Jason Hi

Lets take your scenario of two identical airframes with the two different 20lb static thrust engines.
Lets say that at 200 mph the airframe drag is 20lbs, thrust equals drag and in that steady state my understanding is that the only way you could go faster in this scenario would be more thrust (yeh less drag as well for any pedants but then it wouldn't be the same airframe[:-])
If I understand correctly the issue of EGV and dynamic thrust would come into play in a different scenario....

Scenario two

Two identical airframes this time very very slippery, same 20lb engines as above and as per your scenario the two different exhaust gas velocities 400 & 1050 mph, I suspect that as the two airframes approach 400 mph the engine with the EGV of 400 mph is producing next to no effective (dynamic?) thrust and therefore stops accelerating. The second airframe equipped with the engine with higher EGV will continue to accelerate until either the total drag equals the thrust or until in this mythical case the dynamic thrust is zero, at I assume the EGV of 1050 mph.

Only my thoughts......

Phil.

Phil

I hear what you are saying but surely the amount the engine is pushing the airframe will be reducing from 1mph to 400mph and not go from 20lb at 399mph to 0lb at 400mph. So could we say at 200mph the thrust of the 400mph EGS engine could be 10lb dynamic thrust and the 1050mph engine could be 16lb ish dynamic thrust? I know these figures don't take drag in to account but you get the idea.

jason

Spot on!

Thrust is thrust but thrust is related to area only in the respect that you need to increase efflux velocity on smaller turbines to compensate for reduced area.

EG: Lets say we have a 12 square inch area efflux, our exhaust velocity is 200 mph

This produces thrust of "X"

Our smaller turbine has only 6 square inches, so if the efflux velocity stays the same the thrust = X/2 (half X)
To create the same thrust we will have to double the efflux velocity to 400 mph

Both give the same thrust (static) but once moving the lower efflux velocity loses its 'pushing power' earlier because the air it is pushing against is moving in the same direction.
So in level flight the plane driven by the lower efflux velocity cannot exceed 200mph (ignoring drag).

In the same level flight the higher efflux velocity plane can exceed 400mph (ignore drag again).

Taking drag into consideration, the 200mph version will get closer to 200mph than the 400mph version will to 400mph due to the square rule of induced drag. It will however (on a like for like airframe) exceed the 200mph version by an appreciable figure.

The acceleration of the larger efflux will initially be higher due to the larger area to ‘push against’ but it will drop off much quicker than the smaller one.

Remember, a jet is not a reaction motor such as a rocket; it has to push against something to function, (air) if that air is moving at the same speed as its efflux velocity then there is no effective thrust!
Paul

Sorry to disagree, by your thinking it would seem that a turbojet is more efficient than a turbofan, and we all know that it is not the case.

From Wikipedia: (http://en.wikipedia.org/wiki/Propulsive_efficiency)


Gaspar

Gaspar

so asuming the two planes in the following senario are identical apart from the engine, which do you think would fly the quickest, plane A or plane B or would they both do the same speed?

A. 20lb static thrust with 1050mph exhaust gas velocity

B. 20lb static thrust with 400mph exhaust gas velocity

Jason

I think you better re-think that statement. Our jets ARE reaction engines. The difference is that the jet gets O2 from the air and the rocket carries its own.

http://csep10.phys.utk.edu/astr161/l...wton3laws.html

Force is force.

With the caveat that I am NOT an engineer, and with all due respect to Paul, I simply don't accept this premise. Could somebody help to enlighten (prove or disprove this premise) for my already over-extended brain? Thanks,

If you look at the link I posted you'll find your answer So far as I know, Newton is still right.

Exactly correct. Turbojets are reaction motors and the total thrust is a function of the combination of its exhaust area and exhaust velocity.

Well, at least in a non-relativistic frame of reference.

Well since he's long dead I'm sure he won't mind LOL.

All I know is that I had an AMT NL Mercury HP that was rated at almost 20lbs of thrust and my buddy had a P-80 that was rated at 18lbs of thrust...My engine had to spool up to 151 K to reach max thrust and the P-80 was much lower, about 119 K...With the engines mounted in the EXACT same planes (HotSpots) and weighed the same, my buddy's plane took off much faster and had more vertical...Although both engines were rated about the same I believe that his JetCat put out more volume and mine put out more velocity and that volume beats velocity in this case...The performance difference was amazing...

Kevin

There is a popular misconception that a jet engine has to push against something to give thrust, in fact the thrust is pure reaction felt within the engine, exactly the same as a rocket engine.

If you could give the jet engine a supply of air in space it would produce thrust the same with nothing to push against.

Thrust is efflux mass x velocity.

Rob.

As others have already pointed out Paul, a jet engine is a reaction engine. It does not work by pushing the exhaust against the air outside the back of the engine. The greatest part of the thrust is reacted against the combustion chamber front wall as reaction to the increase in momentum imparted by the heating of the air, the second greatest part of thrust is created by the compressor. Together they create several times more thrust than the net thrust of the engine. The airflow aft of the combustion chamber does not produce thrust except in a small amount if you have the correctly designed divergent pipe, otherwise all flow aft of the combsution chamber front wall produces a rearward force that deducts from the forward force produced in the compressor and combustion chambers. Even an afterburner does not create a forward force as such, its effect is to reduce the rearward force in the jet pipe.
H.

The jet engine simply throws molecules (mass) rearwards (velocity) and the reaction to this is what drives the engine in the opposite direction.

One popular way to describe this is this I learned at school: you stand on a wheeled cart that has a pile of bricks on it. As you start throwing some bricks rearward off the cart as hard as you can, the reaction of force required to throw them will slowly push the cart forward.
To maximize the obtained speed of the cart you can throw more bricks per second or increase your throw velocity or use bricks with more mass.
With this analogy we can clearly relate to a "top speed" being reached for the cart relative early as a person cannot throw that very fast.

So even for a reaction type propulsion there is a definitive top speed just as a DF or prop.

The rpm of the turbine can be left out of the equation, its just a result of other internal design parameters such as blade area and other geometrical dimensions, flow efficiency, sealing between the stages and EGT to name a few. The same goes for fuel consumption i.e efficiency of the total design

my two cents

What would a plot of thrust vs. forward air speed look like for one our motors?

The answer to your question is not so simple but look here for a good explanation, http://home.anadolu.edu.tr/~mcavcar/.../Jetengine.pdf

It's not quite as easy as it sounds, though for the speeds at which we operate models it is fairly straightforward. The net thrust, exhaust velocity minus intake velocity falls in pretty much a straight line from static value at no airspeed to zero thrust where airspeed equals exhaust speed. So for example if static thrust is 20lbs and exhaust speed is 800mph, net thrust at 400mph would be expected to be 10lbs. But, there is another effect, that of the intake ram air effect which does much of the compressor's work thus freeing up much of the energy to be used as thrust instead of driving the compressor, thus increasing the thrust. This starts at zero with zero airspeed and rises exponentially as the airspeed rises. The net thrust is the combination of these two effects. Initially the thrust falls as forward speed reduces it and the ram air effect has little power. As speed rises the thrust contrinues to fall but the ram air effect becomes more powerful and eventually its increase matches the decrease caused by exhaust:airspeed difference and the thrust has reached its minimum point. Above that speed, the ram air effect rising exponentially is going up faster than the straight line decrease from exhaust speed is going down, and thrust rises again back up to and even over the static level.

At slow speeds therefore the reduction in thrust will be almost straight line towards the zero point because the ram air effect is negligible. For most of the speeds at which our models operate then, the reduction in thrust will be fairly straight line, barely moderated by the ram effect.
H

Thanks Bob and Harry for the very useful info.

If ram effect is negligible at the speeds we operate, then it would seem from the above explanations that if motors A and B have the same static thrust, but A has a higher exhaust velocity, then A would have more available thrust than B at the upper end of our operating range - say 200 mph forward air speed (if you fly in the U.S.). Is that a generally true statement?

Yes, but the difference is a small value of pounds, and though it will increase the aircraft speed, at high speed the total drag curve will be dominated by the exponentially rising friction drag curve so the increase in speed will be small, in the magnitude of a handful of mph. A disadvantage is that, all else being equal, the higher exhaust velocity engine will burn fuel more rapidly to get the same thrust, so you get a shorter flight or start with more fuel = more weight = slower top speed!
You will certainly get ram effects at 150 - 200mph, but due to model design they will be less than on full size. Our ducts do not provide 100% of their air to the engine and ram it into the engine! Sometimes the engine is not in a duct at all, just sitting open in the fuselage, the airflow into the intake will be fast but will slow down inside the diverging fuselage and the engine itself sits in a much slower airflow. Even fully ducted installations in models still have quite a bit of bypass air around the engine. Engines out in the open may get more ram effect than an engine which is simply sitting unducted in the fuselage but still at these speeds it is not going to make a marked difference.

Thanks again Harry. Very enlightening.

Well, it would appear that I am wrong to some degree.
There is some ‘true’ reactive force involved, I stand corrected on that point.
However it is still true that higher exhaust velocity will give a higher top speed.

Paul

It is all reactive force Paul
H.

Newton is STILL correct and he's from the UK! Action/Reaction nothing more. In fact the Active force is the propellant and the Reactive force is your plane moving in the other direction. Now it doesn't get any simpler than that