JAA Performance - conflicting feedback

[NineEighteen]

I have some conflicting feedback on a JAA Performance question.

Q. In wet conditions, what extra percentage over the calculated landing distance must be available for a turbojet?

A) 43%
B) 92%
C) 67%
D) 15%One set of feedback says (B) 92% which, I guess, includes the 1.67 factor for a turbojet as well as the 1.15 for wet conditions. I have some conflicting feedback which says the answer is (D) 15%. I was inclined to go for that because the question states what extra percentage over the calculated landing distance suggesting that the calculated landing distance already includes the 1.67 factor and so only 15% needs to be added.

Does anyone have a fresh opinion?

Thanks in advance
0918

[Alex Whittingham]

It depends what is meant by the calculated distance. If the examiner means 'calculated gross distance' the answer is 92%, if he means 'calculated net distance' it's 15%. The majority concensus is 92% but either answer could be correct.

[pugzi]

Actually the clue comes from the words "what must be available". The available distance must be a factor of 1.67 of the calculated landing distance. Remeber you are not allowed to use all the available distance.

[flystudent]

These questions ge tme every time when I practise them, I seem to interpret them wrong . Anyone got any advice (other than not having the CAA translate the exam on a computer !!)

so given the above having 2 possible correct answers would you object to the question ?

Could someone please clarrify how the 92% comes about ?

If I have calculated my landing distance to be 1000m and I get there and its raining which wasnt in my "calculation" then I woudl factor in aother 15%.

where does the 92% come from is it bascially, the calculated /60 X100 then x1.15 which seems to come to 92. so thats's the distance available, however its not really available (to use) ??!!

Thanks in advance

[pugzi]

Remember that most manuals assume paved level and dry. Therefore corrections need to made if these conditons do not apply.
The CAP manual suggests that and extra 15% needs to be added (Factor 1.15) for wet. So take 100ft then multiple by 1.15. This is you "actual" landing distance. However, in the cap manual it states "the landing distance must not exceed 60% of the landing distance available". So multiple again by 1.67 to find out what that available distance is. Now you'll see the number 192 pop into the calculator.
Your actaul landing distance which you have calculated using graphs and wet factor is not allowed to be more than 60% of the available landing distance!!. It does mean it "available to use", it means that there must be some spare runway left when you land in case you hicup or sneeze and balls up the landing. Its for safety.

[Alex Whittingham]

I don't think its as clear as you infer, pugzi. The actual regulation makes no mention of 'calculated' landing distance. It says:

(a) An operator shall ensure that the landing mass of the aeroplane determined in accordance with JAR–OPS 1.475(a) for the estimated time of landing at the destination aerodrome and at any alternate aerodrome allows a full stop landing from 50 ft above the threshold:

(1) For turbo-jet powered aeroplanes, within 60% of the landing distance available; or

(2) For turbo-propeller powered aeroplanes, within 70% of the landing distance available;

And the answer to the question lies in the interpretation of that word. 'Calculated' could mean, as you suggest, the gross LDR but it could equally well mean the net LDR, particularly as all Class A aircraft manuals have the 1.67 safety margin built into the graphs and tables so all values 'calculated' would already include that factor.

To clarify, you are allowed to use all the landing distance available but you are not allowed to plan to use all the LDA.

Flystudent, in short yes, appeal the question. It is ambiguous.

[pugzi]

Alex your right, and I neglected to mentioned that the demonstrated/planned performance must be within 60% of the runway, whether you actually do or not in real life is neither hear nor there, sorry. The joy of the JAA ambiguity.

Steve

[NineEighteen]

Thanks very much for the assistance gentlemen. I think this may be a case of inaccurate feedback. Maybe the actual JAA question is clearer.

I think I'm making progress in getting my head around the concept so hopefully I'll have half a chance in the exam.

Having done some more feedback questions last night, there are a number of identical questions with different answers depending on which FTO you believe. So I'll be back into the books today, trying to understand how that can happen! :*

Thanks again
0918

[Alex Whittingham]

It is sadly true. You can see from the above how different interpretation can lead to differing, and equally viable, views about which is the correct answer. In the end you pays your money and takes your choice. Ask the FTO why they have chosen that answer.

The fault lies with the JAA Central Question Bank, which is not properly validated. The UK CAA do their best but what is clear and unambiguous to one man is often nothing like clear to another. As an example, a recent Ops Procedure question asked the candidates to identify that, during refuelling with passengers on board, smoking is not allowed. It was phrased as 'A no smoking ban must be introduced'. At least one candidate read that as a double negative and ticked another answer.

[NineEighteen]

Thanks again Alex,

I guess it's a case of reading the questions extra carefully and if there's any doubt about what's being asked, raise the query in the normal fashion.

Here's hoping for a fair exam on Monday. Best of luck to all.

Cheers
0918

[NineEighteen]

I have another confusing question:5. Two identical turbojets are at the same altitude and same speed and have the same specific fuel consumption. Plane 1 weighs 130,000kg and the fuel flow is 4,300kg/hr. If plane 2 weighs 115,000kg, what is the fuel flow?

A) 3804kg/hr
B) 4044kg/hr
C) 3364kg/hr
D) 3530kg/hr
One set of feedback gives (A) as the answer and another gives (C). The difference being that to get 3364kg/hr the masses of the aircraft are squared instead of the direct ratio method that I used in getting 3804kg/hr?

Why on earth would I square the aircraft mass? :rolleyes:

Thanks
0918

[paco]

Looks like you could do that with a simultaneous equation

Phil

[Keith.Williams.]

This is because (as know from your previous study of JAR ATPL level POF) the coefficient of induced drag (CDI) and hence the induced drag (DI) are proportional to the square of the coefficient of lift.

You might (or might not) remember the equation CDI = CL Squared / k Pi A

Lift equals weight in level flight so if we increase weight we must increase lift, by increasing CL by the same fraction (assuming speed remains unchanged). This increases CDI and DI by the square of that fraction. So if we double the lift we get 4 times the induced drag. This requires 4 times the thrust which in a jet means 4 times the fuel flow.

So to find the new fuel flow we use:

New FF = Old FF x ((new mass/old mass)squared)

In this case New FF = 4300 x ((115/130)squared)

Which gives New FF = 3364 (option C)

All of this is in fact a bit of a fudge in that the total drag also includes some profile drag. This will not change significantly with changes in mass. But there is insufficint information in the question to include profile drag so option C is (probably) the one they are looking for.

If you ever forget which way up the new mass and old mass go in this equation, just put in some figures and see what comes out. Then apply a quick health check by noting that increasing mass must increase fuel flow. In this case the new mass (aircraft 2) is lower than the old mass (aircraft 1) so the fuel flow goes down.

This string illustrates not only the ongoing problem with JAR questions, but also a weakness in the way schools deal with feedback. Because we are all in competition with each other we do not put enough effort into comparing feedback.

Students commonly exchange feedback and every instructor will be familiar with the experience of a student saying "Ah but my feedback from ........... gives a different answer". Instead of spending a little time considering why the other school has come up with a different answer we all too frequently say something like "I cannot be responsible for what they say at........" or " Ah that just shows how little they know at ........ aren't you lucky you are studying with us?".

Students would be far better served and the question bank far healthier if schools could bring themselves to discuss such questions and point out ambiguities to the CAA.

[Alex Whittingham]

...and, to put the other side of the arguement:

Keith has, I think correctly, assumed that profile drag does not change but has continued the assumption, I think incorrectly, to completely ignore profile drag in the solution.

At VMD we know that profile drag equals induced drag. Assume each have notional 100% values, total drag is 100 + 100 = 200. Reduce induced drag by 10%, now induced drag is 90% of what it was before, profile drag is still 100%, new total drag 100 + 90 = 190, a 5% reduction in total drag, not 10%.

I would solve it like this:

At VMD the profile drag is the same as the induced drag. Not knowing the values we will give each a value of 1 so the total drag is 1+1 = 2.

We know that

SFC = fuel flow ÷ thrust

If the second plane is lighter its thrust must be less because there is less induced drag, but how much less? If the aircraft has constant speed then thrust equals drag. We know that induced drag is proportional to the square of the weight. The weight has reduced by a factor of

115000 ÷ 130000 = 0.884

so the induced drag should have reduced by

0.884 x 0.884 = 0.782

Because the profile drag is unchanged, as it does not change with mass, and the induced drag is decreased to 0.782 of its former value the total drag is now 1 + 0.782 = 1.782 as opposed to 2, a reduction of

1.782 ÷ 2 = 0.891

and therefore the fuel flow should reduce by 0.891

4300 x 0.891 = 3831 kg/hr

This answer is contrived, and will only work from a starting point of VMD where we know the values of profile drag and induced drag are the same. It could be worked in a similar way to give a different answer for 1.32VMD where induced drag is roughly a quarter of the total. The logical extension of this is that at very high speeds nearly all the drag is profile drag and weight changes therefore have very little effect on fuel flow.

As a short cut, for small changes around VMD, the fuel flow changes roughly in proportion to the weight change.

4300 x 0.884 = 3801 kg/hr

And I would go for answer (a)

[malaysian eaglet]

Concerning the question of 2 turbojets, in fact it is a quite classic french question. but some details are incorrectly translated. The two aircraft are not exactly at the same speed but at "holding speed" in the french "énoncé"==> which means that the 2 aircraft are at the same best CL/ Cd ratio ("finesse" in french). Consequently, the drags are in the ratio of the weights but also the thrusts. As the specific consumption of a turbojet can be considered as constant
we get the relationship
FF= Csp x W/ f where f is the french "finesse" (Cl/Cd ratio), Csp the specific consumption (Kg per unit of thrust and per hour, W is the weight .

[Keith.Williams.]

Alex,

Your answer is clearly a logical one, but (I think) exhibits the same weakness as mine. We are both making assumtions about what the examiners have forgotten or have forgotten to tell us.

My answer assumes that they want the student to ignore Profile drag, while yours assumes that they expect the student to go for VMD or 1.32 VMD. My approach will come closer to the truth at low speeds (but will never actually be true) whereas yours will be the more accurate at high speeds(and will be true provided you pick the correct speed).

The closest thing we have to any real clue of what they might expect is the material they publish for use in the exams. In the CAP 697 (yes I know it's for flight planning), Figure 4.4 gives the fuel flow for various masses at holding speed. Making a few quick comparisons I see that they predict the % change in fuel flow to be slightly less than the % change in weight. But it is certainly closer to a 1 to 1 ratio than any square ratio. But does this mean that the PERF examiners are using the same logic???

This string continues to illustrate the truth in my earlier comment. The FTOs would all serve their students far better if we spent more time discussing the feedback we have and what it means.

[NineEighteen]

Well, neither of these questions came up but I'm convinced we could begin another debate on at least half of the exam I just sat. The trouble is that I don't have enough time to memorise that amount of questions for querying purposes.

I found myself torn between feedback answers and my understanding of the subject. One of my nightmare questions came up...For an aircraft flying at the Long Range cruise speed, what happens to (i) specific range and (ii) fuel flow over time?

A) (i) Increases (ii) Decreases
B) (i) Decreases (ii) Increases
C) (i) Increases (ii) Increases
D) (i) Decreases (ii) DecreasesOne set of feedback gives option (A) as correct (I agreed, so I went for this one) and another set said (B) was correct. http://socjopatus.w.interia.pl/19.gif

I have no idea how I've scored in this exam as I'm possibly thinking too hard about questions that, in all likelihood, are designed to produce a simple answer. Thanks for all the replies on this thread but I'm none the wiser so I really hope I've lucked out because I'm tired and I need a rest.

0918 http://socjopatus.w.interia.pl/31.gif

[malaysian eaglet]

Sorry, but this question is a real feedback of a JAR ATPL, ...on the french side. So this question can be in the question bank. (But we can suppose..... correctly translated)

[malaysian eaglet]

The Long Range is 1% less than the Maxi range. So the following which is related to the maxi range can apply to the Long Range
PRIMO the Fuel Flow decreases because we are at at particular point of the CL/CD ratio graph (Cd/ square root of CL) and with the weight decreasing the thrust required reduces as the weight
SECUNDO(we are at high RPM, the SFC variation can be considered as neglectible) and the FF is going to vary as the weight (FF= SFCx W / (Cl/Cd ratio))
TERTIO There is an altitude optimum which is increasing with the reducing weight. However, over a certain altitude the compressibility effects increases the drag with the increase of TAS (we reach the critical Mach)and so the Long Range decreases. (This is reinforced for the Long Range as the Long Range speed is greater than the Maxi range).
As a consequence for a Long Range Flight, the Specific range increases at the beginning then decreases when the critical Mach is reached

[pugzi]

Almost

Specific range when it all comes down to it is TAS/sfc x total thrust.
SFC x Thrust being fuel flow.

Over time weight decreases and therefore the total drag drag curve will move down and left. This means that the IAS for the long range cruise speed will fall and the drag for this speed will fall.
Less drag will require less thrust, so total thrust can be reduced. Therefore fuel flow will fall. However, if IAS is being decreased to maintain the IAS LRC then TAS and MACH NO will fall correspondingly. This allows you to climb slowly, to benefit from reduced sfc. Climbing slowly as the IAS falls allows you to keep the TAS and MACH NO. constant and therefore keep the compressibility effect effect on drag out of the equation.
All in all, SR increases because of reduced thrust required, and if you climb to stay with the optimum altitude the point is further enforced as sfc reduces.

This was a duffers guide, so excuse the symplistics.

[malaysian eaglet]

Sorry, but the answer is not complete
First
Specific Air Range = (TAS/Drag) x (1/SFC)
Secondly
At low altitude the SAR will be found at 1.32 Vimd
Tertio
Over the optimum altitude for the Specific Air range when the compressibility effects appear (Vimd = Mach crit) the aircraft is flown at constant Mach Number. Below the tropopause the TAS will decrease as the LSS will decrease. Over the tropopause the TAS will be constant. For flying the Maxi range, everything will depend on the aircraft characteristics. A fast aircraft for example B747-400 is already flown at constant Mach number at relatively low altitude. Others aircraft will be flown at constant CAS up to the optimum altitude for the SAR then at constant Mach Number when the compressibility effects appear.
Suggests to read SWATTON (Aircraft performance Theory page 117 and followings)
I was forgetting one point: the Long range or the Maxi Range are quite useful after the lost of one engine (or 2 on a big A/C) in such a case even my 747-400 is flown at 1.32 Vimd and not a constant Mach (this Vimd will reduce with the weight in order to maintain the same angle of attack but will be constant for a given weight throughout the whole range of altitude) example B747-400 on 3 engines for 270t .==> 303 kt from FL50 to FL 250 the IAS reduction begin at FL 260 for a Mach Number of .746 where we can suppose the increase of drag becomes significant - typical range of work for the Mach trim) . the Long Range is also useful after depressurisation (Low altitude and low speed...)

[NineEighteen]

Thanks very much to all those who contributed to this thread. I managed to pass the exam despite being somewhat bemused by certain areas of the subject.

Very best of luck to all those taking the exam in the future but beware the dodgy feedback!