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.
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 may be chosen such as mass, or concentrations which can be by weight or can be molar if reactions are important.
Material balances can be based on total mass, mass of dry solids, or mass of particular components, for example protein.
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) / (100 + x) = 0.045
where = x + 0.1 = 0.045(100 + x)
x = 4.6 kg
So the composition of the whole milk is then fat = 4.5%, water = 90.5/104.6 = 86.5 %, protein = 3.5/104.6 = 3.3 %, carbohydrate= 5.1/104.6 = 4.9% and ash = 0.8%
Concentrations can be expressed in many ways: weight/ weight (w/w), weight/volume (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 in kg 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 chapter the term mole means a mass of the material equal to its molecular weight in kilograms. In this chapter percentage signifies percentage by weight (w/w) unless otherwise specified.
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/m3. Calculate the concentration of salt in this solution as a (a) weight fraction, (b) weight/volume fraction, (c) mole fraction, (d) molal concentration.
(a) Weight fraction:
20 / (100 + 20) = 0.167: % weight / weight = 16.7%
(b) Weight/volume:
A density of 1323kg/m3 means that lm3 of solution weighs 1323kg, but 1323kg of salt solution contains
(20 x 1323 kg of salt) / (100 + 20) = 220.5 kg salt / m3
1 m3 solution contains 220.5 kg salt.
Weight/volume fraction = 220.5 / 1000 = 0.2205
And so weight / volume = 22.1%
c) Moles of water = 100 / 18 = 5.56
Moles of salt = 20 / 58.5 = 0.34
Mole fraction of salt = 0.34 / (5.56 + 0.34) = 0.058
d) The molar concentration (M) is 220.5/58.5 = 3.77 moles in 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 of 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 K, 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/m2 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 105Pa, or very nearly 1 atm = 100 kPa.
If air consists of 77% by weight of nitrogen and 23% by weight of oxygen calculate:
(a) the mean molecular weight of air,
(b) the mole fraction of oxygen,
(c) the concentration of oxygen in mole/m3 and kg/m3 if the total pressure is 1.5 atmospheres and the temperature is 25 oC.
(a) Taking the basis of 100 kg of air: it contains 77/28 moles of N2 and 23/32 moles of O2
Total number of moles = 2.75 + 0.72 = 3.47 moles.
So mean molecular weight of air = 100 / 3.47 = 28.8
Mean molecular weight of air = 28.8
b) The mole fraction of oxygen = 0.72 / (2.75 + 0.72) = 0.72 / 3.47 = 0.21
Mole fraction of oxygen = 0.21
c) In the gas equation, where n is the number of moles present: the value of R is 0.08206 m3 atm/mole K and at a temperature of 25oC = 25 + 273 = 298 K, and where V= 1 m3
pV = nRT
and so, 1.5 x 1 = n x 0.08206 x 298
n = 0.061 mole/m3
weight of air = n x mean molecular weight
= 0.061 x 28.8 = 1.76 kg / m3
and of this 23% is oxygen, so weight of oxygen = 0.23 x 1.76 = 0.4 kg in 1 m3
Concentration of oxygen = 0.4kg/m3
or 0.4 / 32 = 0.013 mole / m3
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.
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 0oC and atmospheric pressure. Calculate (a) the mass fraction and (b) the mole fraction of the CO2 in the drink, ignoring all components other than CO2 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
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 / (1000 + 5.9) = 5.9 x 10-3
(b) Mole fraction of carbon dioxide in drink = 0.134 / (1000/18 + 0.134) = 2.41 x 10-3
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: Balance across equipment in continuous centrifuging of milk
If 35,000kg of whole milk containing 4% fat is to be separated in a 6 hour 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 = 35000/6 = 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 / hr and skim milk (5833 – 465) = 5368 kg/hr
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.
Another class of situations which arise are 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.
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: Drying Yield
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 1 000kg potato entering
As 8% of potatoes are lost in peeling, potatoes to drying are 920 kg, solids 129 kg
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.