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Question 248-1 : Propeller blade twist is the ? [ Question security ]

Varying of the blade angle from the root to the tip of a propeller blade

.the pitch angle varies from root to tip to maintain a near constant angle of attack along the blade .the reason a propeller is 'twisted' is that the outer parts of the propeller blades like all things that turn about a central point travel faster than the portions near the hub if the blades had the same geometric pitch throughout their lengths portions near the hub could have negative aoas while the propeller tips would be stalled at cruise speed twisting or variations in the geometric pitch of the blades permits the propeller to operate with a relatively constant aoa along its length when in cruising flight .propeller blades are twisted to change the blade angle in proportion to the differences in speed of rotation along the length of the propeller keeping thrust more nearly equalized along this length exemple 348 Varying of the blade angle from the root to the tip of a propeller blade.

Question 248-2 : A rotating propeller blade element produces an aerodynamic force f that may be resolved into two components . a force t perpendicular to the plane of rotation thrust . a force r generating a torque absorbed by engine power . the diagram representing a rotating propeller blade element during the ?

Diagram 1

. /com en/com080 1001 gif. exemple 352 Diagram 1.

Question 248-3 : A rotating propeller blade element produces an aerodynamic force f that may be resolved into two components . a force t perpendicular to the plane of rotation thrust . a force r generating a torque absorbed by engine power . the diagram representing a rotating propeller blade element during ?

Diagram 2

. /com en/com080 1001 gif. exemple 356 Diagram 2.

Question 248-4 : A rotating propeller blade element produces an aerodynamic force f that may be resolved into two components . a force t perpendicular to the plane of rotation thrust . a force r generating a torque absorbed by engine power . the diagram representing a windmilling propeller is . err a 081 1006 ?

Diagram 4

. /com en/com080 1001 gif. exemple 360 Diagram 4.

Question 248-5 : The variation of propeller efficiency of a fixed pitch propeller with tas at a given rpm is shown in . err a 081 1019 ?

Figure 4

. /com en/com080 1018 jpg.variation of propeller efficiency of a fixed pitch propeller exemple 364 Figure 4.

Question 248-6 : Which statement is correct regarding a propeller .i increasing tip speed to supersonic speed does not affect propeller noise .ii increasing tip speed to supersonic speed increases propeller efficiency ?

I is incorrect ii is incorrect

A propeller must be able to absorb all the shaft power developed by the engine and also operate with maximum efficiency throughout the required performance envelope of the aircraft the critical factor is tip velocity if tip velocity is too high the blade tips will approach the local speed of sound and compressibility effects will decrease thrust and increase rotational drag supersonic tip speed will considerably reduce the efficiency of a propeller and greatly increase the noise it generates .in most installations increasing the number of blades helps to reduce noise cockpit noise comes from a variety of sources engine exhaust slipstream and the propeller vibrations are also perceived as noise in the cockpit a two blade propeller produces an inherent once per revolution vibration that shakes the airframe so a three blade propeller will be inherently smoother and therefore quieter etc exemple 368 I is incorrect, ii is incorrect.

Question 248-7 : Which statement is correct .i at a given rpm the propeller efficiency of a fixed pitch propeller is maximum at only one value of tas .ii a constant speed propeller maintains near maximum efficiency over a wider range of aeroplane speeds than a fixed pitch propeller ?

I is correct ii is correct

.for a fixed pitch propeller the pitch angle cannot be changed it varies from root to tip to maintain near constant angle of attack along the blade at a given rpm the propeller efficiency of a fixed pitch propeller is maximum at only one value of tas when a constant speed propeller maintains near maximum efficiency over a wider range of aeroplane speeds . /com en/com080 1175 png. constant speed propeller pitch control .coarse pitch is large blade angle low rpm and high forward airspeed .fine pitch is small blade angle high rpm and low forward airspeed exemple 372 I is correct, ii is correct.

Question 248-8 : Which statement is correct when comparing a fixed pitch propeller with a constant speed propeller .i a constant speed propeller reduces fuel consumption over a range of cruise speeds .ii a constant speed propeller improves take off performance as compared with a coarse fixed pitch propeller ?

I is correct ii is correct

.a constant speed propeller maintains near maximum efficiency over a wider range of aeroplane speeds than a fixed pitch propeller therefore a constant speed propeller reduces fuel consumption over a range of cruise speeds . /com en/com080 1175 png. pitch control .coarse pitch is large blade angle low rpm and high forward airspeed .fine pitch is small blade angle high rpm and low forward airspeed .a coarse fixed pitch propeller improves cruise performance not take off and climb performances exemple 376 I is correct, ii is correct.

Question 248-9 : Which statement is correct regarding a propeller .i increasing tip speed to supersonic speed increases propeller noise .ii increasing tip speed to supersonic speed decreases propeller efficiency ?

I is correct ii is correct

.a propeller must be able to absorb all the shaft power developed by the engine and also operate with maximum efficiency throughout the required performance envelope of the aircraft the critical factor is tip velocity if tip velocity is too high the blade tips will approach the local speed of sound and compressibility effects will decrease thrust and increase rotational drag supersonic tip speed will considerably reduce the efficiency of a propeller and greatly increase the noise it generates .in most installations increasing the number of blades helps to reduce noise cockpit noise comes from a variety of sources engine exhaust slipstream and the propeller vibrations are also perceived as noise in the cockpit a two blade propeller produces an inherent once per revolution vibration that shakes the airframe so a three blade propeller will be inherently smoother and therefore quieter etc exemple 380 I is correct, ii is correct.

Question 248-10 : Which statement about propeller icing is correct .i propeller icing increases blade element drag and reduces blade element lift .ii propeller icing reduces propeller efficiency ?

I is correct ii is correct

exemple 384 I is correct, ii is correct.

Question 248-11 : A windmilling propeller ?

Produces drag instead of thrust

.a windmilling propeller produces drag roughly equivalent to that of a solid disk of the same diameter as the propeller itself . drag from a windmilling propeller is high it is being driven by the relative airflow and is generating both drag and torque a feathered propeller generates the least drag there is no torque because it is not rotating and the parasite drag is a minimum because the blades are edge on to the relative airflow exemple 388 Produces drag instead of thrust.

Question 248-12 : The correct sequence of cross sections representing propeller blade twist is . err a 081 1283 ?

Sequence 2

Img /com en/com080 16658 jpg.the reason a propeller is 'twisted' is that the outer parts of the propeller blades like all things that turn about a central point travel faster than the portions near the hub exemple 392 Sequence 2.

Question 248-13 : An aeroplane is in a level turn at a constant tas of 300 kt and a bank angle of 45° its turning radius is . given g= 10 m/s² ?

2381 metres

Radius m = v² m/s / g tan bank angle .300 kt = 154 33 m/s.154 3² = 23818 77.radius m = 23818 77 / 10 x tan 45°.radius = 2381 877 m exemple 396 2381 metres.

Question 248-14 : By what percentage does the lift increase in a level turn at 45° angle of bank compared with straight and level flight ?

41%

.load factor = 1 / cos bank angle.load factor = 1 / cos45°.load factor = 1 41.compared with straight and level flight your load factor increases by 1 41 or 41% thus you must increase lift by 41% to counteract and stay in level exemple 400 41%.

Question 248-15 : In a steady level co ordinated turn the load factor n and the stall speed vs will be ?

N greater than 1 vs higher than in straight and level flight

exemple 404 N greater than 1, vs higher than in straight and level flight.

Question 248-16 : Two identical aircraft a and b with the same mass are flying steady level co ordinated 20 degree bank turns if the tas of a is 130 kt and that of b is 200 kt ?

The rate of turn of a is greater than that of b

Img /com en/com080 222 jpg.v in m/s .aircraft a radius = 65²/ 10 x tan 20 = 1160 m .aircraft b radius = 100²/ 10 x tan 20 = 2747 m .aircraft a rate of turn = 65 / 20 x 3 14 x 1160 = 8 9° .aircraft b rate of turn = 100 / 20 x 3 14 x 2747 = 5 8° .notice tas in nm can be divided by 2 to get an approximate speed in meters per second 130 kt / 2 = approximately 65 m/s exemple 408 The rate of turn of a is greater than that of b.

Question 248-17 : The bank angle in a rate one turn depends on ?

Tas

.rule of thumb .angle of bank rate 1 at tas 120 kt = 120 + 120/2 = 18° exemple 412 Tas.

Question 248-18 : The stall speed in a 60° banked turn increases by the following factor ?

1 41

.1 / square root cos 60 = 1 41 exemple 416 1.41.

Question 248-19 : If an aeroplane carries out a descent at 160 kt ias and 1000 ft/min vertical speed ?

Weight is greater than lift

exemple 420 Weight is greater than lift.

Question 248-20 : What is the approximate value of the lift of an aeroplane at a gross weight of 50000 n in a horizontal co ordinated 45 degrees banked turn ?

70000 n

.lift = weight / cos bank angle.lift = 50000 / cos 45 = 70710 n . bjacoby .it is nearer to 80 000 than 70 000. .no it is nearer to 71000 than 80000 exemple 424 70000 n.

Question 248-21 : Assuming zero thrust the point on the diagram corresponding to the value for minimum sink rate is . err a 081 323 ?

Point c

. /com en/com080 323 jpg. exemple 428 Point c

Question 248-22 : Assuming zero thrust the point on the diagram corresponding to the minimum glide angle is . err a 081 324 ?

Point b

. /com en/com080 323 jpg. exemple 432 Point b

Question 248-23 : Which point in the diagram gives the lowest speed in horizontal flight . err a 081 325 ?

Point d

. /com en/com080 323 jpg. exemple 436 Point d

Question 248-24 : What is the correct relationship between the true airspeed for i minimum sink rate and ii minimum glide angle at a given altitude ?

I is less than ii

.the minimum sink rate speed is vmp velocity for minimum power .the minimum glide angle occurs at vmd velocity for minimum drag . 1135.vmp is a lower speed than vmd minimum sink rate is less than minimum glide angle exemple 440 (i) is less than (ii).

Question 248-25 : Why is vmcg determined with the nosewheel steering disconnected ?

Because the value of vmcg must also be applicable on wet and/or slippery runways

.vmcg the minimum control speed on the ground is the calibrated airspeed during the take off run at which when the critical engine is suddenly made inoperative it is possible to maintain control of the aeroplane using the rudder control alone without the use of nosewheel steering as limited by 667 n of force 150 lbf and the lateral control to the extent of keeping the wings level to enable the take off to be safely continued using normal piloting skill .the nosewheel steering is disconnected because the value of vmcg must also be applicable on wet and/or slippery runways .in the determination of vmcg assuming that the path of the aeroplane accelerating with all engines operating is along the centreline of the runway its path from the point at which the critical engine is made inoperative to the point at which recovery to a direction parallel to the centreline is completed may not deviate more than 9 1 m 30 ft laterally from the centreline at any point exemple 444 Because the value of vmcg must also be applicable on wet and/or slippery runways.

Question 248-26 : By what approximate percentage will the stall speed increase in a horizontal co ordinated turn with a bank angle of 45° ?

19%

.the stalling speed of an airplane increases as the angle of bank increases .load factor in a turn is 1/cos bank angle .load factor is 1/cos45° = 1/0 707 = 1 414.stall speed increases with the square root of the load factor .square root 1 414 = 1 189.example .your aircraft has a stall speed of 100 kt in straight and level flight it will stall at 100 kt x 1 189 = 118 9 kt in a horizontal co ordinated turn with a bank angle of 45° .thus we can say that stall speed has gone up by 19% exemple 448 19%.

Question 248-27 : An aeroplane has a stall speed of 100 kt when the aeroplane is flying a level co ordinated turn with a load factor of 1 5 the stall speed is ?

122 kt

.stall speed increases with the square root of the load factor .vs = 100 kt x square root 1 5.vs = 100 kt x 1 22.vs = 122 kt exemple 452 122 kt.

Question 248-28 : An aeroplane has a stall speed of 100 kt at a load factor n=1 in a turn with a load factor of n=2 the stall speed is ?

141 kt

.stall speed increases with the square root of the load factor .vs = 100 kt x square root 2.vs = 100 kt x 1 41.vs = 141 kt exemple 456 141 kt.

Question 248-29 : How does vmcg change with increasing field elevation and temperature ?

Decreases because the engine thrust decreases

.if conditions are hot high and humid the air is less dense this reduces thrust and means when you lose the critical engine on takeoff there is less yawing effect from asymmetric thrust than there would be in denser air .vmcg concerns the calibrated airspeed for minimum control on the ground i e using primary aerodynamic controls only to correct the yawing tendency we know that flying surfaces are more effective the faster air flows over them so if the yaw is less pronounced the required airspeed for counteracting it need not be so high .as elevation and temperature increase hot high and humid vmcg decreases .. increases because at a lower density a larger ias is necessary to generate the required rudder force is incorrect because at high altitudes versus low altitudes you fly at the same ias for same amount of lift only tas changes with air density . increases because vmcg is related to v1 and vr and those speeds increase if the density decreases is incorrect because vmcg will determine v1 . decreases because vmcg is expressed in ias which decreases with constant tas and decreasing density is incorrect because the ias is not decreasing at constant tas and decreasing air density suppose unlimited engine power exemple 460 Decreases, because the engine thrust decreases.

Question 248-30 : The lift coefficient cl of an aeroplane in steady horizontal flight is 0 42 an increase in angle of attack of 1 degree increases cl by 0 1 a vertical up gust instantly changes the angle of attack by 3 degrees the load factor will be ?

1 71

.1° => 0 1.3° => 0 3. 0 42 + 0 3 / 0 42 = 1 71 exemple 464 1.71

Question 248-31 : When an aeroplane performs a straight steady climb with a 20% climb gradient the load factor is equal to ?

0 98

.climb gradient% = climb gradient° / 0 6.climb gradient° = climb gradient% x 0 6.it works for small angles only here is the exact formula .tan alpha = 20 / 100 = 0 2.alpha = cotan 0 2.alpha = 11 309° .angle of climb = 20% = 11 31°.n load factor = lift x cos11 31 / mass.assuming mass = lift in level flight = 1.lift = mass ==> n = cos11 31.n = 0 98

Question 248-32 : What is the approximate diameter of a steady level co ordinated turn with a bank angle of 30 degrees and a speed tas of 500 kt ?

23 km

.radius = tas² / g x tan bank angle .tas in m/s.1 kt = 0 515 m/s.500 kt = 257 5 m/s.radius = 257 5² / 10 x tan30° .radius = 66306 / 5 773 = 11484 m.diameter = 2 x 11484 = 22968 m 22 968 km exemple 472 23 km.

Question 248-33 : Which of the following statements is correct .i vmcl is the minimum control speed in the landing configuration .ii the speed vmcl can be limited by the available maximum roll rate ?

I is correct ii is correct

.vmcl is the minimum ias at which directional control can be maintained with the aircraft in the landing configuration but with the added ability of being able to roll the aircraft from an initial condition of steady flight through an angle of 20 degrees in the direction necessary to initiate a turn away from the inoperative engine s in not more than 5 seconds exemple 476 I is correct, ii is correct.

Question 248-34 : Which statement is correct about an aeroplane that has experienced a left engine failure and continues afterwards in straight and level cruise flight with wings level ?

Turn indicator neutral slip indicator neutral

. 2385. turn and slip indicator .the wings and therefore the instrument and the central portion of the tube are level .the aircraft is not accelerating laterally so the tube isn't being snatched away sidewayds leaving the ball deflected by lagging behind .the balls weight is acting towards the earth as per usual and is opposed by the ball being held up by the bottom edge of the central part of the tube again no unbalanced forces .so there is no mechanism to cause the ball to be deflected so it isn't exemple 480 Turn indicator neutral, slip indicator neutral.

Question 248-35 : Which statement about minimum control speed is correct ?

Vmca depends on the airport density altitude and the location of the engine on the aeroplane aft fuselage or wing

. cs 25 . e vmcg the minimum control speed on the ground is the calibrated airspeed during the take off run at which when the critical engine is suddenly made inoperative it is possible to maintain control of the aeroplane using the rudder control alone without the use of nosewheel steering as limited by 667 n of force 150 lbf and the lateral control to the extent of keeping the wings level to enable the take off to be safely continued using normal piloting skill in the determination of vmcg assuming that the path of the aeroplane accelerating with all engines operating is along the centreline of the runway its path from the point at which the critical engine is made inoperative to the point at which recovery to a direction parallel to the centreline is completed may not deviate more than 9 1 m 30 ft laterally from the centreline at any point vmcg must be established with .1 the aeroplane in each take off configuration or at the option of the applicant in the most critical take off configuration .2 maximum available take off power or thrust on the operating engines .3 the most unfavourable centre of gravity .4 the aeroplane trimmed for take off and.5 the most unfavourable weight in the range of take off weights exemple 484 Vmca depends on the airport density altitude, and the location of the engine on the aeroplane (aft fuselage or wing).

Question 248-36 : Which of the following statements is correct .i when the critical engine fails during take off the speed vmcl can be limiting .ii the speed vmcl is always limited by maximum rudder deflection ?

I is incorrect ii is incorrect

.i when the critical engine fails during take off the speed vmcl can be limiting incorrect vmcl does not apply to take off .vmcl is the minimum control speed during landing/approach with all engines operating it is the calibrated airspeed at which when the critical engine is suddenly made inoperative it is possible to maintain control of the airplane with that engine still inoperative and maintain straight flight with an angle of bank of not more than 5 degrees .ii the speed vmcl is always limited by maximum rudder deflection incorrect vmcl can be limited by maximum rudder deflection in order to maintain control of the airplane with critical engine fails and maintain straight flight with an angle of bank of not more than 5 degrees but this is not always the case exemple 488 I is incorrect, ii is incorrect.

Question 248-37 : Given an initial condition in straight and level flight with a speed of 1 4 vs the maximum bank angle attainable without stalling in a steady coordinated turn whilst maintaining speed and altitude is approximately ?

60°

.stalling speed vs in a steady coordinated turn = stalling speed vs in level flight x squared root of load factor.load factor = 1/cos bank angle.load factor = 1 4/1 ² = 1 96.bank angle = cos^ 1 1/1 96 .bank angle = cos^ 1 0 51 = 59 33°.the question states approximately so answer 60° is the right one .example .stall speed in straight and level flight is 100 kt we are at 140 kt 1 4 vs .whilst maintaining speed and altitude we can turn until a bank angle of 60° approximately without stalling exemple 492 60°.

Question 248-38 : Given .aeroplane mass 50 000kg.lift/drag ratio 12.thrust per engine 28 000n.assumed g 10m/s².for a straight steady wings level climb of a four engine aeroplane the one engine inoperative climb gradient is ?

8 5%

.weight = 50000 x 10 = 500000 newtons.total thrust only 3 engines = 3 x 28000 n = 84000 n..climb gradient % = 100 total thrust/weight 1/ lift/drag ratio .climb gradient % = 100 84000n/500000n 1/12 .climb gradient % = 100 0 168 0 08333 .climb gradient % = 100 x 0 084667 = 8 4667% exemple 496 8.5%.

Question 248-39 : During a steady horizontal turn the stall speed ?

Increases with the square root of the load factor

exemple 500 Increases with the square root of the load factor.

Question 248-40 : For shallow climb angles the following formula can be used gamma = climb angle ?

Sin gamma = t/w cd/cl

.this formula for small angles and for simple calculations gives the relationship between flight path angle thrust weight drag and lift .sin gamma = thrust drag / weight.or.sin gamma = thrust / weigth drag / weight.for small climb angles lift equals weight the formula can be written as .sin gamma = thrust / weigth drag / lift.or.sin gamma = thrust/weigth cd/cl exemple 504 Sin gamma = t/w - cd/cl.


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