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How Fast Does Air Travel Out Of A Turbofan?  
User currently offlineTriebwerk From United States of America, joined Sep 2008, 126 posts, RR: 0
Posted (5 years 6 months 1 week 6 days ago) and read 7367 times:

(Assume that the air is traveling out of the jet portion, not the fan portion. I'm talking about the kind of turbofan you would find on a 777, say, or an A380.

14 replies: All unread, jump to last
 
User currently offlineTdscanuck From Canada, joined Jan 2006, 12709 posts, RR: 80
Reply 1, posted (5 years 6 months 1 week 5 days 22 hours ago) and read 7348 times:



Quoting Triebwerk (Thread starter):
(Assume that the air is traveling out of the jet portion, not the fan portion. I'm talking about the kind of turbofan you would find on a 777, say, or an A380.

I've been told before that it's very close to Mach 1, although I haven't seen any hard numbers to back that up. It has to be higher than the aircraft's forward speed (typically around Mach 0.8-0.89) but, for noise reasons, wouldn't be supersonic so it's a fairly narrow band.

Tom.


User currently offlineTristarSteve From Sweden, joined Nov 2005, 3930 posts, RR: 34
Reply 2, posted (5 years 6 months 1 week 5 days 22 hours ago) and read 7338 times:



Quoting Tdscanuck (Reply 1):
It has to be higher than the aircraft's forward speed (typically around Mach 0.8-0.89)

Well now you made me think. What about when the aircraft is stationary on the ground, is the exhaust speed still M0.9?
What about Concorde? Travelling at M2.2, engine exhaust only M0.9?

I will go away for a think.


User currently offlineTriebwerk From United States of America, joined Sep 2008, 126 posts, RR: 0
Reply 3, posted (5 years 6 months 1 week 5 days 8 hours ago) and read 7223 times:

There has to be enough force from the air exiting the turbofans to propel the plane forward, of course. Perhaps the answer is as simple as finding the momentum of an accelerating plane, then finding out how quickly the displaced air would have to be traveling to equal that momentum.

(Weight of plane x speed) = (weight of air displaced x speed).

I'm no physics major, though, so I figure this question is best left to someone who knows what they're doing.


User currently offlineWPIAeroGuy From United States of America, joined Aug 2007, 240 posts, RR: 0
Reply 4, posted (5 years 6 months 1 week 5 days 7 hours ago) and read 7210 times:

Theoretically you want the air moving as close the forward velocity as possible. In practice it will always be moving faster, because if they were equal your mass airflow through the engine would have to be infinite. Treibwerk you're on the right track with the momentum, but I think you're going to have to account for the aerodynamic drag as well. To find the velocity its pretty simple, its T=mdot(u9-u0) where the u's are exit and inlet(aicraft) velocity and mdot is the mass airflow through the engine. The problem is finding the mass airflow, you need engine specific information to find that out. I'll agree with Tom in that its somewhere short of Mach 1 but I'm not sure.

Most supersonic aircraft use low-bypass turbofans because they need the high-velocity air to reach supersonic speeds. The larger the fan, the more air you move, but you move it slower for the same thrust. From the equation above, your thrust will be negative if your exit velocity is slower than your inlet velocity, therefore supersonic planes need thrust, but they must get it from a high exit velocity rather than greater mass airflow.



-WPIAeroGuy
User currently offlineThegeek From Australia, joined Nov 2007, 2638 posts, RR: 0
Reply 5, posted (5 years 6 months 1 week 4 days 18 hours ago) and read 7117 times:



Quoting Tdscanuck (Reply 1):
I've been told before that it's very close to Mach 1, although I haven't seen any hard numbers to back that up. It has to be higher than the aircraft's forward speed (typically around Mach 0.8-0.89) but, for noise reasons, wouldn't be supersonic so it's a fairly narrow band.

Wouldn't even the bypass air be hotter than the ambient air, so therefore it's supersonic speed would be faster than the aircraft's supersonic speed?


User currently offlineTdscanuck From Canada, joined Jan 2006, 12709 posts, RR: 80
Reply 6, posted (5 years 6 months 1 week 3 days 23 hours ago) and read 6976 times:



Quoting TristarSteve (Reply 2):
What about when the aircraft is stationary on the ground, is the exhaust speed still M0.9?

Should be, yes.

Quoting TristarSteve (Reply 2):
What about Concorde? Travelling at M2.2, engine exhaust only M0.9?

No, Concorde (and any other supersonic aircraft) has supersonic exhaust. It's subsonic at the back of the engine and accelerated to supersonic speed by a con/di nozzle. This is part of the reason you see variable nozzles on supersonic aircraft but almost never on subsonic.

Quoting Triebwerk (Reply 3):
Perhaps the answer is as simple as finding the momentum of an accelerating plane, then finding out how quickly the displaced air would have to be traveling to equal that momentum.

(Weight of plane x speed) = (weight of air displaced x speed).

This is the gist of it, but you really need to do it as (sum of forces) = (change in momentum). You pick up drag in the sum of forces and the change in momentum drives you to delta-speeds, not absolute speeds.

Quoting Thegeek (Reply 5):
Wouldn't even the bypass air be hotter than the ambient air, so therefore it's supersonic speed would be faster than the aircraft's supersonic speed?

Yes, fan air is noticably hotter than ambient, so it has a higher sonic velocity. However, for noise reasons I think you need to stay below ambient sonic, otherwise I think you'd induce shockwaves at the interface between the fan flow and ambient air.

Tom.


User currently offlineUAL Bagsmasher From United States of America, joined Sep 1999, 2143 posts, RR: 10
Reply 7, posted (5 years 6 months 1 week 2 days 16 hours ago) and read 6835 times:

On the CRJ-200, the exhaust speed at maximum thrust is approximately 544 mph measured at 25 feet behind the engine. The temperature of the exhaust at this point is approximately 440 degrees F. These numbers are from the CRJ-200 AMM and the engine in this example is the CF-34.

The same engine at idle thrust produces an exhaust speed of approx. 68 mph at a temperature of approx. 140 degrees F measured at the same location. I hope this helps.


User currently offlineTriebwerk From United States of America, joined Sep 2008, 126 posts, RR: 0
Reply 8, posted (5 years 6 months 1 week 2 days 5 hours ago) and read 6765 times:

544 miles an hour and 440 degrees at 25 feet away? Hope the people at Juliana make sure to keep their distance.  Smile

On that note, does anyone have any exhaust velocity statistics for rockets? The Saturn V, for example?


User currently offlineJetMech From Australia, joined Mar 2006, 2635 posts, RR: 53
Reply 9, posted (5 years 6 months 1 week 2 days 1 hour ago) and read 6733 times:



Quoting TristarSteve (Reply 2):
What about when the aircraft is stationary on the ground, is the exhaust speed still M0.9?

If the engine is at idle, I'd hazard a guess that the exhaust velocity would be somewhat lower than M 0.9 . I was recently involved with the marking of a propulsion assignment, where the students were asked to calculate the exhaust velocities of a given twin spool engine operating as alternatively as a turbofan and turbojet.

Obviously, there was a fair bit of simplification involved, but as far as I could tell, the exhaust gas velocities from the various streams were calculated from temperature differences. For the core flow, it was the temperature difference between the LPT discharge and exhaust nozzle exit (To5 - To9).

Getting back to the aircraft being stationary on the ground, I'd say that the core velocity would change somewhat between idle, and say T/O power settings. If we assume that as the power setting increases, that there is more heat discharged from the LPT - and hence a higher To5 - and that furthermore, To9 is ambient temperature and constant, then in theory, the core velocity should be higher compared to lower power settings.

The formula used by the students to calculate the velocity was v = SQRT [ 2*Cp*(To5-To9) ] , with Cp equal to 1005 J / Kg*K .

Quoting Tdscanuck (Reply 6):
Yes, fan air is noticeably hotter than ambient

The fan air velocities were also calculated in the same manner. This time, IIRC, they worked out the temp rise across the fan (To13-To2), assumed it was the same as the temp drop from fan discharge to the nozzle exit (To13-To9), and again, the used the formulae v = SQRT [ 2*Cp*(To13-To9) ].

Quoting Triebwerk (Reply 3):
(Weight of plane x speed) = (weight of air displaced x speed).



Quoting Tdscanuck (Reply 6):
You pick up drag in the sum of forces and the change in momentum drives you to delta-speeds, not absolute speeds.



Quoting WPIAeroGuy (Reply 4):
To find the velocity its pretty simple, its T=mdot(u9-u0) where the u's are exit and inlet(aircraft) velocity and mdot is the mass airflow through the engine.

I think there are two different, but related thrust formulas. The first is from F=Ma, which upon substitution gives N = kg*(m/s^2), which coverts into the impulse - momentum equation, N.s = kg*(m/s). There is also the mass flow rate - change in velocity formula, N = kg/s*[ V exit (m/s) - V inlet (m/s) ].

I think the impulse-momentum and mass flow rate - change in velocity formulas are most often used for aerospace applications. I have seen the impulse - momentum equation used for rockets, and the mass flow rate - change in velocity used for propellers and turbofans.

Quoting WPIAeroGuy (Reply 4):
but they must get it from a high exit velocity rather than greater mass airflow.



Quoting WPIAeroGuy (Reply 4):
From the equation above, your thrust will be negative if your exit velocity is slower than your inlet velocity, therefore supersonic planes need thrust, but they must get it from a high exit velocity rather than greater mass airflow.



Quoting Tdscanuck (Reply 6):
It's subsonic at the back of the engine and accelerated to supersonic speed by a con/di nozzle.

I'm curious as to whether a high-bypass turbofan would be more efficient if used for supersonic aircraft, compared with the low by-pass devices currently used? As far as I can tell, a CD nozzle works by having a certain upstream stagnation pressure before the throat. If the area ratio of the throat is designed correctly, and exposed to a certain upstream stagnation pressure, the gas should reach sonic velocity at the throat, and then get expanded to supersonic velocity by the divergent part.

In any case, the important thing to me at least, seems to be the large pressure difference across the nozzle that accelerates the gases from sub to super sonic velocities. If this is the case, I wonder if a high-bypass turbofan could be successfully used as the device to supply gasses to a CD nozzle?

Is there something intrinsically unsuitable about a CF-6 for instance that makes it unable to be used for supersonic duties? I understand the large fan of a high bypass device may create drag issues, but if we ignore this, is there anything else making a high-bypass engine unsuitable?

Regards, JetMech



JetMech split the back of his pants. He can feel the wind in his hair.
User currently offlineJetMech From Australia, joined Mar 2006, 2635 posts, RR: 53
Reply 10, posted (5 years 6 months 1 week 2 days 1 hour ago) and read 6730 times:



Quoting Triebwerk (Reply 8):
On that note, does anyone have any exhaust velocity statistics for rockets? The Saturn V, for example?

The Rocketdyne F1 apparently had exhaust gases that had a temperature of about 3,200 degees Celcius ( 5,800 Fahrenheit ). The ehxaust gas velocity for the Space Shuttle Main Engine (SSME) is apparently 4,444m/s ( 14,580 fps )

http://en.wikipedia.org/wiki/F-1_(rocket_engine)
http://en.wikipedia.org/wiki/Space_shuttle_main_engine
http://en.wikipedia.org/wiki/SSME_energy_and_power_relationships

Regards, JetMech



JetMech split the back of his pants. He can feel the wind in his hair.
User currently offlineWPIAeroGuy From United States of America, joined Aug 2007, 240 posts, RR: 0
Reply 11, posted (5 years 6 months 1 week 2 days ago) and read 6719 times:



Quoting JetMech (Reply 9):
I'm curious as to whether a high-bypass turbofan would be more efficient if used for supersonic aircraft, compared with the low by-pass devices currently used? As far as I can tell, a CD nozzle works by having a certain upstream stagnation pressure before the throat. If the area ratio of the throat is designed correctly, and exposed to a certain upstream stagnation pressure, the gas should reach sonic velocity at the throat, and then get expanded to supersonic velocity by the divergent part.

I think theoretically it would work, but the inlet diffuser would be so large it wouldn't work well on todays designs. Maybe along the lines of a supersonic BWB?



-WPIAeroGuy
User currently offlineTdscanuck From Canada, joined Jan 2006, 12709 posts, RR: 80
Reply 12, posted (5 years 6 months 6 days 21 hours ago) and read 6564 times:



Quoting JetMech (Reply 9):
I'm curious as to whether a high-bypass turbofan would be more efficient if used for supersonic aircraft, compared with the low by-pass devices currently used?

Is there something intrinsically unsuitable about a CF-6 for instance that makes it unable to be used for supersonic duties? I understand the large fan of a high bypass device may create drag issues, but if we ignore this, is there anything else making a high-bypass engine unsuitable?

The issue is pressure...a C/D nozzle converts pressure to velocity. You need to feed it with high pressure to get high velocity out.

A big fan moves a lot of air with a low delta-V, which is equivalent to a low-delta P. It's like the difference between a high-volume low-pressure pump and a low-volume high-pressure pump...the former is more efficient at moving fluid around *if* you don't need a big pressure differential. Unfortunately, TANSTAAFL and you need a lot of pressure to feed your C/D nozzle.

Tom.


User currently offlineJambrain From United Kingdom, joined Sep 2008, 251 posts, RR: 0
Reply 13, posted (5 years 6 months 6 days 8 hours ago) and read 6510 times:

If you care to put a few sums into an equation


Fundamentals of Compressible Fluid Dynamics By P. Balachandran section 13.2.4 on turbofans

see on Google books:-
http://books.google.co.uk/books?id=K...wqqUMSNccja0q0MuV-QDJaDA#PPA447,M1

but say Trent 1000
Fan diameter: 112 inches (2.85 m)
Take-off thrust: 75000 lbf (flat-rated to ISA+15C)
Airflow: ~2850lb/s

334090 N Force
1295 kg / s Airflow

F=Ma

At full thrust on ground with no forward velocity
260 m/s (that's average not jet portion)

I haven't done the math for jet velocity but one of the examples gives jet velocity of 670 m/s at static temp of 800K



Jambrain
User currently offlineJetMech From Australia, joined Mar 2006, 2635 posts, RR: 53
Reply 14, posted (5 years 6 months 5 days 3 hours ago) and read 6384 times:



Quoting WPIAeroGuy (Reply 11):



Quoting Tdscanuck (Reply 12):
The issue is pressure...a C/D nozzle converts pressure to velocity. You need to feed it with high pressure to get high velocity out.



Quoting Tdscanuck (Reply 12):
A big fan moves a lot of air with a low delta-V, which is equivalent to a low-delta P.

I see. I was actually thinking along the lines of what you have mentioned. I suppose that an engine used as a motivator for a CD duct would not only need to put out a decent mass flow rate, but also have the ability to develop a good head of stagnation pressure as well. As you have pointed out, a turbojet or low bypass turbofan seem to be the best type of jet engine to provide these requirements.

Regards, JetMech



JetMech split the back of his pants. He can feel the wind in his hair.
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