# Jet engine thermodynamic non sequitur.

1. Jul 5, 2007

### parsec

Hi all,

This has been bugging me all day, so I thought I'd seek the help and advice of the internet.

When a plane is stationary, you’d have to say that its jet engines are doing no useful work. The engines apply a static force on the airframe and could be said to be running at zero efficiency (ignoring the fact that they generate shaft power and compressed air for the plane’s auxiliary systems). I guess this means that the hot exhaust gases ejected from the rear of the engine lose all of their energy through heating up the surrounding atmosphere (entirely plausible).

Now consider a plane taxiing at a slow velocity at the same engine speed (rpm). The engines are doing useful work, and the propulsive power it is producing could be said to be the product of the plane’s velocity and frictional force the plane’s airframe sees; P=FV (in the tyre-tarmac interface and drag).

What causes this dramatic increase in efficiency? Mechanically, the same scenario is easy to visualize and rationalize in a car. When the engine is decoupled from the drivetrain (in neutral), the car revs freely, and not much air and fuel (accelerator pedal excursion) is required to redline the car. Here, the work done by the engine is used purely to overcome piston friction and other mechanical losses. When the car is coupled to the drivetrain, much more air and fuel is needed to speed the engine up as the energy is consumed by accelerating the vehicle.

From the jet engine’s point of view, the only thing that has changed is the velocity of the intake air. When this intake air is no longer stationary, the engine seems to do useful work. It is hard to visualize how this works.

A simpler related problem involves the propulsion of an inflated balloon. If the balloon is held static while it deflates, all of its energy is lost in heat and noise. By simply allowing the balloon to be unrestrained, and hence achieve a non zero velocity relative to the surrounding air, the balloon seems to do useful work in the form of giving itself kinetic energy. How can simply allowing the balloon to be unrestrained so dramatically increase its propulsive efficiency?

Imagine the scenario from the jet engine's or balloon's point of view. All that has changed is the ability to propel yourself forward, but you are still applying the same force you were applying when restrained.

The implication is that when unrestrained, the environment or surroundings aren't as hot as a result of your jet engine or balloon exhaust, because some of this energy has gone into the plane's/balloon's inertia. I can accept this, but I don't see how restraint (or lack thereof) can cause the exhaust to reconfigure itself magically in this fashion.

It would seem at first glance that this lack of restraint allows the jet stream to create less turbulence and more ordered work in the form of propulsive thrust.

2. Jul 5, 2007

### Staff: Mentor

For the purpose of efficiency, the work done by a jet engine is measured from the mass flow rate of air out the exhaust. The plane may not utilize that work, but it is still being generated.

edit: also, it is true that idling engines are typically pretty inefficient. In an idling car engine, for example, all of the energy generated is lost to friction. This isn't any big deal.

Last edited: Jul 5, 2007
3. Jul 5, 2007

### K.J.Healey

Whats your "propulsive efficiency" equation that you're talking about? Is it a specific quantity?
Lets assume an inflated ballon. It has potential energy U. The work done by the system when the balloon is held in place while it deflates would be:

W = W_air + W_balloon

Well W_balloon = F*d, since no displacement == 0.

The AIR on the otherhand, is moving at a a velocity defined such that
W_air = U =>> Turns into complete kinetic energy of the air. The air moves really fast. All the balloon's potential becomes kinetic in the air.

Now DON'T restrain the balloon. What happens?

W = W_air + W_balloon

Now W_balloon isn't zero. Which means that :
U = W_air + W_balloon
W_air = U - W_balloon

As you can see, the work done on the AIR is much less. And this makes sense if you think about it relative to the balloon. If the balloon is sitting still its air displacement can be measure from either an observer or the balloon and its the same. But when the balloon gets moving, the balloon still sees the air moving out just as fast as before, BUT a stationary OBSERVER sees the air move much slower relative to the initial position of the balloon.
Think about the possibility when the velocity of the air leaving the balloon is equal to the balloon's speed.
A stationalry observer would see the air stay still, while the balloon flies off at 2*V, where V is the velocity of the air as witnessed by the balloon.

Work done is always the same as the potential energy created by the system.
It just depends on what you ask the work is being done on. (oh man thats so grammatically errored)

4. Jul 5, 2007

### rcgldr

"Useful work" depends on what you consider as the work output of a jet engine. Even when "staionary", jet engines are performing work by accelerating air and burnt fuel. The result of this work is in increase in kinetic energy (relative to the surrounding air), and the thrust produced as a result of the acceleration of air and burnt fuel (there's also a big increase in temperature).

Taxing on the ground isn't going to be much more efficient that standing still. However a jet engine's efficiency does improve with speed and altitude. At speed, ram air effect supplies the intake with higher pressure air. At higher altitudes, air temperature and density lowers, which also helps.

Most of the losses are heat related, not friction. The heat losses go into heating the cooling systems, and the heated air out of the exhaust.

As I mentioned, the velocity of intake air produces a ram air effect that help, but your point is about useful work done by the thrust of a jet engine. In this case, thrust (force) times distance is the work done. Power would be thrust times air speed. Note that as speeds increase, power increases, as maximum thrust doesn't decrease with speed (within reason). This is just an issue of what you consider to be the work done by a jet engine, the acceleration of air and burnt fuel, or the thrust times the distance moved.

Last edited: Jul 5, 2007
5. Jul 5, 2007

### FredGarvin

The unrestrained scenario simply allows you to use the definition of work with respect to what is outside of the engine's control volume. The engine is always producing work. Whether you utilize that work does not have any bearing on it.

Technically in static conditions, even the definition of thrust horsepower causes one to conclude that there is no work being done. However, we know that there is always the $$\tau \dot{\theta}$$ power being produced. I find it analogous to the person pushing on a wall; they are exerting and using energy, even to the point of exhaustion, but doing no work.

6. Jul 5, 2007

### parsec

Everything everyone has posted makes sense so far. I'd like to quote individual posts but it'd take too much time.

The lengthy post involving the balloon makes sense, but what about at the instant the balloon is released, where the balloon has no forward momentum, and for all intensive purposes is doing the same amount of work accelerating the air as when it was restrained.

Also, does that mean a propulsive device is most efficient when its forward velocity equals the exit velocity of its air/jet stream?

I agree with this sentiment, however I can't get my head around the jet stream changing its properties due to the plane being unrestrained.

consider the following:

Say the engine is supplied with so much fuel with a thermal content of P joules/sec or watts.

In the static scenario, ignoring friction in the engine and the auxillary systems, the jet stream exhausted out of the rear of the engine delivers Q watts of heat to the outside atmosphere. A lot of this is in heat, the rest is noise and the high speed jet stream shearing against the motionless surrounding air. We can say P=Q in this scenario, and all of the power put into the engine results in heating the atmosphere.

Now release the plane's brakes. Some of the power P must be invested in building up the plane's inertia, so Q can no longer equal P. P = Q + U (power invested in moving the plane) From the jet engine's point of view, its jet stream must be losing less energy to the atmosphere. It is still being supplied with P watts of fuel, but now its jet stream has suddenly reconfigured itself so that P - U energy is exhausted into the atmosphere.

Sorry if that's a little scatterbrained. Hope it makes some sense. Clearly I'm slightly confused :)

7. Jul 5, 2007

### rcgldr

Which means work is peformed on the air that is accelerated (ignoring the heat aspect).

I think you're missing the point of my post. Aircraft engines perform work on air, regardless of whether the aircraft is moving with respect to the air or not. So even in a "static" situation, the power of a jet engine, based on the rate of work peformed on the air can be determined, and it's not zero.

The comparason to pushing on a wall is not a good one, because the net result of the forces is to compress the wall and what ever is pushing the wall a small amount, and there's no movement (except for the initial compression reaction to the application of forces). In the case of a jet engine, a force is applied to a mass of air over a distance as the air accelerates, and that translates into real work being done, and in turn the cacpacity to accelerate the air can be converted into power.

Last edited: Jul 5, 2007
8. Jul 5, 2007

### parsec

sure, work is being performed on the air. i agree with this, but surely when the plane is allowed to move, less work is performed on the air as some of the work must be used to actually move the plane.

i don't understand what part or the jet engine or its interaction with the air changes to allow some of this work to be invested in the plane's inertia.

9. Jul 6, 2007

### rcgldr

The jet stream output of a "static" "typical" military type jet engine is around 1300mph. Assuming it remains at around 1300mph, then as the aircraft speed increases, then the acceleration of air to 1300mph represents less work done and there would be less thrust. However military jet engines have nozzles, there's a ram air effect that increases intake air pressure, and at altitude the air is much colder, so a high rate of thrust can be maintained at high speeds and high altitudes. There are some jets that can reach mach 1.5 at 200 feet above sea level (F-111). The SR71 uses very high speed jet exhaust, uses a movable cone in front of the intake to allow the supersonic shock wave to help compress intake air, and also uses partial compressor stage bypass tubes to feed air directly to the combustion chamber to run at Mach 3 (2000+mph), at about 30% "throttle" (top speed is restricted the temperature the aircraft can take due to air friction). Obviously, the engines on a SR-71 have an exhaust speed well over 2000mph.

The point of all of this is that jet engines have the ability to maintain thrust (and work done on the air) at higher speeds because the configurations can be change dynamically (nozzles, inlet valves, bypass pipes, moving cones, ...).

Another wierdness of jet engines, is that the rate of fuel consumed verus distance traveled remains about the same regardless of aircraft speed.

A jet engine accelerates air backwards, and the air responds to this backwards acceleration with a forwards reaction force that is the "thrust" that moves or accelerates the aircraft forwards.

Last edited: Jul 6, 2007
10. Jul 6, 2007

### FredGarvin

That is somewhat true. Actually it is a combination of a little less work performed plus less wasted kinetic energy of the jet. That is why, if you look at a plot of propulsive efficiency vs. airspeed, it pretty much increases up to around 85% (depending on altitude).

I guess all I can say to this is that energy is not being wasted restraining and thus wasting the energy provided by the jet. At zero forward speed, the propulsive efficiency is zero because of the definition of the efficiency. Remember, propulsive efficiency is the ratio of the useful propulsive power divided by the sum of that energy PLUS the wasted or unused kinetic energy of the jet. BY DEFINITION, if the aircraft is at a standstill, you are wasting 100% of the provided kinetic energy of the jet and thus your propulsive efficiency is 0. The jet doesn't really care one way or another what is going on around it. It's going to grab a gulp of air, compress it and throw it out the back. How much of that work we grab and how much is wasted gets rolled into that efficiency number.

$$\eta_{propulsive} = \frac{2}{1+\frac{V_{exhaust}}{V_{aircraft}}}$$ or if you prefer the long winded version...

$$\eta_{propulsive} = \frac{m V_a(V_j-V_a)}{m\frac{\left[V_a(V_j-V_a)+(V_j-V_a)^2\right]}{2}}$$

You can think about it this way when it comes to the propulsive efficiency:

- $$F$$ is maximum when $$V_a =0$$ however $$\eta_p =0$$

- $$\eta_p$$ is maximum at $$\frac{V_j}{V_a}=1$$ but then $$F = 0$$

Last edited: Jul 6, 2007
11. Jul 12, 2007

### parsec

Thanks. I think I understand what I was misunderstanding. The concept seems more intuitive if I think about the useful thrust being generated through a momentum exchange mechanism. I guess that means that when the aircraft is travelling at the same speed as the exhaust, the air remains relatively undisturbed and all of the work goes into propelling the aircraft.

Presumably your propulsive efficiency equation does not apply to the case where the exhaust is slower than the aircraft? (I guess no useful thrust can be produced in this scenario).