Tag: concentration

These are the fundamental equations you need to know, not ALL the equations. Nowhere close (good luck). But as these equations commonly show up, I would recommend reading through them and understanding them

Dimensionless variables

In my previous post we talked about physical variables, but only went through one type: substantial variables. Here we’ll talk about the 2nd type: natural variables (also known as dimensionless variables). Natural variables are variables that are dimensionless, hence have no units. They can occur when the substantial variables used to derive the natural variable cancel each other out, or do not require any units to express magnitude (i.e ratios).

In bioprocessing, dimensionless variables are often used to represent a physical phenomenon that is the result of multiple substantial variables. Since bioprocessing involves many complex processes (reactor configuration, operating conditions, liquid properties), it is usually simpler to represent them as a dimensionless group that is the sum of all these processes. This is called dimensional analysis. We will go through this later on, but right now you just need to know of their existence.

Reynolds Number (Re)

You probably already know of this; if not, you will do very soon! It is likely the most common dimensionless group in the field! Reynolds number is simply the ratio of inertial forces (creating motion) to viscous (reducing motion) forces, and is used to describe the turbulence or “chaos” in the fluid.

Re = inertial forces / viscous forces

In terms of physical variables, this is the equation for Reynolds number in pipes:

Re = ρud/μ

Where:

• ρ = Liquid density (kg/m³)
• u = Liquid velocity (m/s)
• d = Diameter of Pipe (m)
• μ = Liquid viscosity (Pa.s)

And this is the equation for the Reynolds number in tanks:

Re = ρNdi²/μ

Where:

• ρ = Liquid density (kg/m³)
• N = Rotational speed of impeller (in rotations per second, RPS or s-¹)
• di = Diameter of impeller (m)
• μ = Liquid viscosity (Pa.s)

If we are to write down all the SI base units:

[kg/m³] x [m/s] x [m] / m-¹ x kg x s-¹

[kg/m³] x [m/s] x [m] / m-¹ x kg x s-¹

= kg x s-¹ / kg x s-¹ = kg x s-¹ / kg x s-¹  = 0

Proving that the Reynolds number is a dimensionless variable

The Reynolds number is an important dimensionless group as it tells us the turbulence of the fluid, which is linked to how well mixed the fluid is. The Reynolds number tells you the type of flow of the liquid:

• Laminar flow: Occurs when Re<10-⁴. In this type of flow, liquid moves in regular patterns and will not mix.
• Transitional flow: Occurs when Re is between 2100 and 10000. 
• Turbulent flow: Occurs when Re>10⁴. In this type of flow, the fluid is moving fast and in a chaotic manner. Linked to well mixed.

Generally, we want our fluid to possess turbulent flow as it is linked with good mixing (but not always!)

Power number (Po)

This is another common dimensionless variable that you will come across. Power number is different for each impeller design and is related to power consumption of an impeller. We will go through this during Fluid mechanics and Reactor design.

Po = Pug / ρ x N³ x di⁵

Where:

• Pug = Ungassed power consumption
• ρ = Liquid density (kg/m³)
• N = Rotational speed of impeller (in rotations per second, RPS or s-¹)
• di = Impeller diameter (m)

Basic equations you should know

You will have learnt this in high school if you did chemistry, but if you didn’t (like me), here are some of the  basic equations you must know.

Molar Mass/Molecular Weight (g/mol)

The molar mass is the mass, in grams of 1 mole of a molecule. It’s basically means the same thing as moles. To find the molar mass of a molecule, you just add up the individual molar masses of each element in the molecule.

For example:

C₆H₁₂O₆ or Glucose

• C, Carbon, has a molar mass of 12g/mol
• H, Hydrogen, has a molar mass of 1g/mol
• O, Oxygen, has a molar mass of 16g/mol

So to find the total molar mass: 

(12 x 6) + (1 x 12) + (16 x 6) = 180g/mol

Also, remember that a mole is equal to 6.02 x 10²³ of that molecule. I have only used this equation in my chemistry modules but I feel like it’s the basic foundation of chemistry/biochem and you should probably know that if you pursue a career in the field.

Mass (g) = molecular weight (g/mol) x moles (gmol)

You’ll probably know this but just in case you don’t, remember this! It pops up a lot and is one of the fundamental equations. I use the equation triangle to remember it, which I’ve put below:

If you need to work out one of the variables, cover it with your thumb, and the position of the two remaining variables lets you know how to calculate it i.e. if you want mass, you need to multiply molar mass by moles, but if you want moles you must divide mass by molar mass. 

Molar mass is the number of neutrons and protons in an element. It’s given in the periodic table (it’s also called molecular weight), but you should know the 4 important ones:

• Carbon: 12
• Hydrogen: 1
• Oxygen: 16
• Nitrogen: 14

Concentrations

There are 4 main equations for calculating concentration:

1. Amount per volume: N/L³ (units mol/L or M)
   – Concentration (mol/L) = amount /volume or moles/ volume
   – Used for salts, buffer concentrations, etc.
   
2. Mass per volume: M/L³  (units g/L)
   – Concentration (g/L) = mass /volume
   – Used for molecules dissolved in water i.e DNA/ RNA concentrations, protein concentrations, etc.
   
3. Weight per volume (Written as a percentage): M/L³ (Units: %)
   – Concentration (%) = mass /volume x 100
   – Same equation as 2. but is written as a percentage i.e. 30% glucose dissolved in water
   
4. Volume per volume (Percentage): L³/L³ (Units: %)
   – Concentration (%) = volume 1/volume 2 x 100
   – Used for liquids dissolved in other liquids (e.g. detergents,)

Stock Solutions

Stock solutions are used to dilute a fluid so that you can manipulate the concentration until it is what you need. This is commonly used to make buffers.

Concentration of stock x volume of stock = concentration of final solution x volume of final solution

•  Cstock x Vstock = Cfinal x Vfinal

If you know 3 of the variables, you can use this equation to find the remaining value. For example if you want to know how much volume of stock you need to get a solution of desired concentration and volume, you can find it by dividing the right side of the equation by Cstock:

Vstock = (Cfinal x Vfinal) / Cstock

Dilution factors of stock solution

if you want to dilute a solution by a certain factor, you can use this equation:

Vstock = Vfinal/Dstock

It’s similar to the last equation, right? Dfinal is always 1, therefore it is cancelled out. To remember how to calculate each variable, you can use the triangle method that I did for molar masses.

Based on the equations, here are some questions you can try out:

1. I have 2 moles of glucose. How much do I have in grams?
2. What is the molar mass of C₂OH₄?
3. I have 2 litres of buffer containing 0.4 mol/L of C₂OH₄. How much do I have in grams? (this question uses 2 equations)
4. I have 4 nanograms of DNA in a 1 millilitre solution. What is the concentration?
5. If I want a 40% glucose solution, and I have 1 litre of water, how many grams of glucose do I need to add?
6. I have a stock solution of 5 litres containing 2g/L of glucose. If my final volume is 3L, what is the final concentration?

If I haven’t explained this in enough detail, please leave a comment and I will write another post to answer your question and go through the topic more.