UAL747 From , joined Dec 1969, posts, RR: Posted (7 years 11 months 2 weeks 4 days 21 hours ago) and read 10495 times:

I've always wondered if someone could tell me what the speed of the air is coming outside a jet engine, in particular the GE90-115B. I know that the power is measured in lbs of thrust, but what would the actual speed be to power a fully laden 777 down the runway?

N231YE From , joined Dec 1969, posts, RR:
Reply 1, posted (7 years 11 months 2 weeks 4 days 21 hours ago) and read 10490 times:

I don't know the velocity (speed) of the jet exhaust for the GE90-115, but as part of my pilot training here in the U.S., I had to have basic understanding of jet blast. For a 707, I believe it is 300 miles/hour (482.8 Km/hour) from a distance of 50 feet (15.2 m) from the tail. I must add, that the velocity is that it is slower for larger and more modern engines. Although their thrust ranges are different, the concept is simple: Older jet engines such as the PW JT8D move a small amount of air at a high velocity, resulting in higher fuel burn and noise, whereas a modern GE90-115 moves a large amount of air at a lower velocity.

Dw747400 From United States of America, joined Aug 2001, 1258 posts, RR: 1
Reply 2, posted (7 years 11 months 2 weeks 4 days 20 hours ago) and read 10478 times:

777236ER From , joined Dec 1969, posts, RR:
Reply 3, posted (7 years 11 months 2 weeks 4 days 20 hours ago) and read 10470 times:

Thrust is essentially the intake mass flow (and so the outlet mass flow) multiplied by the outlet flow velocity (ignoring pressure thrust and the forward speed of the aircraft).

The outlet mass flow equals the outlet velocity multiplied by the density multiplied by the cross-sectional area of the outlet. Assume that the density is pretty much 1.225 kg/m^3 and that the outlet area is the same as the inlet area (128 inches fan diameter = 3.25 m which leads to an area of pi*3.25^2 / 4 = 8.3 m^2 ish.)

So, the thrust (115000lbs, or 511.5kN) equals 1.225*8.3*v^2, leading to the outlet velocity equaling sqrt(511.5*10^3 / (1.225*8.3)) = 224.3 m/s = 500 mph.

Disclaimer: it's very late and all this is off the top of my coffee-fueled head. Feel free to disagree entirely!

AeroWeanie From United States of America, joined Dec 2004, 1608 posts, RR: 52
Reply 4, posted (7 years 11 months 2 weeks 4 days 19 hours ago) and read 10447 times:

The maximum massflow that you can get through any opening occurs when the flow is M=1 at the opening. This is called choked flow. Engine exits are typically set up to be choked. One exception is hush kits, which use slightly larger exit areas.

MarkC From United States of America, joined Apr 2006, 259 posts, RR: 0
Reply 5, posted (7 years 11 months 2 weeks 4 days 19 hours ago) and read 10435 times:

Don't know about GE.

A 94" 4000 has a velocity of 680 mph measured at 30 feet behind the exhaust.

Good effort! I'm answering this in the morning, so I have an advantage. I think your logic is correct, but in my opinion you've made a couple of incorrect assumptions....

Your assumption that the annulus area at the back of the aircraft is the same as the annulus area at the front will not be true (I don't know what the dimensions are, but they won't be the same. You can tell this just by looking at the engine). Also, you have taken the intake area to be the area of a circle of a diameter equalling the fan diameter. But what about the spinner? The intake area is actually an annulus with an outer diameter equal to the diameter of the fan, and a height equal to the height of the fan blades, significantly less than the fan case diameter. Again, I don't have these figures to hand.

I think the best way of working it out would be to work out the mass flow at the intake (mass flow equals annulus area multiplied by density multiplied by velocity of intake, which is usually designed to be about Mach 0.6), assume that the cabin pressurization bleeds are negligible (obviously some of this mass flow of air that goes in the front won't come out of the back, instead it will end up in the aircraft cabin), and then use your equation for thrust to work out the average velocities of the hot and cold jets (velocity equals required thrust divided by mass flow). Obviously this is true only at take-off (for other conditions thrust won't be 115klbf, density won't be 1.225kg/m^-3, and we would have the aircraft speed to worry about).

Skipping the first part of the above, according to random website (http://www.turbokart.com/about_ge90.htm), GE90-115 mass flow is approx 3000lb/s, which is 1361kg/s. 115lbf in N is 511.545KN. Therefore, average of the hot and cold jet velocities is 511,545/1361 = 375.8m/s, or 838.8 mph.

This speed is only valid while the aircraft is not moving forwards and attempting to create 115Klbf of thrust. If it is moving forwards, add the forward speed of the aircraft to this number. Obviously my value is wrong if the figure of 3000lb/s is incorrect. Also, the cold jet will be moving somewhat slower than the hot jet, which will be significantly faster (Mach 1, I believe, but I don't know the temperature or pressure, so Idon't know what that is in m/s).

F14D4ever From United States of America, joined May 2005, 319 posts, RR: 4
Reply 8, posted (7 years 11 months 2 weeks 3 days 19 hours ago) and read 10149 times:

Quoting TEAtheB (Reply 7): Obviously my value is wrong if the figure of 3000lb/s is incorrect

3000 pps was the figure I always carried in my head, but checking some cycle data, we're low by at least 10%. Speaking of cycle data, it says fan exit velocity is below 1000 ft/sec and core exit velocity is above 1300 ft/sec at nominal takeoff thrust.