At reference or see Flight Planning Manual MRJT 1 Figure 4 4 The final reserve fuel taken from the HOLDING PLANNING table for the twin jet aeroplane is based on the ?
Attainment > AIM
Finish the ENDURANCE/FUEL CALCULATION and determine ATC ENDURANCE for a twin jet aeroplane with the help of the table provided Contingency is 5% of the planned trip fuel and fuel flow for extra fuel ?
Given OAT +5°CDuring climb average head wind component 20 ktTake off from MSL with the initial mass of 3 650 lbsFind Time and fuel to climb to FL75 2464 ?
Maximum allowable take off mass 64 400 kgmaximum landing mass 56 200 kgmaximum zero fuel mass 53 000 kgdry operating mass 35 500 kgestimated load 14 500 kgestimated trip fuel 4 900kgminimum take off fuel 7 400 kgfind maximum allowable take off fuel Pressure altitude aeroplane mass flaps up with minimum drag airspeed. in order to find maximum allowable take off fuel we must consider mtow (maximum take off weight) 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 the lower limitation. 
At reference or use flight planning manual sep 1 table 2 2 3 given fl 70oat 19°clean mixture 2300 rpm find fuel flow in gallons per hour gph and tas err _a_033 536 2 35 gph tas 59 kt. no need calculation img /com_en/com033 536 jpg . 
At reference or see flight planning manual mrjt 1 figure 4 5 3 1 given flight time from top of climb to enroute point in fl280 48 min cruise procedure long range cruise lrc temp isa 5° ctake off mass 56 000 kgclimb fuel 1 100 kgfind distance in nautical air miles nam this leg and fuel consumption err _a_033 542 2 35 gph tas 59 kt. 56000 1100 = 54900 kg on table tas 437 kt adjustments operation at non standard temperatures states to decrease tas 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 3390 (by interpolation) 52900 kg 54900 52900 = 2000 kg see section 5 4 2 method on caa cap697 flight planning manual that kind of questions. 
Planning a flight from paris charles gaulle to london heathrow a twin jet aeroplanepreplanning maximum take off mass 62 800 kgmaximum zero fuel mass 51 250 kgmaximum landing mass 54 900 kgmaximum taxi mass 63 050 kgassume following preplanning results trip fuel 1 800 kgalternate fuel 1 400 kgholding fuel final reserve 1 225 kgdry operating mass 34 000 kgtraffic load 13 000 kgcatering 750 kgbaggage 3 500 kgfind take off mass tom 2 35 gph tas 59 kt. Take off mass = dry operating mass + traffic load + fuel (trip + contingency + alternate + final reserve) contingency fuel 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 a jet aircraft 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 higher of (i) or (ii) below (i) either 5% of planned trip fuel or in event of in flight replanning 5% of trip fuel the remainder of flight (ii) an amount to fly 5 minutes at holding speed at 1 500 ft (450 m) above destination aerodrome in standard conditions note baggage mass already included in traffic load traffic load the total mass of passengers baggage cargo including any non revenue load. Question 88-8
You are to determine maximum fuel load which can be carried in following conditions dry operating mass 2800 kg trip fuel 300 kg payload 400 kg maximum take off mass 4200 kg maximum landing mass 3700 kg 2 35 gph tas 59 kt. img /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 300 kg it means that we can take off with 800 kg (500 kg + 300 kg) then we will land at our maximum landing mass. Question 88-9
At reference or see flight planning manual mrjt 1 figure 4 5 1given brake release mass 62000 kgtemperature isa +15°cthe fuel required a climb from sea level to fl330 err _a_033 549 2 35 gph tas 59 kt. img /com_en/com033 549 jpg . Question 88-10
At reference or see flight planning manual sep 1 figure 2 1 given take off mass 3500 lbsdeparture aerodrome pressure altitude 2500 ftoat +10°cfirst cruising level fl 140oat 5°cfind time fuel and still air distance to climb err _a_033 557 22 min 6 7 gal 45 nam. img /com_en/com033 557 jpg . Question 88-11
At reference or see flight planning manual sep1 figure 2 2 table 2 2 3a flight has to be made with single engine sample aeroplanefor fuel calculation allow 10 lbs fuel start up and taxi 3 minutes and 10 lbs of additional fuel to allow the climb 10 minutes and no fuel correction the descent planned flight time overhead to overhead 03 hours and 12 minutesreserve fuel 30% of trip fuel power setting 23 in hg or full throttle 2300 rpm 20°c lean flight level 70 and oat isa +20°c the minimum block fuel err _a_033 591 22 min 6 7 gal 45 nam. total flight time 3h12 + 3 min + 10 min = 3h25 (3 42h) fuel comsumption at fl70 (isa +20°c) is (68 5+69)/2 = 68 75 pph 3 42h x 68 75 = 235 pph 235 + 10 lbs the climb + (reserve fuel 30% of 235) + 10 taxi = 325 lbs. Question 88-12
A flight has to be made with a multi engine piston aeroplane mep 1 for fuel calculations take 5 us gallons the taxi and an additional 13 min at cruise condition to account climb and descent calculated time from overhead to overhead 1h 47 min powersetting 45% 2600 rpmcalculated reserve fuel 30% of trip fuel fl100 temperature 5°c find minimum block fuel 1506 22 min 6 7 gal 45 nam. total flight time 3h12 + 3 min + 10 min = 3h25 (3 42h) fuel comsumption at fl70 (isa +20°c) is (68 5+69)/2 = 68 75 pph 3 42h x 68 75 = 235 pph 235 + 10 lbs the climb + (reserve fuel 30% of 235) + 10 taxi = 325 lbs. Question 88-13
A flight has to be made with sep for fuel calculation allow 10 lbs fuel start up and taxi 3 min and 1 gallon of additional fuel to allow the climb 10 min and no fuel correction the descent planned flight time overhead to overhead 3h 12 min reserve fuel 30% of trip fuel power setting 25 inch hg or full throttle 2100 rpm 20 °c fl70 and oat 11 °cthe minimum block fuel 1507 22 min 6 7 gal 45 nam. total flight time 3h12 + 3 min + 10 min = 3h25 (3 42h) fuel comsumption at fl70 (isa +20°c) is (68 5+69)/2 = 68 75 pph 3 42h x 68 75 = 235 pph 235 + 10 lbs the climb + (reserve fuel 30% of 235) + 10 taxi = 325 lbs. Question 88-14
A flight has to be made with sep1 for fuel calculation allow 10 lbs fuel start up and taxi 3 min and 1 gallon of additional fuel to allow the climb 10 min and no fuel correction the descent planned flight time overhead to overhead 3h 12 reserve fuel 30% of trip fuel power setting 25 inch hg or full throttle 2100 rpm 20 °c lean fl70 and oat 11 °c the minimum block fuel 1508 22 min 6 7 gal 45 nam. total flight time 3h12 + 3 min + 10 min = 3h25 (3 42h) fuel comsumption at fl70 (isa +20°c) is (68 5+69)/2 = 68 75 pph 3 42h x 68 75 = 235 pph 235 + 10 lbs the climb + (reserve fuel 30% of 235) + 10 taxi = 325 lbs. Question 88-15
Given dry operating mass 33500 kg traffic load 7600 kg maximum allowable takeoff mass 66200 kg taxi fuel 200 kg tank capacity 16100 kg the maximum possible takeoff fuel 22 min 6 7 gal 45 nam. total flight time 3h12 + 3 min + 10 min = 3h25 (3 42h) fuel comsumption at fl70 (isa +20°c) is (68 5+69)/2 = 68 75 pph 3 42h x 68 75 = 235 pph 235 + 10 lbs the climb + (reserve fuel 30% of 235) + 10 taxi = 325 lbs. Question 88-16
Given brake release mass 58 000 kgtemperature isa 15the fuel required to climb from an aerodrome at elevation 4000 ft to fl300 1514 22 min 6 7 gal 45 nam. total flight time 3h12 + 3 min + 10 min = 3h25 (3 42h) fuel comsumption at fl70 (isa +20°c) is (68 5+69)/2 = 68 75 pph 3 42h x 68 75 = 235 pph 235 + 10 lbs the climb + (reserve fuel 30% of 235) + 10 taxi = 325 lbs. Question 88-17
Given distance from departure to destination 500 nmtrue track 090°wind 090°/20kttas 150 ktwhat the distance and time of pet from departure point Distance 283 nm time 3 min. Ground speed out 130 kt ground speed home 170 kt pet = d x gsh / (gso + gsh) pet = 500 x 170 /(130 + 170) = 283 nm. Question 88-18
Use route manual chart e hi 1an aircraft has to fly from glasgow 55°52'n 004°27'w to benbecula 57°29'n 007°22'w cruising at 320 kt tas assuming a headwind of 40 kt and cruise fuel consumption of 2300 kg/h what the forecast fuel used this flight err _a_033 6 Distance 283 nm time 3 min. Ground speed = 320 40 = 280 kt 136 nm/(280/60)= 29 14 minutes 2300/60 = 38 333 kg/min 29 14 x 38 333 = 1117 kg. Question 88-19
Given distance from departure to destination 500 nmsafe endurance 4 htas 140 ktground speed out 150 ktground speed home 130 ktwhat the distance and time of psr from departure point Distance 279 nm time min. Point of safe return (psr) = endurance x homeward gs / (outbound gs + homeward gs) ground speed out = 150 kt ground speed home = 130 kt point of safe return (psr) = 4 x 130 / (150 + 130) point of safe return (psr) = 65000 / 280 point of safe return (psr) = 1 86 h 1 86 x 60 = 111 6 min distance of psr from departure point at a speed of 150 kt 111 6 min x (150/60) = 279 nm. Question 88-20
On a given path it possible to chose between four flight levels fl each associated with a mandatory flight mach number m the flight conditions static air temperature sat and headwind component hwc are given below fl 370 m = 0 80 ts = 60°c hwc = 15 ktfl 330 m = 0 78 ts = 60°c hwc= 5 ktfl 290 m = 0 80 ts = 55°c hwc = 15 ktfl 270 m = 0 76 ts = 43°c hwc = 0the flight level allowing highest ground speed Distance 279 nm time min. Set corresponding mark m(kt) against outside temperature at flight altitude read in front of mach number on outer scale true air speed example 60° mach 0 8 => tas 451 kt fl 370 => tas = 451 kt => gs = 439 kt fl 330 => tas = 440 kt => gs = 435 kt fl 290 => tas = 456 kt => gs = 441 kt fl 270 => tas = 445 kt => gs = 445 kt the flight level allowing highest ground speed fl 270. Question 88-21
Given distance from departure to destination 410 nm safe endurance 3 6 h true track 055° w/v 180/35 tas 120 ktwhat the distance of psr from departure point Distance 279 nm time min. First step find gs out given true track true air speed wind direction velocity required wind correction angle ground speed computer solution a) set true track to true index b) turn indicator to wind direction in this case using black azimuth graduation (the angle being upwind counting anti clockwise) c) shift speed arc corresponding to true air speed so as to coincide with wind speed on indicator d) read wind correction angle at same place read ground speed under center bore from scale on axis of slide setting 055° to true index set indicator to 180° on black azimuth circle (being upwind) adjust speed arc labelled 120 kt of diagram slide to wind speed 35 kt of indicator scale reading under plotted point read wind correction angle 14° under center bore read ground speed 136 kt step two proceed on same way gs home (you will find 97 kt) step three now apply psr formula psr = time x gs out x gs home / (gs out + gs home) psr = 3 6 x 136 x 97 / (136 + 97) psr = 203 8 nm this a 4 points question at exam mathematical calculation on this kind of exercise valid only one right angled triangle which not case here only computer enables you to find good answer. Question 88-22
The surface weather system over england 53°n 002°w err _a_033 22 An occluded front moving east. img /com_en/com022 33a png over england surface weather system an an occluded front img /com_en/com033 22 jpg the arrow attached to occluded front indicates direction speed (20 km/h). Question 88-23
Which describes intensity of icing if any at fl 150 in vicinity of toulouse 44°n 01°e err _a_033 25 An occluded front moving east. its due to presence of cbs around toulouse always assume moderate or severe icing next to them. Question 88-24
Given distance a to b 2050 nmmean groundspeed 'on' 440 ktmean groundspeed 'back' 540 ktthe distance to point of equal time pet between a and b An occluded front moving east. ground speed out (gso) = 440 kt ground speed home (gsh) = 540 kt distance to pet = distance x gsh / (gso + gsh) distance to pet = 2050 x 540 / (440 + 540) distance to pet = 1107000 / 980 = 1130 nm. Question 88-25
Which the heaviest type of precipitation if any forecast bordeaux/merignac at 1000 utc err _a_033 27 An occluded front moving east. img /com_en/com033 27 jpg at 1000 utc if any heaviest type of precipitation forcast light rain. Question 88-26
Given distance from departure to destination 315 nmtrue track 343°w/v 015/15tas 100 ktwhat the distance of pet from departure point An occluded front moving east. under index set true track 343° centre dot on tas 100 kt with rotative scale set wind 015° / 15 kt now drift always measured from heading to track turn to set true heading 345° (343° + 2° left drift) under index you now read a ground speed out of 88 kt proceed in same way to find ground speed home of 112 kt (right drift of 4° true heading of 159°) distance to pet = distance x gsh / (gso + gsh) distance to pet = 315 x 112 / (88 + 112) distance to pet = 35280 / 200 = 176 nm. Question 88-27
What lowest cloud conditions oktas/ft are forecast johannesburg/jan smuts at 0300 utc err _a_033 31 An occluded front moving east. ft0900 prob30 0305 3000 bcfg bkn004 30% chance that temporarily between 0300 utc 0500 utc there will be patches fog with 3000 m visibility 5 to 7 oktas of cloud at 400 ft. Question 88-28
Given distance from departure to destination 210 nmsafe endurance 2 5 htrue track 035w/v 250/20tas 105 ktwhat the distance of psr from departure point An occluded front moving east. point of safe return (psr) = endurance x homeward gs / (outbound gs + homeward gs) on computer under index set true track 035° centre dot on tas 105 kt with rotative scale set wind 250°/20 kt we read a right drift of 5° drift always measured from heading to track so turn to set true heading 030° (035° 5° right drift) under index you read a ground speed out of 122 kt proceed on same way gs home under index set true track 215° centre dot on tas 105 kt with rotative scale set wind 250°/20 kt we read a left drift of 4° drift always measured from heading to track so turn to set true heading 219° (215° + 4° right drift) under index you read a ground speed home of 88 kt point of safe return (psr) = 2 5 x 88 / (122 + 88) point of safe return (psr) = 220 / 210 point of safe return (psr) = 1 04 h distance of psr from departure point at a speed of 122 kt 1 04 x 122 = 127 nm. Question 88-29
Given following d = flight distancex = distance to point of equal timegso = groundspeed outgsr = groundspeed returnthe correct formula to find distance to point of equal time X = d x gsr / (gso + gsr). point of safe return (psr) = endurance x homeward gs / (outbound gs + homeward gs) on computer under index set true track 035° centre dot on tas 105 kt with rotative scale set wind 250°/20 kt we read a right drift of 5° drift always measured from heading to track so turn to set true heading 030° (035° 5° right drift) under index you read a ground speed out of 122 kt proceed on same way gs home under index set true track 215° centre dot on tas 105 kt with rotative scale set wind 250°/20 kt we read a left drift of 4° drift always measured from heading to track so turn to set true heading 219° (215° + 4° right drift) under index you read a ground speed home of 88 kt point of safe return (psr) = 2 5 x 88 / (122 + 88) point of safe return (psr) = 220 / 210 point of safe return (psr) = 1 04 h distance of psr from departure point at a speed of 122 kt 1 04 x 122 = 127 nm. Question 88-30
The flight crew of a turbojet aeroplane prepares a flight using following data flight leg air distance 2 700 nm flight level fl 310 true airspeed 470 kt tailwind component at this level 35 kt initially planned take off mass without extra fuel on board 195 000 kg fuel price 0 28 euro/l at departure0 26 euro/l at destinationto maximize savings commander chooses to carry extra fuel in addition to that which necessary the optimum quantity of fuel which should be carried in addition to prescribed quantity err _a_033 35 The fuel transport operation not recommended in this case. the fuel cheaper at destination the fuel transport operation not recommended in this case. Question 88-31
From which of following would you expect to find information regarding known short unserviceability of vor tacan and ndb The fuel transport operation not recommended in this case. the fuel cheaper at destination the fuel transport operation not recommended in this case. Question 88-32
The wind velocity over italy err _a_033 54 A maximum of kt at fl38. img /com_en/com033 54 jpg a jet crossing france italy at fl 380 with a maximum of 110 kt. Question 88-33
Assuming following data ground distance to be covered 2 000 nmcruise flight level fl 330cruising speed mach 0 82 true airspeed 470 kt head wind component 30 kt planned destination landing mass 160 000 kg temperature isa cg 37total anti ice on pack flow hi time needed to carry out such a flight err _a_033 56 A maximum of kt at fl38. nam = ngm x (tas/gs)= 2000 x (470/440) = 2136 nam img /com_en/com033 56 jpg between 2100 2200 we can find our answer 4 h 43 min 210883 i think that answer wrong because (4 39 + 4 51)/2 = 4 45 h > 4h 27' (4h 43' > 4 75h) so correct answer as i think 4h 27' no times on annex are shown as 4 39 4h39 4 51 4h51 it's written on top below fuel consumed (kg) time (h min). Question 88-34
When calculating fuel required to carry out a given flight one must take into account 1 wind2 foreseeable airborne delays3 other weather forecasts4 any foreseeable conditions which may delay landingthe combination which provides correct statement A maximum of kt at fl38. nam = ngm x (tas/gs)= 2000 x (470/440) = 2136 nam img /com_en/com033 56 jpg between 2100 2200 we can find our answer 4 h 43 min 210883 i think that answer wrong because (4 39 + 4 51)/2 = 4 45 h > 4h 27' (4h 43' > 4 75h) so correct answer as i think 4h 27' no times on annex are shown as 4 39 4h39 4 51 4h51 it's written on top below fuel consumed (kg) time (h min). Question 88-35
Which of following flight levels if any forecast to be clear of significant cloud icing and cat along marked route from shannon 53°n 10°w to berlin 53°n 13°e err _a_033 64 A maximum of kt at fl38. img /com_en/com033 64 jpg there are isolated embedded cbs from below fl100 to fl220 so fl 210 excluded you have cat area n°2 from fl270 to fl400 so fl290 excluded only fl250 forecast to be clear of significant cloud icing cat along marked route from shannon to berlin. Question 88-36
Given distance from departure to destination 215 nmsafe endurance 3 3 htrue track 005°w/v 290/15tas 125 ktwhat the distance of psr from departure point A maximum of kt at fl38. point of safe return (psr) = endurance x homeward gs / (outbound gs + homeward gs) on computer under index set true track 005° centre dot on tas 125 kt with rotative scale set wind 290°/15 kt we read a right drift of 7° drift always measured from heading to track so turn to set true heading 358° (005° 7° right drift) under index you read a ground speed out of 120 kt proceed on same way gs home under index set true track 185° centre dot on tas 125 kt with rotative scale set wind 290°/15 kt we read a left drift of 7° drift always measured from heading to track so turn to set true heading 192° (185° + 7° right drift) under index you read a ground speed home of 128 kt point of safe return (psr) = 3 3 x 128 / (120 + 128) point of safe return (psr) = 422 4 / 248 point of safe return (psr) = 1 70 h distance of psr from departure point at a speed of 120 kt 1 7 x 120 = 204 nm. Question 88-37
Given distance from departure to destination 200 nmsafe endurance 3 htas 130 ktground speed out 150 ktground speed home 110 ktwhat the distance psr from departure point A maximum of kt at fl38. point of safe return (psr) = endurance x homeward gs / (outbound gs + homeward gs) ground speed out = 150 kt ground speed home = 110 kt point of safe return (psr) = 3 x 110 / (150 + 110) point of safe return (psr) = 330 / 260 point of safe return (psr) = 1 2692 h distance of psr from departure point at a speed of 150 kt 1 2692 h x 150 = 190 nm. Question 88-38
What minimum visibility m forecast 0600 utc at london lhr egll err _a_033 71 A maximum of kt at fl38. point of safe return (psr) = endurance x homeward gs / (outbound gs + homeward gs) ground speed out = 150 kt ground speed home = 110 kt point of safe return (psr) = 3 x 110 / (150 + 110) point of safe return (psr) = 330 / 260 point of safe return (psr) = 1 2692 h distance of psr from departure point at a speed of 150 kt 1 2692 h x 150 = 190 nm. Question 88-39
Excluding rvsm an appropriate flight level ifr flight in accordance with semi circular height rules on a course of 180° m A maximum of kt at fl38. the question states excluding rvsm img /com_en/com033 1228 jpg a vfr fligh level flxx5 or flxx5 below fl290 in accordance with semi circular rules a magnetic heading of 180° we need a even level. Question 88-40
Given distance from departure to destination 400 nmsafe endurance 2 5 htas 115 ktground speed out 130 ktground speed home 105 ktwhat the distance of psr from departure point A maximum of kt at fl38. point of safe return (psr) = endurance x homeward gs / (outbound gs + homeward gs) ground speed out = 130 kt ground speed home = 105 kt point of safe return (psr) = 2 5 x 105 / (130 + 105) point of safe return (psr) = 262 5 / 235 point of safe return (psr) = 1 12 h distance of psr from departure point at a speed of 130 kt 1 12 h x 130 = 145 nm.
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