# Introduction to Chemical Engineering Processes/The most important point

## Component Mass Balance

Most processes, of course, involve more than one input and/or output, and therefore it must be learned how to perform mass balances on . The basic idea remains the same though. We can write a mass balance in the same form as the overall balance for each component:

${\displaystyle In-Out+Generation=Accumulation}$

For steady state processes, this becomes:

${\displaystyle In-Out+generation=0}$

The overall mass balance at steady state, recall, is:

${\displaystyle \Sigma {\dot {m}}_{in}-\Sigma {\dot {m}}_{out}+m_{gen}=0}$

The mass of each component can be described by a similar balance.

${\displaystyle \Sigma {\dot {m}}_{A,in}-\Sigma {\dot {m}}_{A,out}+{m}_{A,gen}=0}$

The biggest difference between these two equations is that The total generation of mass ${\displaystyle m_{gen}}$ is zero due to conservation of mass, but since individual species can be consumed in a reaction, ${\displaystyle {m}_{A,gen}\neq 0}$ for a reacting system

## Concentration Measurements

You may recall from general chemistry that a concentration is a measure of the amount of some species in a mixture relative to the total amount of material, or relative to the amount of another species. Several different measurements of concentration come up over and over, so they were given special names.

### Molarity

The first major concentration unit is the molarity which relates the moles of one particular species to the total volume of the solution.

${\displaystyle Molarity(A)=[A]={\frac {n_{A}}{V_{sln}}}}$ where ${\displaystyle n{\dot {=}}mol,V{\dot {=}}L}$

A more useful definition for flow systems that is equally valid is:

 ${\displaystyle [A]={\frac {{\dot {n}}_{A}}{{\dot {V}}_{n}}}}$ where ${\displaystyle {\dot {n}}_{A}{\dot {=}}mol/s,{\dot {V}}_{n}{\dot {=}}L/s}$

Molarity is a useful measure of concentration because it takes into account the volumetric changes that can occur when one creates a mixture from pure substances. Thus it is a very practical unit of concentration. However, since it involves volume, it can change with temperature so molarity should always be given at a specific temperature. Molarity of a gaseous mixture can also change with pressure, so it is not usually used for gasses.

### Mole Fraction

The mole fraction is one of the most useful units of concentration, since it allows one to directly determine the molar flow rate of any component from the total flow rate. It also conveniently is always between 0 and 1, which is a good check on your work as well as an additional equation that you can always use to help you solve problems.

The mole fraction of a component A in a mixture is defined as:

${\displaystyle x_{A}={\frac {n_{A}}{n_{n}}}}$

where ${\displaystyle n_{A}}$ signifies moles of A. Like molarity, a definition in terms of flowrates is also possible:

 Mole Fraction Definition ${\displaystyle x_{A}={\frac {{\dot {n}}_{A}}{{\dot {n}}_{n}}}}$

If you add up all mole fractions in a mixture, you should always obtain 1 (within calculation and measurement error), because sum of individual component flow rates equals the total flow rate:

 ${\displaystyle \Sigma x_{i}=1}$

Note that each stream has its own independent set of concentrations. This fact will become important when you are performing mass balances.

### Mass Fraction

Since mass is a more practical property to measure than moles, flowrates are often given as mass flowrates rather than molar flowrates. When this occurs, it is convenient to express concentrations in terms of mass fractions defined similarly to mole fractions.

In most texts mass fraction is given the same notation as mole fraction, and which one is meant is explicitly stated in the equations that are used or the data given.

 Note: In this book, assume that a percent concentration has the same units as the total flowrate unless stated otherwise. So if a flowrate is given in kg/s, and a composition is given as "30%", assume that it is 30% by mass.

The definition of a mass fraction is similar to that of moles:

${\displaystyle x_{A}={\frac {m_{A}}{m_{n}}}}$ for batch systems

 Mass fraction of Continuous Systems ${\displaystyle x_{A}={\frac {{\dot {m}}_{A}}{{\dot {m}}_{n}}}}$

where ${\displaystyle m_{A}}$ is the mass of A. It doesn't matter what the units of the mass are as long as they are the same as the units of the total mass of solution.

Like the mole fraction, the total mass fraction in any stream should always add up to 1.

 ${\displaystyle \Sigma x_{i}=1}$

## Calculations on Multi-component streams

Various conversions must be done with multiple-component streams just as they must for single-component streams. This section shows some methods to combine the properties of single-component streams into something usable for multiple-component streams(with some assumptions).

### Average Molecular Weight

The average molecular weight of a mixture (gas or liquid) is the multicomponent equivalent to the molecular weight of a pure species. It allows you to convert between the mass of a mixture and the number of moles, which is important for reacting systems especially because balances must usually be done in moles, but measurements are generally in grams.

To find the value of ${\displaystyle {\bar {MW}}_{n}={\frac {g{\mbox{ sln}}}{mole{\mbox{ sln}}}}}$, we split the solution up into its components as follows, for k components:

${\displaystyle {\frac {g{\mbox{ sln}}}{mole{\mbox{ sln}}}}={\frac {\Sigma {m_{i}}}{n_{n}}}=\Sigma {\frac {m_{i}}{n_{n}}}}$

${\displaystyle =\Sigma ({\frac {m_{i}}{n_{i}}}*{\frac {n_{i}}{n_{n}}})=\Sigma (MW_{i}*x_{i})}$

where ${\displaystyle x_{i}}$ is the mole fraction of component i in the mixture. Therefore, we have the following formula:

 ${\displaystyle {\bar {MW}}_{n}=\Sigma (MW_{i}*x_{i})_{n}}$ where ${\displaystyle x_{i}}$ is the mole fraction of component i in the mixture.

This derivation only assumes that mass is additive, which it is, so this equation is valid for any mixture.

### Density of Liquid Mixtures

Let us attempt to calculate the density of a liquid mixture from the density of its components, similar to how we calculated the average molecular weight. This time, however, we will notice one critical difference in the assumptions we have to make. We'll also notice that there are two different equations we could come up with, depending on the assumptions we make.

#### First Equation

By definition, the density of a single component i is: ${\displaystyle {\rho }_{i}={\frac {m_{i}}{V_{i}}}}$ The corresponding definition for a solution is ${\displaystyle \rho ={\frac {m{\mbox{ sln}}}{V{\mbox{ sln}}}}}$. Following a similar derivation to the above for average molecular weight:

${\displaystyle {\frac {m{\mbox{ sln}}}{V{\mbox{ sln}}}}={\frac {\Sigma {m_{i}}}{V_{n}}}=\Sigma {\frac {m_{i}}{V_{n}}}}$

${\displaystyle =\Sigma {\frac {m_{i}}{V_{i}}}*{\frac {V_{i}}{V_{n}}}=\Sigma ({\rho }_{i}*{\frac {V_{i}}{V_{n}}})}$

Now we make the assumption that The volume of the solution is proportional to the mass. This is true for any pure substance (the proportionality constant is the density), but it is further assumed that the proportionality constant is the same for both pure k and the solution. This equation is therefore useful for two substances with similar pure densities. If this is true then:

${\displaystyle {\frac {V_{i}}{V}}={\frac {m_{i}}{m_{n}}}=x_{i}}$, where ${\displaystyle x_{i}}$ is the mass fraction of component i. Thus:

 ${\displaystyle {\rho }_{n}=\Sigma {(x_{i}*{\rho }_{i})}_{n}}$ where ${\displaystyle x_{i}}$ is the mass fraction (not the mole fraction) of component i in the mixture.

#### Second Equation

This equation is easier to derive if we assume the equation will have a form similar to that of average molar mass. Since density is given in terms of mass, it makes sense to start by using the definition of mass fractions:

${\displaystyle x_{i}={\frac {m_{i}}{m_{n}}}}$

To get this in terms of only solution properties (and not component properties), we need to get rid of ${\displaystyle m_{i}}$. We do this first by dividing by the density:

${\displaystyle {\frac {x_{i}}{{\rho }_{i}}}={\frac {m_{i}}{m_{n}}}*{\frac {V_{i}}{m_{i}}}}$

${\displaystyle ={\frac {V_{i}}{m_{n}}}}$

Now if we add all of these up we obtain:

${\displaystyle \sum \left({\frac {x_{i}}{{\rho }_{i}}}\right)={\frac {\Sigma {V_{i}}}{m_{n}}}}$

Now we have to make an assumption, and it's different from that in the first case. This time we assume that the Volumes are additive. This is true in two cases:

1. In an ideal solution. The idea of an ideal solution will be explained more later, but for now you need to know that ideal solutions:

• Tend to involve similar compounds in solution with each other, or when one component is so dilute that it doesn't effect the solution properties much.
• Include Ideal Gas mixtures at constant temperature and pressure.

2 In a Completely immiscible nonreacting mixture. In other words, if two substances don't mix at all (like oil and water, or if you throw a rock into a puddle), the total volume will not change when you mix them. And the total volume in this case will be sum of volume of individual components.

If the solution is ideal, then we can write:

${\displaystyle {\frac {\Sigma {\dot {V}}_{i}}{{\dot {m}}_{n}}}={\frac {{\dot {V}}_{n}}{{\dot {m}}_{n}}}={\frac {1}{{\rho }_{n}}}}$

Hence, for an ideal solution,

 ${\displaystyle {\frac {1}{{\rho }_{n}}}=\sum \left({\frac {x_{i}}{{\rho }_{i}}}\right)_{n}}$ where ${\displaystyle x_{i}}$ is the mass fraction of component i in the mixture.

Note that this is significantly different from the previous equation! This equation is more accurate for most cases. In all cases, however, it is most accurate to look up the value in a handbook such as Perry's Chemical Engineers Handbook if data is available on the solution of interest.