In this screencast we are going to look at a combustion reaction, and solve for not only
the amount of components in the product gas, but also the dew point temperature of the
product or stack gas. Generally the only condensable species in the stack gas is water. So to find the dew point
temperature we are going to use Raoult's Law that says the partial pressure of water is
equal to the saturated pressure of water at the dew point temperature. So in order to
apply this equation what we are going to need to find is that mole fraction of water. So
in this problem we have 10 moles of ethane, and it is going to be combusted with 40 % excess
air, and so what that means is that we will not only have O2 in our product gas. We will
also have N2. In addition the fractional conversion of ethane is 0.75, which means that we will
have ethane in our product gas, and finally the selectivity of the desired product, which
is CO2 versus the undesired product. Which is CO is going to equal 5. So this means that
we have incomplete combustion occurring as well. So lets start with a flow chart of our
process. We will start by figuring out how much oxygen, and therefore, nitrogen comes
into our system. So we are told that there is 40% excess oxygen. In order to be able
to solve for that we have to write the combustion reaction. So we have C2H6, which reacts with
7/2 moles of O2, and that gives us 2 moles of CO2, plus 3 moles of H2O. Note that I haven't
written the incomplete combustion reaction as excess oxygen is always based on the desired
reaction. So for every mole of ethane. We are going to need 3.5 moles of oxygen. So
if we have 10 moles of ethane. We multiply it by 3.5. That would give us our theoretical
amount of oxygen. However since we have 40% excess we multiply this 1.4, and we end up
with 49 moles of O2 coming in. From that we can find the number of moles of our N2 coming
in. Since the relationship is 3.76 times the number of moles of O2 equal the number of
moles of N2, and this is equal to therefore 184 moles N2. It is very important in these
combustion reaction not to forget the nitrogen, because the composition of your product stream
is going to be completely wrong. So now the next thing we are going to do is use the fractional
conversion to find the amount of ethane that is coming out. So we know that 0.75 is equal
to the number of moles of ethane coming in minus the number of moles of ethane coming
out. Divided by the number of moles of ethane coming in, and this allows us to solve for
the number of moles of ethane coming out, which is equal to 2.5 moles. So the next thing
we are going to do is use our atomic species balances to solve for the rest of our components.
We are going to start with our carbon balance. So we have 10 moles of ethane coming in, and
for every mole of ethane we have 2 atoms of carbon, and that is going to equal the moles
of ethane coming out times those 2 carbon atoms. Plus the number of moles of CO coming
out. Plus the number of moles of CO2 coming out. Each of them has 1 carbon atom per mole,
but if you recall from our selectivity the number of moles of CO2 equal 5 times the number
of moles of CO. So we can replace that nCO2 with 5 times nCO. That allows us to solve
for the number of moles of carbon monoxide, which is 2.5 moles, and then to solve for
the number of moles of CO2 we just multiply that by 5, and get the number of moles of
CO2 equaling 12.5 moles of CO2. The next balance I am going to do is the hydrogen balance.
In a combustion reaction I always save the oxygen balance for last, because oxygen is
in so many of the components in a combustion reaction. So for our hydrogen we have our
10 moles of ethane coming in, and there are 6 hydrogen atoms in 1 mole of ethane, and
now this is again going to equal 2.5 times our 6, plus now there are 2 atoms of hydrogen
in every mole of H2O. We can solve for the number of moles of H2O, which is equal to
22.5 moles. And finally we do our oxygen balance. So coming in we have the number of moles of
oxygen. There are 2 oxygen atoms in every mole of O2, and this is going to equal the
22.5 moles from the H2O plus the 2.5 moles from the CO, plus 2 times the 12.5 moles from
the CO2. Since there are 2 oxygen atoms, plus 2 times the number of moles of O2 coming out,
and so what we are left with is that the number of moles of O2 that are coming out are equal
to 24 moles, and since nitrogen is not a reactive species the number of moles of N2 equal the
number of moles that come in. Since we are looking for the mole fraction of water. We
are going to have to find the total moles. So we are going to have to add up all the
products in our stack gas. So when we add this all up we get a total of 248 moles, and
look at the difference we would have gotten if we hadn't included nitrogen. Which is a
substantial amount of the moles in the product gas. The definition of a mole fraction is
the number of moles of H2O, and I am going to call this yH2O. So this is going to equal
our 22.5 moles divided by our 248 moles, and so our mole fraction of water is 0.09. Now
we can find the partial pressure of water. So our y times our pressure is equal to 0.09
times our 2 atmospheres, which is our pressure times 760 mmHg per atm, because generally
the tables that are used for saturation pressures are given in mmHg, and this is going to equal
137.7 mmHg. So when we go to a table of vapor pressure of water as a function of temperature,
and look up this saturated or vapor pressure. We find that the dew point temperature or
the temperature that goes with this saturation pressure is equal to 58.2 degree Celsius.