In this problem what you are looking at is a diagram of a thrust reverser of a turbo
fan engine.
I am showing you 2 views.
This is a view down the front engine air is drown into the fan, and from the side view
you are looking at the fan as well, and air is drawn from the right to left through the
fan.
Through the back of the engine.
When an airplane lands the close doors at the backside of the engine.
Instead of directing thrust straight backwards propel the airplane forward.
It redirects thrust and the air in the engine curves backwards in this direction, and it
pushes the air towards the front of the airplane, and it flows out.
In this problem with some parameters that are able to you.
We want to calculate how big this reverse thrust.
One thing that we know is that the diameter of the fan is 2 m.
We know that the engine is drawing in a mass flow rate of air of 500 kg/s.
Air flows through the engine and is reflected by an angle of 60 degrees from the horizontal,
because the air is heated as it flows through the engine, and expands, and more importantly
the area on the exhaust is some what smaller.
The velocity exiting through the thrust reverse.
We will call it V2, is three times the magnitude of the velocity entering engine, which we
will call V1.
What is a little bit tough to see from this diagram is not only is the air being deflected
this direction.
It is also being deflected 60 degrees part away from you out of the screen part of the
engine into the screen.
We can visualize by maybe looking straight into the engine.
With the engine on the left.
The air is flowing out at you and away from the engine.
at an angle 60 degree.
So the geometry is not all that simple if we look at the side view.
It is not simply moving upward or downward but rather its radially a symmetric problem.
That is convenient in this problem because we don't have to worry about thrust developing
asymmetrically or a thrust developing more in the vertical direction in the y-direction
or the z-direction into our out of the screen.
We only need to worry about thrust being developed in the horizontal x-direction.
So lets begin this problem by drawing a control volume, and I am going to draw it around these
thrust reversers.
As I am drawing it just for convenience to make our lives easy.
I am drawing the control volume every perpendicular to our flow.
So I am going to draw it straight along the fan and come back and close it off in that
direction.
In 3-D control volume you could imagine that it would be.
I am trying to draw it all the way around the engine itself, but in all locations I
am drawing the control volume.
So that it is perpendicular to the flow.
I have brought in the momentum conservation equation in the x-direction, and because I
only care about the x-direction, I have not included the y and z directions.
One thing that is going on is that this problem is at steady state, and under steady state
conditions.
This time derivative would go to 0.
SO what we need to do is brain storm all of the forces acting on the control volume, and
then identify locations where fluid flows into out out of the control volume.
The only force acting on the control volume is the force of the thrust reversers on the
air.
That force will occur at this back region of the engine, and what it does is we got
negative momentum in the x-direction entering, and it leaves the engine.
It as momentum in both the positive and negative y-direction, but more importantly now contains
momentum in the positive x-direction.
So it has gain momentum, and what we would find is that the net force acting on the air,
or acting on the control volume is actually in the positive direction, and that may seem
strange,because the velocity of the airplane is from left to right.
So to slow the airplane down we would expect that there would be a reverse force acting
on the airplane to the left to accelerate in the negative direction, but what we would
find.
When we work with this equation is that we would find a force acting on the air in the
positive direction.
By newtons 3rd law it means that the air is acting on the thrust reversers in the negative
direction.
So lets write it out.
Lets say the force of the reverser acting on the air, and that force is the only force
acting on the air, and note that the sums of the forces is not 0 in this case.
It is not a static problem.
Now lets identify the way air crosses the control volume.
Of course it passes through the fan, and the entrance, and then exits the control volume
due to the thrust reverser.
I have called these locations 1, and location 2.
So lets break this area integral up into two separate integrals.
First one we will integrate over the surface area of the fan.
So lets say mu, rho, V, dot n, dA for the fan, and then surface 2 is the thrust reverser
and we will integrate the same equation over the area of the reverser.
Lets evaluate the dot products of these two surfaces areas.
If we examine the fan again surface area 1 here is the unit normal n1.
Thus V1.
Well becomes the arrows are acting in opposite directions.
It is equal to the negative magnitude of the velocity coming into the engine.
If we examine the second surface area the unit normal on area 2 is acting in the exact
same direction as the velocity.
So V2 dot n2 is equal to the positive magnitude V.
We also need to find the x components of the velocity at location 1 and 2.
In this case the velocity V1 is acting only in the negative x-direction.
So U1 is equal to negative the magnitude of V1.
In area 2 U2 the x component is acting in the positive V2 direction, but we need a cosine
theta in there to account for the fact that only some of the air velocity is acting in
the x direction.
We can also say that because V2 the magnitude of the velocity is 3 times the magnitude of
the velocity entering the fan we can make that substitution right now.
Say that U2 is equal to 3 times the magnitude of V1 cosx.
So lets come back to this integral, because I know the velocity U is uniform over cross
sectional area 1.
So this U can come out of this integral.
The density of air lets say that it is constant.
It can also come out of the integral, and V dot n are also constant, because of the
uniform velocity and they come out of the first integral.
When can say that these, all three of these can come out of this second integral as well.
Evaluating integral 1 over the first surface area is U1, rho times, be careful with the
signs now.
The negative magnitude times A1 plus the second integral.
Is U2, rho, now times the positive magnitude of V2, times A2.
What I can also note is the simplification that the mass flow rate is equal to the density
of the air times the magnitude of the velocity coming into the engine times A1.
All the mass coming into the engine is got to leave to the thrust reversers, and that
is equal to the density of the air times the magnitude of the velocity at V2 times A2.
I am going to make that substitution on both of these terms, and I will come up with an
expression for the thrust of the reverser acting on the air, but I also not that U1
is equal to the negative magnitude of the velocity entering the engine, That is equal
to the negative and I will use Q1 for the volumetric flow rate divided by the area of
that fan.
The volumetric flow rate I can expresses that as the mass flow rate divided by the density
times A1, and I know that U2 us equal to 3 times the magnitude of the velocity entering
the fans, times the cosine of theta.
That then is equal to 3, m dot, over rho, A1, cosine of theta.
So I am going to make this substitution for U1, and this substitution for U2, and I will
come up with a simplified expression for the force of the thrust reversers on the air.
What I find is that the force of the reversers acting on the air scales with the mass flow
rate to the second power, and what is more significant is that it never be negative.
If the thrust reverser is never negative as long as that angle theta is 90 degrees or
less, and in that case if it was 90 degrees for example the force the reversers would
just direct the air vertically.
So finally what is important to note.
If we use newtons 3rd law arguments.
What we are interested in.
Is the force of the reversers on the airplane, because it is the force of the reversers that
attach to the airplane that actually slow it down.
That would be exactly equivalent to the force of the air acting on the reversers and that
is equal to the negative of the force of the reversers acting on the air, which is what
we found in this analysis.
So if I plug in numbers ultimately the force of the reverser on the airplanes equals -200000
N, and this is equivalent to -45000 lb acting in a negative X-direction.