Please explain how.....?in a by-pass turbo-jet engine

[yawskidball]

Reading through some notes on how a bypass turbo jet engine produces thrust and 'hit a brick wall' in a paragraph that stated......"efficient mixing of the two streams in the exaust gas flow gives a cooler and therefore lower velocity jet efflux which improves propulsive efficiency."

Could you please explain how the cooler slower jet efflux improves propulsive efficiency here?

:ok:

[BigHairyBum]

I think the increased efficiency is caused by the reduced shear zone. Slower air from the big bypass causes less shear when meeting the outside ambient air at the back of the engine. Should the bypass not exist, a pure jet will cause rapidly moving air (much hotter too) to shear with the relatively static ambient air. In other words, air flow around the back of the engine is smoother.

Noise is much less on a bypass engine too. Probably for the same reason.

There may of course be other reasons for the increased efficiency, I could only remember this one.

[john_tullamarine]

.. on the noise side, if memory serves me correctly, the relationship is velocity to the eighth power .. hence high bypass engines, for similar thrust ratings, are much quieter than low bypass.

[Mr Levitator]

The efficiency bit comes from the fact that it is more efficient to move a lot of air a little instead of a little air a lot. Therefore if the core exhaust is mixed with the bypass, the bypass exhaust speed is higher and the core ehaust speed lower. Since there is more bypass exhaust than core, more thrust is developed.

Talking about noise, when the core/bypass exhaust is mixed this produces a VERY turbulent flow. Instead of having two shear zones as in a un-mixed exhaust system, there is now only one in a mixed exhaust system. This can actually be louder, it is all in how the two are mixed....you may have seen the wavy exhaust cowls/guides on the newer turbo fans (Trent 700??). These induce a series of votices which help spead the shear zone out over a larger cross section. This increase in shear zone size is what will reduce the noice output.

Mr L.

[pipex]

A turbojet engine produce thrust solely by the high acceleration of air as it passes through and exhausts the core engine.
A turbofan accelerates a larger mass of air than a turbojet,the bypass air generates quite a bit of thrust due to the highly efficient design of the turbofan blades.At lower altitudes more than 80% percent of a high bypass engine's thrust comes from the shrouded fan,resulting in greater fuel efficiency.

[Keith.Williams.]

I think that we need to be a llittle bit careful here. The improved propulsive efficiency in a by-pass engine is not caused by the reduced noise. It is more true to say that the noise is reduced because by-pass engines give less kinetic energy to the air, so there is less energy available to make the noise. The improved propulsive efficiency of by-pass engines is caused by the fact that they give less kinetic energy to the air.

To understand why mixing the exhaust also improves propulsive efficiency we need to look at what propulsive efficinecy is all about and what factors affect it.

An engine takes in chemical energy if its fuel and converts this into thermal energy (heat) and mechanical energy (pressure). The ratio of thermal + mechanical energy out divided by chemical energy in is the thermal efficiency of the engine. None of this has anything to do with propulsive efficiency.

An aircraft engine must then convert this energy into kinetic energy and transfer it into the airframe. Propulsive efficiency is the ratio of energy given to the airframe divided by the thermal and mechanical energy extracted from the fuel. The principal inefficiency in tranferring the energy from the engine to the airfame is caused by the loss of kinetic energy into the exhaust gasses. So to maximise our propulsive efficiency we must minimise the amount of kinetic energy that is left in our exhaust gasses when they leave the engine.

The kinetic energy possessed by the exhaust gas is equal to half of its mass multiplied by the square of its velocity. But in flight, the air is already moving when it enters the engine, so it already has some kinetic energy. This means that the kinetic energy that we have lost to the air is equal to its kinetic energy as it leaves the engine minus the kinetic energy it had when it arrived at the inlet. This means that kinetic energy lost to the air is equal to

1/2 mass airflow x (exhaust velocity squared minus inlet velocity squared)

This means that if we are to minimise the enregy lost to the airflow we must either minimise the mass airflow, or minimise the difference between the inlet velocity and the exhaust velocity.

But the whole point of the exercise to to produce thrust and thrust is equal to mass airflow multpilied by the acceleration that the engine gives to that airflow.

So to maintain any given value of thrust, if we reduce the mass airflow we must increase the acceleration. Or if we reduce acceleration we must increase the mass flow. But increasing acceleration increases the difference between exhaust velocity and inlet velocity. This difference is squared in the kinetic energy loss equation, whereas the mass flow is not. So it is more efficient to use a large mass airflow and give it a small acceleration.

All of the above can be encapsulated in the equation below:

Propulsive efficiency = (2 x TAS) / (TAS + Exhaust velocity) x 100%

For 100% propulsive efficiency the exahust velocity must equal the TAS. But this will give zero thrust, so we can never achieve 100% propulsive efficiency when we are generating thrust.

We can however maximise our propulsive efficiency by arranging for our exhaust velocity to be as close as possible to our TAS. This of course means a corresponding increase in the mass airflow passing through the engine. So for maximum propulsive efficiency we must enusre that our mass airflow is sufficiently large to give us the required thrust using only a small acceleration. It is this process that gives by-pass engines their improved propulsive efficiency.

Now to get back to our original scenario we can examine the efffect of mixing the core and by-pass gas flows.

let us start by assuming the following:

Inlet (total) mass airflow (Minlet) = 2 units per second. This is then divided into 1 unit of core mass flow (Mcore) and 1 unit of by-pass flow (Mby-pass).

Inlet velocity (Vinlet) = 100, Core exhaust velocity (Vcore) = 300 and by-pass exhaust velocity (Vby-pass) = 200.

Thrust = mass x acceleration so:

Core thrust = Mcore x (Vcore - Vinlet) = 1 x (300 - 100) = 200
By-pass thrust = Mby-pass x (Vby-pass - Vinlet) = 1 x (200 - 100) = 100.

So our total thrust = 200 + 100 = 300 units of thrust.

Kinetic energy = 1/2 m x (velocity squared)

So kinetic enrgy in core exhaust = 1/2 x Mcore x (Vcore squared) = 1/2 x 1 x (300 squared) = 45000 units.

And kinetic energy in the by-pass exhaust = 1/2 x Mby-pass x (Vby-pass squared) = 1/2 x 1 x (200 squared) = 20000 units.

So the total kinetic energy in the exhaust gas = 45000 + 20000 = 65000 units.

But when the air entered the engine inlet it had kinetic energy = to 1/2 Minlet x (Vinlet squared) = 1/2 x 2 (100 squared) = 10000 units.

So the kinetic energy lost to the air = 65000 - 10000 = 55000 units.

This means that in generating 300 units of thrust we are losing 55000 units of kinetic energy per second.

Now consider the effect of mixing the core and by-pass gas flows to produce a common mixed exhaust velocity (Vmixed). Using the same mass flow at the inlet (Minlet) = 2 units, which will also be the mass flow of the mixed exhaust (Mmixed), and the same inlet velocity (Vinlet) = 100.

Thrust = mass x acceleration = Mmixed x (Vmixed - Vinlet)

Rearranging this equation gives

Vmixed = (thrust / Minlet) + Vinlet = (300 / 2) + 100 = 250.
So the velocity of our mixed exhaust gas is 250.

This means that the kinetic energy in this mixed exhaust =

KE mixed = 1/2 Mmixed x (Vmixed squared) = 1/2 x 2 x (250 squared) = 62500 ubits.

Minlet and Vinlet have not changed so the kinetic energy of the inlet air is still 10000 units, so the kinetic energy lost = 62500 - 10000 = 52500 units.

This means that we are now generating the same 300 units of thrust but we are now losing only 52500 units of kinetic energy per second compared to our initial loss rate of 55000 units per second. This reduced energy loss equates to an improved propulsive efficiency.

[yawskidball]

:cool: Thank you one and all for your interest and help. I can quite confidently say that I am in a better understanding of the topic. It sure makes learning an enjoyable experience with you guys (:D ;) :p :8 :ooh:) around!

Cheers

[unruly]

Keith,

Just one word....WOW! :ok:

[DoctorA300]

Keith,
Another word, you need to read more playboy and less P&W books. :8 :8 :E
Brgds
Doc

[Keith.Williams.]

Actually I only own one rather old P&W book......

But I have thumbed the pages rather a lot!!!


Now the RR book? Well that is another story....So many pictures!!!

[cribble]

:D
K.W
I saved the page to disk. If I use the argument i will be sure to attribute.