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Can a 747 maintain 5000 ft with 3 eng out?

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Moderator
Registered: May 2009
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As she can climb with 400 tons and 2 engines out at low altitudes, why shouldn't she be able to maintain altitude at a lighter weight with 3 engines out?

Remember, it's not a matter of the power sum. A 100-horsepower car is not necessarily twice as fast as a 50-hp car. The drag rises exponentially.

Has anyone tried this in the sim?


Cheers,

|-|ardy
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Registered: Jan 2013
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Location: YSSY
I suspect it would be random to test this in the real sim as this is not a certification requirement, and as such probably was not tested. You probably know that the "big" simulators are calculating very little physics, they mostly interpolate between large tables of accumulated flight testing data. So it essentially becomes an aerodynamical modelling exercise. Anyone up for it?

Cheers

- Balt
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Registered: May 2009
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Are you sure this is not a certification requirement? As we all know, there are definitely performance data for idle descents, with various idle thrust settings, or with certain engines completely out, or with manual thrust settings above idle. If the sim can compute the descent path under such conditions, it should also know under which conditions the descent path rises to 0 and above. E.g. by applying thrust on #2 and #3, or on all engines, or just on #3 etc. pp. (taking the gross weight and density altitude into account, of course).


Cheers,

|-|ardy
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Registered: Jan 2013
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Location: YSSY
Hi Hardy,

I just had a look in the QRH (CF6 powered ex CLX) and as stated before, the most engines you can lose are two. Lose any more than that, and you've entered the flight testing regime. Here's some data for conditions "maximum mass" and "ISA+10 or below":

1 engine inop: Level off at 26900ft (max mass 393t)
2 engines inop: Level off at 12600ft (max mass 385t)

From simple extrapolation follows: 3 engines out, -1700ft...

Long range cruise ability:
1 engine inop: 25100ft (max 400t)
2 engines inop: 6700ft (max 400t)
I'll spare the extrapolation on this one...

It does get more interesting at lower mass though:

1 engine inop: Level off at 42000ft (max mass 196t)
2 engines inop: Level off at 32300ft (max mass 385t)

From simple extrapolation follows: 3 engines out, 22600ft...

Long range cruise ability:
1 engine inop: 40900ft (max 200t)
2 engines inop: 30700ft (max 200t)

Extrapolating:
3 engines inop: 20500ft

All assuming linear extrapolation is valid... which is (edit) not valid has Hardy mentioned below!

Cheers

- Balt
« Last edit by Balt on Sun, 05 May 2013 23:52:41 +0000. »
Member
Registered: May 2010
Posts: 558
Quote
Simple extrapolation: 3 engines out, -2000ft...


Does lift due to air density increase/decrease linearly with altitude in a standard atmosphere?
(i.e. not including ground effect)
Moderator
Registered: May 2009
Posts: 3289
Hi Balt,

yep, but I think this linear extrapolation is too simplified: As you know, the atmospheric altitude system is not linear, and, secondly, the net thrust sum depending on the number of engines running isn't linear either.

The pressure difference between, say, FL200 and FL300 is not the same as that between sea level and FL100.

A carriage with three horses is not much faster than the same carriage with one horse, the former is significantly faster only if the drag is significantly high. Below LRC speed, at best lift/drag speed, the drag is relatively low.

For these reasons (among many others) PSX uses a complicated system of exponential functions (it's actually impossible to solve anything in the engine/aerodynamics model in a linear way).


Cheers,

|-|ardy
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John H Watson wrote
Quote
Simple extrapolation: 3 engines out, -2000ft...


Does lift due to air density increase/decrease linearly with altitude in a standard atmosphere?
(i.e. not including ground effect)

No. If that was true, the tip of the Sydney Tower would be in vacuum. Edit: This joke goes into the wrong direction, sorry :-) Another picture: If that was true, you wouldn't need oxygen bottles when you're on the top of the Everest (unless your name is Messner).
« Last edit by Hardy Heinlin on Sun, 05 May 2013 23:59:51 +0000. »
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oops. sorry guys, I "fast" posted before running off to a meeting... just updated the post without seeing that there appears considerable interest. Let me put some more thought (and real physics, no linear interpolations) into this and get back to you later!

Cheers

- Balt
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Registered: May 2009
Posts: 740
Location: Sydney, Australia
Isn't it just math...?

If we know how much thrust an engine can produce at altitude we know the zero drag acceleration.

We should know the variables to calculate drag in a standard atmosphere, and with drag we have the lift. ( If we can't get drag, it all falls apart of course!)

To simplify, If thrust => drag, and lift => g, it will maintain altitude.

http://en.wikipedia.org/wiki/Drag_equation

Maybe too simple? I know it doesn't take into account variable drag of other engines, rudder etc, but it's a start?

:P
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Sure, these are the basics. But we are in a very small range now where just a few percent can kick the balance off the scope. And once you've started to drift off the optimal point on the lift/drag curve, you increase the drag more and more with the rising nose. The difference is huge. A fixed drag value is of no help here. Since you have no power reserves, you have no chance to compensate the increasing drag. Just a few knots, or centigrades, or tons may already push you onto the unstable side.

The consequence of my theoretical model (which agrees with 1 and 2 engine out data very well) is so that the lift/drag allows you to maintain altitude with 3 engines out. There are no exact lift/drag data available for this, so it's a plain mathematical, theoretical result.


|-|
Member
Registered: May 2011
Posts: 92
There is the general rule on twins that you lose 70 to 75% of your power (most noticeable in the climb performance, not as much in the level flight). Roughly put, you go from a twinned 747 that already lost 70% of the total power, to another 70% reduction. That leaves you with not,even 10% of the original total power. You might gain some through depressurizing, putting on standby electrical power,etc but the engine still has to power a 747 in other areas as well. You fly unassymetrical, with slip,... Any minor mistake that puts your speed below minimum drag and you are unable to accelerate out of it (no excess power available to do that in level flight)'and you decelerate to your graveyard,... Minor thoughts but put everything into the equation and my personal feeling ends up with a sustained gradual descent. Not even close to level flight. :shock:
Moderator
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Here again, a linear extrapolation is too simple, in my opinion. A climb is something completely different than a cruise. Imagine this: While you are able to push a car forward on an flat street with your hands, you can't do this on a 5 uphill street. Pushing 3 tons forward is something else than pushing 3 tons uphill. The difference is huge. It's not a linear function where 0 is 0% and 90 is 100%. You can't push 3 tons vertically up with your hands, but you can roll 3 tons forward. You need to work with vector trigonometry here, with sine functions etc.

When you look at the empirically measured minimum engine-out cruise altitudes, you see that this 75% rule can only apply to climb situations, not for cruise.

Now, from gravity over to drag:

The second point is: The drag increases with airspeed.

The third point is: The drag increases exponentially (squared)

Another thought: Don't underestimate the power of a 747 engine. They can push 180 tons into the sky, and you can do the same with 400 tons. From those 400 tons take away, for example, 20 tons. 20 tons are a lot when you run uphill. This additional free energy can now be transformed into speed. Reducing the weight further, then finally, at 180 tons, the 747 shoots up like a rocket, you could momentarily pitch over the 30 line, making a flight path of ca. 20. Remember, the force required to compensate the gravity vector doesn't rise linearly. For 20 uphill you need way more than 120% of level flight power.

What I'm trying to say is: These engines have huge power reserves as they have to deal with great differences in gross weight (and density altitudes up to FL450). For example, check the difference of the idle descent path angle when anti-ice becomes active and idle thrust rises by just a few percent. The increase in power is not much on the instruments. But the effect on the flight range is significant. Gedankenexperiment: Instead of setting all four levers from flight idle to approach idle, leave them at flight idle and bring only one lever out of idle. Move it forward until you get the same flight range as with all four in approach idle. Do you need to push that single lever to the max? Probably not. I guess you'll have a lot of power reserves on that single lever. Now push that single lever to the max and wait until the aircraft levels off :-) ... or maybe it will never level off and keeps descending along a 1 glide path, or 1.5, or 0.5. I don't know.

I don't know if level flight is possible. I'm just saying that linear math models cannot be applied here.


Cheers,

|-|ardy


P.S.: On a twin prop, do you need max power on the good engine for level flight at low altitudes? No. Otherwise you couldn't climb with one engine. The energy requirement for a climb is enormous, not to compare with level flight. So there are great power reserves for level flight.
« Last edit by Hardy Heinlin on Mon, 06 May 2013 13:59:20 +0000. »
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Maybe you should turn around the thinking. Although not in concrete numbers, you know the power the engines need to generate. If you have all 4 running, they each provide 25% of that total sum. Turn off 3 and you get 1 engine that needs to provide 100% of that same amount of power. The question is: where do you read that amount of power?

Could you try something for me? The closest indication of power is actually fuel flow if I remember correctly. I have to go dig in the books, so don't shoot me if my memory is playing tricks on me, but on the older 737-200, you could simply double the fuel flow. 210kts clean required 1100kg/hr with two engines. Single engine you got: 210kts required 2200kg/hr fuel flow on the working engine. Now this is an older low by-pass ratio engine. For the fun of it I checked our 777. In a holding we use the number 5-6 tons/hour. QRH gives me the same amount of fuel flow on the running engine for engine out cases.

So the thing you could try: note total fuel flow for 4 engines running in the condition you want to check. Shut down one engine and after stabilization, note total fuel flow for 3 engines. Shut down the second engine and note fuel flow. What is the relation between these values? Is it the same? Does it rise? Does it decrease? Linear? If there is some kind of relationship to be found, try to deduct a required fuel flow for 1 engine running and see if the engine is actually capable of running at that fuel flow.

Just a thought (based on long gone memories... so I could be wrong here...)
« Last edit by IefCooreman on Mon, 06 May 2013 19:37:10 +0000. »
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Below are some screenshots at 250 KIAS, 3000 feet, 300 tons, ISA, with various engines out. The FF numbers (in tons) are:

0 eng out:
2.0 2.0 2.0 2.0 - sum: 8.0

1 eng out:
2.3 2.3 0.0 2.3 - sum: 6.9

2 eng out:
0.0 2.7 0.0 2.7 - sum: 5.4

3 eng out:
0.0 3.5 0.0 0.0 - sum: 3.5

:-)

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|-|
Member
Registered: Jun 2009
Posts: 250
Location: Potsdam, Germany
May I drop a though here...
Only one engine excerts an amount of yaw which will be
compensated with a wing low. This reduces the area of
the wing, results in a higher wingload and this again
increases drag... besides the additional drag of the hull.
We cannot do the calculation without knowing all drag.
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That's right. There are countless factors involved, of course. Also, by the way, for the thing with the FF we need to consider the FF required to get the engine running at all; at which point the engine will not produce useable thrust but a lot of FF already.
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I didn't realise the simplifications I was suggesting were so simple they were not going to help :)

She's a complex beast - amazing stuff!
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My accountant saw the fuel flow numbers decrease with the number of engines, and came with an interesting suggestion.
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Registered: Jul 2009
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Location: Loomis California (near Sacramento)
Some years ago my old airline (UAL) had an incident which ended up with one of our 747 -100s landing at Narita on a single engine! Had about 20,000 lbs of fuel on board but it was all in one tank (no 1 think it was).

I think that the airplane could NOT sustain level flight on the single engine, but it's been awhile and I'm not SURE about that!

I'd bet that incident is documented someplace!

(The crew was not aware that they could have used the jettison system to get fuel to the other engines. Some of the old time flight engineers liked to use the dump system to transfer fuel, and the company did not point out this ability in training . It was in the manual if you knew where to look however).

jj
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Location: Sydney, Australia
http://articles.latimes.com/1988-05-03/news/mn-2148_1_engines-lands-safely

Quote
Boeing spokesman David Jimenez in Seattle, Wash., said he was not surprised that the jet landed safely.

"Almost any aircraft we have is capable of operating on one engine," he said. But Jimenez said he did not know of another 747 losing power in three engines.


Hehe... define 'almost' and 'operating' :)
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Good story, JJ.

Re FF: Not to forget that in the lower FF range there is not much thrust. A certain part of the energy is used just to keep the engine hot and running. If I exaggerate a bit and cut off 1.5 as an operating minimum from the above sample for each engine, the result is this:

0 eng out:
0.5 0.5 0.5 0.5 - sum: 2.0

1 eng out:
0.8 0.8 0.0 0.8 - sum: 2.4

2 eng out:
0.0 1.2 0.0 1.2 - sum: 2.4

3 eng out:
0.0 2.0 0.0 0.0 - sum: 2.0

In either case, I have the feeling that the 3 eng out scenario should require more FF and the airspeed should probably decrease slowly. But it's still just a theory so far.


|-|
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In the absence of drag, lift and thrust models for the 744, this is of course a speculative but nonetheless interesting conversation.

From geometry, one can estimate the tail force needed to maintain lateral control. Obviously the fuselage takes a portion of this, perhaps as much as 30%.

Assuming lateral control is to be retained, and one CF6 engine is producing 100% of its 251.3 kN of thrust produces the following numbers:

Inboard engine yaw moment at 100.0% thrust: 2935.5 kNm
Outboard engine yaw moment at 100.0% thrust: 5235.1 kNm
Assuming straight flight is maintained:
Tail force needed for inboard engine case: 94.7 kN = 37.7% of engine power
Tail force needed for outboard engine case: 168.9 kN = 67.2% of engine power

This does not take dihedral or dropping the wing (which would make things better) or windmilling engines on the dead side (which would make things worse) into account.

This then boils the question down to whether you can keep a 744 aloft using 82 kN or 157 kN of thrust. That should be easy to test. Fly with the following parameters and try to maintain altitude: 82 kN / 4 = 20 kN which is ~8% power. 157 kN / = 39 kN or ~16% power. Will she fly like that? Well, 8% is less than flight idle, not possible to test, but 16% is about 0.7t/h FF, which just about is flight idle. As someone has mentioned before, engine power is directly proportional to the fuel flow.

Cheers

- Balt
« Last edit by Balt on Tue, 07 May 2013 02:51:18 +0000. »
Member
Registered: May 2011
Posts: 92
You won, I got the books out... and the infamous TSFC variable.

TSFC: thrust specific fuel consumption, a very important number when it comes to studying engine efficiency "as a whole" (without the inlet section though, this is testbench stuff). It's basically an indication of how much fuel the engine requires to generate 1 unit of thrust. When you study the number in function of different parameters, you get a good idea of how a jet engine works and what parts become important in what areas.

My memory is not that bad anymore apparently :-). TSFC drops _slowly_ with required engine thrust up to a certain point. The reason is fairly easy: the harder the engine has to work, the higher the turbine inlet temperatures, the higher compressor and turbine efficiencies. In cruise you get this optimum between 95 and 100% N1. At lower speeds the bucket (drop and increase of TSFC) becomes slightly more "pronounced".

I'm actually looking at an actual CF6-50 chart, to give you an idea. Based on the numbers I see on the screenshot, FF evolution seems to be correct although I'm not sure if it's going to be so pronounced. I'll quote some numbers from the charts, but since it's a chart for a cruise level (30000ft), it's only indicative. Pressure altitude doesn't really influence a lot (at low speed reynolds kicks in but not very noticeable), ambient temperature does.

M 0.4, N1 80%, TSFC 0,49
M 0.4, N1 90%, TSFC 0,47

This gives an efficiency increase of about 4%. Lower N1's will give more pronounced efficiency changes.

BTW: accountants are correct if they suggest to shut down one engine on the approach (in cruise it only leads to slower and lower flights so that's where you lose the money) ... but it did lead to engine out taxying becoming a standard procedure. :-)
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Registered: Apr 2012
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And Hardy... those screenshots literally just guaranteed that I have sweet dreams when I go to bed in the next few minutes. You know that meme image?

User posted image

Yeah that's how I'm feeling right about now.
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Registered: Jan 2013
Posts: 47
Location: YSSY
IefCooreman wrote
You won, I got the books out... and the infamous TSFC variable.


Speed does affect things a fair bit: Thrust specific fuel consumption (TSFC) can be approximated by

TSFC/TSFC_ref = 1+0.5*M

Since the Mach number depends on speed and density (which in turn depends on pressure and temperature) you have a whole kittylitter of interdependencies!

That means at 3000ft, 200kts will yield 3.4% less consumption per thrust unit than 250kts! When splitting percentage points, that might make all the difference.

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