Performance questions.

[Mohit_C]

Hi all,

I have quite a lot of doubts on some questions from the Volare question bank, 27 question and I would appreciate if I could get some explanations for them. Thanks in advance.

56) With all engines out, a pilot wants to fly for maximum
time. 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.
->Isn't it true that an minimum angle of descent the a/c flies at maximum (L/D) which correponds to minimum drag? Don't quite understand this one.

59) At constant thrust and constant altitude the fuel flow of
a jet engine
A) increases slightly with increasing airspeed. (Mark scheme answer)
B) is independent of the airspeed.
C) decreases slightly with increasing airspeed.
D) increases with decreasing OAT.

68) How does the thrust of fixed propeller vary during takeoff
run ? The thrust
A) increases slightly while the aeroplane speed builds up.
B) varies with mass changes only.
C) has no change during take-off and climb.
D) decreases slightly while the aeroplane speed builds up. (Mark scheme answer)
->Okay I know that fact but I don't understand why.

30) The climb gradient of an aircraft after take-off is 6% in
standard atmosphere, no wind, at 0 ft pressure altitude.
Using the following corrections:
"± 0,2 % / 1 000 ft field elevation"
"± 0,1 % / °C from standard temperature"
" - 1 % with wing anti-ice"
" - 0,5% with engine anti-ice"
The climb gradient after take-off from an airport situated
at 1 000 ft, 17° C; QNH 1013,25 hPa, with wing and
engine anti-ice operating for a functional check is :
A) 3,9 % (Mark scheme answer)
B) 4,3 %
C) 4,7 %
D) 4,9 %
->My workings: 6%-0.2%-1%-0.5%=4.3% (no temperature correction as at SL it's standard)

32) Two identical turbojet aeroplanes (whose specific fuel
consumption is assumed to be constant) are in a
holding pattern at the same altitude. The mass of the
first one is 95 000 kg and its hourly fuel consumption is
equal to 3100 kg/h. Since the mass of the second one is
105 000 kg, its hourly fuel consumption is:
A) 3787 kg/h
B) 3426 kg/h
C) 3259 kg/h (Mark scheme answer)
D) 3602 kg/h
->My workings: 3100/95000=0.03263*105000=3426.3kg/h

36) The point where Drag coefficient/Lift coefficient is a
minimum is
A) at stalling speed (VS).
B) on the "back side" of the drag curve.
C) the point where a tangent from the origin touches the drag
curve. (Mark scheme answer)
D) the lowest point of the drag curve.
->I think it's the lowest point of the drag curve.

48) An aeroplane carries out a descent from FL 410 to FL
270 at cruise Mach number, and from FL 270 to FL 100 at
the IAS reached at FL 270.
How does the angle of descent change in the first and in
the second part of the descent?
Assume idle thrust and clean configuration and ignore
compressibility effects.
A) Increases in the first part; is constant in the second. (Mark scheme answer)
B) Increases in the first part; decreases in the second.
C) Is constant in the first part; decreases in the second.
D) Decreases in the first part; increases in the second.
->Well from FL410 to FL270 at constant MN, IAS and TAS increase. As angle of descent is equal to R/C divided by TAS, in the first part would decrease. From FL270 to FL100 at constant IAS, TAS decreases leading to an increase in the angle of descent. Isn't this so?

264) The pilot of a jet aeroplane wants to use a minimum
amount of fuel between two airfields. Which flight
procedure should the pilot fly?
A) Maximum endurance.
B) Holding.
C) Long range.
D) Maximum range. (Mark scheme answer)
-> Minimum amount of fuel would mean minimum fuel flow which means maximum endurance no?

265) When V1 has to be reduced because of a wet runway the
one engine out obstacle clearance / climb performance:
A) increases / increases.
B) remains constant / remains constant.
C) decreases / decreases.
D) decreases / remains constant. (Mark scheme answer)
->Well it's obvious at one engine out the obstacle clearance decreases but so does the climb performance, right?

331) The take-off distance of an aircraft is 600m in standard
atmosphere, no wind at 0 ft pressure-altitude.
Using the following corrections:
"± 20 m / 1 000 ft field elevation"
"- 5 m / kt headwind"
"+ 10 m / kt tail wind"
"± 15 m / % runway slope"
"± 5 m / °C deviation from standard temperature"
The take-off distance from an airport at 1 000 ft
elevation, temperature 17°C, QNH 1013,25 hPa, 1% upslope,
10 kt tail wind is:
A) 555 m
B) 685 m
C) 755 m (Mark scheme answer)
D) 715 m
->My workings: 600m+20m+15m+100m=735m (yet again, the temperature at SL is standard)

344) The take-off distance of an aircraft is 800m in standard
atmosphere, no wind at 0 ft pressure-altitude.
Using the following corrections :
"± 20 m / 1 000 ft field elevation "
"- 5 m / kt headwind "
"+ 10 m / kt tail wind "
"± 15 m / % runway slope "
"± 5 m / °C deviation from standard temperature "
The take-off distance from an airport at 2 000 ft
elevation, temperature 21°C, QNH 1013.25 hPa, 2% upslope,
5 kt tail wind is :
A) 810 m
B) 970 m (Mark scheme answer)
C) 890 m
D) 870 m
->My workings: 800m+40m+10m+30m+50m=930m (temperature at SL is ISA+2)

306) For a piston engined aeroplane, the speed for maximum
range is :
A) that which gives the maximum lift to drag ratio. (Mark scheme answer)
B) that which givesthe minimum value of drag.
C) that which givesthe maximun value of lift
D) 1.4 times the stall speed in clean configuration.
-> Doesn't A correpond to the value of minimum drag?

307) Other factors remaining constant, how does increasing
altitude 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.
->My book states that Vx increases and Vy decreases with increasing altitude.

197) To minimize the risk of hydroplaning during landing the
pilot should:
A) make a "positive" landing and apply maximum reverse thrust
and brakes as quickly as possible.
B) use maximum reverse thrust, and should start braking below
the hydroplaning speed.
C) use normal landing-, braking- and reverse technique.
D) postpone the landing until the risk of hydroplaning no longer
exists.
->This question had no answer, but I would think it's B. Is this correct?

189) For a jet transport aeroplane, which of the following is
the reason for the use of 'maximum range speed' ?
A) Minimum specific fuel consumption. (Mark scheme answer)
B) Minimum fuel flow.
C) Longest flight duration.
D) Minimum drag.

174) Given that the characteristics of a three engine turbojet
aeroplane are as follows:
Thrust = 50 000 Newton / Engine
g = 10 m/s²
Drag = 72 569 N
Minimum gross gradient (2nd segment) = 2.7%
SIN(Angle of climb) = (Thrust- Drag) / Weight
The maximum take-off mass under 2nd segment
conditions is:
A) 101 596 kg (Mark scheme answer)
B) 286 781 kg
C) 74 064 kg
D) 209 064 kg
->My workings give me B.

180) Which statement about reduced thrust is correct?
A) In case of reduced thrust V1 should be decreased.
B) Reduced thrust can be used when the actual take-off mass
is less than the field length limited take-off mass. (Mark scheme answer)
C) Reduced thrust is primarily a noise abatement procedure.
D) Reduced thrust is used in order to save fuel.
-> I agree with B, but A made me doubt too. Why would V1 be increased with reduced thrust, if A is not correct?

176) Which of the following set of factors could lead to a V2
value which is limited by VMCA?
A) High take-off mass, low flap setting and high field elevation.
B) Low take-off mass, high flap setting and low field elevation. (Mark scheme answer)
C) Low take-off mass, low flap setting and low field elevation.
D) High take-off mass, high flap setting and low field elevation.
-> I would agree with the low take-off mass and low field elevation but how does a high flap setting increase VMCA (as the question states V2 limited by VMCA)? My books state that VMCA decrease with higher flaps. What do you guys think?

242) For a turbojet aeroplane, what is the maximum landing
distance for wet runways when the landing distance
available 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) = 2070m

248) The speed for maximum lift/drag ratio will result in :
A) The maximum endurance for a propeller driven aeroplane.
B) The maximum angle of climb for a propeller driven
aeroplane.
C) The maximum range for a propeller driven aeroplane. (Mark scheme answer)
D) The maximum range for a jet aeroplane.
-> I agree with C but doesn't it also give maximum angle of climb?

229) The 'climb gradient' is defined as the ratio of
A) true airspeed to rate of climb.
B) rate of climb to true airspeed.
C) the increase of altitude to horizontal air distance expressed
as a percentage. (Mark scheme answer)
D) the increase of altitude to distance over ground expressed
as a percentage.
->My book says: gamma=R/C divided by TAS.

Sorry for the number of questions, perhaps I have confused some main concepts of Performance.

Thanks.

[Pitch+Power]

130

Climb Gradient = (Thrust - Drag) / weight x 100%

( 4 x 75000 ) - ( 150000 x 10 / 14 ) drag= mass x g / 14
----------------------------------
150000 x 10

= 0.1286 x 100% = 12.86 %

351
TurboJET is 60% LDA TurboPROP is 70%

353
There is a minimum climb requirement, from assumed engine failure, must be able to maintain a positive rate of climb at 400', and not less than 0.75% at 1500'

that's all I got time for at the moment...! :8

[RTN11]

46) I think they are trying to say that the take off mass is limited by the speed of the tyres, therefore the downhill slope wouldn't help because the tyres still wouldn't be able to go any quicker.

273) the regulations are that the clearway used cannot be more than 1.5 x take off run available. So even though there is 2000m of clearway, you can only use 1500m of it (3000 x 0.5)

[Edgington]

30) The climb gradient of an aircraft after take-off is 6% in
standard atmosphere, no wind, at 0 ft pressure altitude.
Using the following corrections:
"± 0,2 % / 1 000 ft field elevation"
"± 0,1 % / °C from standard temperature"
" - 1 % with wing anti-ice"
" - 0,5% with engine anti-ice"
The climb gradient after take-off from an airport situated
at 1 000 ft, 17° C; QNH 1013,25 hPa, with wing and
engine anti-ice operating for a functional check is :
A) 3,9 % (Mark scheme answer)
B) 4,3 %
C) 4,7 %
D) 4,9 %
->My workings: 6%-0.2%-1%-0.5%=4.3% (no temperature correction as at SL it's standard)

But you do need temperature correction, at 1000ft temp in I.S.A. would be standard 15 - 2 = 13. Now it is 4 degrees above that.

68) How does the thrust of fixed propeller vary during takeoff
run ? The thrust
A) increases slightly while the aeroplane speed builds up.
B) varies with mass changes only.
C) has no change during take-off and climb.
D) decreases slightly while the aeroplane speed builds up. (Mark scheme answer)
->Okay I know that fact but I don't understand why.

I also don't understand the answer, it says in the powerplants book that a fixed pitch prop is most efficient at cruising speeds. So to me it would seem that thrust would increase to start with.

[Torque2]

The corrections for temperature variation from ISA need to be applied.

If the airport was at sea level, the assumed temperature would be +15C.
The temperature is assumed to drop at a rate of 2C/1000 feet as you know, but from the origin at sea level, so at 1000 ft the assumed temperature should be 13C giving an ISA temperature difference of 17-13 = 4C.

Make that correction and then the rest of your calculation s are correct. :ok:

[Keith.Williams.]

56) With all engines out, a pilot wants to fly for maximum
time. 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.
->Isn't it true that an minimum angle of descent the a/c flies at maximum (L/D) which correponds to minimum drag? Don't quite understand this one.

Flying for maximum time means flying for maximum endurance. An aeroplane in flight uses up energy in order to move forward through the air. This energy is provided by the engines in powered flight.

When the engines stop in flight the aeroplane has a limited store of energy (kinetic energy by virtue of mass and airspeed, plus potential energy by virtue of its weight and height). During the subsequent glide it uses up this energy in moving forward through the air. Glide endurance is the time it takes to use up this store of energy.

The rate of using up energy is the power required. So for maximum glide endurance we must glide an the minimum power required speed Vmp. Gliding at Vmd give maximum still air glide range, but not maximum glide endurance. So there is no correct answer in this question. Options A and C are actually the same speed.


59) At constant thrust and constant altitude the fuel flow of
a jet engine
A) increases slightly with increasing airspeed. (Mark scheme answer)
B) is independent of the airspeed.
C) decreases slightly with increasing airspeed.
D) increases with decreasing OAT.

In steady state conditions the engine fuel control system will maintain a constant fuel/air ratio. If airspeed increases at constant altitude, the mass flow rate of air passing through the engines will increase. The fuel control unit will increase fuel flow to maintain constant fuel/air ratio.

Strictly speaking this will not give constant thrust, so this question is not entirely correct for all possible airspeeds. If we were to accelerate at constant throttle setting, the thrust would decrease slightly up to a certain airspeed then start to increase again. (You need to look at your AGK notes for this). So in the low speed range where thrust is decreasing slightly as airspeed increases, we will need to burn more fuel to maintain constant thrust.



68) How does the thrust of fixed propeller vary during takeoff
run ? The thrust
A) increases slightly while the aeroplane speed builds up.
B) varies with mass changes only.
C) has no change during take-off and climb.
D) decreases slightly while the aeroplane speed builds up. (Mark scheme answer)
->Okay I know that fact but I don't understand why.

Both props and jets create thrust by taking a mass of air and accelerating it rearwards. The thrust is the Newton’s Third Law Reaction to this process.

Newton’s Second Law tells us that the thrust = Mass x Acceleration

For a constinuous process such as an aeroplane propulsion system we modify this slightly to give

Thrust = Mass Airflow x Change in speed of the airflow.

For a prop this gives

Prop thrust = mass airflow x (Propwash speed – TAS)

Before brake release TAS is zero so we have

Thrust = Mass airflow x (Propwash speed – 0)

But as we accelerate after brake release the TAS gradually increases, so the acceleration that we give to the air decreases. This causes the thrust to decrease as the aeroplane speed builds up.

So thrust is maximum before brake release, then it gradually decreases as aeroplane speed builds up.



30) The climb gradient of an aircraft after take-off is 6% in
standard atmosphere, no wind, at 0 ft pressure altitude.
Using the following corrections:
"± 0,2 % / 1 000 ft field elevation"
"± 0,1 % / °C from standard temperature"
" - 1 % with wing anti-ice"
" - 0,5% with engine anti-ice"
The climb gradient after take-off from an airport situated
at 1 000 ft, 17° C; QNH 1013,25 hPa, with wing and
engine anti-ice operating for a functional check is :
A) 3,9 % (Mark scheme answer)
B) 4,3 %
C) 4,7 %
D) 4,9 %
->My workings: 6%-0.2%-1%-0.5%=4.3% (no temperature correction as at SL it's standard)

You are correct with the exception of the temperature.

At 1000 ft the temperature should be 13 degrees C (that is 15 – 2).

ISA deviation is the temperature that we have minus the temperature that we would have in ISA.

In this case temperature deviation = 17 – 13 = +4. This means that we must also subtract 0.4% from our climb gradient to give 3.9%.



32) Two identical turbojet aeroplanes (whose specific fuel
consumption is assumed to be constant) are in a
holding pattern at the same altitude. The mass of the
first one is 95 000 kg and its hourly fuel consumption is
equal to 3100 kg/h. Since the mass of the second one is
105 000 kg, its hourly fuel consumption is:
A) 3787 kg/h
B) 3426 kg/h
C) 3259 kg/h (Mark scheme answer)
D) 3602 kg/h
->My workings: 3100/95000=0.03263*105000=3426.3kg/h

You are correct. If you look at the Volare database you will find two questions of this type. One has the correct answer, but this one does not.

Provided it is a Jet aircraft flying at Vmd and we ignore the slight change in Vmd due to the weight change, we can assume that for any % change in weight we will get the same % change in drag. This enables us to use the equation

Fuel Flow 1 / Weight 1 = Fuel Flow 2 / Weight 2

Which is essentially what you have done.


36) The point where Drag coefficient/Lift coefficient is a
minimum is
A) at stalling speed (VS).
B) on the "back side" of the drag curve.
C) the point where a tangent from the origin touches the drag
curve. (Mark scheme answer)
D) the lowest point of the drag curve.
->I think it's the lowest point of the drag curve.


If Drag Coefficient / Lift Coefficient is minimum then Lift Coefficient / Drag Coefficient is maximum. This occurs at Vmd. This occurs at the lowest point on the drag curve. BUT This is also the point at which a tangent from the origin touches the Power Required Curve. So be careful with this type of question.


48) An aeroplane carries out a descent from FL 410 to FL
270 at cruise Mach number, and from FL 270 to FL 100 at
the IAS reached at FL 270.
How does the angle of descent change in the first and in
the second part of the descent?
Assume idle thrust and clean configuration and ignore
compressibility effects.
A) Increases in the first part; is constant in the second. (Mark scheme answer)
B) Increases in the first part; decreases in the second.
C) Is constant in the first part; decreases in the second.
D) Decreases in the first part; increases in the second.
->Well from FL410 to FL270 at constant MN, IAS and TAS increase. As angle of descent is equal to R/C divided by TAS, in the first part would decrease. From FL270 to FL100 at constant IAS, TAS decreases leading to an increase in the angle of descent. Isn't this so?

In a constant Mach descent in the ISA troposphere the TAS will increase to match the increasing local speed of sound. To achieve this in a glide you must gradually push the nose down. This will increase the angle of descent.

In a constant IAS decent the TAS must decrease. But the rate of change will be much less than that in a constant Mach descent. To achieve the required rate of deceleration we need to set an angle of attack at which the drag is slightly greater than the component of weight that is acting down the flight path. Drag at any given IAS (strictly speaking we should use EAS, but IAS is pretty close) is proportional to angle of attack. So to maintain constant drag as we descend we need to maintain constant angle of attack. If we do this the angle of descent will remain constant.

So the required option in this question is A.


264) The pilot of a jet aeroplane wants to use a minimum
amount of fuel between two airfields. Which flight
procedure should the pilot fly?
A) Maximum endurance.
B) Holding.
C) Long range.
D) Maximum range. (Mark scheme answer)
-> Minimum amount of fuel would mean minimum fuel flow which means maximum endurance no?

If we burn minimum fuel over a given time we will be flying at the speed at which our fuel load will last for the greatest possible time. That means that we will be at our maximum endurance speed.

But if we burn minimum fuel between two points we will be flying at the speed at which our fuel load will give maximum range.



265) When V1 has to be reduced because of a wet runway the
one engine out obstacle clearance / climb performance:
A) increases / increases.
B) remains constant / remains constant.
C) decreases / decreases.
D) decreases / remains constant. (Mark scheme answer)
->Well it's obvious at one engine out the obstacle clearance decreases but so does the climb performance, right?

You are missing the point of this question. Our obstacle clearance and climb performance calculations are always based on the assumption that the critical engine will be seen to fail at V1.

This means that we must accelerate from V1 to V2 with one engine out.
When we reduce V1 for a wet runway we do not reduce V2. This means that we need to do a greater part of our acceleration to V2 with one engine out. This means that decreasing V1 increases the take-off distance to reach screen height. This in turn means that we will be lower when we fly over any obstacles in the climb out. So reducing V1 for a wet runway will reduce our obstacle clearance.

But climb performance depends on thrust, drag and weight. None of these are affected by the value of V1. (In effect we can say that the aeroplane is no longer affected by the runway conditions after it has lifted off.) So reducing V1 for a wet runway will not affect climb performance.


331) The take-off distance of an aircraft is 600m in standard
atmosphere, no wind at 0 ft pressure-altitude.
Using the following corrections:
"± 20 m / 1 000 ft field elevation"
"- 5 m / kt headwind"
"+ 10 m / kt tail wind"
"± 15 m / % runway slope"
"± 5 m / °C deviation from standard temperature"
The take-off distance from an airport at 1 000 ft
elevation, temperature 17°C, QNH 1013,25 hPa, 1% upslope,
10 kt tail wind is:
A) 555 m
B) 685 m
C) 755 m (Mark scheme answer)
D) 715 m
->My workings: 600m+20m+15m+100m=735m (yet again, the temperature at SL is standard)

You have made the same error as you did in question 30. Temperature deviation in this question = +4 degrees C so you need to add 20 m to give the correct solution of 755m


344) The take-off distance of an aircraft is 800m in standard
atmosphere, no wind at 0 ft pressure-altitude.
Using the following corrections :
"± 20 m / 1 000 ft field elevation "
"- 5 m / kt headwind "
"+ 10 m / kt tail wind "
"± 15 m / % runway slope "
"± 5 m / °C deviation from standard temperature "
The take-off distance from an airport at 2 000 ft
elevation, temperature 21°C, QNH 1013.25 hPa, 2% upslope,
5 kt tail wind is :
A) 810 m
B) 970 m (Mark scheme answer)
C) 890 m
D) 870 m
->My workings: 800m+40m+10m+30m+50m=930m (temperature at SL is ISA+2)

You have made the same error again here.

Temperature deviation = 21 – (15 – 4) = +10. So you must add 50m to give the correct answer of 970m.

Remember

Temp deviation = Actual temperature – ISA temperature for that altitude.


306) For a piston engined aeroplane, the speed for maximum
range is :
A) that which gives the maximum lift to drag ratio. (Mark scheme answer)
B) that which givesthe minimum value of drag.
C) that which givesthe maximun value of lift
D) 1.4 times the stall speed in clean configuration.
-> Doesn't A correpond to the value of minimum drag?


Max endurance speed = Vmp (at the bottom of the power required curve)
Max range speed = Vmd which is where a tangent from the origin touches the power required curve.
Vmd is also the speed at which lift/drag ratio is maximum.

This question is bit dubious because Vmd also gives minimum drag and (in another Performance question) is about 1.4 Vs1 for a typical piston prop aircraft.

This question has been successfully appealed in the distant past.



307) Other factors remaining constant, how does increasing
altitude 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.
->My book states that Vx increases and Vy decreases with increasing altitude.

This question and numerous others like it have been successfully appealed many times in the past. This is because the answers depends upon the type of aeroplane )prop or jet) and the type of speed (EAS, IAS, CAS, TAS)

For jets aircraft

EAS Vx is constant Vy decreases.
IAS and CAS Vx increase slightly while Vy decreases.
TAS Vx and Vy both increase.

In all cases Vx is less than Vy below the absolute ceiling and Vx = Vy at the absolute ceiling.


197) To minimize the risk of hydroplaning during landing the
pilot should:
A) make a "positive" landing and apply maximum reverse thrust
and brakes as quickly as possible.
B) use maximum reverse thrust, and should start braking below
the hydroplaning speed.
C) use normal landing-, braking- and reverse technique.
D) postpone the landing until the risk of hydroplaning no longer
exists.
->This question had no answer, but I would think it's B. Is this correct?

If there is a risk of hydroplaning you do not know how well your wheel brakes will work. You should also use reverse thrust to maximise your chance of stopping in time. So you need to start braking as soon as possible. Option A is correct.



189) For a jet transport aeroplane, which of the following is
the reason for the use of 'maximum range speed' ?
A) Minimum specific fuel consumption. (Mark scheme answer)
B) Minimum fuel flow.
C) Longest flight duration.
D) Minimum drag.


Maximum Range Cruise Speed Vmrc gives maximum range for a give fuel load. Maximum possible still air range would be achieved if we were to fly such that we had maximum TAS/Drag ratio (to give best airframe efficiency) and simultaneously have the minimum SCF (to give maximum engine fuel efficiency). This occurs when running at 85% to 95% rpm in cold high-density air. The most important factor for SFc is the rpm band. The optimum cruise altitude (in still air) is that at which the thrust at 85% to 95% rpm equals the drag at Vmrc. Flying at Vmrc at this altitude will give maximum still air range.

Strictly speaking this will not give the absolute minimum SFC because air density at that altitude will be low, but it will be pretty close to it.

Vmrc is about 1.32 Vmd, so none of the other options are true.


174) Given that the characteristics of a three engine turbojet
aeroplane are as follows:
Thrust = 50 000 Newton / Engine
g = 10 m/s²
Drag = 72 569 N
Minimum gross gradient (2nd segment) = 2.7%
SIN(Angle of climb) = (Thrust- Drag) / Weight
The maximum take-off mass under 2nd segment
conditions is:
A) 101 596 kg (Mark scheme answer)
B) 286 781 kg
C) 74 064 kg
D) 209 064 kg
->My workings give me B.

You have probably overlooked the fact that the Second Segment climb requirement is based one the critical engine being inoperative. So this is effectively a twin engine aeroplane. Rework your calculation on this basis and you will probably get the correct answer.


180) Which statement about reduced thrust is correct?
A) In case of reduced thrust V1 should be decreased.
B) Reduced thrust can be used when the actual take-off mass
is less than the field length limited take-off mass. (Mark scheme answer)
C) Reduced thrust is primarily a noise abatement procedure.
D) Reduced thrust is used in order to save fuel.
-> I agree with B, but A made me doubt too. Why would V1 be increased with reduced thrust, if A is not correct?

The main reason for using reduced thrust is to preserve engine life by running them at lower RPM. But you can do this only when you have spare performance available.

Reducing thrust will increase the distance needed to lift-off and reach screen height. This will reduce your field limited TOM. So you can use reduced thrust for take-off only if your actual Tom is less that the field limited TOM. It must also be less than the climb limited TOM and the obstacle limited TOM.

But using reduced thrust will not reduce the required minimum V1. Remember you must be able to see an engine failure at V1 and still accelerate to V2 and screen height within the TODA. With less thrust you will need to be closer to V2 when you see the engine failure at V1.



176) Which of the following set of factors could lead to a V2
value which is limited by VMCA?
A) High take-off mass, low flap setting and high field elevation.
B) Low take-off mass, high flap setting and low field elevation. (Mark scheme answer)
C) Low take-off mass, low flap setting and low field elevation.
D) High take-off mass, high flap setting and low field elevation.
-> I would agree with the low take-off mass and low field elevation but how does a high flap setting increase VMCA (as the question states V2 limited by VMCA)? My books state that VMCA decrease with higher flaps. What do you guys think?

V2 min is limited by

1.13 Vsr for turboprops with 2 or 3 engines and fro jets with no means of reducing their one-engine-out stall speed.
1.05 Vsr for turbo-props with more than 3 engines and for jets with a means of reducing their one-engine-out stall speed.
1.1 Vmca for all aircraft.

This means that the minimum value of V2 cannot be less than the greater of 1.13 or 1.08 of the stall speed or by 1.1 VMCA.

So anything which reduces Vsr or increases Vmca will tend to make Vmca the most limiting factor.

Low mass and high flap angles both reduce Vsr.


Vmca is not related to flap angle. Vmca is determined by the amount of thrust that is available from the live engine. Any factor which increases thrust will increase Vmca. Thrust is proportional to mass airflow rate through the engine. This in turn is proportional to air density. So anything which increases air density will increase thrust and increase Vmca. This will make Vmca more likely to limited the minimum value of V2. Factors which increase air density are low field elevation, low pressure altitude, low air temperature and low humidity.



242) For a turbojet aeroplane, what is the maximum landing
distance for wet runways when the landing distance
available 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) = 2070m

You must remember that you must defactorise distance available or factorise distances used.

In this case we have

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

Your mistake was in multiplying by 1.15 instead of dividing.


248) The speed for maximum lift/drag ratio will result in :
A) The maximum endurance for a propeller driven aeroplane.
B) The maximum angle of climb for a propeller driven
aeroplane.
C) The maximum range for a propeller driven aeroplane. (Mark scheme answer)
D) The maximum range for a jet aeroplane.
-> I agree with C but doesn't it also give maximum angle of climb?

For a simple normally-aspirated, fixed pitch prop aeroplane Vx is a smidgin less than Vmp and Vy is a smidgin more that Vmp. Most of the piston/prop performance questions are based on these figures.

Supercharging and constant speed props move Vy closer to Vmd.


229) The 'climb gradient' is defined as the ratio of
A) true airspeed to rate of climb.
B) rate of climb to true airspeed.
C) the increase of altitude to horizontal air distance expressed
as a percentage. (Mark scheme answer)
D) the increase of altitude to distance over ground expressed
as a percentage.
->My book says: gamma=R/C divided by TAS.

Strictly speaking climb gradient = height gain divided by distance flown over the ground. Option D is closest to this definition.

But if we assume still air conditions then we can also say that climb gradient = height gain / distance flown through the air.

And dividing both height gain and distance flown through the air by time gives

Climb gradient = ROC/TAS


So we could argue that options B, C and D are all correct(ish). You should always appeal any question that appears to have multiple correct options.

[redsnail]

Keith, thank you. :ok:

[Mohit_C]

Thanks everyone, and especially Keith for the time!:ok:

[Mohit_C]

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


Q56) With all engines out, a pilot wants to fly for maximum
time. 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?


307) Other factors remaining constant, how does increasing
altitude 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 aircraft

EAS 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?


242) For a turbojet aeroplane, what is the maximum landing
distance for wet runways when the landing distance
available 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) = 2070m

You must remember that you must defactorise distance available or factorise distances used.

In this case we have

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

Your 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?

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.

Thanks.

[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.

[Mohit_C]

Thanks Keith, that has cleared my doubts.:ok:

[Mohit_C]

Practicing some questions involving annexes I have come across the following three which I can't seem to get the right answer, apart from the fact that they're horribly printed:

Q247) (For this question use annex 032-6569A or Performance
Manual SEP 1 Figure 2.4)
With regard to the landing chart for the single engine
aeroplane determine the landing distance from a height
of 50 ft .
Given :
O.A.T : 27 °C
Pressure Altitude: 3000 ft
Aeroplane Mass: 2900 lbs
Tailwind component: 5 kt
Flaps: Landing position (down)
Runway: Tarred and Dry
A) approximately : 1120 feet
B) approximately : 1700 feet
C) approximately : 1370 feet
D) approximately : 1850 feet (mark scheme answer)

http://img367.imageshack.us/img367/8056/q247wb9.png (http://imageshack.us)
http://img367.imageshack.us/img367/q247wb9.png/1/w748.png (http://g.imageshack.us/img367/q247wb9.png/1/)

Q218) (For this question use annex 032-6581A or Performance
Manual SEP 1 Figure 2.3)
Using the climb performance chart, for the single engine
aeroplane, determine the ground distance to reach a
height of 2000 ft above the reference zero inthe
following conditions:
Given :
O.A.T. at take-off: 25°C
Airport pressure altitude: 1000 ft
Aeroplane mass: 3600 lbs
Speed: 100 KIAS
Wind component: 15 kts Headwind
A) 24 637 ft
B) 18 832 ft
C) 18 347 ft (mark scheme answer)
D) 21 505 ft

http://img371.imageshack.us/img371/7803/q218ao8.png (http://imageshack.us)
http://img371.imageshack.us/img371/q218ao8.png/1/w736.png (http://g.imageshack.us/img371/q218ao8.png/1/)

Q232) (For this question use annex 032-4733A or Performance
Manual MRJT 1 Figure 4.28)
What is the minimum field length required for the worst
wind situation, landing a twin jet aeroplane with the antiskid
inoperative?
Elevation: 2000 ft
QNH: 1013 hPa
Landing mass: 50 000 kg
Flaps: as required for minimum landing distance
Runway condition: dry
Wind:
Maximum allowable tailwind: 15 kt
Maximum allowable headwind: 50 kt
A) 2600 m.
B) 2700 m.
C) 2900 m.
D) 3100 m. (mark scheme answer)

http://img371.imageshack.us/img371/577/q232cv1.png (http://imageshack.us)
http://img371.imageshack.us/img371/q232cv1.png/1/w563.png (http://g.imageshack.us/img371/q232cv1.png/1/)

Thanks.

[Keith.Williams.]

The first point to note is that your Annexes are not up to the current amendment state as used by the UK CAA. But they are the ones that were used for these questions when first used.

The second point tio note is that in many graphical questions you will not get exactly the correct answer. But you should be close enough to select the correct option.

Q247) (For this question use annex 032-6569A or Performance
Manual SEP 1 Figure 2.4)
With regard to the landing chart for the single engine
aeroplane determine the landing distance from a height
of 50 ft .
Given :
O.A.T : 27 °C
Pressure Altitude: 3000 ft
Aeroplane Mass: 2900 lbs
Tailwind component: 5 kt
Flaps: Landing position (down)
Runway: Tarred and Dry
A) approximately : 1120 feet
B) approximately : 1700 feet
C) approximately : 1370 feet
D) approximately : 1850 feet (mark scheme answer)

Using the current version of this graph I get 1790 ft. this is closest to option D 1859 ft.


Q218) (For this question use annex 032-6581A or Performance
Manual SEP 1 Figure 2.3)
Using the climb performance chart, for the single engine
aeroplane, determine the ground distance to reach a
height of 2000 ft above the reference zero inthe
following conditions:
Given :
O.A.T. at take-off: 25°C
Airport pressure altitude: 1000 ft
Aeroplane mass: 3600 lbs
Speed: 100 KIAS
Wind component: 15 kts Headwind
A) 24 637 ft
B) 18 832 ft
C) 18 347 ft (mark scheme answer)
D) 21 505 ft

Putting the data into the graph gives ROC = 1130 fpm

Height gain to get to 2000ft above RZ = 2000 - 50(screen height) = 1950 ft

Dividing height gain by ROC gives a climb time of 1.725663717 minutes

Putting the mean altitude (2000 ft and mean OAT 23 deg C) into the CRP 5 gives TAS = 105 kts

Ground distance = time x ground speed

If we assume nil wind then ground speed is approximately equal to TAS = 105 knots giving distance = 105 kts 101.3 ft/min/knot x 1.725663717 minutes = 18355 ft.

Using the 15 kt headwind unfactored (ground speed = 105 - 15 = 90 knots)
This gives distance = 90 x 101.3 x 1.725663717 = 15732 ft
Note, This is how the CAP 698 state you should do it for the SEP1.

Uisng 50% of the 15 knot headwind (ground speed = 105 - 7.5 = 97.5)
This gives 97.5 x 101.3x 1.725663717 = 17044 ft

All of these answers are lower than the lowest option (which is the correct option). So you can identify 18347 as the correct option.


Q232) (For this question use annex 032-4733A or Performance
Manual MRJT 1 Figure 4.28)
What is the minimum field length required for the worst
wind situation, landing a twin jet aeroplane with the antiskid
inoperative?
Elevation: 2000 ft
QNH: 1013 hPa
Landing mass: 50 000 kg
Flaps: as required for minimum landing distance
Runway condition: dry
Wind:
Maximum allowable tailwind: 15 kt
Maximum allowable headwind: 50 kt
A) 2600 m.
B) 2700 m.
C) 2900 m.
D) 3100 m. (mark scheme answer)

The worst case wind is 15 knot tailwind.
The least distance will be with flap 40

Go into the graph at 50000 Kg mass and move horizontally to the 1000 ft pressure altitude in the top right set of lines (Anti-skid inop)

Then go vertially down through the flpa sacles (bacuse you are using flap 40) and straight through surface condition (because runway is dry) down to 15 kt tailwind.

Floow the sloping lines up to the right to the reference line then go vertically down to come out at 3100 m.

[Mohit_C]

I have a question. The rule of using 50% of a headwind and 150% of a tailwind, is it to be applied whilst doing all the annexes? I have been doing the annexes just with the headwind or tailwind given, not using that factor. Is this correct?

[Keith.Williams.]

In the UK the method for dealing with this type of question is detailed in the CAP698. This book includes the graphs plus worked examples.

For the SEP1 the worked example of the calculation of distance to climb uses the wind without any factorisation. But in a similar worked example for the MEP1 the worked solution includes 50% of the stated headwind.

Logically the factorisation should apply to both. But that is no guarantee that the exam questions have not been produced in accordance with the two different methods shown in the CAP698.

[Mohit_C]

So basically I will just have to see the example in the exam and figure it out. Apart from that, thanks for the previous replies.