Material Balance of a Chemical Engineering Process

Material balance calculation is a necessary thing in chemical engineering. Material balance or mass balance is the calculation of the amount of each component at the inputs and outputs in every process step.

Before material balance calculation, we should have a proper idea about a system, system boundary, and surrounding. A system is a region or a quantity of matter that we have chosen to study. Anything outside of the system is surrounding. The real or imaginary surface that separates the system from its surroundings is the boundary.

General Material Balance Equation

In an industrial process, we insert materials into a system and products are taken out of the system. In material balance calculation, we calculate the percentages or masses of what we input and what we get from the output.

Calculation of material balance for an entire process is a complex task. Therefore, we calculate the material balance for every single process step. The general material balance equation is as follows.

Material in + Generation = Material out + Consumption + Acuumulation

Material balance can be applied to the total mass or total moles of the stream, the mass of a single compound, the mass of an atomic species, the moles of a compound, the moles of an atomic species, etc.

There are processes with or without a chemical reaction and steady and unsteady processes. According to the type of the process, the above equation is changed.

1. An unsteady process with a chemical reaction or a nuclear reaction
   In an unsteady process, the input rate is not equal to the output rate. Therefore, there is a material accumulation in the system. Since there is a chemical reaction or a nuclear reaction, new materials are generated, and some materials are consumed in the system. So, the material balance for such a system would be,

Material in + Generation = Material out + Consumption + Accumulation

2. A steady process with a chemical reaction or a nuclear reaction
   In a steady-state process, there is no accumulation of material inside the system. The input rate is equal to the output rate. Therefore, the accumulation term would be zero. if there is a chemical reaction or a nuclear reaction, new materials are generated and some materials are consumed. So, the material balance for such a system would be,

Material in + Generation = Material out + Consumption

3. Unsteady process without a chemical reaction or a nuclear reaction
   In an unsteady process, there is material accumulation. If there is no chemical reaction or a nuclear reaction, there is no material generation or material consumption. Therefore, those two terms would be zero. In such systems, only physical processes like distillation, adsorption, condensation, etc. The material balance equation for such a system would be as follows,

Material in =Material out + Accumulation

4. Steady process without a chemical reaction or a nuclear reaction
   For a steady process without a chemical reaction, there is no material accumulation, no material generation, or material consumption. We get the same material that we input. But only physical changes have occurred. Therefore, the material balance can be simplified as follows,

Material in =Material out

Material balance problems that do not involve chemical reactions

Let’s consider a recovery process of ethanol from a gas mixture that contains air, water vapor, and ethanol vapor. We are going to separate ethanol from gas. Before calculating the material balance, the first step that should be done is to prepare a flow diagram for the process. The simple flow diagram for the above process is as follows.

In the above process, there are no chemical reactions. Both absorption and distillation are physical separation methods. Therefore, in this process, there is no material generation or consumption. If we consider this process operates at a steady state, there is no material accumulation. For ease of calculation, the overall process can be divided into two systems.

The first system is the absorption column. For the absorption column, there are two inputs and two outputs. The mass percentages of each input and output of the absorption column are as follows.

If the input gas flow rate is 1000 Kg/hr, we can calculate the masses of each component of each stream using material balance. Each component of the system (water, air, ethanol) is applied to the mass balance separately.

A, B, C, and D are the flow rates of each stream. Since this process is not included a chemical reaction and operates at a steady state, the material balance equation would be,

Material in = Material out

A + C = B + D

Air

Ethanol

Water

For the above system input and output flow rates can be summarized as follows.

We take both the distillation column and condenser as the second system. For this system, there are one input and two outputs. The mass percentages of each input and output of this system are,

Material in = Material out

D = E + F

Ethanol

Water

By solving the above two simultaneous equations (eq1 and eq2) we can calculate the flow rates of E and F.

By substitution of the above value to equation 2, we can calculate the value of F.

For system 2, input and output flow rates can be summarized as follows.

Material balance problems that involve chemical reactions

In processes where a chemical reaction is included, new materials are generated, and some materials are consumed in the system. Therefore, output materials are not similar to input materials. So, input masses are not equal to the output masses of materials. In this situation, it is easy to use the mole balance of components.

Let’s consider the H₂SO₄ manufacturing process. The overall process can be simplified as follows. This process includes chemical reactions.

First sulfur is burned in a furnace and results in SO₃. Input sulfur has 70% purity with 30% of inert impurities. The flow rate of input sulfur is 2000 Kg/hr.  

In the furnace, 99% of sulfur is converted to SO₂ and SO₂ + air mixture is passed to the oxidizer. At the oxidizer, SO₂ is converted into SO₃ gas.

At the absorber, SO₃ gas is absorbed by water resulting in 98% of H₂SO₄. The chemical reactions at each step are given below.

At furnace

S + O₂ → SO₂

At oxidizer

2SO₂ + O₂ → 2SO₃

At absorber

SO₃ + H₂O → H₂SO₄

By finding the moles of one component, we can find the moles and the masses needed for all the components. We are dividing the overall process into sub-systems for ease of calculation. All the calculations are done for a one-hour process window.

The furnace has two inputs and two outputs.

Total from the total mols of sulfur, 99% is converted into SO₂. Only 1% is removed with impurities.

43 231.43 mols of SO₂ gas was inserted into the Oxidizer. If we consider the oxidizer is 100% efficient, the total SO₂ will be converted to SO₃.

2SO₂ + O₂ → 2SO₃

According to the stoichiometry, generated SO₃ mols are equal to SO₂ mols.

SO₃ mols = 43 231.43 mol

Water and output air (SO₃ + Air) from the oxidizer was inserted into the Absorbe. At the absorber, SO₃ gas is absorbed by H₂O resulting in 98% of H₂SO₄.

The amount of entering SO₃ mols into the absorber is 43 231.43 mol.

SO₃ + H₂O → H₂SO₄

According to the stoichiometry, mols of SO₃ are equal to the mols of H₂SO₄ mols produced. Therefore,

In the output stream of H, there is only 98% of H₂SO₄. The total mass of the output stream of H per hour is as follows.

In the absorption column, there is no generation of H₂O. Therefore, it would be zero. Consumed mols of water are equal to the consumed mols of SO₃.

Mass balance for H₂O