Hi everyone and welcome to this new class in the engine and propulsion system.
Today we will be discussing the combustion chamber.
As we've seen in previous classes, the air surrounding the aircraft is even taken to the jet engine
through the inlet and then is compressed.
Now that we have the air at high pressure, we will distribute this air to the inside of the combustion
chambers.
Now, this combustion chambers are small cavities, right.
Located on the surrounding of the shaft.
Okay.
And the main idea of a combustion chamber is to mix the air, which is.
Which comes from the surroundings of the aircraft.
Mix it with the fuel that we have in our aircraft.
But how does a combustion chamber work?
Exactly.
So the most important thing to take into account here is that we will take fire and we will introduce
it inside of the combustion chamber.
And also we will produce we will create a way in which the fuel will be in LED or will be taken also
to the inside of the combustion chamber.
So in other words, we need fuel, we need air, and we need them inside of the combustion chamber.
And the ways in which we put the feel and the inside of the combustion chamber are different.
The fuel will simply have a fuel inlet and an injector, which through swirl vanes will be moved or
will be distributed, will be spread to the inside of the combustion chamber.
The air, on the other hand, is actually moves from the.
From the upper part of the combustion chamber and moved throughout the exterior of the chamber itself.
So in other words, the chambers, as you can see, are distributed surrounding the shaft and they are
encased right in these casings.
Now the air will move.
In the inside of the casing, but outside of the structure of the combustion chamber itself.
But because we have some cavities.
We have some holes.
Right.
This air will move to the inside of the combustion chamber, too.
So really the idea is that because we have these holes, the air is inputted towards the interior of
the combustion chamber, as you can see.
This produces.
The mixture between air and the full right.
Two ways in which we approximate how the fuel and the air actually get into the combustion chamber.
But this has been proved to be really effective at mixing the two fuels of the two flows.
Right.
And as you can see, as I mentioned before, they are distributed in this way.
It's not just one combustion chamber, but rather a few combustion chambers distributed in in a way
that they provide the most efficiency for that mixture to happen.
So these are some types of designs, several designs that we can consider.
So for combustion chambers, right, there's not one only solution that works, right?
But rather, it has to be a compromise between many variables.
And as you can see, it's just it.
The most common one that you will find in commercial aircraft particularly, but also for turbojet propulsion
systems, is this type of combustion chamber in which simply as mentioned before, we have an injector
for the fuel.
Through the swirl vein, we are able to distribute this fuel throughout the combustion chamber, the
inside of the combustion chamber and the air which moves at the beginning between the the exterior of
the combustion chamber itself and then moves to the inside.
They will be combined in an effective way to provide the high level energy of the few of the flow that
we actually need.
And this is the geometry of our Rolls-Royce engine used for turbofan engines.
As you can see, it really is the somewhat simple the idea.
But of course, taking into account that we need the materials to withstand all of the adverse conditions,
very high demanding conditions that will be in the inside of the combustion chamber.
In terms of the type of jet engine fuel.
Right.
We need a fuel for aircraft that has actually low freezing temperature points.
That is, of course, because aircraft, when they are flying at a certain altitudes, the air surrounding
that aircraft will be really low.
Right.
The temperature of that air will be really low, meaning that we need to avoid freezing of the fuel
inside of the tanks.
Okay.
And the this is why kerosene is used and not actually just gasoline or diesel fuel.
Okay.
Now, we also, of course, need high energy density fuels, which basically, as it happens in many
engineering fields, trucks is down.
So using hydrocarbons.
We need this type of fuel because it's the most energy dense.
Right.
And if we use another type of fuel, we won't be able to guarantee as much thrust in the sense that
not only are we not able to guarantee that thrust, but rather the thrust output it by using other fuels
would simply be too low to actually provide a thrust enough to an aircraft to fly.
Right.
This is why we are more or less stuck using kerosene for now.
And again, one other interesting consideration, which is not only the only case for historically all
of the aircraft that have been developed is actually to have the capability of auto ignition.
This basically means that we do not need sparks.
We do not need extra components in our aircraft or in our combustion chamber to provide the energy for
that combustion to emerge to happen, but rather because we are already using kerosene and it auto ignites
and the temperature at which the flow is inside of the combustion chamber, it will automatically and
by itself combine with the air and have this chemical reaction.
Okay.
So types in terms of types of the fuels that you will typically see in commercial aircraft, they usually
call the jet a jet A one jet B, all turbo jet propulsion.
We usually have the jet GP one, JB two, and so on.
Okay, now we can get into the detail of what exactly all of these types of fuels guarantee, but usually
they work at different temperatures, they have several different levels of energy density they can
withstand again, or that they are able to produce an amount of thrust to another.
Okay.
And if you want to get into these detail for that, it's pretty easy to to see some tables and to identify
some of the differences between all of these fuels.
Right.
However, we won't get into detail in this class.
And what I wanted to do actually is a quick exercise for you to practice and to really get a bit more
into the engineering mentality of having to actually solve this mathematically.
Right.
Now, as you can see over here, the combustion chamber is capable of intake and the hydrocarbons.
Right.
And also in taking the air on the surrounding of the aircraft right now, as far as you know, air or
the atmosphere at reasonable altitudes is mostly composed by nitrogen.
And to write the dioxide, nitrogen and also oxygen and the balance.
Roughly speaking, from an engineering standard point of view is more or less 3.7, six times the amount
of nitrogen per amount of oxygen.
Okay.
Roughly speaking, the atmosphere is composed of 79% of nitrogen and 21% of oxygen.
And if you do the math, it really comes down to 3.7, six times more nitrogen with respect to the oxygen.
This is why this equation has this this peculiarity.
Okay.
And as you can see, the fuel, which, as we mentioned before, must be a combo.
So the hydrocarbon is combined with the air.
And then therefore, as a consequence, the reaction produces carbon dioxide, which we know is not
the best selling point of aircraft.
It also produces some water, H2O, and it can produce some residual of oxygen and particularly nitrogen.
Okay.
So the question is.
How do we balance this equation?
Right.
If we put more fuel.
Right.
How much more residual would we have?
Okay.
And the other question that we can ask ourselves is, how do we optimize this reaction?
Right.
And so now we will proceed to solve this equation optimally.
Now, the first thing that I did when considering that this has to be optimal is to actually neglect
the oxygen going outside of being a product of the reaction in the sense that the reaction will be optimally
defined if all of the oxygen which is in the air right.
Is able to react with the fuel and to produce thrust.
Okay.
So I just skipped this concept over here.
And we are left with this equation.
Now I'm going to propose that you solve these equations by yourself.
And then you try to think how you can actually solve this.
Considering that need needs to be optimal.
In other words, there's this Z parameter over here A, B, and C, and we need to calculate what these
parameters should be in order for the left side of the equation to be equal to the right half of the
equation.
Okay.
And so I ask you to try to solve this by hand, however, is convenient to you.
And you try to identify what the amounts of oxygen, the amount of air should be with respect to the
amount of fuel.
Right.
For instance.
And you will also see what the byproducts are necessarily because of this reaction.
Okay.
Now, in the next slide, I actually have solved it and I will go through the next slide to present
to you the solution, because I think it's a good exercise for you to understand how it can be solved.
But I really think that it would be really useful to you to actually try to solve it before and see
if there are some mistakes in the way that you approached it or if you already did it.
Fine.
From the beginning.
Right.
And I would encourage you to actually fail, because that's the best way to learn in engineering.
And we need failure.
We need mistakes.
We need to try to solve this equation in many ways and figure out the way that works best.
Okay.
So please try to solve the equation.
Otherwise, I'll just go to the next slide and explain how it is solved.
So this is the equation again.
We have the I hope that you actually tried to do the exercise, right.
And as you can see, there's the hydrocarbon.
This particular number that I simply got right, I'd just like to see 12 and age 26.
It could be another type of kerosene.
Fuel.
I just took this as a as an example to solve this equation.
Right.
And then we have z times the oxygen and nitrogen of the air.
Then eight times the carbon dioxide be times the amount of water that is ejected from the from the aircraft,
from the engine.
And finally, sea times the amount of nitrogen, which, by the way, as you can see, the nitrogen
is not really reacting.
It's just moving from the left side of the equation to the right side of the equation.
And the answer is similar to C and to right.
Okay.
So let's solve it point by point, step by step.
Now, the first equation that I would derive from this set of equations, actually right, is the fact
that we initially have 12 atoms of carbon.
Right?
And that this carbon, it's just the only carbon that actually appears on the left side of the equation
must be equal to all of the carbon appearing to the right side of the equation.
So basically 12 C's right must be transformed or equated to a times C.
Okay.
So in other words, c 12 is equal to a number multiplied by C.
So basically A is equal to 12.
Right.
In a similar way we can actually deduce B to write H 26.
There's no more hydrogens on the left side of the equation, and hydrogens on the right side of the
equation only appear on this product over here inside of the water.
Right.
So basically at 26, H becomes B times two H.
Right.
And so what number must be B?
Right.
In order to satisfy this equation, when it should be 13, 13 times two is 26.
So as you can see, we already balanced the carbons and the hydrogens.
The next part is to actually balance the nitrogen out.
Right.
This is 3.76, which is the ratio of the hour of nitrogen divided by oxygen.
Right.
Times Z because we don't know the amount of air that we will have on as compared to the fuel right will
be equal to the only nitrogen that appears on the right side of the equation, which is C times N2.
So in other words, C is equal to 3.76 times Z.
Now, as you can see, this is not a number.
This is an equation.
This is a ratio between C and Z.
And thus we need another equation to actually solve Z.
And this is where the oxygen comes in.
Right.
This is the most complex equation in the sense that oxygen appears on the left side of the equation,
but also on the right side.
And not only that, but actually we have two main components of the oxygen.
I also added the O2 for illustration purposes.
Right.
These could be also added.
And you would see how all of these variables interact with each other.
And you can make some models with that.
Okay.
But just for simplicity's sake, I consider these zero the O2 to be zero.
Right.
So D equal to zero.
And then we simply have the tosay right because this is dioxide.
Is equal to to A plus B.
And from this.
We know, right?
A is 12 and B is 13.
So 12 times two plus 13.
This is 37, right.
And because we have a two multiplying the Z over here, the two will have to go to the other side dividing.
Right.
So this is 24 plus 13, which is 37 divided by two, 18.5.
This is the value that we obtain from Z, right?
And now because we now see which is the ratio of air with respect to hydrocarbons.
Okay.
We see that we can actually determine, see to see becomes equal to 69.56.
Okay.
And so this is the main equation that ends up solving.
Right.
This documentary reaction, this type of reaction is typically studied in chemistry in order to balance
out and to identify just how much air that we really need for the fuel.
Right.
And as you can see, the amount of air with respect to fuel is massive.
Right.
18.5 times more air than fuel.
Okay.
So this means that the fuel interacts with the air, but the most of the flow is still air.
Right.
And as you can see, this also points out just how much carbon dioxide is produced because of the fact
that we are using kerosene.
Okay.
The fact that it's also this water byproduct and some nitrogen leaving the aircraft.
Okay.
And so this would be all in terms of the combustion chamber.
I also wanted to make this a mathematical exercise for you to get a bit more into the, you know, the
math of engineering, which is clearly fundamental.
And I hope this was productive class.
Okay.
We can now move to the next class.