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Question 253-1 : An operator has a fleet of jet aeroplanes of one type some aeroplanes have steel brakes and others carbon brakes how might the taxi procedures for the two variants vary the aeroplanes with steel brakes should be taxied at a ? [ Explanation maintenance ]
Lower speed using frequent short brake applications whereas the aeroplanes with carbon brakes should be left to accelerate to a higher taxi speed before slowing right down to limit the number of brake applications
Question 253-2 : Why are large aeroplane required to be equipped with a fuel jettisoning system ?
To reduce aircraft mass after take off in order to comply with the climb gradients required under cs 25.
Cs 251001 fuel jettisoning system a a fuel jettisoning system must be installed on each aeroplane unless it is shown that the aeroplane meets the climb requirements of cs 25119 and 25121 d at maximum take off weight less the actual or computed weight of fuel necessary for a 15 minute flight comprised of a take off go around and landing at the airport of departure with the aeroplane configuration speed power and thrust the same as that used in meeting the applicable take off approach and landing climb performance requirements of this cs–25this goes against what is commonly believed and taught the reason for a fuel jettison system is not actually to get below mlm as most people think but instead is to get light enough to fulfil the requirements for minimum climb gradients such as landing climb or oei approach climb if the aircraft can fulfill all these climb requirements at the mtom minus a short 15 minute flight then it does not need a fuel jettison system at all such as the b737 and a320 family the reason is that an overweight landing is not a massive problem the aircraft is checked afterwards but it is not likely to be a problem at the time but the climb performance on a go around needs to be acceptable to avoid terrain as that is very important to avoid a disastrous accidentAn overweight aircraft may damage the runway if the pcn, pavement classification number, is lower than the acn, aircraft classification number. to reduce the mass to maximum landing mass. it will reduce the risk of fire on landing.
Question 253-3 : An aircraft was performing the noise abatement departure procedure 1 nadp 1 when it suffers an engine failure the commander ?
Has the automatic right to ignore the noise abatement procedure.
Easa air opscatopmpa130 noise abatement procedures — aeroplanes a except for vfr operations of other than complex motor powered aeroplanes the operator shall establish appropriate operating departure and arrivalapproach procedures for each aeroplane type taking into account the need to minimise the effect of aircraft noise b the procedures shall 1 ensure that safety has priority over noise abatement and 2 be simple and safe to operate with no significant increase in crew workload during critical phases of flightMust request permission from the appropriate ats unit to abort the noise abatement procedure. must continue the noise abatement departure procedure1 (nadp 1). must continue with noise abatement departure procedure 2 (nadp 2).
Question 253-4 : In a descent an aircraft is landing into a rapidly increasing tailwind due to wind shear what will most likely happen ?
The aircraft will fly below the glideslope.
for these questions involving wind shear and micro bursts thinking in terms of the aircraft's energy is useful a headwind will increase the energy performance and airspeed while a tailwind will decrease the energy performance and airspeedthis scenario can be considered in 3 parts position 1 on the figure an increasing head wind which increases the energy of the aircraft the airspeed increases and the aircraft floats up above the flight profileposition 2 on the figure the aircraft now flies under the wind shear micro burst the previous headwind decreases and the airspeed decreases the aircraft now sinks due to the loss of energy and also the down draft the angle of attack will decrease as the relative airflow is now coming from a higher orientationposition 3 on the figure the aircraft leaves the down draft but now encounters an increasing tail wind this steals more energy and the airspeed decreases and the aircraft sinks furthernote although the figure shows a departing aircraft the scenario order works for this question regarding an aircraft on the approach tooThe aircraft will maintain the glide slope as stability is increased. the aircraft will fly above the glideslope. the aircraft will fly initially above and then below the glide slope.
Question 253-5 : A snowtam identifies a number of hazards which may cause serious consequences for flight safety and airport operations information about would be included in a snowtam and would specifically mitigate against one of these hazards ?
Braking efficiency.
looking at the answer options applicable touchdown techniques when to do soft or hard landings > incorrect the crew uses the information in a snowtam to determine their flying techniques as dictated by the weather surface conditions the probability or risk of a runway excursion > incorrect a snowtam does not include this information but can be used by crew to determine the risk of an excursion from details of the prevailing conditions the landing distances required > incorrect these are type specific and are not included in a snowtam braking efficiency > correct a snowtam includes information on braking efficiency either measured or estimated more information below from icao annex 15icao annex 15 appendix 2 snowtam format9 item h estimated surface friction on each third of the runway single digit in the order from the threshold having the lower runway designation numberfriction measurement devices can be used as part of the overall runway surface assessment some states may have developed procedures for runway surface assessment which may include the use of information obtained from friction measuring devices and the reporting of quantitive values in such cases these procedures should be published in the aip and the reporting made in item t of the snowtam formatthe values for each third of the runway are separated by an oblique stroke without space between the values and the oblique stroke for example 555Applicable touchdown techniques (when to do soft or hard landings). the landing distances required. the probability or risk of a runway excursion.
Question 253-6 : An aircraft is flying from frankfurt germany to quebec canada while in the nat hla an in flight emergency calls for a diversion to keflavik iceland the aircraft lands at maximum landing mass and vacates the runway brake temperature warnings light up the crew becomes concerned about the possibility ?
Stop on the taxiway do not apply the parking brake and request the airport fire service to attend.
Looking at the answer options taxi to the terminal where the crew can check the brakes approaching from the side to avoid injury from possible disc blowout > incorrect to continue taxiing will prevent the brakes from cooling it would be better to stop now on the taxiway besides parking next to the terminal often a large glass covered building full of people may not be a good idea in the event of an explosion also when inspecting over heated brakes approach from the front or rear not the side as the sidewall is the weakest part of the tyre stop on the taxiway apply the parking brake request the airport fire service to attend and evacuate the aircraft > incorrect do not apply the parking brake as this may fuse the brakes also why evacuate for overheated brakes stop on the taxiway do not apply the parking brake and request the airport fire service to attend > correct it is best to stop on the taxiway so clear of the runway no further taxiing means the brakes can start to cool do not set the parking brake as this may fuse the brakes onto the wheelexpedite taxiing to a clear area away from buildings and or other aircraft in case of a disc blowout > incorrect to continue taxiing will prevent the brakes from cooling it would be better to stop now on the taxiway certainly a fast expedited taxi would be a bad idea as it would require further brakingbrake system overheat conditiona braking system works by converting the kinetic energy of a moving aircraft into heat to prevent damage to the tyres and undercarriage structure the heat energy must be dissipated rapidly into the surrounding air if this does not happen and the amount of heat generated becomes excessive as can be the case after an aborted take off or following a landing at an excessive mass andor speed the tyres can overheat and burst consequently brake andor wheel fires are likely to occurthe usual strategies for cooling hot brakes include giving consideration to an appropriate parking area ie use a remote location away from other aircraft buildings parking into the windchocking the nose wheel and releasing the parking brake the brake temperatures may be so high that the brakes may weld together and consequently do not release even after the brakes cool down andusing brake fans when availableTaxi to the terminal where the crew can check the brakes approaching from the side to avoid injury from possible disc blowout. stop on the taxiway, apply the parking brake, request the airport fire service to attend, and evacuate the aircraft. expedite taxiing to a clear area away from buildings and / or other aircraft in case of a disc blowout.
Question 253-7 : What is one main factor to be remembered when considering the ditching of an aircraft ?
On the high seas it is usually best to ditch parallel to and on top of the primary swell system except in high wind conditions.
Refer to figure ditchingthe following is a list of generally accepted considerations and techniques for ditching power on if there is a choice in the matter power on is preferable to power off for ditching use of power allows more control of both the rate of descent and point at which touchdown is made reduce aircraft weight a lighter aircraft allows a lower approach speed and will probably remain afloat higher in the water and for longer thus facilitating occupant evacuation burning off or dumping fuel also has the advantage of increasing buoyancy in some aircraft types by creating a larger air mass held within the fuel tanks configuration gear up is the optimum configuration for ditching most manufacturers recommend the maximum deployment of available slatsflaps is desirable to minimise approach speed in ideal conditions smooth water or very long swells land into the wind this will ensure the minimum possible touchdown speed and help minimise impact damage where the swell is more marked it may be advisable to ditch along the swell accepting a crosswind component and the higher touchdown speed thus minimising the potential for nosing into the face of the rising swell the best touchdown point is on the top of the swell with the second best on the back of the swell aim to remain well clear of the advancing face of the swellIt is usually best to ditch perpendicular to, and on top of, the primary swell system, regardless of the sea state, except in high wind conditions. on the high seas, it is usually best to ditch in any direction considered necessary, regardless of the sea state. when deciding to ditch, in any direction, considerations of the wind may be omitted.
Question 253-8 : Which of the following will shorten the holdover time hot applicable for de icinganti icing procedures 1 high wind velocity 2 light wind velocity 3 ambient temperature above 0 degrees celsius 4 high relative humidity 5 aeroplane skin temperature above the ambient temperature 6 low concentration ?
1 4 and 6.
Looking at the answer options 1 high wind velocity > correct this can disturb the anti icing fluid or even blow it off the aircraft 2 light wind velocity > incorrect light or calm winds will not disturb the protective fluid 3 ambient temperature above 0 degrees celsius > incorrect warm temperatures are good news for hold over times 4 high relative humidity > correct the higher the water content of the air the more contamination on the aircraft as it cools on the surface 5 aeroplane skin temperature above the ambient temperature > incorrect if the aircraft is warmer than ambient this will discourage contamination be careful a cold soaked wing is detrimental to hold over time as an aircraft following a long cruise will have a very cold skin and low temperature fuel in the tanks 6 low concentration of anti icing fluid in the anti icing fluidwater mix > correct the more the anti icing fluid is diluted then the less effective it isicao doc 9640 chapter 4 holdover time hot 42 hot is the estimated time the anti icing fluid will prevent the formation of ice and frost and the accumulation of snow on the protected treated surfaces of an aeroplane these hots are generated by testing fluids under a variety of temperature and precipitation conditions that simulate the range of weather experienced in winter43 numerous factors that can affect the de icinganti icing performance and hots of de icinganti icing fluids have been identified these factors include but are not limited by the following a type and rate of precipitation b ambient temperature c relative humidity d wind direction and velocity including jet blast e aeroplane surface skin temperature andf de icinganti icing fluid type fluidwater ratio temperature cautionowing to the many variables that can influence hots the time of protection may be reduced or extended depending on the intensity of the weather conditions heavy precipitation high moisture content high wind velocity and jet blast can reduce hot below the lowest time in hot guidelines hot may be reduced when aircraft skin temperature is lower than outside air temperature weather conditions for which no hot guidelines exist are referenced in the hot guidelines1, 4, and 5. 2, 3, and 5. 2, 3, and 6.
Question 253-9 : Thunderstorm activity takes place in the vicinity of an airport during the take off roll the crew notices rapid build up of airspeed soon after lift off the ground proximity warning activates and the aircraft loses altitude the most probable cause is a microburst which by the start of the take off ?
On the rwy ahead of the aircraft.
Flying through a microburst1 imagine you're approaching a microburst at 80 knots as you enter the vortex ring you experience turbulence and a rapid increase in ias as you pick up a headwind you start to climb as your performance increases2 as you enter the strongest part of the horizontal shaft your airspeed peaks at 100 knots an increase of 20 knots you're still climbing as you enter the downdraft3 inside the downdraft your headwind starts to switch to a tailwind you're caught in the downdraft sinking quickly toward the ground your airspeed begins to decrease 4 as you exit the downdraft your tailwind increases rapidly the shear drops your airspeed to 60 knots which is 20 knots below your original speed you're at a high angle of attack and still descending 5 you encounter more turbulence as you enter the final vortex if you haven't already hit the ground you may begin to climb again as you exit the vortex the microburst is most likely over the runway and ahead of the aircraft at the start of the take off roll initially the pilots will experience a rapid increase in ias due to headwind Right of the rwy, abeam the threshold. left of the rwy, abeam the threshold. on the rwy, behind the aircraft.
Question 253-10 : What do you understand by aircraft security check ?
It is an inspection of those parts of the interior of the aircraft to which passengers may have had access together with an inspection of the hold of the aircraft in order to detect prohibited articles and unlawful interferences with the aircraft.
Icao annex 17aircraft security check an inspection of the interior of an aircraft to which passengers may have had access and an inspection of the hold for the purposes of discovering suspicious objects weapons explosives or other dangerous devices articles and substancesIt is an inspection of the interior and accessible exterior of the aircraft in order to detect prohibited articles and unlawful interferences that jeopardise the security of the aircraft. it is an inspection done by a flight safety inspector who checks if airworthiness certificate of an aircraft is valid. it is an inspection carried out before flight to ensure that the aircraft is fit for the intended flight.
Question 253-11 : A de icing anti icing procedure is carried out in two steps when does the hold over time start ?
At the beginning of the anti icing step.
Hold over time with a one step process starts at the beginning of de ice anti ice process a one step process is the application of just a single coat of de icing or anti icing fluidwith a two step process starts at the beginning of anti icing process a two step process is an initial coat of de icing fluid to clean the aircraft of contamination followed by a second step of applying anti icing fluid to now protect the clean aircraft from further contamination note in a two step process the hold over time will include the time taken to apply the last step of anti icing fluid complete checks and paperwork and to depart the aerodrome if the hold over time expires prior to take off then the aircraft can not depart until de iced and anti iced again icao doc 9640 chapter 5 holdover times51 holdover time hot is the estimated time the anti icing fluid will prevent the formation of ice and frost and the accumulation of snow on the protected treated surfaces of an aeroplane … 56 the holdover time begins with the start of the final de icinganti icing application and ends after an elapsed time equal to the appropriate holdover time chosen by the pilot in commandnote two step de icinganti icing this process contains two distinct steps the first step de icing is followed by the second step anti icing as a separate fluid application after de icing a separate overspray of anti icing fluid is applied to protect the aeroplane's critical surfaces thus providing maximum anti icing protectionAt the beginning of the de-icing step. at the end of the anti-icing step. at the end of the de-icing step.
Question 253-12 : The angle of attack of an aerofoil section is defined as the angle between the ?
Undisturbed airflow and the chord line.
Img669Local airflow and the mean camber line. local airflow and the chord line. undisturbed airflow and the mean camber line.
Question 253-13 : In a stationary subsonic streamline flow pattern if the streamlines converge in this part of the pattern the static pressure i will and the velocity ii will ?
I decrease ii increase.
Img1466static pressure decreases in a venturi and airflow speed increases(i) increase, (ii) increase. (i) increase, (ii) decrease. (i) decrease, (ii) decrease.
Question 253-14 : The si units of air density i and force ii are ?
I kgm3 ii n.
The use of units follow the international rules and style conventions click on the following link to open a pdf file of units of measurement in a new tab pdf678pdf679(i) kg/m², (ii) kg (i) n/m3, (ii) n (i) n/kg, (ii) kg
Question 253-15 : The units of wing loading i w s and ii dynamic pressure q are ?
I n m² ii n m².
The use of units follow the international rules and style conventions click on the following link to open a pdf file of units of measurement in a new tab pdf679(i) n / m³, (ii) kg / m². (i) kg / m, (ii) n / m². (i) n / m, (ii) kg.
Question 253-16 : The aeroplane drag in straight and level flight is lowest when the ?
Parasite drag is equal to the induced drag.
Total drag is lowest when parasite drag is equal to the induced drag 1084Parasite drag equals twice the induced drag. induced drag is equal to zero. induced drag is lowest.
Question 253-17 : Considering a positive cambered aerofoil the pitch moment when cl=0 is ?
Negative pitch down .
It is because the pressure distribution is producing negative lift at the front and positive lift at the back of the airfoilremember when cl = 0 a cambered section will have a negative alpha 2416other questions you will see on this refer to a symmetrical section it has no pitch and a negative cambered section will pitch nose upInfinite positive (pitch-up). equal to zero.
Question 253-18 : An aeroplane maintains straight and level flight while the ias is doubled the change in lift coefficient will be ?
X 025.
In straight and level flight lift does not change as it is only balancing against weightaircraft is maintaining level flight and therefore lift cannot change lift = cl 12rho v² s rho = density example with s = 2 rho = 10 cl = 1 and ias = 100lift = 1 x 05 x 10 x 100 x 100 x 2 = 100000 and now with s = 2 rho = 10 cl = 1 and ias = 200lift = 1 x 05 x 10 x 200 x 200 x2 = 400000we can't modify s and rho we can only change lift coefficient cl and in order to maintain lift at 100000 we have to multiply cl by 025 X 2.0 x 0.5 x 4.0
Question 253-19 : Which formula or equation describes the relationship between force f acceleration a and mass m ?
F=ma.
Force energy brought to bear which tends to cause a motion or changemass a measure of the amount of material contained in a bodyacceleration rate of change of velocity velocity + time or distance + time²A=f.m f=m/a m=f.a
Question 253-20 : Static pressure is acts ?
In all directions.
Static pressure is atmospheric pressure measured at a point where there is no external disturbance and the flow of air over the surface is perfectly smoothOnly in the direction of the total pressure. only perpendicular to the direction of the flow. only in direction of the flow.
Question 253-21 : Lift is generated when ?
The flow direction of a certain mass of air is changed.
Lift is an upward force whose line of action is at right angles to the relative airflow direction and acts on the centre of pressurelift =cl x 12 rho v² x scl = lift coefficientrho = densityv = tas in ms s = surfaceif an aircraft is providing a lift force upwards there must be something going downwards to react this would be the change of direction of the airflowA certain mass of air is accelerated in its flow direction. a symmetrical aerofoil is placed in a high velocity air stream at zero angle of attack. a certain mass of air is retarded.
Question 253-22 : Consider the steady flow through a stream tube where the velocity of the stream is v an increase in temperature of the flow at a constant value of v will ?
Decrease the mass flow.
The mass flow is density x area x velocityarea and velocity remain constant the only thing that changes is rhowith a temperature increase the density decreases thus the mass flow decreases through the stream tubeIncrease the mass flow. not affect the mass flow. increase the mass flow when the tube is divergent in the direction of the flow.
Question 253-23 : Which one of the following statements about bernoulli's theorem is correct ?
The dynamic pressure increases as static pressure decreases.
Pt total pressureps static pressurepd dynamic pressure 12 x density x tas² bernoulli's theorem is pt = ps + pdthen if pd inscreases ps decreases as pt always remains constantThe dynamic pressure decreases as static pressure decreases. the total pressure is zero when the velocity of the stream is zero. the dynamic pressure is maximum in the stagnation point.
Question 253-24 : If in a two dimensional incompressible and subsonic flow the streamlines converge the static pressure in the flow will ?
Decrease.
Total pressure pt always remains constant and bernoulli's theorem is pt = ps + pd ps = static pressure and pd = dynamic pressure the streamlines converge the speed will increase which increases the dynamic pressure thus static pressure must decrease for the total pressure to remain constantIncrease. not change. increase initially, then decrease.
Question 253-25 : Bernoulli's equation can be written as pt= total pressure ps = static pressure and q=dynamic pressure ?
Pt = ps + q.
Pt total pressureps static pressurepd dynamic pressure 12 x density x tas² bernoulli's theorem is pt = ps + pdPt = ps - q pt = q - ps pt = ps / q
Question 253-26 : Which of the following statements about boundary layers is correct ?
The turbulent boundary layer has more kinetic energy than the laminar boundary layer.
The pressure pattern over the wing goes from high at the leading edge to low at the point of maximum camber and then back to high again at the trailing edge so from the point of max camber backwards the airflow is moving from a low pressure to a high pressure region called an adverse pressure gradient the air would not do this unless it was being pushed by some other factor and this factor is the kinetic energy of the moving air the boundary layer has less kinetic energy than the freestream air and gradually slows down when it stops or reverses it breaks away from the surface or separatesas the laminar type of boundary layer has less kinetic energy than the turbulent type it will slow down quicker and break away earlier so the laminar boundary layer has less kinetic energy and breaks away earlier the laminar layer however causes less dragto try for the highest possible lift at high angles of attack you need to keep the airflow attached as long as possible and this means having a high kinetic energy boundary layer various high lift devices are there simply to re energise a boundary layer that is slowing down never mind the drag it is lift we want at the stallThe turbulent boundary layer is thinner than the laminar boundary layer. the turbulent boundary layer gives a lower skin friction than the laminar boundary layer. the turbulent boundary layer will separate more easily than the laminar boundary layer.
Question 253-27 : On an asymmetrical single curve aerofoil in subsonic airflow at low angle of attack when the angle of attack is increased the centre of pressure will assume a conventional transport aeroplane ?
Move forward.
The pressure created by an aerofoil at any point may be represented by a vector at right angles to its surface whose length is proportional to the difference between absolute pressure at the point and the free stream static pressureall of them can be represented by a single vector acting at a particular point called the centre of pressure 669the centre of pressure is a theoretical point on the chord line through which the resultant of all forces the total reaction is said to actits position is usually around 25% of the way from the leading edge simply because more lift is generated there but it moves steadily forward as the angle of attack is increased until just before the stalling angle when it moves rapidly backwards the centre of pressure's most forward point is just before the stalling angle this is why an aeroplane's nose drops when the wings stall and the centre of pressure moves behind the cgthus when speed is increased in straight and level flight on a positively cambered aerofoil you have to decrease the angle of attack to keep the the total lift force constant and the point where the resultant of all forces are acting the centre of pressure moves aftMove aft. remain matching the airfoil aerodynamic centre. remain unaffected.
Question 253-28 : The cl alpha curve of a positive cambered aerofoil intersects with the vertical axis of the cl alpha graph ?
Above the origin.
Cl = coefficient of liftalpha = angle of attack 2773for a positive cambered aerofoil the curve intersects the vertical axis of the graph above the origin because a positive cambered aerofoil produces lift with a zero pitching moment 0° angle of attack In the origin. below the origin. nowhere.
Question 253-29 : The angle of attack of a two dimensional wing section is the angle between ?
The chord line of the aerofoil and the free stream direction.
680The chord line of the aerofoil and the fuselage centreline. the fuselage centreline and the free stream direction. the chord line and the camber line of the aerofoil.
Question 253-30 : The angle between the airflow relative wind and the chord line of an aerofoil is ?
Angle of attack.
Img680'alpha' angle of attack is the angle between the airflow relative wind and the chord line of an aerofoilGlide path angle. climb path angle. same as the angle between chord line and fuselage axis.
Question 253-31 : The angle between the aeroplane longitudinal axis and the chord line is the ?
Angle of incidence.
Aeroplane's angle of incidence the angle between the longitudinal axis and the wing root chord linethe angle of incidence is a fixed value 681others important angles aeroplane's angle of attack the angle between its speed vector and longitudinal axisaeroplane's pitch angle the angle between its longitudinal axis and the horizontal planeaeroplane's flight path the angle between its speed vector and the horizontal plane 682Climb path angle. angle of attack. glide path angle.
Question 253-32 : With increasing angle of attack the stagnation point will move i and the point of lowest pressure will move ii respectively i and ii are ?
I down ii forward.
1576in red the stagnation point will move downin blue the point of lowest pressure will move forward(i) up, (ii) aft. (i) down, (ii) aft. (i) up, (ii) forward.
Question 253-33 : The aerodynamic centre of the wing is the point where ?
The pitching moment coefficient does not vary with angle of attack.
When upper and lower surface lift act through different points the result is a pitching momentthe aerodynamic centre of the wing is the fixed point on the chord line about which no change in pitching moment is felt when the angle of attack variesthis is not the center of pressure which is a theoretical point on the chord line through which the resultant of all forces the total reaction is said to act the center of pressure moves forward when the angle of attack is increasedChanges of lift due to variations in angle of attack are constant. aerodynamic forces are constant. the aeroplane's lateral axis intersects with the centre of gravity.
Question 253-34 : On a swept wing aeroplane at low airspeed the pitch up phenomenon ?
Is caused by wingtip stall.
Flying at low speed means flying at high angle of attack there is a tendency for the swept wing to develop a strong spanwise flow towards the wingtip when the wing is at high angles of attack com encom080 27jpgstall occurs at the wingtips first resulting in a shift of the center of lift of the wing in a forward direction relative to the center of gravity of the airplane causing the nose to pitch upNever occurs, since a swept wing is a “remedy” to pitch up. is caused by extension of trailing edge lift augmentation devices. is caused by boundary layer fences mounted on the wings.
Question 253-35 : The lift of an aeroplane of weight w in a constant linear climb with a climb angle gamma is approximately ?
Wcosgamma.
Lift = weight x cos climb angle 1526W(1-sin.gamma) w(1-tan.gamma) w/cos.gamma
Question 253-36 : Which one of the following statements about the lift to drag ratio in straight and level flight is correct ?
At the highest value of the liftdrag ratio the total drag is lowest.
Liftdrag ratio is maximum at the speed for minimum total dragThe highest value of the lift/drag ratio is reached when the lift is zero. the lift/drag ratio always increases as the lift decreases. the highest value of the lift/drag ratio is reached when the lift is equal to the aircraft weight.
Question 253-37 : At a load factor of 1 and the aeroplane's minimum drag speed what is the ratio between induced drag di and parasite drag dp ?
Didp = 1.
The minimum drag speed occurs at the speed where the induced drag is equal to the parasitic drag this is the speed at which the best gradient of climb is achievedIt varies between aeroplane types. di/dp = 2. di/dp = 1/2.
Question 253-38 : The correct drag d formula is ?
D= cd 12 rho v² s.
Where cd = drag coefficientrho = densityv = tas in ms s = surfacedrag is the force that opposes the forward motion of a body through the air it's an aerodynamic force on a body acting parallel and opposite to the relative windD= cd 2 rho v² s d= cd 1/2 rho v s d= cd 1/2 1/rho v² s
Question 253-39 : The value of the parasite drag in straight and level flight at constant weight varies linearly with the ?
Square of the speed.
Parasite drag only varies with speed and is directly proportional to v² com encom032 209jpgSpeed. angle of attack. square of the angle of attack.
Question 253-40 : An aeroplane accelerates from 80 kt to 160 kt at a load factor equal to 1 the induced drag coefficient i and the induced drag ii alter with the following factors ?
I 116 ii 14.
Induced drag varies with lift speed and aspect ratio is inversely proportional to aspect ratio and v² so multiply by 1v² and directly proportional to lift²cl²weight²we know that speed is double 80 kt to 160 kt so 12 x rho x v² is multiplicate by 4 to maintain lift l constant you have to divide the lift coefficient cl by 4 lift formula = 12 x rho x v² x s x cl if cl is divide by 4 thus the coefficient of induced drag cdi from cdi = cl² pi x aspect ratio will be divide by 16using now the induced drag formula = 12 x rho x v² x s x cdi knowing that cdi is divide by 16 and 12x rho x v² is multiplicate par 4 it result that induced drag will be divide by 4(i) 1/4 (ii) 2. (i) 1/2 (ii) 1/16. (i) 4 (ii) 1/2.
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