The goal of this screencast is too look into detail at the idea of percent excess of air
in the context of combustion reactions. Here we have the combustion of ethane. Any combustion
reaction involves the reaction of a hydrocarbon, or other fuel source, with oxygen. In this
case we can see ethane is reacting with oxygen to form carbon dioxide in the first reaction.
This is the complete combustion reaction. On the other hand we can also react ethane
with oxygen to form carbon monoxide, which is the partial reaction. In most applications
the oxygen source for combustion is air. Air is 21% oxygen by mol, and 79% Nitrogen. Air
is generally fed to the reaction in excess for several reasons. In general the reason
for combustion is not to generate products of interest, but to harness the energy, here
labeled Q, that's released in these exothermic reactions, for some other purpose. If we look
at the oxygen requirement for complete combustion relative to partial, we can see that the complete
combustion requires more oxygen, so by introducing more oxygen to the reactor we can preferentially
favor the complete combustion reaction. The downside of adding too much excess air is
that all the nitrogen, which is inert in this case, makes its way to the flue gas, so we're
essentially increasing the amount of nitrogen that's in the flue gas. As we introduce more
inert into the reactor, once we consume all the fuel there's more stuff in the reactor
that absorbs that heat, so we're actually transferring less heat out of the reactor,
so there's an optimization to increase the conversion of the fuel, and maximize the heat
transfer across the reactor. If we want to generalize the idea of excess air in the context
of any combustion reaction, the excess air is defined as the amount by which the air
fed to the reactor exceeds the theoretical requirement of air. Mathematically we can
express the percent excess of air as the moles of air fed, minus moles of air theoretically
required to consume all the fuel, divided by the moles of air that are theoretically
required, times 100%. The calculation of the theoretical air requires a couple of things.
First we know that air is primarily nitrogen, which does not react, so the amount of air
that contains the theoretical requirement of oxygen is a first consideration. The theoretical
requirement of oxygen, and therefore air as well, is based on complete combustion of the
fuel. In other words we are assuming that all of the carbon in the fuel is completely
consumed, and all of it generates carbon dioxide via complete combustion. Let's look at an
example in a combustion of ethane. In this case we're feeding 100 moles of ethane to
the reactor, we have the reaction for complete combustion of ethane, as shown earlier, and
we also had the partial reaction. We're feeding air to the reactor, which we know is 21% oxygen,
79% nitrogen, and it's being fed at 100% excess. We want to find the amount of air that's being
fed to the reactor for two conditions, in the case where all the ethane is being converted
to carbon dioxide, or in the case where we have 30% of reacted ethane to form carbon
monoxide. According to the equation we showed earlier the percent excess depends on two
things. The air that we're feeding to the reactor, and the air that we theoretically
need. The theoretical requirement is based on the oxygen content that's required to consume
all the fuel. So we'll start by finding the theoretical moles of oxygen that we need,
and we'll scale that up to the moles of air that we need. If we look at the combustion
reaction for complete combustion we can see that for every mole of ethane that reacts
we need 7.5 moles of oxygen. So if we multiply 100 moles of ethane that are being fed, times
the 7.5 for the stoichoimetric coefficient of oxygen, we can see that we need 350 moles
of oxygen to consume all the ethane. So we need 350 moles of oxygen, and we know that
air is 21% oxygen by mol, so if we divide our 350 moles of oxygen by 0.21, we'll need
1667 moles of air theoretically. So we know the air is supplied at 100% excess, and we
know what we theoretically need, so we have everything we need to find out the mole of
air that's actually fed to the reactor. If we plug into the equation for percent excess
air, 100% is equal to the air that's being fed, minus the theoretical amount of air that
we just calculated as 1667, divided by that theoretical requirement times 100%. So if
we make that calculation, the air that's actually fed to the reactor in this case is 3333 moles
of air. This calculation is assuming complete conversion of methane to form carbon dioxide.
How does this change if we have 30% of the reacted ethane to form carbon monoxide? Take
a minute to jot down how you think the answer is going to change, and what else we need
to make this calculation. The tendency there is to find a new theoretical air requirement
based on 70 moles of ethane reacting to form carbon dioxide, and 30 moles to react to form
carbon monoxide. In actuality the amount of air we need shouldn't change. The percent
excess only depends on complete combustion, it has nothing to do with what actually happens
in the reactor, or how much of the fuel goes to carbon dioxide or carbon monoxide. The
percent excess calculation is based on the ratio between the fuel and the air, assuming
complete conversion and complete combustion only. Let's look at one more example where we have an
equimolar mixture of ethane and propane. On the inlet side we have 50 moles of each. The
air that's fed to the reactor has the same composition, this time it's fed at 25% excess.
So we want to find the air that's being fed to the reactor for a mixture of two fuels,
we've shown that the partial combustion reactions don't matter, right, the percent excess is
only dependent on complete combustion. So here we have the complete combustion reaction
for ethane, and the complete combustion reaction for propane. As we did before, the first thing
we're going to want to find out is the theoretical number moles of oxygen that are required, and in
this case it's going to be the oxygen required to burn 50 moles of ethane, plus 50 moles
of propane. From the reaction for ethane we can see that we need 7.5 moles of oxygen for
every moles of ethane that we're combusting. So if we had 50 moles of ethane that we're
reacting, and we multiply that by 7.5, we have the oxygen requirement to burn all of
the ethane. If we apply the same idea to propane we need 5 moles of oxygen for every mole
of propane. So we multiply our 50 moles of propane that we have by the five moles of
oxygen required per mole of propane. So the theoretical oxygen requirement is the sum
of these two terms. If we add up the oxygen that's required to burn each fuel then we
need 425 moles of oxygen, which again we can multiply by every mole of air, by 0.21 moles
of oxygen to find the theoretical air requirement, so in this case dividing by 0.21 moles of
oxygen per mole there we can find that we need 2024 moles of air theoretically. Now
we know that the air is fed at 25% excess, so if we go back to the calculation of percent
excess we can plug in the theoretical requirement, the percent excess that we're given, to find
the number of moles that are actually being fed. If we scroll down here and show the formula
we know that 25% excess is equal to the number of moles of air that's fed, minus the air
theoretically required, divided by the air that's theoretically required. So
plugging in our value of 25% to the theoretical requirement of 2024 that we found, the number
of moles of air that are being fed to the reactor in this case is 2530 moles of air.
So in this case we found the air that's being fed to the reactor at 25% excess by assuming
complete combustion of each fuel in the mixture, and assuming complete combustion only. So
any time we're looking for the percent excess air whether it's a pure species that's being
combusted, or a mixture, we're looking for complete combustion of all fuel in the mixture.