis is chosen and in such cases a convenient quantity such as the total raw materials into one batch or passed in per hour to a continuous process are often selected. Having selected the basis, then the units can be chosen such as mass, or concentrations which can be weight or molar if reactions are important.

Total mass and composition

Material balances can be based on total mass, mass of dry solids, or mass of particular components, for example protein.

EXAMPLE 2.1. Constituent balance of milk
Skim milk is prepared by the removal of some of the fat from whole milk. This skim milk is found to contain 90.5% water, 3.5% protein, 5.1% carbohydrate, 0.1% fat and 0.8% ash. If the original milk contained 4.5% fat, calculate its composition assuming that fat only was removed to make the skim milk and that there are no losses in processing.

Basis: 100 kg of skim milk. This contains, therefore, 0.1 kg of fat. Let the fat which was removed from it to make skim milk be x kg.

Total original fat     = (x + 0.1 ) kg
Total original mass = (100 + x) kg

and as it is known that the original fat content was 4.5% so

x + 0.1
= 0.045
100 + x

whence x + 0.1            = 0.045(100 + x)
x     =
4.6 kg

So the composition of the whole milk is then
fat =     4.5% ,
  water =
= 86.5 %
  protein =
= 3.3 %
  carbohydrate =
= 4.9%
  and ash   =
=  0.8%


Concentrations can be expressed in many ways: weight/weight fraction (w/w ), weight/volume fraction (w/v), molar concentration (M), mole fraction. The weight/weight concentration is the weight of the solute divided by the total weight of the solution and this is the fractional form of the percentage composition by weight. The weight/volume concentration is the weight of solute in the total volume of the solution. The molar concentration is the number of molecular weights of the solute expressed as moles in 1 m3 of the solution. The mole fraction is the ratio of the number of moles of the solute to the total number of moles of all species present in the solution. Notice that in process engineering, it is usual to consider kg moles and in this book the term mole means a mass of the material equal to its molecular weight in kilograms. In this book, percentage signifies percentage by weight (w/w) unless otherwise specified.

EXAMPLE 2.2. Concentration of salt in water
A solution of common salt in water is prepared by adding 20 kg of salt to 100 kg of water, to make a liquid of density 1323 kg m-3. Calculate the concentration of salt in this solution as a (a) weight/weight fraction, (b) weight/volume fraction, (c) mole fraction, (d) molar concentration.

(a) Weight fraction:

= 0.167
100 + 20

% weight/weight   = 16.7%

(b) Weight/volume fraction:
A density of 1323 kg m-3 means that 1m3 of solution weighs 1323 kg, but 1323 kg of salt solution contains

x 1323 kg salt = 220.5 kg salt m-3.
100 + 20
 and so  1 m3 solution contains 220.5 kg salt.

Weight/volume fraction          =
= 0.2205.

           and so weight/volume           = 22.1%

(c) Mole fraction:
Moles of water         =
             = 5.56.

Moles of salt            =
= 0.34.

Mole fraction of salt  =
5.56 + 0.34

          and so mole fraction =      0.058

(d) The molar concentration (M) is 220.5/58.5 = 3.77 moles in 1 m3.

Note that the mole fraction can be approximated by the (moles of salt/moles of water) as the number of moles of water are dominant, that is the mole fraction is close to 0.34/5.56 = 0.061.
As the solution becomes more dilute, this approximation improves and generally for dilute solutions the mole fraction of solute is a close approximation to the moles of solute/moles of solvent.

In solid/liquid mixtures all these methods can be used but in solid mixtures the concentrations are normally expressed as simple weight fractions.

With gases, concentrations are primarily measured in weight concentrations per unit volume or as partial pressures. These can be related through the gas laws. Using the gas law in the form:

pV = nRT

where p is the pressure, V the volume, n the number of moles, T the absolute temperature, and R the gas constant which is equal to 0.08206 m3 atm mole-1 K-1. The molar concentration of a gas is then

n/V = p/RT

and the weight concentration is then nM/V where M is the molecular weight of the gas.

The SI unit of pressure is the N m-2 called the Pascal (Pa). As this is of inconvenient size for many purposes, standard atmospheres (atm) are often used as pressure units, the conversion being 1 atm = 1.013 x 105 Pa, or very nearly 1 atm = 100 kPa.

EXAMPLE 2.3. Air composition
If air consists of 77% by weight of nitrogen and 23% by weight of oxygen calculate the:
(a) mean molecular weight of air,
(b) mole fraction of oxygen,
(c) concentration of oxygen in mole m-3 and kg m-3 if the total pressure is 1.5 atmospheres and the temperature is 25°C.

(a) Taking the basis of 100 kg of air:
it contains
moles of N2 and
moles of O2

          Total number of moles        = 2.75 + 0.72   = 3.47 moles.

So mean molecular weight =
= 28.8.
Mean molecular weight of air      =               28.8

(b) The mole fraction of oxygen
= 0.21
2.75 + 0.72

Mole fraction of oxygen in air    = 0.21

and this is also the volume fraction


(c) In the gas equation, n is the number of moles present, p is the pressure, 1.5 atm and the value of R is 0.08206 m3 atm mole-1 K-1 and at a temperature of 25°C = 25 + 273 = 298 K, and V = 1 m3

                           pV  =  nRT

and so 1.5 x 1 = n x 0.08206 x 298
n = 0.061 mole
  weight of air in 1m3 = n x mean molecular weight
    = 0.061 x 28.8    = 1.76 kg
and of this 23% is oxygen, weighing 0.23 x 1.76 = 0.4 kg.

Concentration of oxygen = 0.4 kg m-3
= 0.013 mole m-3.

When a gas is dissolved in a liquid, the mole fraction of the gas in the liquid can be determined by first calculating the number of moles of gas using the gas laws, treating the volume as the volume of the liquid, and then calculating the number of moles of liquid directly.

EXAMPLE 2.4. Carbonation of a soft drink
In the carbonation of a soft drink, the total quantity of carbon dioxide required is the equivalent of 3 volumes of gas to one volume of water at 0°C and atmospheric pressure. Calculate (a) the mass fraction and (b) the mole fraction of the C02 in the drink, ignoring all components other than C02 and water.

Basis 1 m3 of water                  = 1000 kg.
Volume of carbon dioxide added = 3 m3.

     From the gas equation     pV = nRT

     1 x 3         =    n x 0.08206 x 273.
     and so  n   =    0.134 mole.

Molecular weight of carbon dioxide             = 44.
and so  weight of carbon dioxide added      = 0.134 x 44 = 5.9 kg.

(a) Mass fraction of carbon dioxide in drink =     5.9/(l000 + 5.9)         = 5.9 x 10-3.

(b) Mole fraction of carbon dioxide in drink  = 0.134/(l000/18 + 0.134) = 2.41 x 10-3

Types of Process Situations

Continuous processes

In continuous processes, time also enters into consideration and the balances are related to unit time. Thus in considering a continuous centrifuge separating whole milk into skim milk and cream, if the material holdup in the centrifuge is constant both in mass and in composition, then the quantities of the components entering and leaving in the different streams in unit time are constant and a mass balance can be written on this basis. Such an analysis assumes that the process is in a steady state, that is flows and quantities held up in vessels do not change with time.

EXAMPLE 2.5. Materials balance in continuous centrifuging of milk
If 35,000 kg of whole milk containing 4% fat is to be separated in a 6 h period into skim milk with 0.45% fat and cream with 45% fat, what are the flow rates of the two output streams from a continuous centrifuge which accomplishes this separation?

Basis 1 hour's flow of whole milk

Mass in
Total mass =
   = 5833 kg.
Fat            = 5833 x 0.04
= 233 kg.

And so water plus solids-not-fat = 5600 kg.

Mass out
Let the mass of cream be x kg then its total fat content is 0.45x. The mass of skim milk is (5833 - x) and its total fat content is 0.0045(5833 - x).

Material balance on fat:
             Fat in = Fat out

    5833 x 0.04 = 0.0045(5833 - x) + 0.45x.
and so     x      =   465 kg.

So that the flow of cream is 465 kg h-1 and skim milk (5833 - 465) = 5368 kg h-1

The time unit has to be considered carefully in continuous processes as normally such processes operate continuously for only part of the total factory time. Usually there are three periods, start up, continuous processing (so-called steady state) and close down, and it is important to decide what material balance is being studied. Also the time interval over which any measurements are taken must be long enough to allow for any slight periodic or chance variation.

In some instances a reaction takes place and the material balances have to be adjusted accordingly.
Chemical changes can take place during a process, for example bacteria may be destroyed during heat processing, sugars may combine with amino acids, fats may be hydrolysed and these affect details of the material balance. The total mass of the system will remain the same but the constituent parts may change, for example in browning the sugars may reduce but browning compounds will increase. An example of the growth of microbial cells is given. Details of chemical and biological changes form a whole area for study in themselves, coming under the heading of unit processes or reaction technology.

EXAMPLE 2.6. Materials balance of yeast fermentation
Baker's yeast is to be grown in a continuous fermentation system using a fermenter volume of 20 m3 in which the flow residence time is 16 h. A 2% inoculum containing 1.2 % of yeast cells is included in the growth medium. This is then passed to the fermenter, in which the yeast grows with a steady doubling time of 2.9 h. The broth leaving the fermenter then passes to a continuous centrifuge which produces a yeast cream containing 7% of yeast, 97% of the total yeast in the broth. Calculate the rate of flow of the yeast cream and of the residual broth from the centrifuge.

The volume of the fermenter is 20 m3 and the residence time in this is 16 h so the flow rate through the fermenter must be  20/16 = 1.250 m3 h-1

Assuming the broth to have a density substantially equal to that of water, i.e. 1000 kg m-3,

Mass flow rate of broth   = 1250 kg h-1

Yeast concentration in the liquid flowing to the fermenter
     = (concentration in inoculum)/(dilution of inoculum)

     = (1.2/100)/(100/2) = 2.4 x 10-4 kg kg-1.

Now the yeast mass doubles every 2.9 h, so in 2.9 h, 1 kg becomes 1 x 21 kg (1 generation).

In 16h there are 16/2.9                 = 5.6 doubling times
    1kg yeast grows to 1 x 25.6 kg = 48.5 kg.
    Yeast in broth leaving               = 48.5 x 2.4 x 10-4  kg kg-1

Yeast leaving fermenter = initial concentration x growth x flow rate
                                   = 2.4 x 10-4 x 48.5 x 1250
                                   = 15 kg h-1

Yeast-free broth flow leaving fermenter  = (1250 - 15) = 1235 kg h-1

From the centrifuge flows a (yeast rich) stream with 7% yeast, this being 97% of the total yeast:

The yeast rich stream is (15 x 0.97) x 100/7        = 208 kg h-1

and the broth (yeast lean) stream is (1250 - 208) = 1042 kg h-1

which contains           (15 x 0.03 )                      = 0.45 kg h-1 yeast and

the yeast concentration in the residual broth       = 0.45/1042 = 0.043%

Materials balance over the centrifuge per hour

Mass in (kg) Mass out (kg)
Yeast-free broth 1235 kg   Broth 1042 kg
Yeast 15 kg   (Yeast in broth 0.45 kg)
      Yeast stream 208 kg
      (Yeast in stream 14.55 kg)
Total 1250 kg   Total 1250 kg

A materials balance, such as in Example 2.6 for the manufacture of yeast, could be prepared in much greater detail if this were necessary and if the appropriate information were available. Not only broad constituents, such as the yeast, can be balanced as indicated but all the other constituents must also balance.

One constituent is the element carbon: this comes with the yeast inoculum in the medium, which must have a suitable fermentable carbon source, for example it might be sucrose in molasses. The input carbon must then balance the output carbon, which will include the carbon in the outgoing yeast, carbon in the unused medium and also that which was converted to carbon dioxide and which came off as a gas or remained dissolved in the liquid. Similarly all of the other elements such as nitrogen and phosphorus can be balanced out and calculation of the balance can be used to determine what inputs are necessary knowing the final yeast production that is required and the expected yields. While a formal solution can be set out in terms of a number of simultaneous equations, it can often be easier both to visualize and to calculate if the data are tabulated and calculation proceeds step by step gradually filling out the whole detail.


Another class of situations which arises includes blending problems in which various ingredients are combined in such proportions as to give a product of some desired composition. Complicated examples, in which an optimum or best achievable composition must be sought, need quite elaborate calculation methods, such as linear programming, but simple examples can be solved by straight-forward mass balances.

EXAMPLE 2.7. Blending of minced meat
A processing plant is producing minced meat, which must contain 15% of fat. If this is to be made up from boneless cow beef with 23% of fat and from boneless bull beef with 5% of fat, what are the proportions in which these should be mixed?

Let the proportions be A of cow beef to B of bull beef.
Then by a mass balance on the fat,

Mass in                     Mass out
A x 0.23 + B x 0.05  =  (A + B) x 0.15.
that is A(0.23 - 0.15) =  B(0.15 -0.05).
         A(0.08)           =  B(0.10).
         A/ B               =  10/8
or      A/(A + B)        =  10/18 = 5/9.

i.e. 100 kg of product will have 55.6 kg of cow beef to 44.4 kg of bull beef.

It is possible to solve such a problem formally using algebraic equations and indeed all material balance problems are amenable to algebraic treatment. They reduce to sets of simultaneous equations and if the number of independent equations equals the number of unknowns the equations can be solved. For example, the blending problem above can be solved in this way.

If the weights of the constituents are A and B and proportions of fat are a, b, blended to give C of composition c:
     then for fat    Aa + Bb = Cc

     and overall    A + B     = C

of which A and B are unknown, and say we require these to make up 100 kg of C then

                   A + B   = 100
B    = 100 - A

and substituting into the first equation

       Aa + (100 - A)b = 100c
or            A(a - b)     = 100(c - b)
or                A         = 100 (c-b)/ (a-b)

and taking the numbers from the example

A =
100 (0.15 – 0.05)
( 0.23 –0.05)
100 (0.10)
   =   55.6 kg
and B    =   44.4 kg

as before, but the algebraic solution has really added nothing beyond a formula which could be useful if a number of blending operations were under consideration.


In setting up a material balance for a process a series of equations can be written for the various individual components and for the process as a whole. In some cases where groups of materials maintain constant ratios, then the equations can include such groups rather than their individual constituents. For example in drying vegetables the carbohydrates, minerals, proteins etc., can be grouped together as ‘dry solids’, and then only dry solids and water need be taken through the material balance.

EXAMPLE 2.8. Drying yield of potatoes
Potatoes are dried from 14% total solids to 93% total solids. What is the product yield from each 1000 kg of raw potatoes assuming that 8% by weight of the original potatoes is lost in peeling.

Basis 1000kg potato entering

As 8% of potatoes are lost in peeling, potatoes to drying are 920 kg, solids 129 kg.

Mass in (kg) Mass out ( kg)
Raw Potatoes:     Dried Product:  
Potato solids 140 kg   Potato solids 129 kg
Water 860 kg   Associated water 10 kg
      Total product 139 kg
      - solids 11 kg
      - water 69 kg
      Water evaporated 781 kg
      Total losses 861 kg
Total 1000 kg   Total 1000 kg

Product yield
= 14%
Notice that numbers have been rounded to whole numbers as this is appropriate accuracy.

Often it is important to be able to follow particular constituents of the raw material through a process. This is just a matter of calculating each constituent.

EXAMPLE 2.9. Extraction of oil fom soya beans
1000 kg of soya beans, of composition 18% oil, 35% protein, 27.1% carbohydrate. 9.4%, fibre and ash, 10.5% moisture, are:
(a) crushed and pressed, which reduces oil content in beans to 6%;
(b) then extracted with hexane to produce a meal containing 0.5% oil;
(c) finally dried to 8% moisture.

Assuming that there is no loss of protein and water with the oil, set out a mass balance for the soya-bean constituents.

Basis 1000 kg

Mass in:
Oil       = 1000 x 18/100    = 180 kg
Protein = 1000 x 35/100 = 350 kg
Total non-oil constituents = 820 kg

Carbohydrate, ash, fibre and water are calculated in a similar manner to fat and protein.

Mass out:
(a) Expressed oil.
In original beans, 820kg of protein, water, etc., are associated with 180 kg of oil.
In pressed material, 94 parts of protein, water, etc., are associated with 6 parts of oil.
Total oil in expressed material 820 x 6/94 = 52.3 kg.
Oil extracted in press    = 180 - 52.3        = 127.7 kg.

(b) Extracted oil.
In extracted meal 99.5 parts of protein, water, etc., are associated with 0.5 parts of oil.
Total oil in extracted meal = 820 x 0.5/99.5 = 4.1 kg.
Oil extracted in hexane    =    52.3 - 4.1      = 48.2 kg.

(c) Water.
In the dried meal, 8 parts of water are associated with 92 parts of oil, protein, etc.
Weights of dry materials in final meal = 350 + 271 + 94 + 4.1 = 719.1 kg.
Total water in dried meal                   = 719.1 x 8/92 = 62.5 kg.
Water loss in drying                         = 105 - 62.5     = 42.5 kg.


Mass in (kg) Mass out ( kg)
      Total oil consisting of 175.9
Oil  180   - Expressed oil  127.7
Protein 350   - Oil in hexane 48.2
Carbohydrate 271   Total meal consisting of: 781.6
Ash and fibre  94   - Protein 350
Water 105   - Carbohydrate 271
      - Ash and fibre  94
      - Water  62.5
      - Oil   4.1
      Water lost in drying  42.5
Total 1000 kg   Total 1000 kg

Material & Energy Balances > ENERGY BALANCES

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Unit Operations in Food Processing. Copyright © 1983, R. L. Earle. :: Published by NZIFST (Inc.)
NZIFST - The New Zealand Institute of Food Science & Technology