Joined: 11 Dec 2005
|Posted: Thu Apr 26, 2012 11:43 am Post subject: Collection of Numerical Problems involved in Performance
RW17 touchdown elevation = 146 feet
TORA = 1400 m
ASDA = 1600 m
TODA = 1800 m
RW 35 touchdown elevation 34 feet
TORA = 1500 m
ASDA = 1700 m
TODA = 1900 m
The slope of RW 35 is: "2.28% up"
146-34 = 112 feet
1500m = 4920 feet
Gradient = 112/4920 x 100 = 2.276%
The determination of the maximum mass on brake release, of a certified turbojet aeroplane with 5o, 15o and 25o flaps angles on take off, leads to the following values, with wind:
Flap angle: 5/15/25 deg
Runway limitation (kg): 66000/69500/71500
Second segment slope limitation: 72200/69000/61800
Head wind: +120 kg/kt
Tail wind: -360 kg/kt
Given that the tail wind component is equal to 5 kt, the maximum mass on
brake release and corresponding flap angle will be:
a) 67700 kg/15 deg <-- Correct
b) 69000 kg/15 deg
c) 72200 kg/5 deg
d) 69700 kg/25 deg
Winds do not affect the climb limited take-off mass so wind correction is required only for field length limit
Thus reducing 1800Kg from runway limited mass we have:
Runway Limits of 64200 / 67700 / 69700
2nd Segment Limits of 72200 / 69000 / 61800
To meet both the field length limit and climb limit our max takeoff weight will be the lowest value for each flap configuration.
For flaps 5 takeoff our max take off weight will be 64200 Kg.
For flaps 15 takeoff our max take off weight will be 67700 Kg.
For flaps 25 takeoff our max take off weight will be 61800 Kg.
Thus the maximum mass on brake release with which we can take off is the highest of the above three values.
It is 67700 and for that weight flaps required will be 15 deg.
Putting in 16500 litres of fuel with an SG of 780 kg/m3, and writing 16500 kg of fuel on the load sheet will result in:
a) TOD and ASD decreasing, and the calculated V2 being too fast <-- Correct
b) TOD increasing and ASD decreasing, and the calculated V2 being too fast
c) TOD and ASD remaining constant, if the calculated speeds are used
d) TOD and ASD increasing, if the calculated speeds are used
16500 litres with sp gravity of .780 is
16500 x 0.78 = 12870 Kg
Thus 12870 Kg of fuel is put and 16500 Kg is shown on paper.
So the all the calculations will be for a heavy weight whereas the aircraft will actually be light.
Thus actual TOD and ASD will be less and calculated V2 will be too fast, because calculations were done for a higher weight.
During certification test flights for a turbojet aeroplane, the actual measured take off runs from brake release to a point equidistant between the point at which VLOF is reached and the point at which the aeroplane is 35 feet above the take off surface are:
- 1747 m all engines operating
- 1950 m, with the critical engine failure recognised at V1, the other factors remaining unchanged.
Considering both possibilities to determine the take off run (TOR). What is the correct distance?
a) 2009 m <-- Correct
b) 2243 m
c) 2096 m
d) 1950 m
All engine distance is factored so it is 1747 x 1.15 = 2009
Engine Fail Distance is unfactored so it remains 1950.
Considering both possibilities to determine the take off run (TOR), the correct distance will be the higher of these figures i.e. 2009m
In the engine failure case, the distances used are gross distances (the distances used by the average aircraft without additional safety factors).
Thus if an engine does fail at VEF and the decision to go is made at V1 there is a 50:50 chance of making the screen height at 35ft (it may be above that or may be below). Similarly, if the decision is made to stop then there is a 50:50 chance of stopping by the end of the ASDA.
Why safety factor are not applied is because it is extremely unlikely that engine will fail exactly at VEF on a field length limiting runway.
The failure at this precise time is assessed as having a probability of less than 1:1,000,000
If an event is extremely unlikely in itself then the difference of net and gross reduces to zero.
So no further safety factors are used.
Since the probability of an all engines takeoff is very high, there is a safety factor involved.
The net distance to 35 ft is 1.15 times the gross distance.
If we compare the distances for engine out takeoff on dry and wet runways and all engine takeoff, then obvioulsy the gross all engines distance to 35ft will be shorter than the other two.
However after applying the safety factor of 1.15, net distance of an all engine takeoff might actually be longer and therefore limiting.
Given that the characteristics of a three engine turbojet aeroplane are as follows:
Thrust = 50000 Newton/Engine
g = 10 m/s2
Drag = 72569 N
Minimum gross gradient (2nd segment) = 2.7%
SIN (Angle of climb) = (Thrust – Drag)/Weight
The maximum take-off mass under second segment conditions is?
a) 101596 kg <-- Correct
SIN (Angle of climb) = (Thrust – Drag)/Weight
Weight = Thrust - Drag/ Sin (angle of climb)
Put in the values and remember to put thrust for 2 engines value as we have to consider one engine out for the second segment calculations.
Weight = 100,000 - 72569 / 2.7%
Weight = 27431/.027
Weight = 1015962 Newtons
Weight is the force on an object due to gravity and is therefore measured in Newtons.
Weight = m x g
Mass = weight/g = 1015962/10 = 101596 Kg
A climb gradient required is 3.3%. For an aircraft maintaining 100 kt true airspeed, no wind, this climb gradient corresponds to a rate of climb of approximately:
CLimb gradient x TAS = Rate of Climb
3.3 x 100 = 330 ft/min
On a segment of the take off flight path an obstacle requires a minimum gradient of climb of 2.6% in order to provide an adequate margin of safe clearance. At a mass of 110000 kg the gradient of climb is 2.8%. For the same power and assuming that the sine of the angle of climb varies inversely with mass, at what maximum mass will the aeroplane be able to achieve the minimum gradient.
a) 118455 kg <-- Correct
b) 106425 kg
c) 102150 kg
d) 121310 kg
Sine of the climb angle = Thrust-Drag /Weight
i.e. Sine of the climb angle varies inversely with mass.
A linear relationship can be used.
Reduction of climb angle (2.6/2.8 = 0.928) will increase the weight limit (2.8/2.6 = 1.077)
110,000 kg x 1.077 = 118,470kg.
Closest option is 118455 Kg.
An aircraft with a mass of 110000 kg is capable of maintaining a grad of 2.6%. With all the atmospheric variables remaining the same, with what mass would it be able to achieve a gradient of 2.4?
a) 119167 kg <-- Correct
Same as above except that the question does not mention the fact that "sine of the climb angle varies inversely with mass".
110000kg x (2.6/2.4) = 119167kg.
In the climb an aircraft has a thrust to weight ratio of 1:4 and a lift to drag ratio of 12:1. While ignoring the slight difference between lift and weight in the climb, the climb gradient will be:
Sine of Climb angle = Thrust - Drag / Weight
For small angles Adjacent is nearly the same as Hypotenuse and Opposite/Adjacent = Sin relation.
So we get an approximate formula for climb gradient, which is:
Climb Gradient = Thrust - Drag / Weight
Just put in the values.
L/D is 12/1. Considering lift almost equal to weight. Weight = 12 and Drag = 1
Thrust to Weight ration is 1:4
If weight is 12 then Thrust is 12 x 1/4 = 3
Climb Gradient = 3-1/12 = 1/6 = 16 = 0.166
Climb Gradient % is 0.166 x 100 = 16.66%
An aircraft with a climb gradient of 3.3%, flying at an IAS of 85 kts. At a pressure ALT of 8500 ft with a temperature of +15 deg C will have a ROC of:
a) 334 ft/min <-- Correct
b) 284 ft/min
c) 623 ft/min
d) 1117 ft/min
Convert IAS into TAS
ROC = TAS x Gradient = 100 x 3.3
Multiply IAS of 85 x 3.3 and you get into the trap set at option (b)
A four jet engined aeroplane (mass = 150000 kg) is established on climb with all engines operating. The lift-to-drag ratio is 14. Each engine has a thrust of 75000 Newtons. The gradient climb is: (given: g=10m/s?):
Climg gradient = Thrust - Drag / Weight
= (75000 x 4) - Drag / Weight
Weight = m x g = 150,000 x 10 = 150,000,0
L/D = 14 so Drag = L/14
For small angles Lift is almost equal to Weight
So Drag = 150,000,0/14 = 107143
Climb gradient = 300,000 - 107143 / 150,000,0 = 0.128
Climb gradient % = 0.128 x 100 = 12.8%
(Refer to figure 032_8-9)What is the maximum vertical speed of a three engine turbo jet aeroplane with one engine inoperative (N-1) and a mass of 75000 kg? Using the following:G = 10 m/s2 1 kt = 100 ft/min SIN (Angle of climb) = (Thrust - Drag)/Weight
Answer: +1267 ft/min
Rate of Climb = Gradient x TAS
From the figure:
TAS = 190
Gradient = (Thrust-Drag)/Weight
Max rate of climb is with max difference between thrust available and thrust required.
From the figure Thrust available for all engines = 240
With one engine out thrust available = 240-80 = 160
Thrust required = 110
Weight = m x g = 750000
gradient = 160,000 - 110,000 / 750,000 = 0.066
ROC = gradient x TAS = 0.066 x 190 = 12.66
given 1kt = 100 ft/min
ROC = 12.66 x 100 = 1267 ft/min
If the conversion factor is not given then before multiplying TAS with gradient we need to convert TAS from nm/hr to ft/min i.e.
190/60 x 6080 = 19253
19253 x 0.066 = 1270 ft/min
Following a take off limited by the 50 ft screen height, a light twin climbs on a gradient of 5%. It will clear a 160 m obstacle in relation to the runway (horizontally), situated at 5000m from the 50 ft point with an obstacle clearance margin of:
a) 105 m <- Correct
b) 90 m
The important point in this simple problem is to cater for the 50 ft point. If you dont you'll end up with the wrong answer of 90m which is present in the options.
The obstacle is at 5000m or 16400 feet from the 50 ft point and is 160m or 525 ft high.
5% gradient means the aircraft is climbing 5 feet for 100 feet horizontal distance.
Therefore for 16400 ft horizontal distance it will be at a height of:
5/100 x 16400 = 820
Since the starting point was at 50 ft above the ground the correct aircraft height will be 820+50 = 870 feet
The difference between aircraft at 870 feet and obstacle at 525 feet is 345 ft or 105m.
If the TAS is 100 kts on the glide slope of 3 deg, what is the Rate of Descent?
Ans: 500 ft/min
Based on trigonometry
Tan of angle = Opposite / Adjacent
Tan of angle = Rate of Descent / Horizontal Distance
Rate of Descent = Tan of angle x Horizontal Distance
Tan of Angle = Glide sLope Angle
Horizontal Distance = Ground Speed
Since we want Rate of Descent in feet per minute, we will have to convert Ground Speed from nm/hr to ft/min by divinding it by 60 and multiplying by 6080.
Rate of Descent (ft/min) = Tan of glide slope angle x GS/60 x 6080
A Rule of Thumb for 3 deg glide slope
Multiply your GS by 5 and that will give a rough rate of descent.
Half the speed and add a Zero
e.g. If GS is 140 then
140/2 = 70 and add a zero (i.e. multiplying by 10) to make it 700 ft/min.
If glide slope is not 3 deg
For every 0.25 of a degree difference above or below the standard 3 deg glide slope, add or subtract respectively 10kts to your Ground speed before using it in the rule of thumb calculation.
e.g. If the GS is 140 and glide slope is 3.5 deg then GS to be used for rule of thumb calculation is 140 + 20 = 160.
(For this question use figure 032_6.2) Using the Power Setting Table, for the single engine aeroplane, determine the cruise TAS and fuel flow (lbs/hr) with full throttle and cruise lean mixture in the following conditions: Given:
OAT: 13 deg C
Pressure altitude: 8000 ft
Answer: 160 kt and 69.3 lbs/hr
The tables provided are for ISA and ISA+20, whereas we are at ISA +14
This involves interpolation.
At ISA, Fuel flow at 8000 feet is 71.1
At ISA+20, Fuel flow at 8000 feet is 68.5
For a difference of 20 (ISA to ISA+20) the difference in fuel flow is 2.6 (71.1-68.5)
For ISA+1 the change in fuel flow is 2.6/20 = 0.13
For ISA+14 ---- 14 x 0.13 = 1.82
The fuel flow is reducing from ISA to ISA+20 (from 71.1 to 68.5)
Thus 71.1 - 1.82 = 69.28
69.3 pph will be the fuel flow at ISA+14
TAS does not require interpolation
Two identical turbojet aeroplanes (whose specific fuel consumptions are considered to be equal) are at holding speed at the same altitude. The mass of the first aircraft is 130000 kg and its hourly fuel consumption is 4300 kg/h. The mass of the second aircraft is 115000 kg and its hourly fuel consumption is:
Ans: 3804 Kg/hr
The rule of thumb is that for small changes, the fuel flow changes roughly in proportion to the weight change.
When the aircraft is light the fuel consumtion will be less.
The reduction factor = 115000 / 130000 = 0.885
So 4300 x 0.884 = 3805 Kg/hr
A jet aeroplane equipped with old engines has a specific fuel consumption of 0.06 kg per Newton of thrust and per hour and, in a given flying condition, a fuel mileage of 14 kg per Nautical Mile. In the same flying conditions, the same aeroplane equipped with modern engines with a specific fuel consumption of 0.035 kg per Newton of thrust and per hour, has a fuel mileage of:
Ans: 8.17 kg/NM
With old engines SFC was 0.06
With new engines SFC is 0.035
Thus aircraft's fuel consumption has improved. That means the fuel mileage will be better (less) than 14 Kg per nautical mile.
The reduction factor = 0.035/.06 = 0.583
14 x 0.583 = 8.16