In this screencast we're going to go over some of the concepts related to combustion,
which is a specific type of reaction that involves the reaction of fuel with oxygen,
as shown in the diagram here on the right. The output from a typical combustion reactor
is generally called the flue gas, it can also be called the stack gas. The goals for this
screencast are to provide an overview of combustion chemistry, looking at a specific example of
methane, as well as go over some specific concepts related to combustion, which are
the percent excess air and the wet and dry basis. Examples of fuel sources that are used
in combustion often include many sources of fossil fuels. For example, oil's used in a
variety of different combustion reactions. Generally oil consists of high molecular weight
hydrocarbons. Alternative fuel is CH4, which is methane, and that is a primary component
of natural gas. Also we have coal, which is mostly carbon, also includes some hydrogen
content as well as sulfur. All of these can be reacted with oxygen to form a variety of
different products, which we'll look at here in a second. The primary source of oxygen
is air, which we know has a composition of 21% oxygen and 79% nitrogen. In the case that
air is used as an oxygen source we can see that nitrogen is a primary component of air,
and it and inert species in this particular class of reaction. In this case all of these
fuels are only reacting with oxygen, so the nitrogen content of air simply passes through
the reactor. When these fuels react with oxygen combustion can either be complete or partial.
The significance of complete combustion all the carbon content of the fuel is converted
to carbon dioxide. The hydrogen in the fuel is converted to water, and if there is sulfur
in the fuel the sulfur is converted to sulfur dioxide.On the other hand, in the case of
partial combustion, the carbon content in the fuel is converted to carbon monoxide.
We still have the case that all the hydrogen in the fuel source is converted to water.
So based on the combustion chemistrys we can take a look at what we would expect to see
in the flue gas. On the product side we're going to see carbon dioxide from complete
combustion, carbon monoxide from partial, and water in either case. If we're feeding
the air in excess, which is common, then we should also expect to see oxygen and nitrogen
in the flue gas, and any un-reacted fuel. So in this case we can see that the products
are not the primary reasons for running these reactions. The real reason for combusting
fossil fuels is to generate the energy that they release upon reaction. These reactions
are highly exothermic and release a lot of heat that can be utilized for other purposes.
Let's look at an example of these ideas in the case of the combustion of methane. Below
we can see two independent reactions for the reaction of methane with oxygen. The first
is a reaction of methane with oxygen to form carbon dioxide, which is the complete combustion
reaction. The second reaction is the combustion of methane with oxygen to form carbon monoxide,
which is the partial reaction. In the example flow chart below the moles of our fuel and
our air are both specified, as well as all the products. Because we can see that there
is carbon dioxide and carbon monoxide in the stack gas we know both of these reactions
are in play. We also have oxygen and nitrogen, as well as some methane in the stack gas,
so the air must be fed in excess, and the conversion is less than 100%. Based on the
moles of each species that are specified we want to find the percent excess air. The calculation
of percent excess air is shown here on the right. We can calculate the excess air as
the moles of air that are fed to the reactor minus the air that's theoretically required
to consume all of the fuel, divided by the theoretical moles of air times 100%. This equation
shows us that the calculation of percent excess air is only related to the amounts of the
fuel and the air that are going into the reactor. Has nothing to do with extent of reaction
or conversion of the fuel inside the reactor, anything about the composition of the output.
It is only related to how much of the fuel we have, how much of the air is being fed
with the fuel, and how much air we actually need to consume all of the fuel. In this case
we know that there is 1047 moles of air going into the reactor, which represents the moles
of air that's fed. If we want to find the theoretical moles of air that's required we
need to look at only the complete combustion reaction. Partial combustion does not go into
the consideration of the oxygen requirement when we're calculating percent excess air,
we are only looking at the primary complete combustion reaction. So based on 100 moles
of methane that are being fed we need to find the theoretical moles of oxygen that are required.
Looking at the complete combustion stoichiometry we know that if we have 100 moles of methane
to react all of it we need 2 moles of oxygen for every moles of methane that we have, so
our theoretical requirement of oxygen should be 200 moles. In this example we're not reacting
methane with pure oxygen, so we should need more air to supply the required amount of
oxygen. We can find the necessary amount of air if we consider that air is composed of
21% oxygen. By multiplying our 200 moles of oxygen that we need by a mole and the composition
of oxygen underneath, which is 21%, we find that we need 952 moles of air theoretically.
Now we know what we have fed to the reactor, as well as what we need, and we can use the
percent excess formula to figure out what the percent excess is in this case. So plugging
in the percent excess is equal to the 1047 moles of air that are fed minus the 952 moles
of air that are theoretically required, divided by that theoretical requirement of 952, and
when we multiply by 100%, we find that we are feeding air at 10% excess. The last thing
we want to talk about that's specific to combustion reaction is the idea of a wet and dry basis.
The wet and dry basis is specifically representing the composition of the stack gas. If the composition
of the stack gas is expressed on a wet basis, that implies that the moles of water is included.
On the other hand if it's expressed on a dry basis as you'd expect, that implies that the
composition of every species is expressed excluding water from the overall composition
and overall mole content. In this example we want to find the mole fraction of carbon
dioxide in the stack gas on both a wet and a dry basis, so we'll start with the wet basis
first. We want to find the mole fraction of carbon dioxide on a wet basis we need the
total number of moles of the entire stack gas including water, so we can simply sum
up the number of moles of every species that we have. In this case for example we have
70 moles of carbon dioxide, we're adding 20 moles of carbon monoxide, and we're adding
the 180 moles of water that we have. If we add oxygen, nitrogen, and methane as well
we have a total of 1157 moles. If we want to find the moles fraction of carbon dioxide
we simply divide the 70 moles of carbon dioxide that we have by the total number of moles
on a wet basis, and we find that the moles fraction is 0.0605. If we want to find the
mole fraction of carbon dioxide on a dry basis we need the total number of moles in the stack
gas, excluding water. This should be the same as the sum of all the species minus the 180
moles of water, or we can simply add every species that's in the stack, except for water.
In this case if we take our total of 1157 moles on a wet basis, subtract 180 moles of
water, we have 977 moles of stack gas on a dry basis. We want to find a mole fraction
of carbon dioxide here, we take our 70 moles, divide it by our 977 moles on a dry basis,
and we find that the mol fraction in this case is 0.0716, so we can see that the mol
fraction increases a little bit on a dry basis because there is no water in the overall molar
content. Expressing the stack gas on a wet and dry basis is unique to combustion. It's
important to be able to go back and forth between the wet and dry basis depending on
the information that's available. For examples of converting wet basis to dry and visa versa
check out another screencast on these type of calculations.