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The flight crew of a turbojet aeroplane prepares a flight using the following ? [ Exam pilot ]

Question 90-1 : 34 430 kg 30 440 kg 32 480 kg 28 720 kg

.nam = ngm x tas/gs .nam = 2500 x 470/440 .nam = 2670 nam..212 800 kg is corresponding to 8083 nam in the table . /com en/com033 99 jpg..8083 2670 = 5413 nam..5413 nam is corresponding to 180 329 kg in the table..fuel = 212 800 180 329 = 32 471 kg..fuel consumption with total anti ice 0n = 32 471 x 1 06 = 34 419 kg . renfernandes i still did not understand what do de values on top of the columns 0 2 4 in this case 8 mean i couldn't choose a column unless i knew what they mean .thanks ..mass at brake release is 212 800 kg ==> 212 ton + 0 8 ton .if mass at brake release was 213 600 kg ==> 212 ton + 1 6 ton . miguel .would you be so kind to explain why 5413 are exactly 180 329 kg because i'm trying to calculate it via interpolation but i cannot find a successful result thank you ..180400 180200 = 200 kg.5419 5402 = 17 nam.17 nam for 200 kg.1 nam = 11 8 kg.5419 5413 = 6 nam.6 x 11 8 kg = 71 kg.180400 71 = 180329 kg

The purpose of the decision point procedure is ?

Question 90-2 : To reduce the minimum required fuel and therefore be able to increase the traffic load to reduce the landing weight and thus reduce the structural stress on the aircraft to increase the safety of the flight to increase the amount of extra fuel

Admin .reduced contingency fuel rcf procedure decision point procedure .this is a technique for increasing the traffic load by reducing the minimum fuel required you reduce the contingency figures by using it only from the decision point to the destination .if an operator's fuel policy includes pre flight planning to a destination 1 aerodrome commercial destination with a reduced contingency fuel procedure using a decision point along the route and a destination 2 aerodrome optional refuel destination the amount of usable fuel on board for departure shall be the greater of 2 1 or 2 2 below .2 1 the sum of . a taxi fuel and. b trip fuel to the destination 1 aerodrome via the decision point and. c contingency fuel equal to not less than 5% of the estimated fuel consumption from the decision point to the destination 1 aerodrome and. d alternate fuel or no alternate fuel if the decision point is at less than six hours from the destination 1 aerodrome are fulfilled and. e final reserve fuel and. f additional fuel and. g extra fuel if required by the commander .2 2 the sum of . a taxi fuel and. b trip fuel to the destination 2 aerodrome via the decision point and. c contingency fuel from departure aerodrome to the destination 2 aerodrome and. d alternate fuel if a destination 2 alternate aerodrome is required and. e final reserve fuel and. f additional fuel and. g extra fuel if required by the commander exemple 194 To reduce the minimum required fuel and therefore be able to increase the traffic load.To reduce the minimum required fuel and therefore be able to increase the traffic load.

At reference or see flight planning manual mrjt 1 figure 4 7 2.for the purpose ?

Question 90-3 : M/kias 74/330 lrc m/kias 74/290 m/kias 70/280

. /com en/com033 120 jpg. exemple 198 M/kias .74/330M/kias .74/330

Using the power setting table for the single engine aeroplane determine the ?

Question 90-4 : 134 kt and 55 7 lbs/h 131 kt and 56 9 lbs/h 125 kt and 55 7 lbs/h 136 kt and 56 9 lbs/h

Admin .first step search for isa temperature .oat is +3°c isa at 6000 ft = 15°c 2 x 6 = +3°c.we are in standard condition isa .for standard day isa ktas is the same 134 kt .step two search for fuel flow .for standard condition isa fuel flow is 55 7 pph pound per hour or lbs/h exemple 202 134 kt and 55.7 lbs/h.134 kt and 55.7 lbs/h.

For turbojet engine driven aeroplane given .taxi fuel 600 kg.fuel flow for ?

Question 90-5 : 77 800 kg 76 100 kg 80 500 kg 79 200 kg

Admin .fuel flow for cruise 6h x 10000 kg = 60000 kg.contingency fuel 5% x 60000 = 3000 kg.taxi fuel 600 kg.alternate fuel 10200 kg.30 minutes fuel for holding at 8000kg/h = 4000 kg.total = 60000 + 3000 + 600 + 10200 + 4000.total = 77800 kg exemple 206 77 800 kg.77 800 kg.

The following apply .temperature isa +15°c.brake release mass 62000 kg.trip ?

Question 90-6 : 13500 kg 13000 kg 13200 kg 13800 kg

Admin . 1194.enter the graph at 5 33 5h40 exemple 210 13500 kg.13500 kg.

At reference or see flight planning manual mrjt 1 figure 4 3 1c. for a flight ?

Question 90-7 : A 17600 kg b 6 hr 50 min a 16200 kg b 6 hr 20 min a 17000 kg b 6 hr 10 min a 20000 kg b 7hr 00 min

.first set the red lines on the graph . /com en/com033 146 jpg.reach the ref lines first and after join the red lines .above landing mass on the right part of the graph you have to use the dashed line it is for high pressure alitude flights exemple 214 (a) 17600 kg (b) 6 hr 50 min.(a) 17600 kg (b) 6 hr 50 min.

At reference or see flight planning manual mrjt 1 figure 4 3 5. for a flight of ?

Question 90-8 : A 18100 kg b 7hr 20 min a 15800 kg b 6hr 00 min a 21800 kg b 9hr 25 min a 19000 kg b 7hr 45min

. /com en/com033 150 jpg.you always need to go first to the ref line and then apply the condition mass temperature wind exemple 218 (a) 18100 kg (b) 7hr 20 min(a) 18100 kg (b) 7hr 20 min

Mark the correct statement .if a decision point procedure is applied for flight ?

Question 90-9 : The trip fuel to the destination aerodrome is to be calculated via the decision point the trip fuel to the destination aerodrome is to be calculated via the suitable enroute alternate a destination alternate is not required the fuel calculation is based on a contingency fuel from departure aerodrome to the decision point

.the decision point procedure permits aircraft to carry less contingency fuel than in the standard case operators select a point called the decision point along the planned route at this point the pilot has two possibilities . reach a suitable proximate diversion airport taking into account the maximum landing weight limitation . continue the flight to the destination airport when the remaining fuel is sufficient exemple 222 The trip fuel to the destination aerodrome is to be calculated via the decision point.The trip fuel to the destination aerodrome is to be calculated via the decision point.

An aeroplane has the following masses .estimated landing weight 50 000 kg.trip ?

Question 90-10 : 4 300 kg 6 185 kg 9 000 kg 6 400 kg

Admin .trip fuel is trip fuel in the operational flight plan trip fuel is 4300 kg exemple 226 4 300 kg.4 300 kg.

In a flight plan when the destination aerodrome is a and the alternate ?

Question 90-11 : 30 minutes holding 1 500 feet above aerodrome b 30 minutes holding 2 000 feet above aerodrome b 15 minutes holding 2 000 feet above aerodrome a 30 minutes holding 1 500 feet above aerodrome a

exemple 230 30 minutes holding 1,500 feet above aerodrome b30 minutes holding 1,500 feet above aerodrome b

A jet aeroplane has a cruising fuel consumption of 4060 kg/h and 3690 kg/h ?

Question 90-12 : 8120 kg 7380 kg 1845 kg 3500 kg

Admin .isolated aerodrome if acceptable to the authority the destination aerodrome can be considered as an isolated aerodrome if the fuel required diversion plus final to the nearest adequate destination alternate aerodrome is more than .for aeroplanes with reciprocating engines fuel to fly for 45 minutes plus 15 % of the flight time planned to be spent at cruising level or two hours whichever is less or. for aeroplanes with turbine engines fuel to fly for two hours at normal cruise consumption above the destination aerodrome including final reserve fuel .2 x 4060 kg = 8120 kg exemple 234 8120 kg.8120 kg.

Given .fuel density = 0 78 kg/l .dry operating mass = 33500 kg .traffic load = ?

Question 90-13 : 17 350 kg 22 100 kg 17 550 kg 21 900 kg

Admin .22500 x 0 78 = 17550 kg.minus 200 kg for taxi we obtain 17350 kg .we also need to check if our maximum allowable take off mass is not exceeded .mtow > dry operating mass + traffic load + fuel.mtow > 33500 + 10600 + 17550.mtow > 61650 kg.that's ok maximum possible take off fuel is 17350 kg exemple 238 17 350 kg.17 350 kg.

Planning an ifr flight from paris to london for the twin jet aeroplane ?

Question 90-14 : 1000 kg 1500 lbs 1100 kg 1000 lbs

Admin . 1224.wind has influence on ground distance but not on the time to reach the top of climb 11 min or fuel flow

Provided that flight conditions on the leg gamma to delta remain unchanged and ?

Question 90-15 : 4475 kg 4745 kg 4635 kg 4250 kg

Admin .actual flight time between beta and gamma 35 minutes .actual fuel used 5285 4970 = 315 kg.fuel flow from beta to gamma 315 / 35 = 9 kg/min.fuel used from gamma to delta 55 min x 9 kg = 495 kg.on arrival overhead delta fuel on board will be .4970 495 = 4475 kg exemple 246 4475 kg.4475 kg.

The required time for final reserve fuel for turbojet aeroplane is ?

Question 90-16 : 30 minutes 45 minutes 60 minutes variable with wind velocity

Admin .an operator shall ensure that the pre flight calculation of usable fuel required for a flight includes .1 taxi fuel and.2 trip fuel and.3 reserve fuel consisting of . i contingency fuel and. ii alternate fuel if a destination alternate aerodrome is required this does not preclude selection of the departure aerodrome as the destination alternate aerodrome and. iii final reserve fuel see appendix below and. iv additional fuel if required by the type of operation e g etops and.4 extra fuel if required by the commander ..appendix . final reserve fuel which shall be . a for aeroplanes with reciprocating engines fuel to fly for 45 minutes or. b for aeroplanes with turbine engines fuel to fly for 30 minutes at holding speed at 1 500 ft 450 m above aerodrome elevation in standard conditions calculated with the estimated mass on arrival at the destination alternate aerodrome or the destination aerodrome when no destination alternate aerodrome is required exemple 250 30 minutes.30 minutes.

A flight has to be made with a single engine aeroplane .for the fuel ?

Question 90-17 : 276 lbs 265 lbs 208 lbs 250 lbs

Admin .total flight time .2h37 + 3 min + 10 min = 2h50 2 83h .fuel comsumption at fl50 isa is . 69 4+71 7 /2 = 70 55 pph.2 83h x 70 55 = 200 pph.200 + 6 lbs for the climb + reserve fuel 30% of 200 + 10 for taxi = 276 lbs exemple 254 276 lbs.276 lbs.

See flight planning manual mrjt 1 figure 4 5 2 and 4 5 3 1.given .distance c d ?

Question 90-18 : 14 500 kg 14 200 kg 17 800 kg 17 500 kg

For 55000 kg we have a tas of 431 kt temperature is isa +12 we have to increase tas by 1 kt per degrees above isa so tas = 431 + 12 = 443 kt .nam = ground distance x tas/gs .nam = 3200 x 443/493 = 2875 nam .at line 55000 the cruise distance nautical air miles is 4151 nautique air miles substract 2875 you find 1276 nam .what is the mass for 1276 nm . /com en/com033 298 jpg.we find 40600 kg this is our end mass .55000 40600 = 14400 kg .last step we have to increase fuel required by 0 6 percent per 10 degrees above isa .14400 x 0 6% = 14486 kg ..see section 5 4 2 method on caa cap697 flight planning manual for that kind of questions exemple 258 14 500 kg.14 500 kg.

At reference or see flight planning manual mrjt 1 figure 4 5 4.planning an ifr ?

Question 90-19 : 273 kg 320 kg 210 kg 263 kg

. /com en/com033 317 jpg.no need for calculation our fuel consumption will be between 270 kg and 275 kg .wind has influence on ground distance but not on the descent time 19 min or fuel flow exemple 262 273 kg.273 kg.

At reference or see flight planning manual mrjt 1 paragraph 2 1 and figure 4 ?

Question 90-20 : Fmp 4% fmp 0% fmp 1% fmp 10%

. /com en/com033 324 jpg. exemple 266 Fmp 4%Fmp 4%

The final reserve fuel for aeroplanes with turbine engines is ?

Question 90-21 : Fuel to fly for 30 minutes at holding speed at 1500 ft 450 m above aerodrome elevation in standard conditions fuel to fly for 45 minutes at holding speed at 1500 ft 450 m above aerodrome elevation in standard conditions fuel to fly for 45 minutes at holding speed at 1000 ft 300 m above aerodrome elevation in standard conditions fuel to fly for 60 minutes at holding speed at 1500 ft 450 m above aerodrome elevation in standard conditions

exemple 270 Fuel to fly for 30 minutes at holding speed at 1500 ft (450 m) above aerodrome elevation in standard conditions.Fuel to fly for 30 minutes at holding speed at 1500 ft (450 m) above aerodrome elevation in standard conditions.

The fuel burn of an aircraft turbine engine is 220 l/h with a fuel density of 0 ?

Question 90-22 : 235 l/h 206 l/h 220 l/h 176 l/h

Admin .volume changes if temperature changes mass consumption will not change .the fuel burn is 220 l/h in mass it is 176 kg/h it will remain the same mass if temperature increases but the volume will be higher .220 x 0 8 / 0 75 = 235 l/h exemple 274 235 l/h.235 l/h.

At reference or see flight planning manual mep1 figure 3 1.a flight is to be ?

Question 90-23 : 6 us gallon 9 us gallon 12 us gallon 3 us gallon

Isa temperature at fl110 is .15°c 2°c x 11 = 7°c.outside air temperature at fl110 is 10°c + 7°c = 17°c . /com en/com033 358 jpg.fuel to climb from 3000 ft to 11000 ft = 9 5 3 5 = 6 us gallon exemple 278 6 us gallon.6 us gallon.

When using decision point procedure you reduce the ?

Question 90-24 : Contingency fuel by adding contingency only from the burnoff between decision point and destination contingency fuel by adding contingency only from the burnoff between the decision airport and destination reserve fuel from 10% down to 5% holding fuel by 30%

.the decision point procedure is a specific set of fuel planning rules which you apply in the pre flight planning phase it makes use of an en route alternate and it tells you to calculate the fuel requirement in two different ways and take the higher of those two fuel figures ..you will need approval from the authority to use this procedure and the overall result in this method of planning will be a reduction of the minimum fuel requirement through a reduction of contingency fuel exemple 282 Contingency fuel by adding contingency only from the burnoff between decision point and destination.Contingency fuel by adding contingency only from the burnoff between decision point and destination.

A jet aeroplane is to fly from a to b the minimum final reserve fuel must allow ?

Question 90-25 : 30 minutes hold at 1500 ft above destination aerodrome elevation when no alternate is required 20 minutes hold over alternate airfield 30 minutes hold at 1500 ft above mean sea level 15 minutes hold at 1500 ft above destination aerodrome elevation

Caestudent06 .if no alternate is required 15 minutes extra reserve fuel must be taken above the 30 minutes . .you are right final reserve fuel for a jet is always 30 minutes at 1500 ft above the aerodrome elevation at holding speed .15 minutes extra reserve fuel additional fuel is mandatory if there is no alternate at destination above the final reserve fuel of 30 minutes exemple 286 30 minutes hold at 1500 ft above destination aerodrome elevation, when no alternate is required.30 minutes hold at 1500 ft above destination aerodrome elevation, when no alternate is required.

At reference or see flight planning manual mrjt 1 figure 4 5 1. given brake ?

Question 90-26 : 1138 kg 1040 kg 1238 kg 1387 kg

. /com en/com033 411 jpg.extrapolate fuel for a mass of 57000 kg . 1150 + 1100 / 2 = 1125 kg.extrapolate fuel for a mass of 57500 kg . 1150 + 1125 / 2 = 1137 5 kg.wind has an effect on ground distance only not for fuel consumption exemple 290 1138 kg1138 kg

At reference or see flight planning manual sep 1 figure 2 2 table 2 2 3.using ?

Question 90-27 : 160 kt and 69 3 lbs/h 158 kt and 74 4 lbs/h 160 kt and 71 1 lbs/h 159 kt and 71 7 lbs/h

.first step search for isa temperature .oat is +13°c isa at 8000 ft = 15°c 2 x 8 = 1°c.we are in isa +14°c .for standard day isa and isa +20°c ktas is the same 160 kt .seconde step search for fuel flow .for standard day isa fuel flow is 71 1 pph.for isa +20°c fuel flow is 68 5 pph.interpolation for isa +14°c is . 71 1 68 5 x 6/20 = 0 8.fuel flow is 68 5 + 0 8 = 69 3 pph exemple 294 160 kt and 69.3 lbs/h.160 kt and 69.3 lbs/h.

At reference or see flight planning manual mep 1 figure 3 3.a flight has to be ?

Question 90-28 : 91 us gallons 76 us gallons 118 us gallons 86 us gallons

.from the table fuel flow in us gallons per hour is 23 3 gph.2h37 + 13min = 2h50 = 2 83 hours .23 3 x 2 83 = 65 93 usg..reserve fuel is 30% of the trip fuel .65 93 x 1 30 = 85 7 usg..minimum block fuel 85 7 + 5 usg for taxi = 90 7 usg . moonen .i found 118 us gallons the power setting is 65% and if i read the table 23 3 gph is for 45% and this is the good response can you explain to me thank you ..we are looking for fuel flow not manifold pressure . /com en/com033 456 jpg. exemple 298 91 us gallons.91 us gallons.

Integrated range' curves or tables are presented in the aeroplane operations ?

Question 90-29 : To determine the fuel consumption for a certain still air distance considering the decreasing fuel flow with decreasing mass to determine the flight time for a certain leg under consideration of temperature deviations to determine the still air distance for a wind components varying with altitude to determine the optimum speed considering the fuel cost as well as the time related cost of the aeroplane

exemple 302 To determine the fuel consumption for a certain still air distance considering the decreasing fuel flow with decreasing mass.To determine the fuel consumption for a certain still air distance considering the decreasing fuel flow with decreasing mass.

At reference or see flight planning manual sep 1 figure 2 4 .using the range ?

Question 90-30 : 865 nm 739 nm 851 nm 911 nm

. /com en/com033 491 jpg. exemple 306 865 nm.865 nm.

At reference or see flight planning manual mrjt 1 figure 4 4. the final reserve ?

Question 90-31 : Pressure altitude aeroplane mass and flaps up with minimum drag airspeed pressure altitude aeroplane mass and flaps down with maximum range speed pressure altitude aeroplane mass and flaps up with maximum range speed pressure altitude aeroplane mass and flaps down with minimum drag airspeed

exemple 310 Pressure altitude, aeroplane mass and flaps up with minimum drag airspeedPressure altitude, aeroplane mass and flaps up with minimum drag airspeed

Finish the endurance/fuel calculation and determine atc endurance for a twin ?

Question 90-32 : Atc endurance 04 07 atc endurance 04 12 atc endurance 03 37 atc endurance 03 52

. /com en/com033 509 jpg.contingency is 5% of the planned trip fuel = 290 kg.final reserve for a jet aeroplane is 30 minutes .minimum take off fuel = 5800 + 290 + 1800 + 1325 = 9215 kg.fuel flow for extra fuel is 2400 kg/h so 585 kg gives 14 minutes 38 seconds exemple 314 Atc endurance: 04:07Atc endurance: 04:07

Given .oat +5°c.during climb average head wind component 20 kt.take off from ?

Question 90-33 : 9 min 3 3 usg 10 min 3 6 usg 7 min 2 6 usg 9 min 2 7 usg

Admin . 2465 exemple 318 9 min. 3,3 usg9 min. 3,3 usg

Maximum allowable take off mass 64 400 kg.maximum landing mass 56 200 ?

Question 90-34 : 11 100 kg 11 400 kg 14 400 kg 8 600 kg

Admin .in order to find the maximum allowable take off fuel we must consider the mtow maximum take off weight and the mlw maximum landing weight ..mtow = dry operating mass + estimated load + maximum take off fuel.64400 = 35500 + 14500 + maximum take off fuel.maximum take off fuel = 14400 kg.mlw = dry operating mass + estimated load + maximum landing fuel.56200 = 35500 + 14500 + maximum landing fuel.maximum landing fuel = 6200 kg.4900 kg trip fuel + 6200 kg = 11100 kg.11100 kg is the lower limitation exemple 322 11 100 kg.11 100 kg.

At reference or use flight planning manual sep 1 table 2 2 3 .given .fl 70.oat ?

Question 90-35 : 12 35 gph tas 159 kt 73 90 gph tas 159 kt 12 35 gph tas 151 kt 11 95 gph tas 160 kt

.no need for calculation . /com en/com033 536 jpg. exemple 326 12.35 gph, tas: 159 kt.12.35 gph, tas: 159 kt.

At reference or see flight planning manual mrjt 1 figure 4 5 3 1. given flight ?

Question 90-36 : 345 nam. 2000 kg 350 nam. 2000 kg 345 nam. 2100 kg 437 nam. 2100 kg

56000 1100 = 54900 kg on table tas is 437 kt .adjustments for operation at non standard temperatures states to decrease tas by 1 knot per degree celsius below isa .isa 5°c > tas = 432 kt..tas x 48/60 = 345 6 nam..on table 54900 > 3736 346 = 3390.we can read for 3390 by interpolation 52900 kg.54900 52900 = 2000 kg .see section 5 4 2 method on caa cap697 flight planning manual for that kind of questions exemple 330 345 namxsx 2000 kg345 namxsx 2000 kg

Planning a flight from paris charles de gaulle to london heathrow for a twin ?

Question 90-37 : 51 629 kg 55 765 kg 51 425 kg 52 265 kg

Take off mass = dry operating mass + traffic load + fuel trip + contingency + alternate + final reserve .contingency fuel is the higher of 5% of trip or 5 minutes holding at 1500ft .5% of trip = 90 kg.5 minutes holding at 1500 ft = 204 kg.the question states holding fuel final reserve 1 225 kg .holding fuel for a jet aircraft is 30 minutes at holding speed at 1500 ft above aerodrome elevation .contingency = 1225/30 x 5 = 204 kg .34000 + 13000 + 1800 + 204 + 1400 + 1225 = 51629 kg .. fuel policy .contingency fuel which shall be the higher of i or ii below . i either .5% of the planned trip fuel or in the event of in flight replanning 5% of the trip fuel for the remainder of the flight . ii an amount to fly for 5 minutes at holding speed at 1 500 ft 450 m above the destination aerodrome in standard conditions .note baggage mass is already included in traffic load .traffic load is the total mass of passengers baggage and cargo including any non revenue load exemple 334 51 629 kg.51 629 kg.

You are to determine the maximum fuel load which can be carried in the ?

Question 90-38 : 800 kg 1000 kg 700 kg 500 kg

. /com en/com033 548 jpg.we can take off with 1000 kg but we can land only after having burn 500 kg if not we will be over weight .the question states that trip fuel is 300 kg it means that we can take off with 800 kg 500 kg + 300 kg and then we will land at our maximum landing mass exemple 338 800 kg.800 kg.

At reference or see flight planning manual mrjt 1 figure 4 5 1.given .brake ?

Question 90-39 : 1700 kg 1650 kg 1750 kg 1800 kg

. /com en/com033 549 jpg. exemple 342 1700 kg.1700 kg.

At reference or see flight planning manual sep 1 figure 2 1. given .take off ?

Question 90-40 : 22 min 6 7 gal 45 nam 24 min 7 7 gal 47 nam 16 5 min 4 9 gal 34 5 nam 23 min 7 7 gal 50 nam

. /com en/com033 557 jpg. exemple 346 22 min, 6.7 gal, 45 nam22 min, 6.7 gal, 45 nam


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