Tag: mass balance

Fermentation Mass Balancing: Part 5 – Worked example

Finally! We’ve done all the waffling about the topic, now we can get straight to it! It’s hard to explain mass balancing with words, so I think working through an example will be the most effective way of explaining a mass balance.

Note: This exercise was taken from Doran’s book: Bioprocess Engineering Principles. I HIGHLY recommend getting this book, it’s basically the bible of bioprocessing. If you want to buy it, click the photo to be taken to the amazon page:

Bioprocess Engineering Principles by Pauline M. Doran: Click on photo to get amazon link!

Continuous Acetic Acid Fermentation

C2H5OH (ethanol) + O2 -> CH3COOH (acetic acid) + H20

The production target is 2kg of acetic acid per hour. However the maximum acetic acid tolerated by cells is 12% mass acetic acid (only 12% of the outlet stream can be acetic acid). Air is pumped into the fermenter at 200gmol per hour.
What minimum amount of ethanol is required?
What minimum amount of water must be used to dilute the output stream to 12% acetic acid?
What is the composition of the outlet gas?

First, lets define the assumptions:

  • Steady state
  • No system leakage
  • 100% conversion of ethanol to acetic acid
  • O2 and ethanol available to all cells
  • Inlet air and outlet gas are dry
  • Feed has the density of water: 1kg/m3
  • Other feed components are negligible
  • Inlet air has composition of 21% O2 and 79% N2

Define mass balance equations:

We know N2 is not involved in any reactions by looking at the chemical equation. So for the total mass of the process and N2:

Mass in = Mass out

We can see from the chemical equation that ethanol and oxygen are converted to water and acetic acid, so the equation for these will be:

Mass in + Mass generated = Mass out + Mass consumed

Define basis:

Since this is a continuous process, we have to decide a point for where we will do the mass balance. To make things more simple, let’s say that we’ll do this mass balance for an hour of fermentation – then we can define the input and output as a constant amount. Also, we should stick to one unit to make it easier to double check our answers. Let’s choose kilograms since the acetic acid is given in kg.

Make a flow sheet

I don’t know why this is important (besides visualising the process) but you will have to do it in your exams, so here is what the flow sheet for this process will look like:

Process flow sheet for continuous acetic acid fermentation (made with draw.io)
Process flow sheet for continuous acetic acid fermentation

Calculations!

Now we know that there is 200gmol of inlet air. As mass = MW x moles, we can find out the mass of the inlet air by determining the mass of O2 and N2. Since we assume the composition of air is 21% O2 and 79% N2, we can find the molar mass:

  • Moles of O2: 21% of 200 = 0.21 x 200 = 42 moles
  • Moles of N2: 79% of 200 = 0.79 x 2oo = 158 moles

To find the mass in grams:

Mass = MW x moles

  • O2: 42 x (16 x 2) = 42 x 32 = 1344g = 1.344kg
  • N2: 98 x (14 x 2) = 158 x 28 = 4424g = 4.424kg

Now that we know the mass of O2 and N2, we can find the total mass of the inlet air:

1.344 + 4.424 = 5.768kg

So the total mass of the inlet air is 5.768kg!

Now, we want 2kg/hr of acetic acid, which is also 12% of the product stream. To find the total weight of the product stream is then just the mass of acetic acid divided by the percentage/100:

Mass of product stream = 2 / (12/100) = 2/ 0.12 = 16.67kg

So now we know the total mass of the product stream. Since we assumed that the rest of the media is negligible, the only other component of the product stream is water. So the mass of water would be:

Total mass – mass of acetic acid = 16.67kg – 2kg = 14.67kg

And voila! Now we know the total mass of 2 streams.

Stoichiometric mass balances

Now lets find the mass of everything in the equations!

We know that there is 2kg of acetic acid. With this and the chemical equation, we can find the masses of every other molecule in the equation.

C2H5OH (ethanol) + O2 -> CH3COOH (acetic acid) + H20

From the chemical equation we can see that for 1 mole of acetic acid produced, there is 1 mole of ethanol and oxygen used and 1 mole of water produced. Lets convert our 2kg of acetic acid in to moles:

2kg = 2000g
Moles = Mass/ MW = 2000g / (12+3+12+16+16+1)
= 2000/ 60 = 33.3 gmol

So for every mole of acetic acid produced, 33.3 moles of ethanol and oxygen are used, and 3.33 moles of water is formed. Let’s convert them into mass:

Mass = Moles x MW

  • Ethanol: 33.3 x 46 = 1582g = 1.582kg
  • Oxygen: 33.3 x 32 = 1065g = 1.065kg
  • Water: 33.3 x 18 = 599.4g = 0.599kg

We know how much water there is in the product stream. So, if 0.599kg of water is produced for 2kg of ethanol, then the water that was added in the feed stream must be the mass of water in the product stream – mass of water produced in the process

14.67kg – 0.599kg = 14.07kg water in feed stream

We also now know the amount of ethanol required to make 2kg of acetic acid. Thus to find the total amount of the feed stream we only need to add the mass of acetic acid and the mass of water in the feed stream together:

14.07 + 1.582 = 15.652kg (total mass of feed stream)

Now for the gas! We know that 1.065kg of oxygen is used to make 2kg of acetic acid. Therefore, to find the amount of oxygen leaving the system (that hasn’t been used), we only need to subtract the oxygen used from the amount of oxygen that entered the reactor in the inlet air:

1.344 – 1.065 = 0.279kg O2 in outlet air

Since nitrogen is not involved in the equation, the mass of nitrogen is the same going in and going out. Therefore the total mass of the outlet gas is:

4.424 + 0.279 = 4.703kg mass of outlet gas

And were done! Lets put it in a table to see if we have filled everything in:

ININOUTOUT
FEED STREAMINLET AIRPRODUCT STREAMOUTLET GAS
WATER14.07X14.67X
ETHANOL1.582X0X
ACETIC ACIDXX2X
OXYGENX1.344X0.279
NITROGEN X4.424X4.424
TOTAL15.6525.76816.674.703

Since the total mass balance for the process is mass in = mass out, we can check if we’ve done our mass balance properly:

Feed stream + inlet air = Product stream + outlet gas
= 15.652 + 5.768 = 16.67 + 4.703
21.4kg = 21.4kg

Everything adds up!

And thats how you do a mass balance! If you have any questions, please leave a comment and I will get back to you ASAP! Also, I know how difficult it is to understand this, so I will go through more examples in the future since thats the best way to learn mass balances in my opinion. If you have any of your own mass balance problems please let me know so I can go through them!

Fermentation Mass Balancing: Part 4 – Assumptions

In bioprocessing it’s usually difficult to calculate things exactly due to the complexity of biological processes. Therefore we tend to make a lot of assumptions.

In your exams or even at work you will need to put all the assumptions you make when doing a mass balance, so I thought I’d write a list of all the common ones that you can use.

Common Assumptions

  • No system leakage (best assumption!!)
  • Inlet air and outlet gas are dry (liquid does not leave system)
  • Cell broth has density of water: 1kg/m3
  • Complete conversion of input material (nutrients to product)
  • Composition of inlet air is 21% mol O2, 79% N2
  • Mass of cells inoculated at start of process is negligible (mainly for batch processes)
  • CO2 leaves in outlet gas
  • Other media components can be ignored
  • O2 and nutrients are available to all cells (vessel broth is homogenous)

There are probably more examples of assumptions you can make, but these are generally applicable to all processes. If you have any questions or want me to add anything, leave a comment and I will get to it ASAP!

Fermentation Mass Balancing: Part 3 – Stoichiometry (Part 1)

So we’ve gone through what mass balancing is and the mass balancing equations. Here is one of the harder parts: calculating the mass balance when there are reactions occurring. This is called stoichiometric mass balancing. This will be basic stoichiometric mass balancing (aka what I learnt in year 1). There is also elemental mass balancing, which is more cell-based and a bit more complex so I’ll go through that in another post.

Stoichiometry (Textbook Definition): The chemical and biochemical reactions involved in the rearrangement of atoms and molecules to form new groups.

Stoichiometry is basically calculating the amount of a reactant or product based on the related chemical equation.
To do this you must know the mass of at least one of the molecules, the moles of each molecule and the molecular weight of each molecule.

The best way to teach this is by example, so here is a chemical equation:

the chemical equation for alcoholic fermentation. 1 mole of glucose is broken down into 2 moles of ethanol and 2 moles of carbon dioxide.
Chemical equation for alcoholic fermentation.

This is the chemical equation for alcoholic fermentation. In this equation, 1 mole of glucose (left) is broken down into 2 moles of ethanol (right) and 2 moles of carbon dioxide (right). Now, imagine we have 300 grams of glucose. How much ethanol would we produce?

Here are the 3 main rules:

  • Matter cannot be created or destroyed. Therefore, the total mass on the left of the equation must be the same as the total mass on the right.
  • Elements cannot be created or destroyed in a chemical equation. Therefore, both sides of the equation will have the same amount of each element.
  • Stoichiometric balances work with moles, not mass – you must convert mass into moles when doing stoichiometric balances.

Firstly, since stoichiometry is done in moles, let’s convert our 300g of glucose into moles. From the basic calculations and equations post, we know that moles = mass / molecular weight. To find the molecular weight of glucose, we must add the molecular weight of all the elements in glucose together. If you google the periodic table (or just google the molecular weight of each element) you will find that:

  • Carbon (C): Molecular weight of 12
  • Hydrogen (H): Molecular weight of 1
  • Oxygen (O): Molecular weight of 16

However, since there is more than 1 of each element (as indicated by the subscripts in the equation) we must multiply the molecular weight of each element by how much there is in glucose.

(C x 6) + (H x 12) + (O x 6) = (12 x 6) + (1 x 12) + (16 x 6) = 180

Hence the total molecular weight of glucose is 180. To find the molar mass:

Moles = Mass / MW = 300/180 = 1.67 gmol

Now that we know the molar mass of glucose, we can determine the molar mass of ethanol. If you look at the equation, you can see that there is 2 moles of ethanol produced for every 1 mole of glucose (if there is no number in front of a molecule it means there is only 1 mole). Therefore, the molar mass of ethanol would be 2 times the molar mass of glucose. So:

Molar mass of ethanol = molar mass glucose x 2
= 1.67 x 2 = 3.33 gmol ethanol

From this, we can find the mass of ethanol using the MW of ethanol and the molar mass calculated above:

Mass = MW x moles
= 46 x 3.33 = 153g

Where 46 is the molecular weight of ethanol. And voila! There we have the amount of ethanol produced with 300 grams of glucose. If you were to find the mass of carbon dioxide produced (which is roughly 147, you can calculate it yourself if you want practice). you’ll find that:

300 grams glucose = 153 grams ethanol + 147 grams carbon dioxide

Both sides of the equation equal 300 grams! This is a good way of checking if you’ve done the mass balance right, since the mass must be the same on both sides of the equation.

Now, lets try a harder one…

Chemical equation for the conversion of glucose to L-glutamic acid.
Chemical equation for the conversion of glucose to L-glutamic acid.

This LOOKS complicated, but the same method applies. How would we work out how much oxygen is needed to produce 40g of glutamic acid (C5H9NO4)?

Firstly, lets find the molecular weight:

(C x 5) + (H x 9) + (N x 1) + (O x 4)
= (12 x 5) + (1 x 9) + (14 x 1) + (16 x 4) = 147

Therefore the molar mass of L-glutamic acid is:

Molar mass = Mass/ MW
= 40/147 = 0.272 gmol

From the chemical equation, we know that 1.5 moles of oxygen is needed to make 1 mole of L-glutamic acid. Therefore we need 1.5 x more of oxygen:

0.272 x 1.5 = 0.408 gmol

To find the mass of oxygen needed:

Mass = MW x moles
= 0.408 x (16 x 2) = 0.408 x 32 = 13.06 grams

Therefore to make 40g of L-glutamic acid, we need 13.06 grams of oxygen.

And there you have it! It sounds complicated, but once you understand it its actually pretty simple! However I don’t know if its just simple to me, so if you have any questions about anything please leave a comment below and I will get back to you!

Fermentation Mass Balancing: Part 2 – Rules

In my previous post I explained (briefly) what mass balancing was. Here I will go over the rules and equations used for mass balancing.

Mass Balancing: General equation

The most important equation that covers all aspects of mass balancing is this:

[Mass in + mass generated] – [Mass out + Mass consumed] = Mass accumulated

Pretty self-explanatory, right? Mass in is the amount of what you put into the system, mass generated is the mass of anything generated inside the system, mass out is the mass of what leaves the system, mass consumed is the mass of what is consumed in the reactor and mass accumulated is the mass of anything that is still inside the system.
This equation can apply to the whole unit operation (the mass of everything all together) or to individual components (water, nutrients, product).

However, depending on the type of process or material you have, you can simplify the equation.

Change in Process variables

A fermentation can be split into 2 states:

  • Steady state: In this state, process variables do not change with time. This means that there is also no change in mass for each component over time. This state generally applies to continuous fermentation, where input, output and the system are the same over time.
  • Unsteady state: Process variables change over time. This is the case for batch or semi-batch operations, where material is not added or removed from the reactor at a constant rate. Due to this, the mass of different components may differ at different stages of the fermentation.

Depending on the state of the process, the mass balance equation will change!

Mass Balance: Simplified equations

Steady state

Since mass does not change over time, there is no accumulation of mass within the reactor/ system. Therefore, the equation can be simplified to these terms:

Mass in + Mass generated = Mass out + Mass consumed

If there are no reactions occurring in a steady state process, meaning the molecules in that component do not interact with each other (i.e. cell mass), there is no mass generated or consumed. Then, the equation can further be simplified to:

Mass in = Mass out

Both these equations can be applied to either the process as a whole (if the above is applicable) or to individual components.

Unsteady state

In an unsteady state process, we should use the general equation. However, we tend to assume (at least at the bachelor degree level) there is no accumulation by assuming total conversion of input to output. Why? This is simpler, and in a batch or semi-batch process the unconverted nutrients tends to be a very small value. We also only do a mass balance of the whole process, not random time points, since this is more useful and much easier.

Therefore, if we assume no accumulation:

Mass in + Mass generated = Mass out + Mass consumed

No reaction

For some components of the process, there is no reaction occurring between them and other components, resulting in no accumulation, generation or consumption. This applies to elements (hydrogen, oxygen, nitrogen) since they can’t be broken down. It can also apply to other components but always double check that they aren’t reactive! In this case, the amount entering should be the same as the amount leaving, in which case:

Mass in = Mass out

Overall process

Mass cannot be created or destroyed, it can only be transferred. Because of this, for the whole process, regardless of whether its steady, unsteady, batch or continuous, the mass of everything entering the reactor will always be the same as the mass of everything leaving the reactor. This is because reactors are closed systems, and do not allow free exchange of mass in or out of the vessel. Therefore, the mass balance for the whole process will be:

MASS IN = MASS OUT



Those are all the rules! My advice would be to use the general equation for each component, and if there is no accumulation/ consumption / generation, just cross them out of the equation. Usually though, the only equations you need are:

MASS IN = MASS OUT

for components with no reaction, and

MASS IN + MASS GENERATED = MASS OUT + MASS CONSUMED


for components with reactions

If you want to know whether you can use the mass in = mass out equation, you can use this guide:

Does mass in = mass out?

Material

Without reaction With reaction
Total mass Yes   Yes
Total number of moles
Yes  
No
Mass of the total amount of molecules
Yes
No

Number of moles of a molecule
 Yes No
Mass of an element Yes Yes
Number of moles of an element Yes Yes

This is complicated and hard to explain with words, so I will do some working out of example questions in later posts to help you understand it better. However, if you have any questions about these rules, please leave a comment and I will get back to you ASAP!

Fermentation mass balancing: Part 1 – Intro

Mass balancing is a fundamental part of bioprocessing that is used to determine the amount of a component coming in or out of a system. A process stream is usually made up of multiple components. For example, feed entering the reactor will contain nutrients, water, growth hormones, etc. whilst the stream leaving the reactor will contain product, cells, waste (ammonia, carbon dioxide) and many other components!

We want to know how much of each component is entering, leaving or inside the reactor or system so that we can characterise the process and understand it better. However, with so many reactions occurring in biological processes, we won’t always know how much of each component there is off the top of our head! This is where mass balancing comes in.

Mass balancing is the calculation of what is going in or out of the system based on what information we have. For example, if we know x amount of cells and x amount of feed are going into the reactor, what is the amount of product and waste produced?
Remember: we need to know one side of the equation (in or out) to figure out the other!

In broader terms, mass balances refer to the balancing of mass going in and out of a system. A system can be anything that has a clearly defined boundary that separates it from the external environment. For example, a cell has a cell membrane, a water bottle has plastic walls, a bioreactor has metal walls. There are 3 main types of systems:

  • Open: Allows energy and matter exchange freely with external environment
  • Closed: Only allows energy exchange freely with external environment
  • Isolated: Does not allow energy or matter exchange with external environment

In most bioprocesses, the system (bioreactor) is closed – matter cannot be exchanged freely with the surroundings. When matter is exchanged in the reactor, it is because we intentionally add or remove components.

a simple diagram of a system. things go in, things come out.
A simple diagram of a system.

Remember: matter cannot disappear! It can only be transferred. Therefore when we add a component to a system we expect to either stay in the system, come out or be transformed into another component.

Now that we know what mass balances are, lets move on to how to do them! If this post wasn’t clear, please leave a comment and I will try to explain it in simpler terms.

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