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172_driver
26th Sep 2008, 05:14
Any kind soul able to explain the advantages of having a convergent-divergent nozzle at the back of the jet engine? I understand that the convergent section extracts the final potential energy (pressure) in the flow, accelerates the air and leaves it at ambient static pressure.

Then the divergent part, I don't quite understand. Does it have anything to do with the nozzle is choked at normal thrust levels?

Thanks in advance :ok:

Mark1234
26th Sep 2008, 06:54
Only thing I know on the subject is that convergent-divergent nozzles are of benefit when the gas flow is supersonic (wrt the local speed of sound for the hot gas).. something to do with the different behaviour of supersonic gas flow vs subsonic.

I'll get back to reading that bit later :)

Brian Abraham
26th Sep 2008, 07:13
See if this answers your question De Laval nozzle - Wikipedia, the free encyclopedia (http://en.wikipedia.org/wiki/De_Laval_nozzle)
A handy site to explore for these sort of questions Short Index of Propulsion Slides (http://www.grc.nasa.gov/WWW/K-12/airplane/shortp.html)

172_driver
26th Sep 2008, 23:16
Yepp... it did give some insights. Thank you!
But I understand this is actually quite advanced mathematics and nothing a humble pilot is ever taught. :8

Practically, do turbofan engines (i.e. CFM56) today utilize the convergent-divergent nozzle, and is the flow supersonic out of the engine supersonic at anytime during the flight?

Milt
27th Sep 2008, 02:34
If the air comes out the back at the same speed as it comes in at the front then there is no thrust - or am I missing something?

Brian Abraham
27th Sep 2008, 05:02
172_driver, this should put you in the picture Propelling nozzle - Wikipedia, the free encyclopedia (http://en.wikipedia.org/wiki/Propelling_nozzle)

Keith.Williams.
27th Sep 2008, 17:04
I'm afraid that this is rather long (as the bishop said the actress) but it is much easier on a board.

When discussing the behaviour of air flowing at low speeds through convergent and divergent ducts we usually assume that.

a. The air is flowing in a steady streamline manner.
b. The mass flow rate of air in Kg/second or lb/second is constant at all points along the tube.
c. The density of the air is constant.
d. The total pressure (Static pressure + Dynamic pressure) is constant at all points along the tube.

The assumption of constant density means that to maintain constant mass flow rate at all points along the tube, we must maintain constant volume flow rate.

Volume rate at any point along the tube is equal to the cross-sectional area of the tube multiplied by the velocity of the air.

So if for example the cross-sectional area of the tube becomes half its initial value, the velocity of the air must become twice its initial value. And if the cross-sectional area become ľ of its initial value, the velocity must become 4 times its initial value.

Now imagine that we have a tube in which the cross-section of the inlet is constant and the cross-section of the outlet can be varied. We have a variable speed fan driving air into the tube and we have instruments at each end to measure the air velocity, static pressure and temperature.

Letís assume that we want to carry out some experiments to see how these variables change, when we adjust the velocity of the incoming air and the size of the outlet. In conducting these experiments we will adjust the fan rpm to give the required air velocity at the tube inlet, and we will adjust the cross-sectional area of the outlet end to control the velocity of the air at the outlet. Now letís suppose that in each experiment we want the outlet airspeed to be 50 units higher than the inlet airspeed.

We adjust the fan so that the airspeed at the inlet of the tube is 50. To have an increase of 50 units as the air flows through the tube, the airspeed at the outlet must be 100 units. This means that the airspeed at the outlet is twice that at the inlet. So we adjust the cross-sectional area of the outlet to Ĺ that of the inlet. If our instruments are reasonably accurate they will show us that we have the required 50 units of velocity at the inlet and 100 units of velocity at the outlet.

If for the next experiment we increase the fan rpm to give us 100 units of airspeed at the inlet, we will find that because the area of the outlet is still Ĺ of the area of the inlet, the airspeed at the outlet is 200 units. We actually want only a 50 unit airspeed increase giving 150 units of airspeed at the outlet, so the outgoing air is now too fast. To reduce it to the required 150 units we increase the cross-sectional area of the outlet and watch the airspeed at the outlet gradually reduce. When we have achieved the required 150 units of airspeed at the outlet, we will find that the cross-section of the outlet is 2/3 that of the inlet.

Provided the density of the air and the mass flow rate both remain constant, we will find that the ratio of the inlet airspeed to the outlet airspeed is the inverse of the ratio of the inlet cross-sectional area to the outlet cross-sectional area. To double the speed (from 50 to 100) we halved the area. And to increase the speed by a factor of 3/2 (from 100 units to 150 units) we reduced the outlet area by to 2/3 of the inlet area.

This suggests that for any inlet airspeed, if we want to accelerate the air flowing through a tube, the outlet area must always be less than the inlet area. For example with 1 000 000 units of inlet airspeed to obtain 1 000 050 units of outlet airspeed, the outlet area must be 1 000 000 / 1 000 050 of the inlet area.

But if we continued to carry out the experiments described above, gradually increasing the inlet airspeed by 50 units, we would find that this assumption begins to break down. If in each experiment we adjusted the area of the outlet in accordance with the inverse ratio law described above, we would find that the airspeed at the outlet of the tube would be higher than expected. To compensate for this we would need to open the tube out a bit more than we had predicted.

This is because the density of the air is not constant. Air is compressible and any decrease in static pressure will allow it to expand. This will reduce its density and increase the cross-sectional area that is required to maintain constant mass flow through a tube.

When the velocity of air increases, its dynamic pressure also increases. This causes its static pressure to decrease to maintain constant total pressure. This decrease in static pressure causes the air to expand, thereby decreasing its density. Because this accelerated air has become bigger, it requires a greater cross-sectional area to maintain constant mass flow rate along the tube.

The rate at which the air expands is determined by the rate of decrease in static pressure. This in turn is determined by the rate of increase in dynamic pressure. Dynamic pressure is proportional to the square of the velocity. So the rate of change increases exponentially as velocity increases. For a given increase in velocity at low and of the velocity range there will be very little change in volume and density. But at the high end of the velocity range the same increase in velocity will give a much bigger change in volume and density.

You can demonstrate this by sketching the curve for dynamic pressure (on the vertical scale) against velocity (on the horizontal scale). Mark off the velocity scale into small sections of equal length. The rate at which dynamic pressure increases is approximately equal to the rate at which the static pressure decreases and the volume of the air increases. So the curve gives an approximation of the rate of expansion of the air. The rate of expansion is very small at the low velocity end, but increases at an increasing rate as you move towards the high velocity end.

So for PPL studies it is OK to assume that density in a venturi is constant because the changes are very small at low speeds. But if we want to study airflows at much higher speeds we must take account of the increasing expansion rates.

If we continued our experiments we would reach a stage at which we needed a parallel tube to give the required 50 units increase in airspeed. In theory this will occur when the air is flowing at the local speed of sound. In this condition the rate of expansion exactly matches the rate of acceleration, so that although the air is accelerating, its increasing volume and decreasing density mean that the mass flow rate is constant.

But as the air continues to accelerate it will start to expand at a rate that is greater than its rate of acceleration. This means that to enable it to accelerate we must open out the tube making it divergent, in order to accommodate the increasing volume of the air.

If we were to carry out these experiments over a wide speed range we would find that.

a. A convergent duct will accelerate air only up to the local speed of sound.
b. A divergent duct is required to accelerate air at supersonic speeds.

In theory a parallel duct will accelerate sonic air because the rate of expansion will exactly match the rate of acceleration. But as soon as sonic air accelerates, it becomes super-sonic and requires a divergent duct in order to accelerate.

In conventional jet engines we create high pressure in the jet pipe to accelerate air through a convergent propelling nozzle. As the air accelerates through the nozzle its static pressure decreases. Under ideal circumstances this process ensures that the static pressure of the exhaust gas is equal to ambient pressure as it emerges from the end of the nozzle. This means that all of the excess pressure in the jet pipe has been used up in accelerating the air and producing thrust.

If we open the throttle further we will increase the pressure in the jet pipe. This in turn will increase the velocity of the air coming out the nozzle. But a convergent duct can accelerate air only up to sonic speed. If we then increase jet pipe pressure even more, the outlet velocity will remain sonic, but the air will still have some residual excess pressure when it emerges from the nozzle. This will then be wasted as the air expands in all direction behind the aircraft.

If we want to use this excess pressure to accelerate the air to supersonic speeds, we must use a divergent nozzle immediately behind the convergent nozzle. Provided we have sufficient excess pressure in the jet pipe, the air will be accelerated up to sonic speed in the convergent section and it will continue to accelerate up to supersonic speed in the divergent section.

In considering all of this we should also note that because of the high temperature of the exhaust gas, the local speed of sound in the propelling nozzle is much higher than that in the surrounding atmosphere.

172_driver
27th Sep 2008, 21:19
Keith, that was excellent. Think I got it down now. :)

Brian, I do appreciate the links as well. They put me in the right direction but did not answer the specific questions I had in my mind. Thank you anyway.

dazdaz
27th Sep 2008, 21:33
Might this link help.......
HTWPhysics (http://www.aoxj32.dsl.pipex.com/NewFiles/HTWPhysics.html)