# An In Depth Guide To Material Balances

## What Is A Material Balance?

Material balances also called mass balances, is an application of the law of conversation of mass, which states matters is neither created nor destroyed. Material balances are vital for chemical engineers as they are the basis of process design and allow the design of units themselves as they determine the quantities of raw materials required for the quantity of product required.

The material balance can be seen as the accounting for materials that enter and leave a process and accounts for all materials even if they undergo chemical reactions, separation, heating, cooling, mixing, drying or any other operation (excluding nuclear reactions) that occurs in the system.

Material balances are a really useful tool for studying plant; operations, troubleshooting, checking actual plant performance versus the design performance (what the plant works at against what it was designed for), can extend the amount of data from plant instrumentation, checks if instruments are calibrated correctly and can help locate sources of material loss.

## Key Definitions

A list of definitions that must be understood when doing material balances:

• Steady-state – Conditions at all points within the process are constant with time.
• Unsteady state – Conditions within the process vary with time.
• System boundary – the imaginary box is drawn around the system that is being analysed.
• Open system (or flow system) – Materials crosses the system boundary during the process.
• Closed system – no materials cross the system boundary except the beginning and end of the process, and changes take place within the system and heat may be transferred.

## System Boundary

The system or region needs to be defined by an imaginary closed box which is called the system boundary, which should always be drawn to avoid making any mistakes (figure 1). A system can be one single process unit, a collection of process units or an entire process.

Figure 1: How to define a system boundary on a system

## Material Balance Equation

As stated previously the material balances obey the law of conversation of mass, thus mass in is equal to mass out. The equation that needs to be remembered by every chemical engineer at all levels is:

Accumulation = (Mass in – Mass out) + Generation – Consumption

• Accumulation – The change in quantities of materials inside the system
• Mass in – Materials entering the system that cross the system boundary
• Mass out – Materials leaving the system that cross the system boundary
• Generation – Materials produced by a chemical reaction that takes place within the system
• Consumption – Materials used by a chemical reaction that takes place within the system

## Simplification For The Material Balance Equation

In certain cases, the material balance equation will be affected:

• No net accumulation – as the process is in steady-state or when you have a batch vessel which starts empty and then ends empty, so no accumulation within the process, the equation becomes:

In + Generation = Out + Consumption

• No reaction – No materials are used up such as when blending, washing or drying. Thus, no generation of a new product of consumption of starting materials, the equation becomes:

Accumulation = In – Out

• No net accumulation or reaction – nothing is consumed, and no new products are created, and no materials remain in the process, the equation becomes:

In = Out

## Choosing A Basis

A basis is a reference point that you can choose when wanting to solve a problem where the information gave is limited or purposely missed out such as an exam type question. A good choice of basis makes material balances much easier to solve, and can be a flow rate, a unit of time or an amount.

When choosing your basis, always state your basis clearly to remove any issues that could arise from colleagues not understanding where your values came from, and for exam markers to be able to get the marks given for choosing a basis.

Other things that need to be considered are; the units to be used, the most convenient basis value to be used and what the question is asking. Having this checklist will make creating a basis straightforward and minimise mistakes.

In a scenario where one stream has enough data, then use that stream as it has the most data and if no data then assume a basis for a stream with known components, and when mass fractions are known then choose either the total mass or the mass flow rate as the basis. Similarly, if mole fractions are given then choose the total number of moles or the molar flow rate.

When having to choose a basis for a continuous steady-state process a sensible choice for the basis is based on the amount of material entering or leaving the system within a period i.e. Kg/hr. For a batch or semi-batch process then base the basis on the inlet amount at the feed initially or the output at the end of the process i.e. Kg/batch.

## Procedure For Solving A Material Balance

• Carefully analyse the situation now what you are being asked to find or work out.
• Label all quantities on the block flow diagram (BFD) or process flow diagram (PFD) using the units given and doing any conversions to make sure all units are the same.
• Choose your basis, keep this simple by using easy numbers i.e. 100 kg or 1 hour.
• Draw a calculations table to make it easier to see the values you have worked out
• Check your solution to make sure it makes sense!

## Material Balance Example

A centrifuge is used to separate particles of a diameter of 0.5 – 50 mm from a liquid.  Yeast cells are recovered from a broth (a liquid mixture containing cells) using the centrifuge. Find out the amount of cell-free discharge if per hour if 1000 L-1 is the feed stream that contains 500 mg cells L-1 with a product stream that contains 50 wt% cells. Assume a density of 1 g cm 1.

Try to do this question without looking at the answer, if you can do this first time without looking at the answer it will be very impressive, but if you cannot don’t worry getting used to material balance questions takes time.

Hint: Use the procedure to solve this question and only use the information you need to be careful of red herrings!

The diameter of 0.5 – 50 mm, is information that isn’t required to answer this question, this is the red herring.

There is no accumulation as we are given a flow term and as this is a separation there are no chemical reactions, thus no consumption or generation

• Material balance: Mass in = Mass out
• Only one unit is mentioned so, it a one-unit process system.
• There are two outlet streams as on is the product and the other one which is the cell-free discharge contains no cells.

Basis:

Basis has been given as per hour, so 1 hour.

• As mass in = mass out then steady-state
• Two components
• No cells in the discharge stream
• No chemical reactions

Draw the process diagram, label what you have and the system boundary:

Set-up calculation table:

For this scenario:

• no outlet streams in the mass in section
• no inlet streams in the mass out section
• no cells in the cell-free discharge stream
 Basis = 1 hour Mass in Mass out Components Total Components Total Cells Liquid Cells Liquid Feed stream 0 0 0 Product stream 0 0 0 Cell-free discharge 0 0 0 0 Total

Put in identifiers:

• Cells – C
• Liquid – L
• Inlet – subscript in
• Outlet – subscript, Out
• Cell-free discharge – D
 Basis = 1 hour Mass in Mass out Components Total Components Total Cells Liquid Cells Liquid Feed stream Cin Lin Cin+Lin 0 0 0 Product stream 0 0 0 Cout Lout Pout  (Cout+Lout) Cell-free discharge 0 0 0 0 Dout Dout Total Cin Lin Cin+Lin Cout Lout+Dout Pout+Dout
• Find Cin and the mass of the feed stream Pin

1000 L-1 is the feed stream that contains 500 mg cells L-1

To make it easier for later on calculations convert mg to g, ss the basis is in calculations table.

To convert from litre to g you need to use the density, as you cannot go from litre to grams.

Density = 1 g cm-3

1000 cm3 = 1 litre

• Find Lin

Simply the mass of cells subtracted from the total mass of the stream

 Basis = 1 hour Mass in Mass out Components Total Components Total Cells Liquid Cells Liquid Feed stream (g) 500 999500 106 0 0 0 Product stream (g) 0 0 0 Cout Lout Pout  (Cout+Lout) Cell-free discharge (g) 0 0 0 0 Dout Dout Total (g) 500 999500 106 Cout Lout+Dout Pout+Dout
• Find Cout

The product cell-free discharge has no cells, thus Cin will be the same as Cout.

 Basis = 1 hour Mass in Mass out Components Total Components Total Cells Liquid Cells Liquid Feed stream (g) 500 999500 106 0 0 0 Product stream (g) 0 0 0 500 Lout Pout (Cout+Lout) Cell-free discharge (g) 0 0 0 0 Dout Dout Total (g) 500 999500 106 500 Lout+Dout Pout+Dout
• Find Pout and Lout

We are told that Cout is 50wt% of the product stream, thus Lout will be the other 50wt% and Pout will be the sum of these two components masses

 Basis = 1 hour Mass in Mass out Components Total Components Total Cells Liquid Cells Liquid Feed stream (g) 500 999500 106 0 Product stream (g) 0 500 500 1000 Cell-free discharge (g) 0 0 Dout Dout Total (g) 500 999500 106 500 500  +Dout 1000  +Dout
• Find Dout

As stated previously mass in = mass out from the material balance

${10}^{6}=1000+{D}_{out}$

 Basis = 1 hour Mass in Mass out Components Total Components Total Cells Liquid Cells Liquid Feed stream (g) 500 999500 106 0 0 0 Product stream (g) 0 0 0 500 500 1000 Cell-free discharge (g) 0 0 0 0 999000 999000 Total (g) 500 999500 106 500 500  + 999000 106

Final checks:

Do a check on your solutions to make sure the values make sense, and the material balance satisfies a steady-state process.

Mass in = Mass out

thus, the material balance is satisfied.