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 microorganisms are related
to the combined carbon in the feed; or the essentially inert nonoil constituents
of the oilseeds in an oilextraction 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
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 m^{3} 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:

20

=
0.167


100
+ 20

% weight/weight
= 16.7%
(b) Weight/volume
fraction:
A density of 1323 kg m^{3} means that 1m^{3} of solution
weighs 1323 kg, but 1323 kg of salt solution contains

20

x
1323 kg salt = 220.5 kg salt m^{3}.


100
+ 20


and so 1
m^{3} solution contains 220.5 kg salt. 

Weight/volume
fraction =


220.5

=
0.2205.


1000

and
so weight/volume =
22.1%
(c) Mole fraction:

Moles
of water =

100

=
5.56. 

18


Moles
of salt =

20

=
0.34.


58.5


Mole
fraction of salt = 
0.34



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 m^{3}.
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 m^{3} 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 10^{5} 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 
77

moles
of N_{2} and

23

moles
of O_{2}



28

32

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

So
mean molecular weight = 
100

=
28.8.


3.47


Mean
molecular weight of air = 28.8 
(b) The mole fraction of oxygen

= 
0.72

=

0.72

=
0.21



2.75
+ 0.72

3.47


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 m^{3}
atm mole^{1} K^{1} and at a temperature of 25°C
= 25 + 273 = 298 K, and V = 1 m^{3}
pV =
nRT
and
so 
1.5
x 1 
=
n x 0.08206 x 298 

n

=
0.061 mole 

weight
of air in 1m^{3} 
=
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}
or 
0.4

=
0.013 mole m^{3}.


32

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 C0_{2} in the drink, ignoring all
components other than C0_{2} and water.
Basis 1 m^{3}
of water =
1000 kg.
Volume of carbon dioxide added = 3 m^{3}.
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 = 
35,000

=
5833 kg.


6


Fat
= 5833 x 0.04 
=
233 kg.

And so water plus
solidsnotfat = 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
(socalled 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 m^{3} 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 m^{3} and the residence time in
this is 16 h so the flow rate through the fermenter must be 20/16
= 1.250 m^{3} 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 2^{1} kg
(1 generation).
In 16h there are
16/2.9 =
5.6 doubling times
1kg yeast grows to 1 x 2^{5.6} 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}
Yeastfree 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) 
Yeastfree 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.
Blending
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 straightforward 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
or
B =
100  A
and substituting
into the first equation
Aa + (100  A)b = 100c
or
A(a  b) = 100(c  b)
or
A = 100 (cb)/
(ab)
and taking the numbers
from the example

A
= 
100
(0.15 – 0.05)


(
0.23 –0.05)


= 
100
(0.10)


(0.18)


=
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.
Layout
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 



Losses: 




Peelings 




 solids 
11 kg 



 water 
69 kg 



Water evaporated 
781 kg 



Total losses 
861 kg 
Total 
1000
kg 

Total 
1000
kg 
Product
yield 
139

=
14%


1000

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 soyabean constituents.
Basis 1000 kg
Mass in:
Oil
= 1000 x 18/100 = 180 kg
Protein =
1000 x 35/100 = 350 kg
Total nonoil 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
BALANCE. BASIS 1000 kg SOYA BEANS ENTERING
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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