SlamClick From United States of America, joined Nov 2003, 10062 posts, RR: 66
Reply 2, posted (11 years 6 months 3 weeks 2 days 10 hours ago) and read 16870 times:
The ratio should be just about exactly the ratio of the local air pressure to standard at sea level.
Standard atmosphere table says where sea level is 1 atmosphere or about 14.7 PSI,
at 10000' it is 0.6877 of sea level
at 20000' it is 0.4595 of sea level
at 30000' it is 0.2970 of sea level
at 36089' it is 0.2234 of sea level ("standard" height of tropopause)
at 40000' it is 0.1851 of sea level
So I would expect a 24K engine to be producing just under 5400 pounds at the tropopause. There might be some other factors but I think these numbers would be pretty close.
Happiness is not seeing another trite Ste. Maarten photo all week long.
Bd5jdave From United States of America, joined Oct 2004, 5 posts, RR: 0
Reply 3, posted (11 years 6 months 2 weeks 3 days 14 hours ago) and read 16385 times:
That depends totally on the aircraft the engine is in. It's all a question of drag. If your 10K engine was in a heavy, fat airplane, drag would limit the airplane's speed. Your engine would produce, say, 2K thrust at 35,000 at max speed (Mach .85 for this example). Now, put this same engine a sleek, small fighter, and your aerodynamic top speed is much faster- say Mach 1.5. Now you have more air entering the inlet to combust, producing more thrust. You have more room to work with the inlet and exhaust, so you can cheat more thrust out of it that way as well.
The above is a simplified example, but a sound one.
It's just not cut and dried. That's why Aero Engineering is so tough. It's a game of give and take. Jets are typically designed for a narrow range in its "cruise speed" because an airplane has to be designed for certain operating conditions. This will take in account the total drag coefficient, design operating altitude, and engine performance. You design for one thing, and then have to change something else. It's an excercise in compromise.
For instance a 22k CFM56-5B5 (very common on A319s) will deliver max 5,020 lbs at FL350.
Other versions, e.g. the -5B4 and -5B6 with 27k and 23.5k TO thrust, have the same max 5,020 lbs at FL350. I would guess that the -5B5 and -5B6 are basically identical to the -5B4, but with derated TO thrust for longer on wing time and reduced maintenance requirements.
So there is not a direct relationship between max thrust and ambient air density. But it is pretty close.
BTW, did you know that the max thrust of a P&W J-75 in a Lockheed U-2R way up in coffin corner is just some 700-800 lbs?
[Edited 2004-10-22 00:29:11]
[Edited 2004-10-22 00:30:11]
[Edited 2004-10-22 00:31:04]
Always keep your number of landings equal to your number of take-offs
IDAWA From Italy, joined Aug 2004, 306 posts, RR: 0
Reply 6, posted (11 years 6 months 2 days 18 hours ago) and read 15720 times:
SlamClick is correct, at least as a general theory. My university knowledge doesn't go deeper than SlamClick's post, so I assume that other posts are correct!
I'd like to add something for piston engines: instead of providing thrust, they provide power (thrust*speed), and power is reduced with the same ratio as thrust (current altitude air density vs. sea level air density).
To mitigate this effect, some piston engined are fitted with a compressor, that allows power to be unchanged by altitude from sea level to the so-called "ciritcal altitude". Above such altitude, power decreases as if the engine hadn't been fitted with the compressor.