The goal of this screencast is to provide an example of a combustion problem that includes
both complete and partial combustion. In this case the combustible species is ethane. We'll
touch on basic reaction concepts, such as degrees of freedom analysis for multiple reactions.
We'll also touch on specific ideas to combustion which are the percent excess of air and the
wet and dry basis of the stack gas. The reaction stoichiometry for ethane is shown, the first reaction we
have the complete combustion of ethane, where carbon is converted to carbon dioxide. The
second reaction shows the partial combustion of ethane, were carbon is converted to carbon
monoxide. In this problem we're given a couple of pieces of information, we're going to assume
a basis of 100 moles of ethane, which is labeled on the process flow chart. We know that there is
50% excess air that's fed to the reactor, so we don't know how much but we know the
ratio relative to the feed. We also know the composition of air is always 21% oxygen, 79%
nitrogen. With this information we want to find the ratio of water to the dry gas in
the stack gas, which is the reactor output. If we look at the output, the stack gas, we
can find all the components that could possible make up the output. So the first three species
are all reaction products. We also have unreacted oxygen, the inert nitrogen from air, and any
unreacted ethane that might be present. Based on the information that's been given let's
do the degrees of freedom analysis using both molecular species balances and atomic species
balances to see if we have enough information to fully characterize the stack gas and calculate
the ratio that's been asked for. The degrees of freedom is equal to the number of unknowns
minus the number of equations that we can write to relate them.
This analysis is important, we want 0 degrees of freedom so that we can assure ourselves that
we have enough equations that we can write to solve for all the unknowns in the process.
We can apply this analysis to both the molecular species as well as the atomic species, so
we'll start with molecular. Looking at the flow chart we have 6 variables that represent
the species in the stack gas, and we also don't know how much air is fed to the reactor,
so we have a total of 7 unknowns. In this case there are 5 molecular species that are
reactive. We're not including nitrogen that's inert, but we can also write a mole balance
on nitrogen. We have one other equation, which is the percent excess, that relates how much
air is fed relative to the ethane feed. Taking the difference between equations and unknown
here shows 0 degrees of freedom, but we have not accounted for the degrees of freedom associated
with each reaction. Each independent reaction adds 1 degrees of freedom to the process,
which essentially accounts for the extent of reaction. For each reaction we don't know
the extent to which the reaction is proceeded. So if we tally up our unknowns and our equations
that we can use to relate them in this case we have 2 degrees of freedom which implies
that we need 2 more things to be specified before we can calculate for all the variables
in this chart. Let's take a look at how that changes if we look at the atomic species.
In this case we still have 7 unknowns that we're trying to solve for and we can write
3 atomic species balances, which include carbon, hydrogen, and oxygen. We also have the inert
balance for Nitrogen, and we still have our one other equation which represents the percent
excess. In this case we don't have any degrees of freedom for the atomic species because
atomic species are conserved. There is no extent of reaction or consumption or generation
to account for when we're talking about atomic species. So in this case we have 2 degrees
of freedom as well. So either way we look at this problem we need 2 more pieces of information
before we can find what's asked for, which is the ratio of water to dry gas in the reactor
output. In this case we already know the percent excess of air, and we'll also assume a fractional
conversion of ethane of 0.9. Of the ethane that reacts 30% is going to form carbon monoxide.
We can start by using the definition of % excess of air to find out how much air is fed to
the reactor per 100 moles of ethane. It's important to keep in mind when we are finding the percent
excess of air it is only based on complete combustion. By definition the percent excess
is equal to the moles fed to the reactor, minus the moles that are theoretically required
to consume all of the fuel, divided by the theoretical requirement. From the complete
combustion chemistry we can first find out the theoretical number of moles of oxygen
that are required to react with all the ethane. From the complete combustion stoichiometry
we can see that for 100 moles of ethane we're going to need 3.5 times that number of moles
of oxygen to completely react. So if we multiply our 100 moles by 7/2 we get 350 moles of oxygen
that are theoretically required. We can find the theoretical amount of air that's required
to deliver our 350 moles of oxygen if we take into account that a mole of air contains 0.21
moles of oxygen. So multiplying 350 by a mole of air per 0.21 moles of oxygen give us 1667
moles of air that are theoretically required to deliver the necessary oxygen. Plugging
in these numbers to the equation for percent excess if we have 50% excess that's equal
to the amount of air that we're actually feeding to the reactor minus our theoretical requirement
of 1667 divided by 1667. So in this case our flow rate of air is equal to 2500 moles. Next
we can use the specified fractional conversion to find how much ethane is leaving the reactor.
The fractional conversion represents for the number of moles that go into the reactor,
how many of them have reacted? Here the fractional conversion is represented with the variable
f. Fractional conversion can be calculated by taking the difference between what goes
in and what goes out, which represents the number of moles that were reacted and relating
that to the number of moles that go in. Here we've specified our conversion is 0.9, and
we have 100 moles going in, our variable for the ethane leaving is n6, so this calculation
shows that n6 is equal to 10 moles of ethane. This also shows us that 90 moles of ethane
have reacted, which we can use to find the number of moles of carbon monoxide leaving
the reactor. The moles of carbon monoxide leaving the reactor is given the variable
n2, and we know that it's equal to 30% of the number of moles of ethane that have been
reacted, which in this case is 90. If we look at the reaction stoichiometry though we know
that for every mole of ethane that is consumed, 2 moles of carbon dioxide are generated, so
we need to multiply by that molar ratio. This calculation shows that we have 54 moles of
carbon monoxide leaving the reactor. Scrolling down here a little bit I've re-labeled our
flow chart to include the values of moles of air, ethane, and carbon monoxide that we
previously found. We can see here that we still have 4 unknowns to solve for, which
means we are going to have to set up a system of mole balances to solve for the remaining
unknowns. We can either do this via molecular species balances, or atomic species balances.
It's important to keep in mind each independent reaction has it's own extent of reaction.
In this case we have two reactions so we have two extents, so these essentially represent
additional variables that would have to be solved for. Any time there is more than one
reaction it is generally easier to set up a system of atomic species balance, which
is what we'll do here. The principle of an atomic species balance is that atomic species
are conserved, so the balances can be written in the form of in=out. We can write that for
every atom that's involved in the reactions. In this case that includes carbon, hydrogen,
and oxygen. We'll start with carbon, and the number of carbon atoms that are entering the
reactor. We have 100 moles of ethane, and 2 moles of carbon for every mole of ethane,
so 2 times 100 must be equal to the carbon leaving. We have 1 mole of carbon for every
mole of carbon dioxide, so we have n1 plus the moles of carbon monoxide, which also has
1 carbon, so plus 54, plus 2 moles of carbon for every mole of ethane leaving, which is
2 times 10. We can repeat this process for hydrogen and oxygen. Counting up the atoms of hydrogen
and oxygen entering and leaving the reactor should lead to the following system of equations.
In this case we can solve these very easily and sequentially, which can be used to calculate
n1, n3, and n4. The only thing that we haven't solved for is n5, which represents the amount
of nitrogen leaving the reactor. In this case nitrogen is inert, so the nitrogen that
enters the reactor with air passes through un-reacted, so we can simply write a mole
balance, also in the form in=out. So in this case n5 is equal to 0.79 times the molar flow
rate of air, which is 2500. So we should have 1975 moles of nitrogen that are leaving the
reactor. In the original problem statement we are asked to find the ratio of water to
dry gas in the stack gas, and we now know the molar amounts of every component in the
stack gas, so we should be able to find this ratio. The stack gas is often expressed on
a dry basis or a wet basis. So if we're looking for the dry gas content it essentially represents
everything in the stack gas without water. So we're essentially being asked to find the
moles of water related to the moles of everything in the stack gas except for water. From the
atom balances we found that we had 270 moles of water in the stack gas. If we want to find
the total amount of dry gas we have we need to find the total moles of stack gas and subtract
the amount of water. If we add up the total number of moles we find that we have 2672
moles in the stack gas, and taking out the 270 moles of water gives us 2402 moles of
dry gas. If we take the ratio of 270 to 2402 we find that the ratio is 0.112, which is
essentially a metric of the humidity content on a dry air basis.