In a given configuration the endurance of a piston engine aeroplane only ? [ Exam pilot ]

Question 80-1 : Altitude speed mass and fuel on board altitude speed and mass speed and mass speed mass and fuel on board

Oszklarska .the fuel on board weight is contained in airplane mass is it not .therefore the 'altitude speed and mass' answer is equivalent to 'altitude speed mass and fuel' . .no you need to know the quantity of fuel on board to be able to calculate endurance to stay airborne as longer as possible

At a higher gross mass on a piston engined aeroplane in order to maintain a ?

Question 80-2 : The airspeed must be increased and the drag will also increase the airspeed must be decreased and the drag will increase the lift/drag ratio must be increased the airspeed must be increased and the drag will remain unchanged

exemple 184 The airspeed must be increased and the drag will also increase. — Admin .at a higher gross mass we need to increase lift if we consider to maintain same altitude .lift = cl x 1/2 rho v² x s.cl = lift coefficient.rho = density.v = tas in m/s .s = surface.since density surface and lift coefficient e g change in angle of attack do not change only speed can changed .therefore the airspeed must be increased and the drag will also increase The airspeed must be increased and the drag will also increase.

On a reciprocating engine aeroplane to maintain a given angle of attack ?

Question 80-3 : An increase in airspeed and power is required a higher coefficient of drag is required an increase in airspeed is required but power setting does not change requires an increase in power and decrease in the airspeed

Admin .at a higher gross mass we need to increase lift if we consider to maintain same altitude .lift = cl x 1/2 rho v² x s.cl = lift coefficient.rho = density.v = tas in m/s .s = surface.since density surface and lift coefficient e g change in angle of attack do not change only speed can changed .therefore power required increases airspeed will increased and the drag will also increase .notice whatever the propulsion system prop or jet it doesn't play a role as seen by its absence in the lift formula

An aeroplane with reciprocating engines is flying at a constant angle of attack ?

Question 80-4 : Remains unchanged but the tas increases increases at constant tas decreases and the cas decreases too because of the lower air density remains unchanged but the the cas increases

exemple 192 Remains unchanged but the tas increases. — Admin .with increasing altitude density rho decreases angle of attack mass and configuration remain constant to maintain lift tas has to increase .drag = cd x 1/2 x rho x v² velocity = tas .v increases while rho decreases drag remains constant Remains unchanged but the tas increases.

On a reciprocating engine aeroplane with increasing altitude at constant gross ?

Question 80-5 : Increases and the tas increases by the same percentage increases but tas remains constant decreases slightly because of the lower air density remains unchanged but the tas increases

exemple 196 Increases and the tas increases by the same percentage. — Admin .this question compares an aircraft at a certain weight and angle of attack for a first altitude and then in the same configuration at a second altitude .you have to understand that the aircraft is in straight and level flight it is not climbing .our aircraft has to produce the same lift .lift l = 1/2 rho cl s v².for a higher altitude density reduces cl and s remain unchanged thus only tas v can and must increase .for that reason the power required must increase Increases and the tas increases by the same percentage.

At reference or see performance manual sep 1 figure 2 4 .with regard to the ?

Question 80-6 : Approximately 1850 feet approximately 1700 feet approximately 1370 feet approximately 1120 feet

exemple 200 Approximately : 1850 feet. — Admin . 1091 Approximately : 1850 feet.

Performance manual sep 1 figure 2 4 .with regard to the landing chart for the ?

Question 80-7 : Approximately 1930 feet approximately 950 feet approximately 1400 feet approximately 750 feet

exemple 204 Approximately : 1930 feet. — Admin . 1128 Approximately : 1930 feet.

For this question use annex 032 005 or performance manual sep 1 figure 2 4 ?

Question 80-8 : Approximately 1800 feet approximately 1300 feet approximately 2000 feet approximately 1450 feet

exemple 208 Approximately: 1800 feet. — Img /com en/com032 295 jpg. oat isa +15°c so oat is +30°c at 0 ft pressure altitude .runway wet grass .landing factor for short grass is 1 15 and landing factor for wet runway is 1 15.1275 ft x 1 15 = 1466 ft.1466 ft x 1 15 = 1686 ft.approximately 1800 feet is our answer Approximately: 1800 feet.

With regard to the take off performance chart for the single engine aeroplane ?

Question 80-9 : 3200 lbs 3000 lbs 2900 lbs > 3650 lbs

exemple 212 3200 lbs. — Admin .complete the graph with the data . 1093.you will find 3200 lbs 3200 lbs.

At reference or see performance manual sep 1 figure 2 2 .with regard to the ?

Question 80-10 : Approximately 2050 ft approximately 1150 ft approximately 2450 ft approximately 1260 ft

exemple 216 Approximately: 2050 ft. — Complete the graph with the data . /com en/com032 297 jpg.you find 1900 ft the closest answer is 2050 ft Approximately: 2050 ft.

At reference or see performance manual sep 1 figure 2 1 .with regard to the ?

Question 80-11 : 71 and 82 kias 65 and 75 kias 73 and 84 kias 68 and 78 kias

exemple 220 71 and 82 kias. — Img /com en/com032 298 jpg. 71 and 82 kias.

At reference or see performance manual sep 1 figure 2 2 .with regard to the ?

Question 80-12 : Approximately 3960 ft approximately 3680 ft approximately 4150 ft approximately 5040 ft

exemple 224 Approximately: 3960 ft. — Img /com en/com032 299 jpg.do not forget to apply 'grass.correction factor' .3300 ft x 1 2 = 3960 ft Approximately: 3960 ft.

With regard to the climb performance chart for the single engine aeroplane ?

Question 80-13 : 1290 ft/min 1370 ft/min 1210 ft/min 1150 ft/min

exemple 228 1290 ft/min. — Admin . 1129.youssef92 .for me i find that the answer is 1370ft/min . .if you start at 15°c instead of 30°c you will find around 1350 1370 ft but the question states o a t isa + 15°c .it means an outside temperature of 30°c at sea level 1290 ft/min.

At reference or see performance manual sep 1 figure 2 2 .with regard to the ?

Question 80-14 : 2275 ft 1900 ft 1600 ft 2000 ft

exemple 232 2275 ft. — Img /com en/com032 3587 jpg..1750 ft x surface factor 1 3 = 2275 ft 2275 ft.

Using the landing diagram for single engine aeroplane determine the landing ?

Question 80-15 : 1350 ft 1550 ft 1020 ft 880 ft

exemple 236 1350 ft. — Admin . 1097.1320 ft close to 1350 ft 1350 ft.

The pilot of a single engine aircraft has established the climb performance ?

Question 80-16 : Degraded improved unchanged unchanged if a short field take off is adopted

An extract of the flight manual of a single engine propeller aircraft is ?

Question 80-17 : 465 m 615 m 540 m 395 m

exemple 244 465 m. — Admin . 1099.1270 kg = 2800 lbs .convert 1520 ft to meters .1520 x 0 3048 = 463 m 465 m.

With regard to the landing chart for the single engine aeroplane determine the ?

Question 80-18 : Approximately 1700 feet approximately 1150 feet approximately 1500 feet approximately 920 feet

exemple 248 Approximately : 1700 feet. — Admin . 1102.you will find 1760 ft close enough to 1700 feet Approximately : 1700 feet.

With regard to the take off performance chart for the single engine aeroplane ?

Question 80-19 : Approximately 2470 feet approximately 1440 feet approximately 2800 feet approximately 2200 feet

exemple 252 Approximately : 2470 feet. — Admin . 1131 Approximately : 2470 feet.

For this question use reference or performance manual sep 1 figure 2 1.with ?

Question 80-20 : 3240 lbs 3000 lbs 2900 lbs greater than 3650 lbs

exemple 256 3240 lbs. — Complete the graph with the data . /com en/com032 359 jpg.you will find 3240 lbs 3240 lbs.

Consider the graphic representation of the power required versus true air speed ?

Question 80-21 : Maximum specific range maximum endurance maximum thrust critical angle of attack

exemple 260 Maximum specific range. — Admin . 1085.maximum specific range is reached at minimum drag speed vmd for a piston engined aeroplane . 1135.vmd is the speed for maximum range in a prop aircraft .vmd is the speed for maximum endurance in a jet aircraft Maximum specific range.

For this question use reference sep 1 figure 2 3 .using the climb performance ?

Question 80-22 : 15640 ft 18909 ft 20109 ft 16665 ft

exemple 264 15640 ft. — At 5000 ft isa t° is 5°c ..1500/1080= 1 39 minute.ground speed 100 + 5 = 105 kt.. 105/60 x 1 39 = 2 43 nm..2 43 = 4 5 km = 14722 ft . /com en/com032 362 jpg.closest answer is 15640 ft normally we should use the climb gradient since the question gives a speed in ias but rate of climb can also lead closely to the correct answer. ebbr1000 .tas= 108 kt.height difference x 100/ climb gradient= 1500 x 100 / 9 9 = 15151 ft. 15151 x gs / tas= 15151 x 113 / 108 = 15852 ft 15640 ft.

For this question use reference .using the climb performance chart for the ?

Question 80-23 : 1140 ft/min and 10 6% 1030 ft/min and 8 4% 1170 ft/min and 9 9% 1140 ft/min and 8%

exemple 268 1140 ft/min and 10.6%. — Isa at 3000 ft .15° 2° x 3 = +9°c.for a given ias the true airspeed is about 2% higher than ias per 1000ft of altitude above sea level . /com en/com032 363 jpg.rate of climb 1140 ft/min .gradient of climb 10 6% . atplea .figure cap698 figure 2 3 1140 ft/min and 10.6%.

At reference or use performance manual sep 1 figure 2 1.airport characteristics ?

Question 80-24 : 1500 ft 2000 ft 1750 ft 1350 ft

exemple 272 1500 ft. — Img /com en/com032 405 png. 1500 ft.

Unless otherwise specified in the afm for a performance class b aeroplane ?

Question 80-25 : 1 05 1 03 1 15 1 1

exemple 276 1.05 — Ecqb04 october 2017..the factor to be applied for each 1% of downslope is 5% or 1 05 1.05

Unless otherwise specified in the afm for a performance class b aeroplane ?

Question 80-26 : 1 05 1 03 1 01 1 15

exemple 280 1.05 — Ecqb04 october 2017..the factor to be applied for each 1% of uphill slope is 5% or 1 05 1.05

The following conditions are observed at an airport runway 13 wind 140° for ?

Question 80-27 : 5 kt 18 kt 15 kt 10 kt

exemple 284 5 kt. — Admin .vt = sin 10° x 30 = 5 kt 5 kt.

Maximum crosswind demonstrated is equal to 0 2 vs0 and the following conditions ?

Question 80-28 :

exemple 288 — Admin .350 300=50°.crosswind sin50x20=15 3kt.demonstrated crosswind 70x0 2=14kt

Which margin above the stall speed is provided for the landing speed reference ?

Question 80-29 : 1 30 vso 1 05 vso 1 10 vso 1 2 vmca

With regard to the graph for the light twin aeroplane will the accelerate and ?

Question 80-30 : No the performance will be worse than in the chart it does not matter which take off technique is being used the chart has been made for this situation performance will be better than in the chart

exemple 296 No, the performance will be worse than in the chart. — Admin .on the associated condtions you can read 'full power before brake release' .if you release the brake before the engine is in full power your acceleration will be slower a longer runway distance is required to accelerate until v1 No, the performance will be worse than in the chart.

The critical engine inoperative ?

Question 80-31 : Increases the power required and the total drag due to the additional drag of the windmilling engine and the compensation of the yaw moment does not affect the aeroplane performance since it is independent of the power plant decreases the power required and increases the total drag due to the additional drag of the windmilling engine and the compensation of the yaw moment increases the power required and decreases the total drag due to the windmilling engine

exemple 300 Increases the power required and the total drag due to the additional drag of the windmilling engine and the compensation of the yaw moment. — When an engine failure occurs in a multi engine aircraft asymmetric thrust and drag produce the following effects on the aircraft's axes of rotation . pitch down along the lateral axis loss of accelerated slipstream over the horizontal stabilizer produces less negative lift . roll down toward the inop engine along the longitudinal axis wing produces less lift on side of failed engine due to loss of accelerated slipstream . yaw toward the inop engine along the vertical axis loss of thrust and increased drag from the windmilling propeller .to compensate for these effects a pilot must add additional back pressure deflect the ailerons into the operating engine and apply rudder pressure on the side of the operating engine .a loss of one engine on a multi engine results in a loss of 50% of all available power and up to 80% of the aircraft's excess power and climb performance due to increased drag from the inoperative engine Increases the power required and the total drag due to the additional drag of the windmilling engine and the compensation of the yaw moment.

Following a take off determined by the 50ft 15m screen height a light twin ?

Question 80-32 : 115 m 100 m 85 m it will not clear the obstacle

exemple 304 115 m. — Climb gradient = change in height / distance travelled.change in height = gradient x distance travelled.change in height = 10 x 10000 = 100000.divide 100000 by 100 to express climb gradient as percentage .100000 / 100 = 1000 m.add 15m screen height = 1015 m..the aircraft will clear a 900 m high obstacle with an obstacle clearance of 115 m 115 m.

A runway is contaminated with 0 5 cm of wet snow .the flight manual of a light ?

Question 80-33 : Increased unchanged reduced substantially decreased

Following a take off limited by the 50 ft screen height a light twin climbs on ?

Question 80-34 : 105 m 90 m 75 m it will not clear the obstacle

exemple 312 105 m. — A 5% climb on 5000 m means a 250 m gain in height 5% > for 100 m horizontal distance you gain 5 m in height + 50 ft 15 m it gives 265 m .265 160 = 105 m 105 m.

The pilot of an aircraft has calculated a 4000 m service ceiling based on the ?

Question 80-35 : Higher than 4000 m less than 4000 m unchanged equal to 4000 m only a new performance analysis will determine if the service ceiling is higher or lower than 4000 m

exemple 316 Higher than 4000 m. — Higher than 4000 m.

The flight manual of a light twin engine recommends two cruise power settings ?

Question 80-36 : An increase in speed fuel consumption and fuel burn/distance same speed and an increase of the fuel burn per hour and fuel burn/distance an increase in speed and fuel burn/distance but an unchanged fuel burn per hour same speed and fuel burn/distance but an increase in the fuel burn per hour

exemple 320 An increase in speed, fuel consumption and fuel-burn/distance. — An increase in speed, fuel consumption and fuel-burn/distance.

At a given mass the reference stall speed of a twin engine turboprop aircraft ?

Question 80-37 : 123 kt 115 kt 125 kt 120 kt

Given .oat 25°c .pressure altitude 3000 ft .rwy 24l .wind 310°/20kt .take off ?

Question 80-38 : 3750 ft 4300 ft 4000 ft 3350 ft

exemple 328 3750 ft. — Admin .runway 24l 240° wind from 310° it is a headwind .headwind component .20 kt x cos angle between the wind and the runway .20 x cos 70° = 7 kt . 1106.accelerate and stop distance = 4000 ft .but heavy duty brakes are installed we must reduce distance by 7% .4000 x 0 93 = around 3750 ft 3750 ft.

Use performance manual mep1 figure 3 2.with regard to the graph for the light ?

Question 80-39 : Longer than the graphical distance shorter than the graphical distance the same as the graphical distance because both techniques are accounted unaffected because all take off techniques are accounted

exemple 332 Longer than the graphical distance. — On the associated condtions you can read 'full power before brake release' .if you release the brake before the engine is in full power your acceleration will be slower a longer runway distance is required to accelerate until v1 Longer than the graphical distance.

For this question use annex ecqb 032 016 v2015 10 or performance manual mep 1 ?

Question 80-40 : 175 ft/min 1250 ft/min 375 ft/min 625 ft/min

exemple 336 175 ft/min. — Img /com en/com032 387 jpg. 175 ft/min.

~

Exclusive rights reserved. Reproduction prohibited under penalty of prosecution.

3159 Free Training Exam