Documentation:CHBE Exam Wiki/2.3 - Solving Reactive Balances

2.3 – Solving Reactive Balances In-depth

2.3.0 – Learning Objectives

By the end of this section you should be able to:

Solve simple reactive balances using the molecular and atomic reactive balances.

2.3.1 – Introduction

We will solve the example in the diagram 4.71 using the molecular species and atomic species balance. Python will be implemented to perform calculations. This section goes in-depth of the logic used to solve the problem that is addressed before.

2.3.2 – Problem statement

Recall the block diagram in figure 4.7-1

$C_{2}H_{6}\longrightarrow C_{2}H_{4}+H_{2}$


What are the mol flows of the exit streams of ethane $C_{6}H_{6}$ and ethylene $C_{4}H_{4}$?

2.3.3 – Solving using the molecular species balance

First, we identify our molecular species; for us, this is Hydrogen, Ethane, and Ethylene.

Next, we write out their respective balance equations. Note how the diagram 4.7-1 only shows the output variables (${\dot {n_{1}}},{\dot {n_{2}}}$)

$C_{2}H_{6}balance:100{\frac {kmol}{min}}_{input}={\dot {n_{1}}}{\frac {kmol}{min}}_{Output}+C_{2}H_{6Consumed}$

$C_{2}H_{4}balance:C_{2}H_{4generated}={\dot {n_{2}}}{\frac {kmol}{min}}_{Output}$

$H_{2}balance:40{\frac {kmol}{min}}_{generated}=40{\frac {kmol}{min}}_{Output}$

To further relate these equations, use the stoichiometrey of the dehydrogenation reaction which gives a relationship between the consumed and generated terms:

$C_{2}H_{6}\rightarrow C_{2}H_{4}+H_{2}$

Since there is a 1:1 ratio between the Generation of Hydrogen to $C_{2}H_{4}$:

$40{\frac {kmol}{min}}H_{2generated}*{\frac {1{\frac {kmol}{min}}C_{2}H_{4}}{1{\frac {kmol}{min}}H_{2}}}=40{\frac {kmol}{min}}C_{2}H_{4generated}$

Since there is a 1:1 ratio between the generation of hydrogen and the consumption of Ethane, $C_{2}H_{6consumed}=40{\frac {kmol}{min}}$.

Substituting the $C_{2}H_{6Consumed}$ and the $C_{2}H_{4generated}$ into their respective equations, we can see that the outputs of ethane and ethylene $({\dot {n_{1}}},{\dot {n_{2}}})$ are $60{\frac {kmol}{min}}$ and $40{\frac {kmol}{min}}$ respectively.

$C_{2}H_{6}balance:100{\frac {kmol}{min}}_{input}={\dot {n_{1}}}{\frac {kmol}{min}}_{Output}+40{\frac {kmol}{min}}C_{2}H_{6Output}$

${\dot {n_{1}}}=60{\frac {kmol}{min}}_{Output}$

$C_{2}H_{4}balance:C_{2}H_{4generated}={\dot {n_{2}}}{\frac {kmol}{min}}_{Output}=40{\frac {kmol}{min}}C_{2}H_{4generated}$

2.3.4 – Solving using the Atomic species balance

Let's recall the Atomic balances:

$Carbon(C){\text{ }}balance:Input=Output$

$Hydrogen(H){\text{ }}balance:Input=Output$

First the Atomic balances are broken down to collect any molecular species containing the atomic species (Carbon and Hydrogen).

The carbon balance becomes: $C_{2}H_{6input}=C_{2}H_{6output}+C_{2}H_{4output}$

The hydrogen balance becomes:

$C_{2}H_{6input}=C_{2}H_{6output}+C_{2}H_{4output}+H_{2output}$

We then isolate the atomic species from the molecular species. Another way of thinking is: How many of N atoms are there in this molecule? For example, there are 2 carbon atoms for every 1 ethane molecule.

The carbon balance becomes:

$C_{2}H_{6input}*{\frac {2C{\frac {kmol}{min}}}{1C_{2}H_{6}{\frac {kmol}{min}}}}=C_{2}H_{6output}*{\frac {2C{\frac {kmol}{min}}}{1C_{2}H_{6}{\frac {kmol}{min}}}}+C_{2}H_{4output}*{\frac {2C{\frac {kmol}{min}}}{1C_{2}H_{6}{\frac {kmol}{min}}}}$

The hydrogen balance becomes:

$C_{2}H_{6input}*{\frac {6H{\frac {kmol}{min}}}{1C_{2}H_{6}{\frac {kmol}{min}}}}=C_{2}H_{6output}*{\frac {6H{\frac {kmol}{min}}}{1C_{2}H_{6}{\frac {kmol}{min}}}}+C_{2}H_{4output}*{\frac {4H{\frac {kmol}{min}}}{1C_{2}H_{6}{\frac {kmol}{min}}}}+H_{2output}*{\frac {2H{\frac {kmol}{min}}}{1H_{2}{\frac {kmol}{min}}}}$

Note from diagram 4.7-1, the input and output values are substituted for their respective integers and variables. Remember that ethane and ethene outputs are defined as (${\dot {n_{1}}},{\dot {n_{2}}}$)

rewriting the balances

The carbon balance becomes: $200{\frac {kmol}{min}}={\dot {2n_{1}}}+{\dot {2n2}}$

The hydrogen balance becomes:

$600{\frac {kmol}{min}}=6{\dot {n_{1}}}+4{\dot {n_{2}}}+80{\frac {kmol}{min}}$