Performance questions.

[Keith.Williams.]

Having looked through the questions, I have four last doubts:
Quote:

Q56) With all engines out, a pilot wants to fly for maximumtime. Therefore he has to fly the speed corresponding to:A) the minimum drag. (Mark scheme answer)B) the critical Mach number.C) the minimum angle of descent.D) the maximum lift.

->If the actual speed we're looking for is that corresponding to minimum power required, isn't it true that due to the shapes of the graphs, the speed that gives minimum angle of descent is practically the same as the speed giving maximum rate of climb (statement from my book), which in turn correponds to minimum power required?

You appear to be confusing glide endurance with glide range.
This question is asking for the speed that gives maximum glide time (endurance). That is not the same as the speed that gives the maximum glide distance (and the minimum glide angle).

Let’s suppose that the speed for maximum glide time (Vmp) was 100 knots.
If you flew at this speed into a 100 knot headwind you would still get maximum glide time, but you would be descending vertically. The smoking hole in the ground would be directly below the point at which the engines failed. So your glide endurance would be maximum but your glide range would be zero and your glide angle would be 90 degrees. This is (hopefully) an artificial scenario, but it illustrates the difference between gliding for range and gliding for endurance.

Quote:
307) Other factors remaining constant, how does increasingaltitude affect Vx and Vy:A) Vx will decrease and Vy will increase.B) Both will increase. (Mark scheme answer)C) Both will remain the same.D) Both will decrease.For jets aircraftEAS Vx is constant Vy decreases.IAS and CAS Vx increase slightly while Vy decreases.TAS Vx and Vy both increase.

->How would the speed of Vx and Vy vary if it was for a propeller driven aircraft?
The (most probable) reason that the CAA do not ask questions about the way in which Vx and Vy for prop aeroplanes vary with altitude is that there are too many variables. Having a supercharger and/or a constant speed prop would change not only the values of Vx and Vy but also change their behaviour.

In all cases Vx would be less than Vy as long as the aeroplane was below its absolute ceiling. And Vx would be equal to Vy at the absolute ceiling. This is because there is only one speed at which steady state flight is possible at the absolute ceiling.

But the actual values of Vx and Vy would depend the type of engine (normally aspirated or supercharged) and the type of prop (fixed pitch or constant speed).

For a simple normally aspirated engine and fixed pitch prop at low altitude Vx would be slightly less than Vmp and Vy would be slightly more than Vmp (we are only talking about a knot or two either way). At the absolute ceiling they would both be slightly less than Vmp. To determine how these speeds varied as altitude increased we would need to consider the type of speed (EAS, IAS, CAS TAS) just as we did with the jet.

The good news is that there is very little risk that this subject will come up in your Perf exam.

Quote:
242) For a turbojet aeroplane, what is the maximum landingdistance for wet runways when the landing distanceavailable at an aerodrome is 3000 m?A) 2 070 m.B) 1565 m. (Mark scheme answer)C) 1800 m.D) 2609 m.->My workings: 60% of 3000= 1800m * 1.15% (as it's wet) = 2070mYou must remember that you must defactorise distance available or factorise distances used.In this case we haveLDA = 3000mDivide by 1.67 for regulatory factor (this is 1/0.7)Then divide by 1.15 for wet surface.This gives 1562mYour mistake was in multiplying by 1.15 instead of dividing.

->I still don't understand why the maximum landing distance is less for a wet runway than for a dry runway. In JAR-OPS it states it must increase 15%. So if, in dry conditions, the aircraft has to stop in 60% of the runway, it would be 1800m. If it's wet it has to stop in 1565m?

You are getting confused with the processes of factorising and defactorising distance. To overcome this problem just remember the golden rule of performance planning…..Be pessimistic and you will not be disappointed.

If we intend to land or take of at a certain mass we go into the graphs and get the graphical distance. We then increase this distance by multiplying it by the various factors to give us the “Distance Required”. This must not be greater than the distance available. In this case we acted pessimistically by increasing the distance we expected to use.

If we want to know the maximum distance at which we can land or take off we work in the opposite direction. We start with the distances available and divide them by the various factors to get the maximum graphical distance. We then put that into the graphs to get the filed limited mass. In this case we are acting pessimistically by reducing the distance that we have available.

So in this question the method is as follows
LDA = 3000m
Divide by 1.67 for regulatory factor (this is 1/0.7)
Then divide by 1.15 for wet surface.
This gives 1562m

We would then put 1562m into our graph, together with wind, pressure altitude and temperature, to get the maximum graphical LD and ensure that our actual LD was not greater than this figure.

Getting back you your question “So if, in dry conditions, the aircraft has to stop in 60% of the runway, it would be 1800m. If it's wet it has to stop in 1565m?”

No, you must still stop within 1800. But to stop within 1800 on a wet surface our mass must be reduced such that we could stop within 1562 if it were dry.

What we are saying is that we must expect to use a greater distance if the runway is wet. But this must still not be more than 60% of the LDA. So we must reduce the distance that we would use on a dry surface to ensure that our distance on a wet surface is still not greater than 60% of the LDA.

Q349)A jet aeroplane is flying long range cruise. How does the
specific range / fuel flow change?
A) Decrease / decrease.
B) Increase / decrease. (Mark scheme answer)
C) Increase / increase.
D) Decrease / increase.
->Is this answer incorrect? I would go for D.

Jet fuel flow is proportional to thrust and is cruise flight thrust = drag. So fuel flow is proportional to drag.

As time passes in the cruise the weight of the aeroplane decreases (because we are burning fuel). This reduces the drag. So we need less thrust (and hence less fuel flow) to maintain speed. So our fuel flow decreases.

Specific Range = nautical miles flown per Kg of fuel used. In this case we have reduced our fuel flow, so we will use less fuel per nautical mile. So our specific range has increased and our fuel flow has decreased.

But be careful here…..our range has decreased because we now have less fuel left on board.