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Boeing published engine hazard areas as shown below:

Danger area

I am curious as to how the danger area is affected when the aircraft is in flight as the jet engine is still vulnerable to foreign object damage.

  1. Assuming idle power conditions at 10000 ft (or just airborne in general), are the danger zones the same as on the ground? If not, would there be a way to calculate this?

  2. Knowing the danger zone for both idle and full power, would it be possible to calculate the danger zone distance for different power settings? E.g. 50% N1.

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  • $\begingroup$ It is doubtful that the area in front of the engines is significant given that the aircraft will probably be outrunning the inlet stream. The area behind the engines will probably combine with the wake of the aircraft, and would probably be difficult to separate the two (or at least not useful since the combined effect is what somebody following the aircraft would be concerned about). If you are asking if something ahead of the aircraft but not directly in the intake stream can get ingested, the answer, I'm guessing, is probably not. $\endgroup$ – Ron Beyer Oct 6 '17 at 20:14
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    $\begingroup$ The "danger zones" that you reference only have meaning when the aircraft is on the ground. Once you're flying, everything in front of the aircraft is in danger of being either sucked into an engine, or else struck by some part of the aircraft structure. How far in front of the aircraft you'd define this "danger zone" would be based as much on assumed reaction time (to get out of the way of a 200? 300? 400? knot aircraft) as on anything else. You can't extrapolate the posted figures into anything comparable for "in flight" conditions -- the dangers (to skydivers?) just aren't the same! $\endgroup$ – Ralph J Oct 7 '17 at 5:39
  • $\begingroup$ @RonBeyer I'm guessing this would depend on the weight of the object being ingested as well? E.g. a small, light RC plane not positioned directly in the intake stream could still be sucked in if close enough to the engine? $\endgroup$ – Zack Oct 7 '17 at 9:52
  • $\begingroup$ I think you should precise what you understanding of "danger area". It is primary defined for airport workers on the ground, but as you speak of suction area, you may define what you understand by being in or out of this area. $\endgroup$ – Manu H Oct 7 '17 at 11:38
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    $\begingroup$ @CarloFelicione Bird strikes have been reported at altitudes above 30,000 feet. $\endgroup$ – Ron Beyer Oct 7 '17 at 22:47
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Let's assume the A320 is powered by V2500 engines, one of the options for it. This engine has an air mass flow of 355 kg/s. At sea level, air density is 1.225kg per cubic m. Hence, 355 kg/s = 355/1.225 = 290 cubic m/s.

Now, let's assume the aircraft is still at sea level, but now at Mach 0.8 (unrealistic, but we'll correct that next). The fan diameter is 1.6 m, which gives an intake area of 2.0 sq.m. Also, Mn 0.8 at 0ft is 272m/s. Hence, every second, the intake sweeps a volume of 272 x 2.0 = 544 cubic m. But, the engine only needs 290 cubic/m of air, per second assuming air density in the intake is the same as the air surrounding the aircraft. Therefore, the diameter of a streamline tube of the intake air will actually be smaller than the diameter of the engine inlet. It will be the area that sweeps 290 cubic m/s at 272m/s = 290/272 = 1.07 sq. m., or a diameter of 1.17m, not 1.6m.

Basically, the engine takes the airflow it wants, not what the intake area x forward speed provides. If the engine wants more (such as when the aircraft speed is low, or stationary, but the engine rpm is high, such as at the start of the takeoff roll), the engine will draw air from a large area in front of the engine (as per the max takeoff conditions diagram). Conversely, when the aircraft is at high speed, and the engine throttled back, the intake will spill the excess air it is providing (causing spillage drag).

enter image description here

Now, let's correct the fact the aircraft can't do Mn 0.8 at sea level. Let's re do the calculation at 35,000ft (10,700m). Here, air density is 0.38 kg per cubic m, and pressure and temperature are 3.46 psi and 219 Kelvin (-54C), compared with 14.7psi and 288 Kelvin (15C) at sea level. Thus our 355 kg/s, which is actually a corrected airflow, is a physical (real) air flow of 95.8 kg/s at 35,000ft since theta = 219/288 = 0.76 and delta = 3.46/14.7 = 0.235. Now 95.8 kg/s at 0.38 kg/ cubic m = 252 cubic m per second. Also, Mn 0.8 at 35,000ft is now 237m/s, not 272 as it was at seal level. Hence, every seond the intake area of 2 sq.m is sweeping through 2 x 237 = 474 cubic m per second. But we only want it to sweep 252, hence we need to find the diameter of a streamline tube that will cause this, at 237m/s. Therefore, we need an area of 252/237 = 1.06 sq. m., which occurs with a diameter of 1.16m. This compares with the physical inlet diameter of 1.6m.

Thus, at Mn 0.8, 35,000ft, max power, the green area at the engine intake is now a tube, of 1.16m dia, that extends in front of the aircraft. For how far this extends, depends as suggested, on what reaction time is required for the aircraft to manoeuvre away from an obstacle in this region, or for the object (a bird?) to manoeuvre out of the aircraft's way.

The exact figure of 1.16 is questionable, as the assumption of air density in the intake not changing from the surrounding air is not entirely realistic. But the general result, is I believe, reasonable.

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  • $\begingroup$ Should the diameter of the streamline tube at sea level not be 1.17m as opposed to 1.47m? (290/272 = 1.07 sq. m., diameter = 2 * √(1.07/3.14) ) $\endgroup$ – Zack Oct 8 '17 at 19:11
  • $\begingroup$ Would it be correct in saying that if the engine intake diameter is less than the fan diameter (I.e. 1.6m) then the result would be a streamline tube extending in front of the aircraft? And, in the case that the engine intake diameter exceeds the fan diameter (1.6m), then the engine would draw air from a large area like in the diagrams? $\endgroup$ – Zack Oct 8 '17 at 19:28
  • $\begingroup$ Yes, I made a mistake with calculating the diameter from the area. It is a dia of 1.17m, as you say. I have corrected the answer - thanks. $\endgroup$ – Penguin Oct 12 '17 at 8:59
  • $\begingroup$ With regard to your second comment, yes, I would agree in general with what you say. Note in the first case, the engine doesn't have to suck, so it's not drawing the air in, just swallowing a portion of what the intake provides. In the second case, rather the engine taking air from a cylinder or tube in front of it, it would probably be drawing it from a shape more like a cone or a hemisphere, I expect. Sorry I missed your comments till now. The only odd thing with my answer, is that even in the extreme sea lavel case, the intake seems way oversized. $\endgroup$ – Penguin Oct 12 '17 at 9:05
  • $\begingroup$ This makes sense. However, could the forward distance of this cone/hemisphere be calculated? For example, if the diameter of the cone/hemisphere is 1.1 metres, would the cone/hemisphere also be extending 1.1 metres forward? $\endgroup$ – Zack Nov 17 '17 at 17:27

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