In this problem we've got a 50,000 kilogram airplane its powered by a turbojet engine
and the engine draws in fuel at 2 kilograms per second, it draws in air at 150 kilograms
per second and as the air passes through the compressor the pressure is increased by a
factor of 20, the fuel has a heating value of 40,000 kilojoules per kilogram and the
question is asking us based on this information what is the initial acceleration of the airplane
if it starts from a rest and full power is immediately applied to the engine and we're
to assume the engine operates on an ideal jet propulsion cycle and we're given the ambient
temperature and pressure of 27 C and 100 kilopascal and it says assume constant specific heats
for air of this value and a heat capacity ratio of 1.4, in a turbojet engine there's
effectively five points, there's the inlet to the compressor, there's the exhaust or
exit of the compressor, which enters the burner section and then the exit of the burner which
enters the turbine, and the exit of the turbine which enters the nozzle and then finally point
five, the exit of the nozzle, so to solve this problem what we want, we want the acceleration
of the airplane and by newton's law we just say the acceleration is equal to the force
of the thrust divided by the mass of the airplane so if we knew the force of the thrust, we
know the mass of the airplane, we'd have the problem solved, if we did a momentum balance
around the engine we'd find that the force of the thrust is equal to the mass flow rate
of air times the difference in velocity V5 the velocity exiting the nozzle minus the
velocity entering the front of the engine and if we make the assumption that the velocity
of the air entering the engine is very small compared to the velocity exiting the back
of the engine we can say that V1 is about equal to zero and the force or thrust is just
mdot times the velocity exiting the nozzle, the thermodynamics we need to work with will
allow us to calculate the velocity at V5, we know the mass flow rate of air is given
to us and therefore we can find the force and hence the acceleration the key to solving
this problem is to do a series of energy balances and some isentropic considerations, one energy
balance we could do is an energy balance around the compressor section of this engine, so
we've got a mass flow of air coming in, a mass flow of air leaving and there's shaft
work entering the compressor and the amount of shaft work entering the compressor which
is used to compress the air by a factor of 20 so P2 is 20 times bigger than P1, and this
shaft work comes from the work out put by the turbine, now lets just do an energy balance
around this section first. So here's our general energy balance equation, the compressor is
adiabatic, there's no significant heat transfer in or out of it, the potential energies are
about the same there is no work leaving the compressor so work out is zero and the velocities
entering and leaving the compressor, although they're different, the differences are really
small. So what we're left with is the amount of shaft work entering the compressor plus
the rate at which enthalpy is entering the compressor is equal to the rate at which enthalpy
is leaving the compressor and if we use a constant heat capacity we can calculate this
relationship for the amount of work coming in and the temperature difference T2 minus
T1 and because the amount of work entering the compressor is a positive value we would
expect the temperature at 2 to be greater than the temperature at 1 so shaft work is
going into the air, heats up the air by the time it leaves the compressor, and although
it's heated up, we don't know how much but because this is an ideal compressor its isentropic,
it's adiabatic and it's reversible we can use the relationship for an ideal gas that
isentropic compression because the two entropies are the same T2 is equal to T1 multiplied
by the pressure ratio and this factor for K and if I plug in values for this, using
the ambient temperature pressure ratio of 20 and this factor of 1.4 for K I find that
the exit of the compressor, it's not even reached the flames yet, but that the temperature
has been raised from 300 Kelvin up to a little over 700 Kelvin. We now know something about
the state of the air and point 2, let's turn our attention to the burner and let's do an
energy balance around the burner, so around the burner there is no shaft work coming into
or out of the burner, there is no heat leaving the burner itself and any differences in kinetic
and potential energies are both negligibly small. So what we're left with is the rate
at which heat is entering the burner, plus the rate at which enthalpy is entering the
burner is equal to the rate at which enthalpy leaves the burner, so again we'd expect, if
you look at it for an ideal gas, we say Qin is equal to mCp times the increase in temperature
as the air exits the burner itself, and in that case if we know Qin we can calculate
T3 the temperature at the exit of the burner just by rearranging this algebraically, we
do know Qin because Qin is the rate at which the engine is consuming fuel multiplied by
the heating value of the fuel, so here is the chemical energy associated with burning
the fuel itself and if I plug in numbers for this there is 80,000 kilojoules per second
of heat entering the burner and that heat is used to heat the air from point 2 to point
3, and we come up with a temperature exiting the burner of a little over 1200 Kelvin. So
we're working our way though this problem, we know the temperature leaving the burner,
lets focus attention on the turbine itself and lets do an energy balance around that,
so here's the simplified form of the general energy balance equation, no work in, no heat
in, differences in kinetic and potential energy are negligibly small so we've got enthalpy
entering the turbine is equal to the rate at which shaft work is leaving the turbine
plus the rate at which enthalpy is leaving the turbine in the simplified form we can
say that the rate at which shaft work is leaving the turbine is equal to the mass flow rate
of air times it's heat capacity times it's difference in temperature, so we expect a
positive value for the rate at which work is leaving and in that case we would expect
T3 to be greater than T4, we'll see a temperature drop across the turbine as shaft work is being
extracted from it, and another thing to consider is the fact that all of the shaft work leaving
the turbine is going into the compressor itself, we can use the relationship the rate at which
work is leaving the turbine equals the rate at which work is entering the compressor and
plugging in relationships, again energy balances on both, we'll come up with the differences
in temperature based on the mass flow rate of air and the heat capacity, and if we plug
in numbers for these, solve for T4 the temperature leaving the turbine at point 4 and I can calculate
a value because I know all the upstream temperatures I can calculate the temperature leaving the
turbine is a little over 800 Kelvin, and this I might add is a very important thing to recognize
for jet propulsion cycle. Another important thing to recognize is that the expansion through
both the turbine and the nozzle is isentropic, so we've got a perfect turbine and effectively
a perfect nozzle, so the expansion between 3 and 4 is isentropic expansion results in
shaft work leaving it and the expansion between 4 and 5 results in the conversion of enthalpy
at point 4 into kinetic energy at point 5, which we're really interested in for the design
of this engine, so what that says is that the entropy at state 5 is equal to the entropy
at state 3 and for an ideal gas we can use this relationship so find the temperature
at point 5 as a function of the pressure ratio so plugging in numbers we get a final value,
the temperature at point 5 of the exhaust is 526 Kelvin, so we see a temperature drop
between 3 and 4 because shaft work is being removed and a temperature drop between 4 and
5 because we're generating kinetic energy at the expense of internal energy at point
4. Even though the expansion is isentropic between 4 and 5 I didn't use a relationship
there because we don't know the pressure at point 4, all we know is that the pressure
at point 5 is 20 times less than the pressure at point 3, so we do know that pressure ratio
so that allows us, based on the fact that we know the temperature at point 3 to find
the temperature at point 5. Now lets do an energy balance around the nozzle, there is
no shaft work entering or leaving the nozzle, the nozzle, lets say it's adiabatic, there's
no heat entering or leaving, let's say that the differences in potential energy are zero,
so we know the velocity leaving the nozzle is a very high velocity and lets say the velocity
at point 5 is so much greater than the velocity at point 4 that the velocity at point 4 is
essentially zero, so there's very little kinetic energy at point 4 as compared to the kinetic
energy at point 5, so here's the simplified form of the energy balance, note that the
mass flow rates of air cancel out of this expression converting kinetic energy at the
expense of enthalpy, so here I'm solving for V5, and if I plug in numbers I come up with
a velocity close to 800 meters per second, and note the conversion here of 1000 joules,
this seems to be a common mistake, students will forget the fact that this is kilojoules
per kilogram Kelvin we need to convert to joules. So we're almost done with the problem,
we do know the force of the thrust is equal to the mass flow rate of air times the velocity
at point 5, and plugging in numbers we come up with the force of 117 kilonewtons or a
little over 26000 pounds, which is a pretty typical force for a turbojet engine, so finally
we'll find the acceleration, this 117,000 Newtons divided by the mass of the airplane,
we come up with an acceleration of a little over 2 meters per second squared and for airplanes
you often hear the acceleration in terms of G-forces so I'll take this value and divide
by gravity and I come up with an acceleration in the forward direction of 0.24 Gs.