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Question 253-1 : Why are large aeroplane required to be equipped with a fuel jettisoning system ? [ Explanation maintenance ]
To reduce aircraft mass after take off in order to comply with the climb gradients required under cs 25.
Cs 25.1001 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 25.119 and 25.121 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–25....this 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 accident.
Question 253-2 : 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 ops..cat.op.mpa.130 noise abatement procedures — aeroplanes.. a except for vfr operations of other than complex motor powered aeroplanes, the operator shall establish appropriate operating departure and arrival/approach 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 flight.
Question 253-3 : 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 airspeed...this 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 profile...position 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 orientation...position 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 further...note although the figure shows a departing aircraft, the scenario order works for this question regarding an aircraft on the approach too.
Question 253-4 : 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 15.....icao annex 15..appendix 2. snowtam format..9. item h estimated surface friction on each third of the runway single digit in the order from the threshold having the lower runway designation number...friction 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 format...the values for each third of the runway are separated by an oblique stroke / , without space between the values and the oblique stroke , for example 5/5/5.
Question 253-5 : 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 ?
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 wheel.expedite 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 braking.brake 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 and/or speed, the tyres can overheat and burst. consequently, brake and/or wheel fires are likely to occur.the usual strategies for cooling hot brakes include giving consideration to an appropriate parking area i.e. use a remote location, away from other aircraft, buildings , parking into the wind.chocking 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 available.
Question 253-6 : 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 slats/flaps 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 swell.
Question 253-7 : Which of the following will shorten the holdover time hot applicable for de icing/anti 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 ?
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 fluid/water mix > correct. the more the anti icing fluid is diluted, then the less effective it is.icao doc 9640.chapter 4. holdover time hot 4.2 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 winter.4.3 numerous factors that can affect the de icing/anti icing performance and hots of de icing/anti 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 icing/anti icing fluid type, fluid/water 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 guidelines.
Question 253-8 : 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 ?
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 increases.2. 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 downdraft.3. 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.
Question 253-9 : 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 17..aircraft 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 substances.
Question 253-10 : 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 fluid.. with 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 times..5.1 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. …..5.6 the holdover time begins with the start of the final de icing/anti icing application and ends after an elapsed time equal to the appropriate holdover time chosen by the pilot in command...note two step de icing/anti 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 protection.
Question 253-11 : The angle of attack of an aerofoil section is defined as the angle between the ?
Undisturbed airflow and the chord line.
Img669
Question 253-12 : 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.
Img1466.static pressure decreases in a venturi, and airflow speed increases.
Question 253-13 : The si units of air density i and force ii are ?
I kg/m3, 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..pdf678..pdf679
Question 253-14 : 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
Question 253-15 : 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. 1084
Question 253-16 : 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 airfoil...remember when cl = 0 a cambered section will have a negative alpha.. 2416.other questions you will see on this refer to a symmetrical section, it has no pitch and a negative cambered section will pitch nose up.
Question 253-17 : An aeroplane maintains straight and level flight while the ias is doubled. the change in lift coefficient will be ?
X 0.25
.in straight and level flight, lift does not change as it is only balancing against weight...aircraft is maintaining level flight and therefore lift cannot change.lift = cl 1/2rho v² s rho = density.. example with.s = 2, rho = 10, cl = 1 and ias = 100..lift = 1 x 0.5 x 10 x 100 x 100 x 2 = 100000.. and now with.s = 2, rho = 10, cl = 1 and ias = 200..lift = 1 x 0.5 x 10 x 200 x 200 x2 = 400000..we 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 0.25.
Question 253-18 : Which formula or equation describes the relationship between force f , acceleration a and mass m ?
F=m.a
.force energy brought to bear, which tends to cause a motion or change...mass a measure of the amount of material contained in a body...acceleration rate of change of velocity velocity + time or distance + time².
Question 253-19 : 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 smooth.
Question 253-20 : 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 pressure..lift =.cl x 1/2 rho v² x s..cl = lift coefficient.rho = density.v = tas in m/s.s = surface..if 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 airflow.
Question 253-21 : 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 velocity..area and velocity remain constant, the only thing that changes is rho...with a temperature increase, the density decreases, thus the mass flow decreases through the stream tube.
Question 253-22 : Which one of the following statements about bernoulli's theorem is correct ?
The dynamic pressure increases as static pressure decreases.
.pt total pressure.ps static pressure.pd dynamic pressure 1/2 x density x tas²..bernoulli's theorem is.pt = ps + pd..then, if pd inscreases, ps decreases, as pt always remains constant.
Question 253-23 : 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 constant.
Question 253-24 : Bernoulli's equation can be written as. pt= total pressure, ps = static pressure and q=dynamic pressure ?
Pt = ps + q
.pt total pressure.ps static pressure.pd dynamic pressure 1/2 x density x tas²..bernoulli's theorem is.pt = ps + pd
Question 253-25 : 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 separates...as 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 drag...to 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 stall.
Question 253-26 : 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 pressure..all of them can be represented by a single vector acting at a particular point, called the centre of pressure. 669.the centre of pressure is a theoretical point on the chord line through which the resultant of all forces the total reaction is said to act..its 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 cg...thus, 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 aft.
Question 253-27 : 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 lift.alpha = angle of attack. 2773.for 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.
Question 253-28 : 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.
. 680
Question 253-29 : 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 aerofoil.
Question 253-30 : 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 line..the angle of incidence is a fixed value.. 681.others important angles.aeroplane's angle of attack the angle between its speed vector and longitudinal axis..aeroplane's pitch angle the angle between its longitudinal axis and the horizontal plane..aeroplane's flight path the angle between its speed vector and the horizontal plane.. 682
Question 253-31 : 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.
. 1576.in red, the stagnation point will move down..in blue, the point of lowest pressure will move forward.
Question 253-32 : 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 moment..the 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 varies..this 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 increased.
Question 253-33 : 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 en/com080 27.jpg..stall 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 up.
Question 253-34 : The lift of an aeroplane of weight w in a constant linear climb with a climb angle gamma is approximately ?
W.cos.gamma
Lift = weight x cos climb angle.. 1526
Question 253-35 : 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 lift/drag ratio the total drag is lowest.
.lift/drag ratio is maximum at the speed for minimum total drag.
Question 253-36 : 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 ?
Di/dp = 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 achieved.
Question 253-37 : The correct drag d formula is ?
D= cd 1/2 rho v² s
.where cd = drag coefficient.rho = density.v = tas in m/s.s = surface..drag 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 wind.
Question 253-38 : 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 en/com032 209.jpg
Question 253-39 : 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 1/16 ii 1/4.
.induced drag varies with lift, speed and aspect ratio, is inversely proportional to aspect ratio and v² so multiply by 1/v² , and directly proportional to lift²/cl²/weight²...we know that speed is double 80 kt to 160 kt , so 1/2 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 = 1/2 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 16...using now the induced drag formula = 1/2 x rho x v² x s x cdi, knowing that cdi is divide by 16 and 1/2x rho x v² is multiplicate par 4, it result that induced drag will be divide by 4.
Question 253-40 : What is the effect of high aspect ratio of an aeroplane's wing on induced drag ?
It is reduced because the effect of wing tip vortices is reduced.
.aspect ratio is defined as the square of the wingspan divided by the area of the wing planform.. /com en/com080 38.jpg.induced drag varies with lift, speed and aspect ratio, is inversely proportional to aspect ratio and v² so multiply by 1/v² , and directly proportional to lift²/cl²/weight².
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