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Windmill driven hydraulic pump

Started by Hardy Heinlin, Tue, 26 Apr 2011 16:09

Hardy Heinlin

Moin,

I wonder what kind of relationship exists between N2 RPM and EDP generated hydraulic pressure.

At normal N2 idle RPM, hydraulic pressure is 3000 psi.

With engine out and, say, IAS 160 knots, N2 may drop to 30%. How much will the hydraulic pressure drop then? Down to 2000 psi?

There must be a graph or a table, I think.

Any ideas?


Cheers,

|-|ardy

Mundyas

Hi
I looked in the book Handling the big Jet by D Davies talking about the 747 Classic.

EDP's "They produce enough Hydraulic Power provided they are kept windmilling fast enough. This will be achieved if 1.3Vs (stall speed) with a minimun of 160 knots is maintained. " He continues "keep demand low controls should not be waggled"
I expect this doesn't really answer the question and maybe where you got 160 knots from.
Appreciate I might be trying to comment on something I don't know much about!

Hardy Heinlin

#2
Thanks for the tip, Andrew.

By "enough Hydraulic Power" he probably means a value between ca. 1500 and 3000 psi.

Below 1400 psi a low press warning occurs, so that would be "not enough" anymore.


Cheers,

|-|ardy

skin

hi
thats what i´ve found ...
http://www.pprune.org/archive/index.php/t-93138.html
seems like 3% N2 are enough for 3000 psi !

QAVION jumping in towards the end of the thread - who is this guy ?  ;)

Hardy Heinlin

I see. Makes sense. It's like electric impedance. You may have 10000 V at the poles, but if you connect a consumer, it may drop to 0.3 V.

The next question then is what did Davies mean by "controls should not be waggled" (to keep the demand low).

When I permanently move the aileron, say, like in a stormy approach, with all engines' N2 RPM windmilling at 30%, how much would the two systems for the aileron depressurize? And how much, if only one system is feeding the aileron?


Cheers,

|-|ardy

Mundyas

Hi |-|

The full quote from the book in relation to an all engines flame out on a classic 747 -  " ... To keep the demand low the controls should not be waggled unnecessarily and other hydraulic services - flaps for example - should not be used."

One example where all engines flamed out was 747-236B 24 June 1982 near Jakarta due to ash ingestion, BA Captain Moody was flying that night at FL370!

Davies again (re hydraulic systems)  "a full analysis shows that the worst double failures are as follows ;-
(a)For pitch control - failure of systems 1 and 2 (or 3 and 4) resulting in the loss of one outboard and the opposite inboard elevators
(b) For lateral control - failure of systems 2 and 3, resulting in the loss of all outboard roll spoilers and reduced rate on all ailerons."

For anyone interested this book has an excellent section on 747 classics especially the hydraulics systems etc.

Have you a copy of the book?  |-| (mine is third edition - Handling the Big Jets D.P Davies). I ask as there is a very good discussion on effect of hydraulic failures on 747 Classic.

Good flying everyone but keep those engines going otherwise glide in using the controls gingerly please all pilots budding or otherwise!

A

Hardy Heinlin

Quote from: MundyasHave you a copy of the book?
Yes, but it's somewhere in my basement at the moment :-) Bought it 20 years ago. Very interesting details and very good writing style, in my opinion.


Cheers,

|-|ardy

Hardy Heinlin

#7
Is there a hydraulic expert aboard?

Let's take the aileron as an example:

The engine is windmilling slowly, we have 3000 psi as long as the aileron doesn't move.

When I move the aileron control on the flight deck to max deflection, the pressure drops to, say, 1000 psi.


What will the aileron surface on the wing do now?

a) Move to max position as usual, but extremely slowly

b) Move promptly as usual, but will never reach max position

c) A mix of a) and b)


What do you think? :-)


Cheers,

|-|ardy



P.S.: As we all know, the stab trim runs at half speed when only one instead of two hydraulic systems move it. Could this be taken as a hint towards answer a)? Or is the half stab trim speed a result of electronic system control?

Mundyas

Hi
Certainly no expert!
..........
Here is a bit more from the book!
Roll Control
Four Ailerons
Left Outboard - Systems 1 and 2
left Inboard - Systems 1 and 3
Right Inboard - 2 and 4
right outboard - 3 and 4
Ten Spoilers  -2,3 and 4
........
So "Single surfaces with a double hydraulic supply will still operate effectively in the event of a single hydraulic failure. The following are the design rates for those primary services with two hydraulic supplies
(Considering only ailerons)

Inboard ailerons - 2 hydr rate 40 down, 45 up in degrees / second - with one hydraulic system 27 down, 35 up in degrees / second
Outboard ailerons - 2 hydr rate 45 down, 55 up in degrees /second - with one hyd system 22 down, and 45 up"

I expect you know most of this already!
........

I would image the aileron would move very slowly (with all engines out etc ) until pressure was low then would stop as is.  Then maybe as pressure increased assuming it does! aileron would move to requested position.

But Davies does say "to keep demand low the controls should not be waggled unnecessarily and other hydraulic services - flaps for example -should not be used"

The strategy presumably if all engines out and EDP's windmilling would be to establish a glide profile.

As I say no expert!

Perhaps search the basement though book doesn't seem to answer yout point.

Wedding to watch or not!
Regards
Andrew

Jeroen Hoppenbrouwers

#9
No expert either, but know a bit from general hydraulics.

Hydraulics use pressure and displacement. If P is 3000 psi and you open the valve to the aileron actuator, at first nothing happens. The 3000 psi reaches the actuator and displaces it 1 mm. This leads to pressure drop. Only if the pump can supply enough to fill in the displacement gap, the pressure will remain up.

A jackscrew actuator such as that of the stabiliser basically needs the same pressure all the way to maximum deflection. A bit of pressure leads to slow displacement, but it screws slowly but surely and eventually will reach the end stop as long as the pump can suck new fluid from somewhere.

Two pumps deliver more fluid flow, and can keep up better with the large volume demand of a jackscrew.

A pure piston actuator (no screw) pushes until the counterpressure (air load) equals the pump pressure (or the end stop makes the counterpressure infinite). No matter how much fluid the pump can suck in, if the pressure isn't high enough, the actuator won't move further.

In principle jackscrew and piston actuators both need a minimum pressure to work, but the pressure does not increase much during jackscrew extension. With a constant air load, a piston works the same. Only with increasing load with extension, the actual pump pressure comes into significant play.

So it all depends on the type of actuator of the ailerons.


Jeroen

Hardy Heinlin

#10
So it's rather a matter of motion speed, not motion length. Answer c)

Quote from: MundyasThe following are the design rates for those primary services with two hydraulic supplies
(Considering only ailerons)

Inboard ailerons - 2 hydr rate 40 down, 45 up in degrees / second - with one hydraulic system 27 down, 35 up in degrees / second
Outboard ailerons - 2 hydr rate 45 down, 55 up in degrees /second - with one hyd system 22 down, and 45 up"

I expect you know most of this already!
I didn't know the exact rates. Interesting. Guess it's time to get the book from the basement :-)


Cheers,

|-|ardy





John H Watson

#12
Quote from: HardyP.S.: As we all know, the stab trim runs at half speed when only one instead of two hydraulic systems move it. Could this be taken as a hint towards answer a)? Or is the half stab trim speed a result of electronic system control?

Each hydraulic system drives a separate motor. The mechanical output of the motors drives a differential gearbox attached to a single screwjack. i.e. the speed is mechanically summed.

In addition to this, there are different flow rate selections. Different solenoids are energised, to select flow rates. The 747-400 varies the rate according to airspeed ranges.

Rgds

J>-

Hardy Heinlin

#13
Thanks, J>-
Sounds like the stab trim with its special system is no good example to get a hint regarding the aileron question. Or is this principle applied to all flight surfaces that are moved by a pair of hydraulic systems?

At least the aileron rates have no exact "x2" factor:
QuoteInboard ailerons - 2 hydr rate 40 down, 45 up in degrees / second - with one hydraulic system 27 down, 35 up in degrees / second
Outboard ailerons - 2 hydr rate 45 down, 55 up in degrees /second - with one hyd system 22 down, and 45 up

Anyway, I'm trying to create a model that is N2 dependent. The lower the N2, the lower the available max hydraulic force. I'll probably reduce both the motion rate as well as the motion length. Say, at 15% N2, the left inboard aileron moves down at a rate of only 5°/sec, and during the motion the pressure drops into the amber range, causing the aileron surface to stop after, say, 5°.

Now here's a dilemma.

When I return my control to neutral, pressure should rise again so that I regain enough energy to move the aileron surface further. On the other side, returning the controls to neutral also commands the surface to move to neutral, i.e. this alone causes the pressure to drop, also.

So I guess the pressure rises whenever the controls are not moved, no matter at which angle they currently are. This leads me to the assumption that ... :-) ... the surface will never fully stop, it might get slower, whereby the pressure rises, whereby the motion continues, whereby the pressure rises etc.

Perhaps there's a "pressure wave" oscillation? It might have a frequency of 0.3 Hz, or 13 Hz?


Cheers,

|-|ardy

Phil Bunch

I fear I may not be following this thread properly, but isn't the fundamental issue to know the total system mechanical efficiency of the hydraulic+mechanical system(s)?  In other words, if one's windmilling engines can put out X amount of energy per second and the systems are using Y amount of energy per second, only by avoiding extracting more energy such as by wiggling the ailerons or other gadgets will one maximize the longevity of the backup-system-operated hydraulics.

Again, please excuse me if I've not followed the discussion (or if my comments are not helpful).
Best wishes,

Phil Bunch

Hardy Heinlin

#15
Not sure I've understood your sentence completely, Phil :-) Perhaps it agrees with my new way of thinking: I've come to the conclusion that the hydraulically moved object will always reach its target as long as pressure is higher than drag*. It's just a matter of time.

At 3000 psi the full travel will take 0.5 sec. At 3 psi (1 hair above friction force*) it will take 2 weeks.

Since pressure drops by permanent wiggling, the pilot should move the controls slowly. This way the pressure remains pretty constant and there's no big delay between commanded and actual surface angle.


Cheers,

|-|ardy


* Or whatever force pushing against the hydraulic force.

Jeroen Hoppenbrouwers

#16
Make that (friction force plus actual current load on the object to be pushed away) and I'm in.

If you try to hydraulically push up a pallet with bricks, it is not just friction force that you need to overcome. If you add more bricks, you need to push harder (more pressure) but not further (same displacement and no fluid losses).

If there's some kind of gearing between the piston and the load to be moved, which in an extreme way is the case when a hydraulic motor (turbine) drives a gearbox or jackscrew, you need just a little above friction but a whole lot of volume.

Just as with a pulley system. With enough pulleys, you need very little 'pressure' above the friction, but you need to 'push' a whole lot.

Jeroen

Hardy Heinlin

#17
Yes.

And that means: The higher the airspeed, the more pressure is required for max surface deflection.

Of course, the actual deflection required for a normal flight will be limited by the aileron lockout (highspeed mode) and the pilot's inputs; nobody would set max deflection on all aileron panels at Mach .8.

Anyway, this leads me to a simple vector formula:
H = hydraulic force vector ("psi")
A = Ram air force vector ("airspeed")

H = A means: Aileron motion freezes at current position
H < A means: Aileron moves to neutral
H > A means: Aileron moves away from neutral


|-|ardy

Pierre Theillere

#18
Hi Hardy!

Looks like quite a clever summary! I guess that the motion speed towards "neutral" could be directly proportional to H-A due to friction.
Oooh, and when the airspeed is low, "neutral" balance position (for ailerons and elevator) is "fully down" due to weight, as it can be seen on (hydraulically) unpowered aircrafts, maybe the transition for the balancing position for ailerons and elevator between "middle" and "down" could be something progressive?
For vertical stabilizer, rudder's "neutral" position without hydraulics could also be related to wind's direction...
Pierre, LFPG

Hardy Heinlin

#19
Agreed, Pierre.

But, the idea with the wind ... I think that needs to be a storm then.


Cheers,

|-|ardy