Principles of material (mass) balance

Principles of material (mass) balance ACCUMULATION = INPUT - OUTPUT + GENERATION - CONSUMPTION Input – enters through the boundary of the system Output – leaves through the boundary of the system Generation – produced within the system Consumption – consumed within the system Accumulation – build-up within the system

General approach to material balance The general approach for carrying out the material balance on a chemical process can be summarised into the following steps: Step 1

Draw a flow diagram and fill in all the given variables

Step 2

Label unknown stream variables with symbols

Step 3

Decide on a “basis” of calculation (do you need one?)

Step 4

If there are mixed mass and mole units, convert to the same units

Step 5

Write material balance equations using all available information.

Write the balance with the fewest unknowns first

Step 6

Check if the equations are solvable or not. You can Degree of Freedon analysis (DOF) to check if the equations are solvable or not. DOF is a powerful tool and used by many chemical engineers.

Step 7

Solve for the unknowns progressively Substitute values on the flow diagram and into other equations as they are found

Step 8

Scale the final values as necessary. Make sure the units are consistent.

Step 9

Check your final answer using dependent equations

Note: Based on the nature of the problem, some of these steps can be avoided.

Step 3 Choice of Basis A basis of calculation is an amount (mass or mole) or flowrate (mass or molar) of one stream or stream components in a process.

Once a basis is chosen then all unknown variables are then determined to be consistent with the basis Examples of basis – 100 kg of feed for liquids and or solids if weight analysis is given 100 moles is a good choice for gases Tips for choosing a basis (1) For any non-reacting system work in the units that are given in the problem (2) In reacting systems it is easier to use moles as the chemical reactions work in moles (3) Often convenient to work with, for example, 100 kg (or kmol) of feed, even if you are given a flow in some other number (e.g. 50 kg). You can always scale up or down later.

Example If a fuel is 80% C and 20% H by weight, what is the C/H ratio in moles?

Choose a basis of 100kg of fuel and so this means you have 80kg of C and 20 kg of H. You then need to determine the number of moles of C you have in 80kg and the number of moles of H you have in 20kg. The ratio is then just no of moles of C divided by number of moles of H. Answer: C/H = 0.335 Example What is the average molecular weight of a gas which contains 10mol% CH4, 60mol% C2H6 and 30mol% H2?

Choose a basis of 100kmol of gas. So you now have 10 kmol of CH4, 60kmol C2H6 and 30kmol H2. You can then calculate (using the molecular weights of each component) the mass in kg of each component.

The total mass of the gas is the sum of the mass of each component which equals 2020kg. The average MW is 2020kg/100kmol which is 20.2kg/kmol Step 5 Write material balance equations using all available information.

Mass conservation can be applied to all material in the process (total balance) or just one component (component balance) Overall Balance Input + Generation – Output – Consumption = Accumulation Component Balance Inputi + Generationi – Outputi – Consumptioni = Accumulationi Special Cases (1) If process reaches steady state (2) If there is no chemical reaction

ACCUMULATION = 0 GENERATION = CONSUMPTION = 0

Therefore in a steady state no reacting system, the material balance becomes

INPUT = OUTPUT Step 6 Check if the equations are solvable or not A frequent problem facing chemical engineers is to solve equations for a process. This is a fairly simple task when you are dealing with a simple process. However, in complicated processes this is not an easy task, and you may spend hours with no result at the end. Not all problems have solutions, and not all problems have unique solutions. Problems in which there is insufficient information are underspecified. Problems with too much information are overspecified. A problem with just the right amount of information is uniquely specified. You can use the Degree of Freedom (DoF) to see if the problem is solvable or not. Degree of freedom determines whether a process is uniquely specified or not.

To perform a DoF, add up the number of unknown variables you’ll need to calculate. Then add up the number of independent equations relating the unknown variables.

Independent equations can come in different flavours - you need to account for all of them to do the degree of freedom analysis properly. The equations may include material balances, information from the problem statement not already listed on the flowchart, physical properties and laws (e.g. ideal gas equation to relate P, V and T; tables of density to interconvert mass and volume), physical constraints (such as the sum of all mass or all molar fractions must add up to 1) and, for a reactive process, stoichiometric constraints derived from the chemical reactions taking place.

For a given problem, where:

DoF = Nunknowns – Nindependent equations

DoF = degrees of freedom Nunknowns = number of unknowns Nindependent equations = number of independent equations • DoF = 0: A numerical value for each unknown can be obtained. The problem is uniquely specified and solvable.

• DoF > 0: The problem is underspecified (underdetermined). There is not enough information (i.e. not enough independent equations) to provide a solution.

• DoF < 0: The problem is over-specified (over-determined). The problem is mathematically and possibly physically inconsistent.

Sources of equations relating to unknown process streams: (1) (2) (3) (4) (5)

Material Balances Energy Balances Process Specifications Physical properties and laws Physical constraints

Step 8 Scale the final values as necessary A balanced process (for materials) means that the material balance is satisfied. So for a steady-state, process with no reactions the material balance is satisfied if

∑ INPUTS

i

= ∑ OUTPUTSi

1kg HNO3 2 kg HNO3=H2O

Mixer 1 kg H2O 1kg HNO3 + 1kg H2O = 2kg (HNO3 = H2O) The system is balanced.

Scaling is the process of changing the values of all stream amounts or flowrates by a proportional amount while leaving the stream compositions unchanged.

Scaling UP – final stream quantities > original Scaling DOWN – final stream quantities < original

1kg HNO3 Mixer

2 kg

1 kg H2O X 20

20 kg HNO3 Mixer 20 kg H2O

40 kg