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Static Thrust?  
User currently offlineBryan Becker From United States of America, joined Mar 2001, 333 posts, RR: 0
Posted (13 years 7 months 4 weeks 1 day 9 hours ago) and read 10693 times:

When they say that a engine produces a certian amount of thrust standing still,lets ay 105lb's of thrust static,does the engine produce less thrust when set up to move something,or does it produce the same? Smile


-BryanBecker Big thumbs up

12 replies: All unread, jump to last
 
User currently offlineAerotech From United States of America, joined Jul 2000, 259 posts, RR: 2
Reply 1, posted (13 years 7 months 4 weeks 1 day 8 hours ago) and read 10658 times:

The actuall thrust produced by the engine is highest at sea level at zero speed. If, for instance the engine is on a plane travelling at 600M.P.H., there is a sufficient amout of RAM air, that cancels out a little of the thrust.

User currently offlineEssentialPowr From United States of America, joined Sep 2000, 1820 posts, RR: 2
Reply 2, posted (13 years 7 months 4 weeks 22 hours ago) and read 10641 times:

Aerokid,

Your response is wrong...As always, here's the math:

F=ma

For a jet, thrust (F) = mass flow rate x jet velocity. At 100% thrust, jet velocity is roughly constant. Mass flow rate varies directly with velocity and inversely with altitude (till you get to space...no atmosphere = no mass flow).

Therefore, up to about 50,000', max thrust is obtained at low altitudes and high speeds...certainly greater than static thrust, which is simply a ref point.

Your UPT interview will be very brief if you miss these concepts...

Best of luck-
Esp


User currently offlineEssentialPowr From United States of America, joined Sep 2000, 1820 posts, RR: 2
Reply 3, posted (13 years 7 months 4 weeks 19 hours ago) and read 10626 times:

Sorry - Aerotech



User currently offlineSabenapilot From Belgium, joined Feb 2000, 2728 posts, RR: 46
Reply 4, posted (13 years 7 months 4 weeks 11 hours ago) and read 10610 times:

the only correct formula for thrust in its simplest form is:
F = Dm x (Vjet - TAS)

Try to answer the above question for yourself with the help of this formula....



User currently offlineAerotech From United States of America, joined Jul 2000, 259 posts, RR: 2
Reply 5, posted (13 years 7 months 4 weeks 7 hours ago) and read 10595 times:

I will do just fine at the board. I know what I'm talking about. Your formula doesn't account for gravity. I guess I explained it wrong-the point I was making was the same as yours-up at high alt.-you can't get your maximum output. No offense, but I don't need your critiquing. And if you critque, make sure you have ALL the facts.The REAL formula looks like this:

Weight of air in pounds per second X velocityThrust ---------------------------------------------------
32.2 (normal acceleration due to gravity, in feet per second)

Imagine an aircraft standing still, capable of handling 215 pounds of air per second. Assume the velocity of the exhaust gases to be 1,500 feet per second. The thrust would then be:

215 lbs of air per second
Thrust --------------------------- X 1,500 feet per second= 6.68 X 1,500 =
32.2 feet per second2

Thrust = 10,020 lbs
If the pressure at the exit plane is not the same as the atmospheric pressure and the aircraft were not standing still, the formula would be somewhat different.




User currently offlineEssentialPowr From United States of America, joined Sep 2000, 1820 posts, RR: 2
Reply 6, posted (13 years 7 months 4 weeks 4 hours ago) and read 10591 times:

Aerotech,

Gravity?? What does gravity have to do with it? Ever heard of thrust @ "0" g? Absolutely.

Your response was that the highest thrust produced by a (assumably a jet @ 600 mph) jet is at sea level, speed = 0.

That is completely wrong. An F100 in an F16 at 1000' MSL and 500 kts is producing MUCH MOre thrust than the same engine in static conditions. This idea is the basic concept of a jet; to miss it is a huge conceptual error.

The equation is Thrust = mdot V, or mass flow rate x velocity, as stated. Reread my post with regard to velocity and altitude.

As for knowing what you're talking about...
Don't tell em a KC10 has ballast weights on the main gear bogies (boogies?) either...


User currently offlineEssentialPowr From United States of America, joined Sep 2000, 1820 posts, RR: 2
Reply 7, posted (13 years 7 months 4 weeks 3 hours ago) and read 10583 times:

Sabenapilot,

thrust = mass flow rate x jet velocity is correct for low, subsonic speeds.



User currently offlineSkwpilot From United States of America, joined Feb 2001, 60 posts, RR: 1
Reply 8, posted (13 years 7 months 4 weeks 3 hours ago) and read 10589 times:

Now I'm a little out of my element here (being a turbo-prop guy and all) so feel free to correct me if I am wrong.

I think that Sabena has the right idea relating jet speed to TAS. The faster the TAS, the faster the Vjet must be to compensate to give the same thrust. Let's look at this in terms of EPR (engine pressure ratio). This concept is non-linear, however gives a good indication of thrust at any speed. With the same EPR, a stationary engine (lets say on top of a 1000 ft mountain just for comparison) will yield a higher EGT temp stationary than one which is inflight at the same altitude and temp. This is because inflight, the ram air causes a slight backup of air at the compressor inlet causing a slightly higher P1 pressure due to "ram rise". The engine inflight has essentially more O2 to burn through higher inlet density. However, the engine rotation will increase, because more air must be "pumped" through the engine to make up for the increased TAS requiring increased Vjet. The engine will create the same power until the N1 and N2 speeds become maxed. Then total power will be lost. This is rarely demonstrated in operational conditions, because aircraft climb too fast for this to occur. This makes our engines "temp out" (EGT) before we reach max compressor speed.

Happy Flying Smile

Skwpilot


User currently offlineSabenapilot From Belgium, joined Feb 2000, 2728 posts, RR: 46
Reply 9, posted (13 years 7 months 3 weeks 6 days 11 hours ago) and read 10577 times:

to Skwpilot-
thank you.

To those starting a fight over something very basic-
the formula I gave you is the only one that can explain the most simple graph in the study of turboreactors. i.e. the Thrust vs Velocity graph.
please note that this formula is only valid in subsonic airflows, so don't use the F16 as an example in this...
Have you ever seen this graph?
Good. If not, I'll try to describe it to you.
It looks like this:
If you set the thrust on the vertical axis and the speed on the horizontal exis, you'll find the representation of thrust produced by an engine running in a constant regime at a given altitude to be a more or less horizontal line sloping slightly downwards if you accelerate (move to the right in the graph), till around 250kts or M.4 where the thrust line will once again start to move upwards to reach around the same value it had at zero speed!

Why?
from the formula I gave you:
F = Dm x (Vjet-TAS)
or:
F = Dm X vjet - Dm x TAS
if your TAS becomes bigger, you substact more in the formula from Dm x vjet, so the thrust gets smaller.
As from 250kts the compressability of the air starts to take effect and although the Dm x TAS still gets bigger all the time, so does the DM x vjet thanks to an increase in Dm, resulting in an small overall increase in thrust to around the initial value.

In general it can be said that an engine running at a constant altitude (for instance MSL) is producing a same specific ammount of thrust for any given power setting at two different speeds.
For maximum thrust (which is also a specific amount of thrust, isn't it?) these two different speeds are:
TAS = 0kts or M.0
TAS = between M0.65 and M0.90 depending on the engine.
This is ideal since the first speed corresponds with the take-off phase (a lot of thrust needed).
The second speed is made by the engine manufactorer to correspond with the cruising speed. Since we normally don't need the maximum available thrust in this case, it allows for the selection of a lower thrust setting and thus a lower fuel flow.
However, in emergency situation like an engine failure, it gives you the chance to get a lot of thrust (called max continuous thrust then due to some EGT limitations) which will allow you to drift down to a lower altitude at only a very slow rate of descent giving you the biggest possible distance to evercome any mountaneous obstacles on you drift down path.

You see, both of you were more or less correct in a way. If two people have a different idea it does not automatically mean one of them must be wrong you know...





User currently offlineAerotech From United States of America, joined Jul 2000, 259 posts, RR: 2
Reply 10, posted (13 years 7 months 3 weeks 6 days 9 hours ago) and read 10569 times:

OOPS! I see where I put down "0" as the speed. I meant to say "600"-to show that you can get better power at low level than high level. I meant to say sealevel at 600 vs. 60,000ft at 600. Type-o. My fault. And I didn't mean gravity like weight-I meant it as it's used in the formula-acceleration due to gravity-32.2 fps. You can't deny that one. And the landing gear thing,- well, it was a joke but appearently you took it seriously. I'm well aware of the hydraluics-I have beem for as long as I can remember. My dad was an E-6 who worked on pneumatics for various AC the Air Force. I have about a thousand books showing all about various landing gear. I just thought I'd be funny-hence the word "Boogie"!  Smile

User currently offlineEssentialPowr From United States of America, joined Sep 2000, 1820 posts, RR: 2
Reply 11, posted (13 years 7 months 3 weeks 6 days 8 hours ago) and read 10563 times:

Sabenapilot,

Friction and compressibility effects are greater as speed nears mach. The equation I have provided is correct for this reason, dur to the fact that at low speeds, compressability effects are negligible.

The F16 example is as valid as any other jet a/c, flying at subsonic velocities less than around .70 mach.





User currently offlineEssentialPowr From United States of America, joined Sep 2000, 1820 posts, RR: 2
Reply 12, posted (13 years 7 months 3 weeks 6 days 7 hours ago) and read 10561 times:

Areotech,

There is still no gravity component required to produce thrust. Mass exists w/o gravity...

You answered the bogie tilt assembly thread rather directly as requiring ballast weights...

Good luck at UPT!


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