Bsergonomics From United Kingdom, joined Jan 2002, 462 posts, RR: 0 Posted (10 years 7 months 3 weeks 3 days 12 hours ago) and read 11847 times:
One of the guys in the office is thinking about getting his multi-engine rating. This led to a discussion on engine-out scenarios. Needless to say, the dreaded '7 engine landing' in the BUFF got thrown in, which led to this question:
How many engines can the BUFF lose and still maintain level flight? The assumption was that all stores would be jettisoned immediately, so we're talking an unladen aircraft. For argument's sake, we talked about no other failures and a 50% fuel load.
There are two scenarios that were brought up: symmetric and asymmetric engine losses.
Does anyone have any idea?
The definition of a 'Pessimist': an Optimist with experience...
AirRyan From United States of America, joined Mar 2005, 2532 posts, RR: 5
Reply 2, posted (10 years 7 months 3 weeks 3 days 5 hours ago) and read 11738 times:
Current BUFFS have Eight Pratt & Whitney engines TF33-P-3/103 turbofan engines each producing up to 17,000 pounds of thrust - I'd like to see the BUFF house 2 GE-90-115's. That's a max thrust of 136,000lbs boosted up to 230,000lbs, even if you could get them on the wings, that might already be a little too much thrust. Maybe 4 GE CF6-80C2 at 60k lbs. thrust from the AF1 and the new C-5M would be nice?
KC135TopBoom From United States of America, joined Jan 2005, 12319 posts, RR: 51
Reply 3, posted (10 years 7 months 2 weeks 6 days 3 hours ago) and read 11302 times:
Under your symmetric thrust scenario, the B-52D/E/F/G (all with J-57 engines) could fly with two engines out, on each side (for a total of 4 engines out). The B-52H could do the same with four TF-33s out.
In an asymmetric thrust condition, all models would only be able to continue to fly with two engines out on one side, and one on the other. Three (or more) engines out on one side would most likely mean the loss of the aircraft.
Remember, the B-52 has a very small rudder, and no airolons. It turns using the large speed brakes on top of the wing. There is only so much drag the speed brakes can add to the high thrust side before the airplane will become uncontrollable.
The speed brakes are hydraulically controlled. With that many engines out, a B-52C/D/or E might have a problem powering the hydraulic packs. On the B-52 F/G/H they have no hydraulic packs, but do have engine driven hydraulic pumps. So as long as you have at least two hydraulic pumps operating for each hydraulic system, there should be enough pressure, at least until it comes time to lower the landing gear.
Blackbird From , joined Dec 1969, posts, RR:
Reply 5, posted (7 years 1 month 3 weeks 3 days 22 hours ago) and read 10471 times:
Well, the early B-52's had inboard-ailerons and 6-spoiler panels per wing for asymmetric roll-use. The later models (The -G and -H) had the inboard ailerons removed which had to do with increasing the wing's fuel capacity and possibly aero-elastic properties and 7-spoilers per wing (I guess to make up the roll-rate, at least as best as possible).
Flexo From St. Helena, joined Mar 2007, 406 posts, RR: 0
Reply 8, posted (7 years 1 month 3 weeks 3 days 3 hours ago) and read 10138 times:
Quoting KC135TopBoom (Reply 7): The B-52 has a glide ratios of around 3 or 4:1. So, if you started this at FL400, in a "light weight aircraft" you could glide about 160,000' ( just under 26.4nm) for a straight in appro
Is this on two engines or no engines at all? I was wondering how far you could get with just two engines left.
If that was the case, then no B-52 would ever have been able to lift off the ground with all eight engines at max power.
The glide ratio (lift to drag ratio - L/D) is likely to be roughly 20:1 - a little better than most airliners due to the slimmer fuselage, the quite low wing loading and rather small (and therefore low drag) nacelles. But it will be rather different with engines at flight idle or dead engines windmilling.
At FL450 a B-52H cruises beautifully with all engines running. At that altitude max power will be reduced by some 80% due to the thin air - 3-4,000 lbs instead of 17,000 per engine. From that we can assume that at sea level it will fly straight and level even at max weight on two symmetric engines, at least with the other six at idle.
But whether we can choose two symmetric engines and still have sufficient hydraulic and electric power available for controlled flight, that's another story.
As described in earliers posts, the B-52 is a terrible thing with asymmetric thrust. It could never be certified to carry pax with that unusual configuration. It is simply not safe enough with asymmetric thrust. But the unusual configuration saves a lot of weight. And all crew members enjoy the comfort of an ejection seat. Makes it able to make moon craters further away from home.
Assuming MTOW = 488,000 lbs and L/D = 20:1, then thrust needed is 488,000 divided by 20 = 24,400 lbs. So yes, it will theoretically fly on two engines doing 17,000 each, until you lower flaps and landing gear.
Remember, a twin airliner shall do a lot more than just fly on one engine. It must take off at MTOW after engine shut down at V1 and climb out at 2.75 degrees with flaps and slats extended and gear down. It must do so even at some airport elevation and on quite hot days when it shall not most of the time be operated with severe weight restrictions. That also means that any twin airliner must be able to cruise at sea level in clean configuration on one third or one fourth of max power on one engine only. At MTOW. And much less than that without payload and with low fuel load.
But lose an engine at high altitude, then you come down!
Always keep your number of landings equal to your number of take-offs