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sam711
8th Mar 2003, 05:16
For the sake of those of us preparing for interviews who haven't any REAL experience flying heavy equipment, could some of our senior statesmen please take the time to provide some straightforward and logical answers to any or all of the following. Instead of starting 100 different threads, I thought I'd just post 100 different questions right here:)

In a typical Boeing or Airbus, is the centre of gravity always in front of the centre of pressure?

What effect does an increasing IAS have on EPR?

Can someone explain how to show where max range and endurance are located on a drag curve for a JET aircraft?
(is that one max range at min drag, and max endurance at 78% (one over the fourth root of 3) min drag?)

(technically speaking) Why is AVTUR used in jet engines?

A simple explanation of Dutch Roll?

What are Krueger and Fowler flaps?

What is a super stall?

What are the purpose of vortex generators?

How does CofG effect Vmca?

What happens to stall speed at very high altitudes?

Where does a swept wing stall first?

What happens to Mach no. during constant TAS climb? Constant IAS?

Is Vmcg always lower than Vmca?

What is the difference between dry and wet V1?

What does it mean if BRW is limited by an obstacle in the 2nd segment?

That should get the ball rolling

;)

Captain Stable
8th Mar 2003, 14:00
That's an awful lot of complex questions, sam.

Ask ten pilots one question and you will get twenty answers.

Do you have a copy of "Handling The Big Jets"? If not, I suggest you try to lay your hands on a copy.

john_tullamarine
10th Mar 2003, 20:21
... although you will get a bunch of answers here, most of these questions are well represented in the archives ....you might find that some time spent running a few searches will give you more information than you will get in this one thread.

regards ...

411A
10th Mar 2003, 20:29
Sam711

Yes, indeed, have (correct) answers to ALL of your tech questions...ground school starts next Monday at KVCV, and my guy will collect fees as you enter.

OTOH, do you not think that all of those questions is a bit much for one entry? Many folks here are more than glad to help...but will only give so much away...:rolleyes:

FlapsOne
10th Mar 2003, 22:16
If you are preparing for interview, you presumably have a license, which presumably required passing exams containing the answers to your questions.

If you need this much of a refresher, I strongly recommend you invest a few pounds/dollars in a couple of books.

Keith.Williams.
11th Mar 2003, 19:29
Okay Sam,

I think you have probably had enough grief by now. Your questions sound more like ATPL theory exam material, but here are some answers.

Yes, C of G is usually in front of the C of P of the wings. But when in trim (with no residual pitching moments) the C of G coincides with the C of P of the entire aircraft.

The EPR question probably relates to the take-off run (it usually does). In this case the EPR decreases as airspeed increases. This is why take-off EPR must be set before about 80 Knots.

If we take a simplistic look at the subjects of range and endurance then we need to remember that maximum endurance requires minimum fuel flow. Fuel flow in a propeller aircraft is proportional to power, while that in a jet is proportional to thrust. In straight and level flight, thrust equals drag. So prop aircraft give best endurance at the minimum power speed (Vmp) and jets give best endurance at the minimum drag sped (Vmd).

For best range (in still air) we need the speed that gives the best trade-off between fuel flow and TAS. You can find this by drawing a tangent from the origin to touch the power required curve for a propeller aircraft and the drag curve for a jet. The required speeds turn out to be Vmd for a prop and about 1.35 Vmd for a jet. In reality of course the matter is much more complex and involves flying at the optimum speed using the optimum RPM at the optimum altitude.

To understand why piston engines and jets use different fuels we must remember that piston engine combustion is intermittent, whereas that in a jet is constant. The intermittent combustion in a piston engine means that ignition must be achieved very often, very quickly and very reliably. To achieve this we use very volatile fuels with low flash points such as petrol.

In a jet engine the constant combustion means that a high temperature flame is available to ignite the incoming fuel. We do not therefore require a highly volatile fuel, so it is safer and much cheaper to burn kerosine-type fuels such as AVTUR. Jet engines can in fact burn virtually any flammable material that can be pumped into them. As with everything else, we can make the question much more complicated by considering such things as detonation and Octane rating, but the above explanation is essentially correct (but incomplete).

Dutch role is a phenomon whereby an aircraft performs a series of yaw-roll-yaw-roll manoeuvres. If an aircraft is laterally stable it will roll away from sideslip. If it is directionally stable it will yaw into any sidelsip. But yawing tends to cause roll and rolling tends to cause yaw. The overall result depends upon the balance between directional stability and lateral stability. If directional stability is stronger than lateral stability, a sideslipping aircraft will yaw into the sideslip to a greater extent than it rolls away from it. This yaw into sideslip will cause roll into the sideslip, which will in turn cause greater yaw. The result of this process will be spiral instability which causes the aircraft to enter a spiral dive.

If the lateral stability is stronger than the directional stability, the aircraft will roll away from the sideslip to a greater extent than it yaws into it. This will cause the aircraft to yaw in the other direction. This yawing will then cause sideslip from the other side, which will then cause the process to be repeated. Given the necessary (unhealthy) balance between the two modes of stability, the aircraft will perform a repeated yaw-roll-yaw-roll manoeuvre.

Krueger falps are a very primitive form of leading edge flap. They are fitted below the leading edges and are hinged at the front. When deployed their trailing edges are pushed down into the airflow. The overall effect is an increase in leading edge camber, which increases lift. But they are not very efficient, so they stall at a fairly low angle of attack. They are therefore used at the wing roots, with something more efficient such as slats outboard. This ensures that the wing roots stall before the tips.

(order of questions is reversed here to aid explanation)

The main reason for using swept back wings is to increase Mcrit. But swept back wings cause the airflow to migrate outboard. This means that the air takes a longer path over the wings, which causes it to slow down. The result is a very slow moving low energy boundary layer over the wing tips. As angle of attack is increased this low energy boundary layer separates easily causing the wing tips to stall before the roots.

But the wing tips are further back than the wing roots, so tip stall causes the C of P to move forward. This in turn causes the aircraft to pitch nose up in the stall. This pitch up increases the angle of attack still further, thereby making the stall more severe. The loss of lift then makes the aircraft start to descend, which increases angle of attack yet again.

The overall result of this sequence is that the aircraft moves deeper and deeper into the stall. This phenomenon is called deep stall or superstall.

Vortex generators are one of the methods commonly used to reduce spanwise flow in swept wing aircraft. They are small thin vertical aerofoils, which are glued to the upper surfaces of the wings. They are typically about one or two inches long and protrude through the boundary layer into the higher energy free stream air above. Their purpose is to energise the boundary layer by mixing it with higher energy free stream air. The turbulent boundary layer produced by this process is far more able to resist spanwise migration and separation. This process is also beneficial in reducing the effects of shock stall at high speeds.

Vmca is the minimum speed at which it is possible to maintain control following the failure of the critical engine in the air in the take-off configuration. By its legal definition it assumes that the C of G is at its most unfavourable (usually the most most aft) allowable position. So the actual position of the C of G does not affect the legally defined Vmca. A forward C of G does however make it easier to maintain control at low speeds.

When the effect of increasing altitude on indicated stalling speed
was an old CAA question, the answer was that it was constant at all altitudes. The logic then was that the same 1/2RhoVsquared was used in the lift equation and the airspeed indicator, so altitude had no effect. But since the introduction of the JAR exams, we have had to take account of the effects of compressibility. The result is a slight increase in indicated stalling speed at high altitude.

Mach number is the TAS expressed as a fraction of the local speed of sound. The local speed of sound in air is equal to 38.94 times the absolute temperature. As altitude increases up to the tropopause, the temperature and the local speed of sound both decrease. So any given TAS becomes a greater fraction of the local sped of sound as altitude increases. So mach number increases in a constant TAS climb.

The easiest way to answer this type of question is to consider a fan shape of three lines joined at the lower ends. Starting at the left, label these lines CAS, TAS and MACH. To identify what happens in a climb or descent below the tropopause, draw the fan so that the constant factor (CAS, TAS or MACH) line is vertical. The slope of the other lines then indicates how they behave, using slope to the left to mean decreasing values and slope to the right to mean increasing values. Move up the lines to simulate a climb and down to simulate a descent.

There is no regulation or physical law which requires that Vmcg must be more or less than Vmca.

BRW is the brake release weight. Increasing weight decreases the maximum climb gradient that an aircraft can achieve after take-off. The second segment of the take-off path is from the point at which the gear is retracted to the point at which the aircraft attains 400 ft. In the case of a close-in abstacle it might be necessary to reduce the take-off (BRW) weight in order to ensure that the climb gradient is sufficient to clear the obstacle safely.

sam711
12th Mar 2003, 03:24
Thank you Keith Williams for taking the time to offer some very precise and logical answers to my Q's:)

Sorry to those that thought the original post was a bit much, and yes I did cop a flogging and was about to delete it.:O

Just to clarify this in my own mind now:

With respect to range / endurance considerations I was led to believe that for a prop best range was where the tangent from the origin touched the power curve (which turns out to be the fourth root of 3 times the min power speed. Which is roughly 1.32 times the min power speed I think). Whereas best endurance was at min power.

But for a jet I thought that best range was at min drag, but best endurance was actually on the back side of the drag curve (one over the fourth root of 3 times the min drag speed. Which turns out to be about 87% of the min drag speed I think)

The difference is significant. Anyone care to comment there?

And the difference between V1, and a V1 for a contaminated rwy too please:D

P.S. Yes I do have a licence, but it has been some time since I've had the study books out, and I have never flown a transport jet aircraft so I was just seeking clarification from some friendly folks that have;)

john_tullamarine
12th Mar 2003, 03:52
Sam,

Please don't delete the initial post ... that deletes the whole thread which is very frustrating for anyone following the story.

There was nothing wrong at all with your post and Keith has gone to some trouble to address your questions. What we were trying to suggest was that a search of the archives is always good practice .. even with the benefit of Keith's words you will find even more detail in the archives .....

Peace, brother ....

Keith.Williams.
12th Mar 2003, 18:27
Sam,

I'm note sure where you got your figures for endurance and range from, but my explanation is what you need to know for the JAA ATPL POF exam. (Which probably means it is not what you need for your interview!!!)

To understand the effects of contamination on V1 we must first remember what V1 is. In effect it is the speed at which we change from being obliged to stop, to being obliged to go in the event of a single engine failure. (I know its a bit more complex than that, but you did ask for simple explanations). This means that at V1 it must be both possible to continue the take-off safely within the remaining take-off distance available or to stop within the remaining accelerate-stop distance available.

Surface contamination has two contradictory effects. Firstly the reduced friction coefficient makes it harder to stop, so V1 must be reduced. But as the depth of contamination increases, it also increases the rolling resistance of the weels being pushed through the contaminant. This tends to offset, but not entirely negate the effects of reduced friction. The overall effect of these factors is that V1 must be reduced by a significant margin when the runway is wet or lightly contaminated. But the degree of reduction required decreases as contaminant depth increases. If you still have a copy of the CAP 698 Performance Manual, you can see these effects on pages 72 and 73.

JOHN,

My comments regarding Sam getting grief were not aimed at you. I agree that a lot of valuable information and excellent explanations can be found by searching this forum. But anyone who goes to the trouble of asking a question here deserves an answer and he did not appear to be getting many.

john_tullamarine
13th Mar 2003, 03:53
All's well, Keith. I would have ventured in with some answers except that I have been up to my ears with work in recent weeks and just didn't have the time ...... your wading in was most fortunate ....