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Which engine is considered critical in the event of an engine failure during ? [ Course landing ]

Question 85-1 : The left engine the right engine the left engine during ground run afterward the right engine both engines are equally critical

Ecqb03 july 2016 ..on a jet engines aeroplane you don't have the additional 'critical engine' notion added to your engine failure .explanation .clockwise rotation as viewed from the pilot's seat the critical engine will be the left engine . /com en/com080 341 jpg.multi engine aeroplanes are subject to p factor just as single engine aeroplanes are the descending propeller blade of each engine will produce greater thrust than the ascending blade when the aeroplane is operated under power and at positive angles of attack the descending propeller blade of the right engine is also a greater distance from the center of gravity and therefore has a longer moment arm than the descending propeller blade of the left engine as a result failure of the left engine will result in the most asymmetrical thrust adverse yaw as the right engine will be providing the remaining thrust exemple 185 The left engine.The left engine.

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

Question 85-2 : 1 15 1 45 0 65 1 05

Ecqb03 july 2016

A pilot is flying a twin engine piston aeroplane with all engines operating ?

Question 85-3 : 150 ft 50 ft 115 ft 70 ft

Ecqb03 july 2016..multi engine class b .multi engine class b aircraft have a requirement to clear obstacles by 50 ft from the end of the tod up to 1500ft using net performance after which the aircraft is considered to be en route .a an operator shall ensure that the take off flight path of aeroplanes with two or more engines determined in accordance with this sub paragraph clears all obstacles by a vertical margin of at least 50 ft . but performance class b multi engined aircraft piston required a minimum of 4% climb gradient at take off all engine operating .at 2000 m or 6560 ft we are already at a height of 35 ft .4% of 6560 ft = 262 ft .35 + 262 = 297 ft will be our minimum height 2000m after tod .obstacle height = 644 ft 500 ft = 144 ft .297 ft 144 ft = 153 ft exemple 193 150 ft.150 ft.

What is the minimum obstacle clearance above obstacle .given perf class b.cloud ?

Question 85-4 : 215 ft 815 ft 50 ft 235 ft

Admin .perf class b .failure of the critical engine is assumed to occur where visual reference is lost .the gradient to engine failure height is the all engine gradient x 0 77 giving net gradient . 1124.15000 ft x 0 3048 = 4572 m.101 kt x 1 852 = 187 km/h.to climb 250 ft at 1830ft/min x 0 77 it takes .250 / 1830x0 77 .0 177 minute.at 187 km/h the distance to climb 250 ft during 0 177 min will be .187 / 60 minutes = 3 12 km/min.3 12 x 0 177 = 0 552 km or 552 m .it remains 4572 552 m before the obstacle .4020 m .the time taken for a distance of 4008 m is .4 020 km / 3 12 km/min = 1 29 min.during this time you will climb of .1 29 min x 400 ft/min = 516 ft .300 + 516 = 816 ft.816 600 = 216 ft exemple 197 215 ft.215 ft.

During take off the third segment begins ?

Question 85-5 : When acceleration to flap retraction speed is started when landing gear is fully retracted when acceleration starts from vlof to v2 when flap retraction is completed

The first segment starts at 'reference zero' and ends when the gear comes up .the second segment lasts until levelling off for flap retraction .the third segment ends when ready for the enroute climb .it is usually a level burst at 400 ft during which acceleration is made to climb speed flaps are retracted and power is reduced to max continuous exemple 201 When acceleration to flap retraction speed is started.When acceleration to flap retraction speed is started.

What is the maximum vertical speed of a three engine turbojet aeroplane with ?

Question 85-6 : +1267 ft/min 1267 ft/min 0 ft/min +3293 ft/min

Calculation for the climb gradient .aircraft weiht in newton 750000 n.thrust 2 engines = 300000 x 2 = 600000 n..sin angle of climb = thrust drag / weight.sin angle of climb = 600000 553000 / 750000 = 0 0626.in percent it is 6 26%.rate of climb = climb gradient x tas.rate of climb = 6 26 x 202 = 1265 ft/min exemple 205 +1267 ft/min.+1267 ft/min.

During the certification flight testing of a twin engine turbojet aeroplane the ?

Question 85-7 : 1779 m 1978 m 1547 m 1720 m

Cs 25 113 take off distance and takeoff run . . 2 115% of the horizontal distance along the take off path with all engines operating from the start of the take off to the point at which the aeroplane is 11 m 35 ft above the take off surface .all engine take off distance is 1547 x 1 15 = 1779 m .one engine take off distance is 1720 m.1779 m is the greatest distance exemple 209 1779 m.1779 m.

For a turboprop powered aeroplane a 2200 m long runway at the destination ?

Question 85-8 : 1339 m 1771 m 1540 m 1147 m

2200/1 15 x 0 7 = 1339 m..notice .0 7 turboprop .0 6 turbojet ..factor 1 15 for a wet runway . brudef .isn't it asked the 'dry runway' landing distance in that case 1 15 shouldn't be taken into account ..the runway at destination is expected to be 'wet' you must be able to stop your aeroplane in the wet required landing distance even if at the time of arrival the runway is dry exemple 213 1339 m.1339 m.

Characteristics of a three engine turbojet aeroplane are as follows .thrust = ?

Question 85-9 : 101 596 kg 74 064 kg 209 064 kg 286 781 kg

Sin angle of climb = thrust drag / weight .or .weight = thrust drag / sin angle of climb .weight = 50000x2 72569 / 0 027.weight = 1015960 n.divide newtons by g = 101596 kg .as we are calculating maximum take off mass for a path to avoid obstacles we must assume one engine out unless the question states otherwise our available thrust is only 2x50000 n . amassa .i'm missing something with the formula explanation .it says .weight = thrust drag / sin angle of climb .weight = 50000 x2 72569 / 0 027.but sin angle of climb is 0 047 .0 027 is angle of climb / 100..2 7% express as a decimal is 0 027 .angle = arctan 0 027 = 1 55°.sin 1 55° = 0 027 exemple 217 101 596 kg.101 596 kg.

Minimum control speed on the ground 'vmcg' is based on directional control ?

Question 85-10 : Primary aerodynamic control only primary aerodynamic control and nose wheel steering primary aerodynamic control nose wheel steering and differential braking nose wheel steering only

exemple 221 Primary aerodynamic control only.Primary aerodynamic control only.

Which of the following represents the maximum value for v1 assuming max tyre ?

Question 85-11 : Vr vmca v2 vref

Admin . 1459.note vmca minimum control speed in the air is located between v1 and vr but vmca is not the maximum value for v1 .vr is the speed at which the rotation of the airplane is initiated to takeoff attitude exemple 225 Vr.Vr.

During certification flight testing on a four engine turbojet aeroplane the ?

Question 85-12 : 3050 m 2938 m 3513 m 2555 m

.you multiply the all engine performance by 1 15 and then compare to the one engine inoperative distance you take the higher figure .all engines operating distance = 2555 x 1 15 = 2938 m .3050 m distance with failure of the critical engine recognised at v1 is the highest distance adopted for the certification file exemple 229 3050 m.3050 m.

In which of the following distances can the length of a stopway be included ?

Question 85-13 : In the accelerate stop distance available in the one engine failure case take off distance in the all engine take off distance in the take off run available

exemple 233 In the accelerate stop distance available.In the accelerate stop distance available.

At which minimum height will the second climb segment end ?

Question 85-14 : 400 ft above field elevation 35 ft above ground when gear retraction is completed 1500 ft above field elevation

.the first segment starts at 'reference zero' and ends when the gear comes up .the second segment lasts until levelling off for flap retraction .the third segment ends when ready for the enroute climb it is usually a level burst at 400 ft during which acceleration is made to climb speed flaps are retracted and power is reduced to max continuous exemple 237 400 ft above field elevation.400 ft above field elevation.

How does tas vary in a constant mach climb in the troposphere under isa ?

Question 85-15 : Tas decreases tas increases tas is constant tas is not related to mach number

Admin .for those questions use the very simple 'ertm' diagram . 1039.the mach line is vertical because the question states in a constant climb mach . ertm for e as/ r as rectified air speed or cas / t as/ m ach .mach number = tas / local sound speed .the velocity of sound is decreasing as temperature decreases to maintain constant mach as the velocity of sound reduces tas has to reduce

The optimum long range cruise altitude for a turbojet aeroplane ?

Question 85-16 : Increases when the aeroplane mass decreases is always equal to the powerplant ceiling is independent of the aeroplane mass is only dependent on the outside air temperature

exemple 245 Increases when the aeroplane mass decreases.Increases when the aeroplane mass decreases.

How does the specific range change when the altitude increases for jet ?

Question 85-17 : First increases then decreases decreases does not change increases only if there is no wind

.specific air range = tas / fuel flow.as altitude increases tas increases therefore specific air range increases .but as for jet aircraft maximum range in still air is achieved at maximum tas/drag ratio and approximately 95% rpm engine efficiency the speed for maximum still air range occurs at 1 32 times the speed of minimum drag as you climb past a certain height the engines go past their best efficiency so initially you increase specific fuel consumption then it decreases

At reference .assuming constant l/d ratio which of the diagrams provided ?

Question 85-18 : C a b d

The curve is the total drag on the 'thrust required curve' or 'drag or thrust required against airspeed' .with less mass you need less lift ==> less lift = less induced drag .induced drag will decrease displacing the total drag curve downwards and to the left .the lowest point on the curve is vmd eas for minimum drag vmd velocity minimum drag decreases exemple 253 CC

Long range cruise is a flight procedure which gives ?

Question 85-19 : A specific range which is approximately 99% of maximum specific range and a higher cruise speed a 1% higher tas for maximum specific range an ias which is 1% higher than the ias for maximum specific range a specific range which is approximately 99% of maximum specific range and a lower cruise speed

exemple 257 A specific range which is approximately 99% of maximum specific range and a higher cruise speed.A specific range which is approximately 99% of maximum specific range and a higher cruise speed.

A commercial flight is planned with a turbojet aeroplane to an aerodrome with a ?

Question 85-20 : 1 440 m 1 250 m 1 090 m 1 655 m

.eu ops 1 515 landing dry runways . a an operator shall ensure that the landing mass of the aeroplane determined in accordance with eu 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 . 2400 x 0 6 = 1440 m exemple 261 1 440 m.1 440 m.

At the destination aerodrome the landing distance available is 3000m .the ?

Question 85-21 : 1565 m 2070 m 1800 m 2609 m

.eu ops 1 515 landing dry runways allows a full stop landing from 50 ft above the threshold . 1 for turbo jet powered aeroplanes within 60% of the landing distance available .landing wet and contaminated runways a the landing distance available is at least 115% of the required landing distance determined in accordance with eu ops 1 515 .3000 m x 0 6 = 1800 m.1800 m / 1 15 = 1565 m exemple 265 1565 m.1565 m.

With zero wind the angle of attack for maximum range for an aeroplane with ?

Question 85-22 : Lower than the angle of attack corresponding to maximum endurance equal to the angle of attack corresponding to maximum endurance equal to the angle of attack corresponding to zero induced drag equal to the angle of attack corresponding to maximum lift to drag ratio

Admin . 1100.for a jet aeroplane maximum endurance is achieved at a speed corresponding to the maximum l/d ratio which is vmd where the gap between power required and power available is greatest at this speed for a conventional aerofoil the angle of attack is about 4° .the speed for maximum range occurs at 1 32 times the speed of minimum drag vmd in level flight as speed goes up the angle of attack goes down angle of attack is less than 4° typically at around 2 5° .example .at 10° you are at the maximum range aoa .at 16° you are at the maximum endurance aoa . 2416 exemple 269 Lower than the angle of attack corresponding to maximum endurance.Lower than the angle of attack corresponding to maximum endurance.

Two identical turbojet aeroplane whose specific fuel consumptions are ?

Question 85-23 : 3804 kg/h 3578 kg/h 3365 kg/h 4044 kg/h

115/130 x4300 = 3803 kg/h exemple 273 3804 kg/h.3804 kg/h.

A jet aeroplane equipped with old engines has a specific fuel consumption of 0 ?

Question 85-24 : 8 17 kg/nm 11 7 kg/nm 10 7 kg/nm 14 kg/nm

14 x 0 035/0 06 = 8 1666 kg/nm exemple 277 8.17 kg/nm.8.17 kg/nm.

The determination of the maximum mass on brake release of a certified turbojet ?

Question 85-25 : 67700 kg / 15° 69000 kg / 15° 72200 kg / 5° 69700 kg / 25°

.cltom climb limited take off mass.flltom field length limited take off mass..the climb limited take off mass is based on still air so the wind correction is only applied to the runway limitation .tailwind component is 5 kt correction is 360 x 5 = 1800 kg..field length limited take off mass .runway limit for flap 5° = 66000 1800 = 64200 kg .runway limit for flap 15° = 69500 1800 = 67700 kg .runway limit for flap 25° = 71500 1800 = 69700 kg ..the climb limited take off mass remain unchanged..so maximum take off mass is 67700 kg and flap 15° exemple 281 67700 kg / 15°.67700 kg / 15°.

During certification test flights for a turbojet aeroplane the actual measured ?

Question 85-26 : 2009 m 1950 m 2096 m 2243 m

.the certificated value of the take off run is the greater of the 'all engine distance x 1 15' or 'engine out distance' ..all engine distance x 1 15 = 1747 x 1 15 = 2009 m .engine out distance = 1950 m exemple 285 2009 m.2009 m.

For a twin engine turbojet aeroplane two take off flap settings 5° and 15° ?

Question 85-27 : 56 000 kg 53 000 kg 52 000 kg 70 000 kg

Admin .we'll gonna perform a flap 15° take off . 1815.you have to be very precise when you are drawing the line otherwise you will always find something closer to 55 000kg than 56 000 kg exemple 289 56 000 kg.56 000 kg.

The lowest take off safety speed v2 min is ?

Question 85-28 : 1 13 vsr for two and three engine turbo propeller and turbojet aeroplanes 1 20 vsr for all aeroplanes 1 15 vsr for all turbojet and turbo propeller aeroplanes 1 20 vsr for all turbo propeller aeroplanes

Cs25 .v2min in terms of calibrated airspeed may not be less than . 1 1 13 vsr for . i two engined and threeengined turbo propeller powered aeroplanes and. ii turbojet powered aeroplanes without provisions for obtaining a significant reduction in the one engine inoperative power on stall speed .. 2 1 08 vsr for . i turbo propeller powered aeroplanes with more than three engines and. ii turbojet powered aeroplanes with provisions for obtaining a significant reduction in the one engine inoperative power on stall speed and. 3 1 10 times vmc established under cs 25 149 .vsr reference stall speed exemple 293 1.13 vsr for two and three engine turbo-propeller and turbojet aeroplanes.1.13 vsr for two and three engine turbo-propeller and turbojet aeroplanes.

Which of the following three speeds of a jet aeroplane are basically identical ?

Question 85-29 : Holding maximum climb angle and minimum glide angle maximum drag maximum endurance and maximum climb angle maximum range minimum drag and minimum glide angle maximum climb angle minimum glide angle and maximum range

.holding speed for a jet is at the vmd speed vmd means velocity minimum drag this is the speed for minimum fuel consumption .for a jet aeroplane the maximum climb angle is achieved at a speed corresponding to the maximum cl/cd ratio which is vmd where the gap between power required and power available is greatest .minimum glide angle speed permits to fly the longest ground distance without wind it is achieved at a speed corresponding to the maximum cl/cd ratio which is vmd exemple 297 Holding, maximum climb angle and minimum glide angle.Holding, maximum climb angle and minimum glide angle.

The lift coefficient decreases during a glide with constant mach number mainly ?

Question 85-30 : Ias increases aircraft mass decreases tas decreases glide angle increases

.tas and ias increase during a descent at constant mach number thus we must decrease our angle of attack otherwise the lift will increase lift = 1/2 rho s v² cl and the descent may be stopped .so it is the increase in tas and ias which lead to decreasing the lift coefficient exemple 301 Ias increases.Ias increases.

During a descent at constant mach number the margin to low speed buffet will ?

Question 85-31 : Increase because the lift coefficient decreases remain constant because the mach number remains constant increase because the lift coefficient increases decrease because the lift coefficient decreases

.during a descent at constant mach number your tas increases .with increasing tas ias is increasing thus from the lift formula the lift coefficient decreases .the gap between your speed and the stall speed increases the margin to low speed buffet will increase because the lift coefficient decreases exemple 305 Increase, because the lift coefficient decreases.Increase, because the lift coefficient decreases.

A jet aeroplane is climbing at a constant ias and maximum climb thrust how will ?

Question 85-32 : Reduce / decrease reduce / remain constant remain constant / decrease remain constant / become larger

.to maintain a constant ias while climbing you have to reduce the climb angle power available is decreasing with an increase of altitude .and a reduction of the climb angle is done by reducing the pitch angle exemple 309 Reduce / decrease.Reduce / decrease.

A jet aeroplane is flying long range cruise how does the specific range / fuel ?

Question 85-33 : Increase / decrease increase / increase decrease / increase decrease / decrease

Jjansen .isn't it 'decrease/increase' long range cruise = 1% less range specific range decrease 4% faster fuel flow increase ..the question doesn't compare long range cruise and maximum range cruise .specific range is given as 'distance covered per unit of fuel'.as you are flying the aircraft mass decreases for the same long range cruise speed the fuel flow decreases .thus the specific range increases along the flight since for a same distance covered your fuel consumption is decreased exemple 313 Increase / decrease.Increase / decrease.

During a glide at constant mach number the pitch angle of the aeroplane will ?

Question 85-34 : Decrease increase increase at first and decrease later on remain constant

Admin .during a glide the aircraft is descending .descent + constant mach number => tas increase.see ertm diagram . 1039.the mach line is vertical because the question states a glide at constant mach number .if tas increase => pitch angle decrease to remain at constant mach exemple 317 Decrease.Decrease.

During a cruise flight of a jet aeroplane at constant flight level and at the ?

Question 85-35 : Decrease / decrease increase / decrease increase / increase decrease / increase

To maintain flight at max range speed which is the tangent to the drag curve and 1 32 vmd we must reduce speed as mass decreases to maintain 1 32 vmd the mass reduction as a consequence of fuel burn means less induced drag the total drag curve moves down and left and takes 1 32 vmd with it exemple 321 Decrease / decrease.Decrease / decrease.

An aeroplane descends from fl 410 to fl 270 at its cruise mach number and from ?

Question 85-36 : I increases ii remains constant i increases ii decreases i remains constant ii decreases i decreases ii increases

Descending at constant mach ias and tas will increase .total drag is proportional to v² drag will increase while descending and in order to maintain a constant mach number you must increased descent angle at idle thrust .at fl270 now we perform descent with a constant ias tas will decrease and density will increase and drag will stay constant our angle of descent will also stay contant exemple 325 (i) increases (ii) remains constant.(i) increases (ii) remains constant.

With a jet aeroplane the maximum climb angle can be flown at approximately ?

Question 85-37 : The maximum cl/cd ratio 1 1 vs the maximum cl/cd² ratio 1 2 vs

Admin . 1100.for a jet aeroplane the maximum climb angle is achieved at a speed corresponding to the maximum cl/cd ratio which is vmd where the gap between power required and power available is greatest exemple 329 The maximum cl/cd ratio.The maximum cl/cd ratio.

What happens to the drag of a jet aeroplane if during the initial climb after ?

Question 85-38 : The drag remains almost constant the drag increases considerably the drag decreases the drag increases initially and decreases thereafter

Admin .during initial climb if ias is maintained constant tas increases see ertm graph below and density decreases .you can maintain a constant angle of attack to produce the same lift and if lift does not change the drag remains almost constant . 1960.the eas/ias line is vertical because the question states constant ias is maintained exemple 333 The drag remains almost constant.The drag remains almost constant.

Which of the following sequences of speed for a jet aeroplane is correct . from ?

Question 85-39 : Vs maximum angle climb speed maximum range speed vs maximum range speed maximum angle climb speed maximum endurance speed maximum range speed maximum angle of climb speed maximum endurance speed long range speed maximum range speed

exemple 337 Vs, maximum angle climb speed, maximum range speed.Vs, maximum angle climb speed, maximum range speed.

If a flight is performed with a higher 'cost index' at a given mass which of ?

Question 85-40 : A higher cruise mach number a lower cruise mach number an increased maximum range an increased long range performance

Cost index is the ratio of time related costs crew salaries maintenance etc to fuel cost as one of the independent variables in the speed schedule computation . cost index ci = flight time related cost/fuel cost ..the cost index allows airlines to weight time and fuel costs based on their daily operations .a higher 'cost index' will result in a higher cruising speed . jomargra .but at a higher cruise mach number the aircraft will burn more fuel and the ci will reduce ..no we are talking about fuel cost not quantity .cost index is a made up figure which when input into the fmc is used to calculate econ speed the higher the cost index number the faster the aircraft flies basically the company decides on the cost of keeping the aircraft in the air and includes all sorts of costs such as crew aircraft operating costs fuel etc they then decide whether they want the aircraft to fly faster or slower and adjust the cost index as needed exemple 341 A higher cruise mach number.A higher cruise mach number.

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