Introduction to Chemical Engineering Processes/Atom balances

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The concept of atom balances[edit]

Let's begin this section by looking at the reaction of hydrogen with oxygen to form water:

 H_2 + O_2 \rightarrow H_2O

We may attempt to do our calculations with this reaction, but there is something seriously wrong with this equation! It is not balanced; as written, it implies that an atom of oxygen is somehow "lost" in the reaction, but this is in general impossible. Therefore, we must compensate by writing:

 H_2 + \frac{1}{2}O_2 \rightarrow H_2O

or some multiple thereof.

Notice that in doing this we have made use of the following conservation law, which is actually the basis of the conservation of mass:


The number of atoms of any given element does not change in any reaction (assuming that it is not a nuclear reaction).

Since by definition the number of moles of an element is proportional to the number of atoms, this implies that  \dot{n}_{A,gen} = 0 where A represents any element in atomic form.

Mathematical formulation of the atom balance[edit]

Now recall the general balance equation:

 In - Out + Generation = Accumulation

In this course we're assuming  Accumulation = 0 . Since the moles of atoms of any element are conserved,  generation = 0 . So we have the following balance on a given element A:


For a given element A,

 \Sigma \dot{n}_{A,in} - \Sigma \dot{n}_{A,out} = 0

Note:
When analyzing a reacting system you must choose either an atom balance or a molecular species balance but not both. Each has advantages; an atom balance often yields simpler algebra (especially for multiple reactions; the actual reaction that takes place is irrelevant!) but also will not directly tell you the extent(s) of reaction, and will not tell you if the system specifications are actually impossible to achieve for a given set of equilibrium reactions.

Degree of Freedom Analysis for the atom balance[edit]

As before, to do a degree of freedom analysis, it is necessary to count the number of unknowns and the number of equations one can write, and then subtract them. However, there are a couple of important things to be aware of with these balances.

  • When doing atom balances, the extent of reaction does not count as an unknown, while with a molecular species balance it does. This is the primary advantage of this method: the extent of reaction does not matter since atoms of elements are conserved regardless of how far the reaction has proceeded.
  • You need to make sure each atom balance will be independent. This is difficult to tell unless you write out the equations and look to see if any two are identical.
  • In reactions with inert species, each molecular balance on the inert species counts as an additional equation. This is because of the following important note:

Note:
When you're doing an atom balance you should only include reactive species, not inerts.

Example
Example:

Suppose a mixture of nitrous oxide ( N_2O ) and oxygen is used in a natural gas burner. The reaction  CH_4 + 2O_2 \rightarrow 2H_2O + CO_2 occurs in it.

There would be four equations that you could write: 3 atom balances (C, H, and O) and a molecular balance on nitrous oxide. You would not include the moles of nitrous oxide in the atom balance on oxygen.

Example of the use of the atom balance[edit]

Let's re-examine a problem from the previous section. In that section it was solved using a molecular species balance, while here it will be solved using atom balances.

Example
Example:

Consider the reaction of Phosphene with oxygen:  4PH_3+8O_2 \rightarrow P_4O_{10} + 6H_2O

Suppose a 100-kg mixture of 50%  PH_3 and 50%  O_2 by mass enters a reactor in a single stream, and the single exit stream contains 25%  O_2 by mass. Assume that all the reduction in oxygen occurs due to the reaction. How many degrees of freedom does this problem have? If possible, determine mass composition of all the products.

For purposes of examination, the flowchart is re-displayed here:

Phosphate Rxn Flowchart.PNG


Degree of Freedom Analysis[edit]

There are three elements involved in the system (P, H, and O) so we can write three atom balances on the system.

There are likewise three unknowns (since the extent of reaction is NOT an unknown when using the atom balance): the outlet concentrations of  PH_3, P_4O_{10}, H_2O

Therefore, there are 3 - 3 = 0 unknowns.

Problem Solution[edit]

Let's start the same as we did in the previous section: by finding converting the given information into moles. The calculations of the previous section are repeated here:

  •  \dot{m}_{out} = \dot{m}_{in} = 100 \mbox{ kg}
  •  \dot{n}_{PH_3,in}= 0.5*(100\mbox{ kg})* \frac{1 \mbox{ mol}}{0.034 \mbox{ kg}} = 1470.6 \mbox{ moles PH}_3 \mbox{ in}
  •  \dot{n}_{O_2,in} = 0.5*(100\mbox{ kg})* \frac{1 \mbox{ mol}}{0.032 \mbox{ kg}} = 1562.5 \mbox{ moles O}_2 \mbox{ in}
  •  \dot{n}_{O_2,out} = 0.25*(100\mbox{ kg})* \frac{1 \mbox{ mol}}{0.032 \mbox{ kg}} = 781.25 \mbox{ moles O}_2 \mbox{ out}

Now we start to diverge from the path of molecular balances and instead write atom balances on each of the elements in the reaction. Let's start with Phosphorus. How many moles of Phosphorus atoms are entering?

  • Inlet: Only  PH_3 provides P, so the inlet moles of P are just  1*1470.6 = 1470.6 \mbox{ moles P in}
  • Outlet: There are two ways phosphorus leaves: as unused  PH_3 or as the product  P_4O_{10} . Therefore, the moles of  PH_3 out are  1*n_{PH_3,out} + 4*n_{P_4O_{10},out} . Note that the 4 in this equation comes from the fact that there are 4 Phosphorus atoms in every mole of  P_4O_{10} .

Therefore the atom balance on Phosphorus becomes:


Phosphorus

 1*n_{PH_3,out} + 4*n_{P_4O_{10},out} = 1470.6

Similarly, on Oxygen we have:

  • Inlet:  2*n_{O_2,in} = 2*1562.5 = 3125\mbox{ moles O}_2
  • Outlet:  2*n_{O_2,out} + 10*n_{P_4O_{10},out} + 1*n_{H_2O,out} = 1562.5 + 10*n_{P_4O_{10},out} + 1*n_{H_2O,out}


Oxygen

 1562.5 + 10*n_{P_4O_{10},out} + 1*n_{H_2O,out}= 3125

Finally, check to see if you can get the following Hydrogen balance as a practice problem:


Hydrogen

 2*n_{H_2O,out} + 3*n_{PH_3,out} = 4411.8

Solving these three linear equations, the solutions are:


 n_{PH_3,out}=1080 , n_{O_2,out}=586, n_{P_4O_{10},out} = 97.66 moles

All of these answers are identical to those obtained using extents of reaction. Since the remainder of the solution to that problem is identical to that in the previous section, the reader is referred there for its completion.

Example of balances with inert species[edit]

Sometimes it's more difficult to choose which type of balance you want, because both are possible but one is significantly easier than the other. As an example, lets consider a basic pollution control system.

Example
Example:

Suppose that you are running a power plant and your burner releases a lot of pollutants into the air. The flue gas has been analyzed to contain 5%  SO_2 , 3%  NO_2 , 7%  O_2 and 15%  CO_2 by moles. The remainder was determined to be inert.

Local regulations require that the emissions of sulfur dioxide be less than 200 ppm (by moles) from your plant. They also require you to reduce nitrogen dioxide emissions to less than 50 ppm. You decide that the most economical method for control of these for your plant is to utilize ammonia-based processes. The proposed system is as follows:

  1. Put the flue gas through a denitrification system, into which (pure) ammonia is pumped. The amount of ammonia pumped in is three times as much as would theoretically be needed to use all of the nitrogen dioxide in the flue gas.
  2. Allow it to react a specified amount of time.
  3. Pump it into a desulfurization system. Nothing new is injected here, it just has a different catalyst than the denitrification, and the substrates are at a different temperature and pressure.

The reactions that occur are:

  1.  2NO_2 + 4NH_3 + O_2 \rightarrow 3 N_2 + 6H_2O
  2.  H_2O + 2NH_3 + SO_2 \rightarrow (NH_4)_2SO_3

If your plant makes  130\frac{ft^3}{s} of flue gas at  T = 900K and  P = 2\mbox{ atm} , how much ammonia do you need to purchase for each 8-hour shift? How much of it remains unused? Why do we want to have a significant amount of excess ammonia?

Assume that the flue gas is an ideal gas. Recall the ideal gas law,  PV = nRT , where  R = 0.0821 \frac{L*atm}{mol*K} .

Step 1: Flowchart[edit]

Flowcharts are becoming especially important now as means of organizing all of that information!

Air Pollution Example.PNG

Step 2: Degrees of Freedom[edit]

Let's consider an atomic balance on each reactor.

  • Denitrification system: 9 unknowns (all concentrations in stream 3, and  \dot{n}_2 .) - 3 atom balances (N, H, and O) - 3 inert species ( CO_2, SO_2, inerts ) - 1 additional info (3X stoichiometric feed) = 2 DOF
  • Desulfurization system: 15 unknowns - 4 atom balances (N, H, O, and S) - 5 inerts ( CO_2, O_2, NO_2, N_2, inerts ) = 6 DOF
  • Total = 2 + 6 - 8 shared = 0 DOF, hence the problem has a unique solution.

We can also perform the same type of analysis on molecular balances.

  • Denitrification system: 10 unknowns (now the conversion  X_1 is also unknown) - 8 molecular species balances - 1 additional info = 1 DOF
  • Desulfurization system: 16 unknowns (now the conversion  X_2 is unknown) - 9 balances = 7 DOF.
  • Total = 1 + 7 - 8 shared = 0 DOF.

Therefore the problem is theoretically solvable by both methods.

Step 3: Units[edit]

The only weird units in this problem (everything is given in moles already so no need to convert) are in the volumetric flowrate, which is given in  \frac{ft^3}{s} . Lets convert this to  \frac{moles}{s} using the ideal gas law. To use the law with the given value of R is is necessary to change the flowrate to units of  \frac{L}{s} :

 130 \frac{ft^3}{s} * \frac{28.317\mbox{ L}}{ft^3} = 3681.2\frac{L}{s}  P\dot{V} = \dot{n}RT \rightarrow 2*3681.2 = \dot{n}_1(0.0821)(900)

 \dot{n}_1 = 99.64\frac{moles}{s}

Now that everything is in good units we can move on to the next step.

Step 4: Devise a plan[edit]

We can first determine the value of  \dot{n}_2 using the additional information. Then, we should look to an overall system balance.

Since none of the individual reactors is completely solvable by itself, it is necessary to look to combinations of processes to solve the problem. The best way to do an overall system balance with multiple reactions is to treat the entire system as if it was a single reactor in which multiple reactions were occurring. In this case, the flowchart will be revised to look like this:

Air Pollution Example Overall System.PNG

Before we try solving anything, we should check to make sure that we still have no degrees of freedom.

Atom Balance

There are 8 unknowns (don't count conversions when doing atom balances), 4 types of atoms (H, N, O, and S), 2 species that never react, and 1 additional piece of information (3X stoichiometric), so there is 1 DOF. This is obviously a problem, which occurs because when performing atom balances you cannot distinguish between species that react in only ONE reaction and those that take part in more than one.

In this case, then, it is necessary to look to molecular-species balances.

Molecular-species balance

In this case, there are 10 unknowns, but we can do molecular species balances on 9 species  (SO_2, NO_2, NH_3, N_2, O_2, CO_2, H_2O, (NH_4)_2SO_3, inerts) and have the additional information, so there are 0 DOF when using this method.

Once we have all this information, getting the information about stream 3 is trivial from the definition of extent of reaction.

Step 5: Carry Out the Plan[edit]

First off we can determine  \dot{n}_2 by using the definition of a stoichiometric feed.

 \dot{n}_{NO_2,in} = 0.03*99.64 = 2.9892 \frac{mol}{s}

The stoichiometric amount of ammonia needed to react with this is, from the reaction,

 \frac{4 \mbox{ moles NH}_3}{2 \mbox{ moles NO}_2} * 2.9892 = 5.96 \frac{ moles \mbox{ NH}_3}{s}

Since the problem states that three times this amount is injected into the denitrification system, we have:


 \dot{n}_2 = 17.88 \frac{moles}{s}

Now, we are going to have a very complex system of equations with the 9 molecular balances. This may be a good time to invest in some equation-solving software.

See if you can derive the following system of equations from the overall-system flowchart above.


 NH_3: \dot{n}_4*x_{NH_3,4} = 17.88 - 4*X_1 - 2*X_2
 SO_2: \dot{n}_4*2*10^{-4} = 0.05 * 99.64 - X_2
 NO_2: \dot{n}_4*5*10^{-5} = 0.03*99.64 - 2*X_1
 N_2: \dot{n}_4*x_{N_2,4} = 3*X_1
 O_2: \dot{n}_4*x_{O_2,4} = 0.07*99.64 - X_1
 H_2O: \dot{n}_4*x_{H_2O,4} = 6*X_1 - X_2
 CO_2: \dot{n}_4*x_{CO_2,4} = 0.15*99.64
 (NH_4)_2(SO_3): \dot{n}_4*x_{(NH_4)_2SO_3,4} = X_2
 Inerts: \dot{n}_4*(1 - 2*10^{-4} - 5*10^{-5} - x_{NH_3,4} - x_{N_2,4} - x_{O_2,4} - x_{H_2O,4} -  x_{CO_2,4} - x_{(NH_4)_2SO_3,4} )  = 0.7*99.64

Using an equation-solving package, the following results were obtained:


 X_1 = 1.492\mbox{ moles}
 X_2 = 4.961\mbox{ moles}
 \dot{n}_3 = 105.62 \frac{mol}{s}
 x_{NH_3, 4} = 0.01884
 x_{N_2,4} = 0.04238
 x_{O_2,4} = 0.05191
 x_{H_2O,4} = 0.03778
 x_{CO_2,4} = 0.1415
 x_{(NH_4)_2SO_3,4} = 0.04697
 x_{I} = 1 - \Sigma \mbox{ (other components) } = 0.6606

Stream 3[edit]

Now that we have completely specified the composition of stream 4, it is possible to go back and find the compositions of stream 3 using the extents of reaction and feed composition. Although this is not necessary to answer the problem statement, it should be done, so that we can then test to make sure that all of the numbers we have obtained are consistent.

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