In order to calculate the performance
of an aircraft we need to solve the general equations of motion. The general point mass equations of motion
for two dimensional symmetric flight consist of many variables. Unfortunately it is not possible
to solve a set of two equations with more than two unknowns. In the last video we determined that it is
possible to express the aerodynamic drag as a function of airspeed
under the assumption that lift equals weight. Along similar lines I would like to try
to express the propulsive force as a function of airspeed in order to simplify
these equations of motion even further. Before we can do that, I have to
refresh your memory a little bit about what you have learned in terms of propulsion
and I have to introduce a few new definitions. Let's start with the fundamental
equation for thrust. Any air breathing engine, as indicated here,
regardless whether it is a propeller, a turbofan, pure jet or a helicopter rotor,
functions according to the same principle. A mass of air is accelerated to a higher velocity
and as a result a force is created according to the momentum equation for steady flow
derived from Newton's second law. Therefore I can represent all propulsion types
with this simple schematic representation. In steady conditions,
the inflow velocity of air equals the flight speed of the aircraft
and the velocity behind the aircraft equals what we call the jet velocity. The mass flow of air through the engine
is indicated by the term mdot. In most propulsion systems used for aviation,
fuel is burnt and thus the mass flow at the exit is the summation of the mass flow of air
and the fuel mass flow. Based on the momentum equation
we can state that the thrust equals the change in momentum rate,
which is the mass flow of air + the mass flow of fuel multiplied with
the jet velocity minus the mass flow of air multiplied with the flight speed. Strictly speaking, this result only holds
if the pressure at the exit of the propulsion system equals the atmospheric pressure. However, the pressure term is in practice
often much smaller than the momentum terms. Furthermore, the mass of the fuel
is much smaller than the mass of air and can therefore be
assumed zero. So we have one simple equation to express thrust. In essence there are two fundamental options
to create the same amount of thrust. One can take a little bit of air
(mdot in the equation) and accelerate it to a high jet velocity. This is typically done by jet engines. Or one can take a large mass of air
and only give it a small acceleration, which is typically done by propellers. So which of these options is better? For that we need to define efficiency. Let us start with the main
objective of an aircraft, which is to fly from A to B
at a specific velocity V. To do so, thrust needs to be
created by the engine. In a time interval delta t, the aircraft
will then have traveled delta x meters. Hence, the energy needed to travel
this distance can be determined from the principle of work
which states that the energy needed equals the distance multiplied with the force. The energy needed per second,
which is power, then equals thrust times delta x
divided by the time delta t. If we take the time interval small enough,
it becomes Thrust times Velocity. This power, we call power available. But in order to create thrust we need to provide
power, for example by burning fuel. This power is called thermal power Q
and for an engine powered by fuel, this equals the mass flow of fuel
multiplied with the energy that is contained in each kg of fuel, a constant
chemical value we call H. The total efficiency of the propulsion system
can be defined as the ratio between power available, which is our end goal,
to fly from a to b, and the thermal power, so essentially the energy we put
into the process is Q. Unfortunately this efficiency
is always smaller than 100%. But why is this the case? Since the engine accelerates air,
kinetic energy is left behind in the atmosphere, which will eventually dissipate into heat. Furthermore, the air leaving the engine
has a higher temperature and thus heat is also left in the atmosphere. The increase in kinetic energy of the flow
per second is what we call jet power. It is the kinetic energy rate behind the engine
minus the kinetic energy rate in front of the engine. If we look back at the total efficiency equation,
we can also express it as Power available divided by thermal power,
multiplied with jet power divided by jet power. This is a little trick in order to separate
different energy losses and to get more insight into the physics behind the process. The total efficiency can then be written
as propulsive efficiency multiplied with thermal efficiency. Let's have a closer look
at the propulsive efficiency. Propulsive efficiency, eta_j, is defined as
power available divided by jet power. Now if we write out these terms
then power available, of course, is thrust times airspeed and jet power
is a half mdot Vj squared, which is the rate of kinetic energy
behind the engine, minus a half mdot times the airspeed squared,
so the speed of the aircraft. But, of course we know that thrust
can also be written as mdot times Vj minus V and then of course we still have to
multiply it with the airspeed V. And what you see in the denominator
is that we can take a half mdot to the left hand side and multiply it
with Vj squared minus V squared. Now we can take the term
you see over here and rewrite it slightly, so if we keep what is on the
numerator the same (which is still mdot times Vj
minus V multiplied with V). Then we can divide it by a half mdot times,
and now I separate some variables, I can say this is Vj minus V
multiplied with Vj plus V). That is interesting, because the term
we have over here is identical to this term, so we can remove it
both in the numerator and the denominator. At the same time we can take
the mdot over here and the mdot over there and remove them. That greatly simplifies our equation
because what we then get, if we take the half and put it on the top,
we get 2 V divided by Vj plus V and if I simplify this even a bit further
and I try to move this V to the denominator then we obtain the final result which is:
2 divided by 1 plus Vj divided by V. And thus our efficiency we started of with
is only a function of the jet velocity and the airspeed of the aircraft. So the propulsive efficiency purely depends
on the ratio of the jet velocity and the flight velocity. If we would make a graph of this equation
with propulsive efficiency on the y-axis and the ratio of airspeed
and jet velocity on the x-axis, you can observe that as the ratio increases,
propulsive efficiency decreases. In addition, I should mention that
Vj is always larger than the airspeed of the vehicle when the propulsion system
delivers a positive force. So, propulsive efficiency is
always smaller than 100%. This is easy to understand
because in order to create thrust we need to accelerate air and leave kinetic
energy in the atmosphere behind us, once the aircraft has passed. This equation gives me an opportunity
to make an interesting observation through a small example calculation. What I'm going to assume is that
I have some kind of a propulsion system which delivers a thrust of 1000 Newton. This is just a purely hypothetical situation. This propulsion system I'm assuming has a
mass flow of air of 10 kg/s flowing through it. So based on our thrust equation
which is mdot times the velocity difference given to the air, we find that in order to
create 1000 N of thrust, Vj minus V must be equal to 100 m/s. So we have to accelerate the air this amount. Now let's have a look at two separate situations. Situation A is an example aircraft,
which is flying at a velocity of 100 m/s, which in aircraft terms is relatively small. This aircraft is delivering 1000 N of thrust
and with this information we can calculate the propulsive efficiency. We know the equation for propulsive efficiency
and this tells us that it's 2 divided by 1 plus Vj (and Vj in our case has to be 100 larger
than the airspeed, so it's 200 m/s) divided by the airspeed (which is 100 m/s). This is in fact 2 divided by 3
and thus you can express that as saying this is an efficiency of 66%. Now let's take another situation, B,
where everything is the same but now we have an aircraft
which is flying at 200 m/s and it's still delivering 1000 N of thrust
by accelerating the same amount of air. From this we can derive
that our jet velocity has to be 100 m/s larger, so it should be 300 m/s now. If we once again calculate propulsive efficiency
then we find that it is 2 divided by 1 plus 300 divided by 200 and if you figure out
what the propulsive efficiency is then we find that that is about 0.8 (which is 80%) So we have completely different efficiencies
but realize that we are generating exactly the same thrust
and we're also having the same mass flow. That's an interesting observation,
only by flying faster the efficiency of the propulsive system
changes by quite a large extent. Based on this, we can conclude
that engines with a high jet velocity have a low efficiency at low speed flight
but can perform quite efficient at high velocities. Amongst other reasons,
such as the engine size and weight per unit of thrust,
this explains why jet engines are used at high speed flight
and propellers for low speed applications.