In this screencast we're going to look at an example of a combustion reaction. Here
we have the combustion of methane, which is CH4. Methane is the main component in natural
gas, and anytime we're talking about the combustion of hydrocarbons there's two competing reactions
that can occur. The first reaction shows the example of complete combustion of methane.
In complete combustion, all carbon is converted to carbon dioxide. The second reaction shows
partial combustion, and in partial combustion all carbon contents are converted to carbon
monoxide. In this example we're give a couple of pieces of information. We're going to assume
a basis of 100 moles of methane that are fed to the reactor, and we know that air is fed
at 25% excess. Inside the reactor we have the complete conversion of methane, and no
partial combustion. Based on this information we want to find the moles of air that are
fed the reactor, and the moles of each component in the stack gas. To accomplish this we're
going to have to set up a system of molecular species balance on each species in the process.
Before we start setting up and solving the system of equations it's important to have
a fully labeled process diagram. Based on the information that's been specified, take
the time to set up a fully labeled process flow chart. An example of a labeled diagram
is shown at the bottom. The first two pieces of information that are given are telling
us something about the feed of the reactor. Here we have our basis of 100 moles of methane,
and we have an unknown variable, n(air), to represent the feed to the reactor. We don't
know how much air there is, but we know it's at 25% excess, and we also know the composition,
we know that air is 21% oxygen, and 79% nitrogen by mol. The last two pieces of information
that are given are telling us something about what's happening inside the reactor, and influences
what we would expect to see in the stack gas. For example, we know there is complete conversion
of methane, so we know we won't find any methane in the output. We also know there is no partial
combustion, so we also wouldn't expect to see any carbon monoxide in the output. So
this is implying that all methane is combusted via complete combustion, and the reaction
is shown below. The complete combustion reaction shows us that we should expect to find carbon
dioxide in the outlet as well as the other product, which is water, and because air is
fed at excess we know that all the oxygen is not going to be consumed in the reaction.
Air is primarily nitrogen, which is inert in this case, so we're also going to have
nitrogen in the output as well. So with all the relevant species accounted for we should
be able to start setting up our system of equations. We'll start by using the definition
of the percent excess of air to find out how much air is being fed to the reactor. The
definition of percent excess of air is the moles of air that are fed to the reactor minus
the moles of air that are theoretically required to consume all the fuel, divided by the theoretical
moles of air. The moles of air that are fed is what we're trying to find out. The moles
of air that are theoretically required are based on our 100 moles of methane that are
fed, as well as the reaction stoichiometry. Here we can see that for every mole of methane
that is consumed, we need two moles of oxygen. We can find the amount of oxygen we need by
taking our 100 moles of methane and multiplying it by the molar ratio that oxygen is consumed
for every mole of methane. That gives us 200 moles of oxygen that we need to completely
consume all of the methane. However, we're not feeding pure oxygen to the reactor, so
we need to find the theoretical amount of air that is required to supply 200 moles of
oxygen. If we take into account that every mole of air has 0.21 moles of oxygen, we can
multiply by that ratio to get the 952 moles of air that we theoretically need in this
case. With the amount of air that we theoretically need accounted for, we can use the equation
for the percent excess to find out how much air is actually being fed. We know that air
is being supplied at 25% excess. We're trying to find out the moles of air that are being
fed, and we just calculated that we theoretically need 952 moles, so if we plug into the equation
we should find that the moles of air that we're feeding to the reactor is equal to 1190
moles in this case. Now that we know how much air we're feeding to the reactor, we should
be able to set up a system of molecular species balance to find all the amounts of each component
in the stack gas. If we're running a molecular species balance on a particular component,
we're relating the moles out to the moles in through the extent of reaction. The moles
out is equal to the moles in plus the stoichiometric coefficient for that species i, times the
extent of reaction, and we can write that for every reactant and product in the reaction.
For example if we apply that to methane we know that there is no moles of methane that
are leaving the reactor. We have 100 moles that are going in, the stoichiometric coefficient
is -1 because it is a reactant, times the extent of reaction. From this equation we
can see that the extent of reaction is equal to 100 moles. If we apply this to the product
carbon dioxide, the amount of carbon dioxide leaving is represented by the variable n(1).
We have no carbon dioxide entering the reactor. The stoichiometric coefficient is +1, because
it's a product, and we're multiplying that by the extent of reaction, and we can apply
that to water and oxygen as well. The amount of water leaving is given the variable n(2).
We had no water going in, we're generating 2 moles of water for every mol of methane
that reacts, times the extent of reaction. For oxygen the variable is n(3). The input
of oxygen is 0.21 times the total moles of air, which we calculated as 1190. This is
also a reactant, so we have a -2 for the stoichiometric coefficient, times the extent of reaction.
If we insert 100 moles for the extent of reaction into these balances, we can find the flow
rates for carbon dioxide, water, and oxygen. The last thing that we need to find is the
moles of nitrogen that are leaving. We don't need to write a molecular species balance
in terms of the extent of reaction, because in this case nitrogen doesn't react. It's
inert, which means that whatever goes in is coming right back out, so we can write a balance
of the form in=out. In this case what's going in is 0.79 times the total moles of air that
we found, again that's 1190, and that must be equal to n(4). And if we make that calculation
we'll find that n(4) is equal to 940 moles of nitrogen. So in the example of the complete
combustion of methane we've shown that we can relate the amount of air that's being
fed to the reactor for a given amount of fuel given the percent excess, and we can relate
the reactor output to what's being fed through stoichiometry, the extent of reaction, and
molecular species balances.