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The correct ANSWER is (a) INCREASES gliding FLIGHT time

Explanation: The benefit of flying at minimum power SPEED is increase in gliding flight time. The minimum sink rate is attained by flying at a relative airspeed of U/\(\sqrt[4]{3}\). At this airspeed the minimum power speed of aircraft is attained. Flying at this speed will maximize the time of gliding flight.

The CORRECT option is (a) u/\(\sqrt[4]{3}\)

Explanation: The MINIMUM SINK rate is attained by flying at a relative airspeed of u/\(\sqrt[4]{3}\). At this airspeed the minimum power speed of aircraft is attained. Flying at this speed will MAXIMIZE the time of gliding flight.

Correct ANSWER is (a) True

To elaborate: The use of partial climbs is to calculate rate of CLIMB. The other method for CALCULATING climb rate is LEVEL accelerated method. Level accelerated is best suitable for thrust producing engines. Partial climb is best suitable for power-producing engines.

The correct OPTION is (a) True

Explanation: At the back SIDE of the drag curve the RATE of change of drag is negative. During this CASE the flight path gradient cannot be alone HANDLED by the elevator control. If the aircraft is flying with airspeed greater than minimum drag speed then the flight path gradient of descent can be increased by increasing airspeed.

The correct answer is (a) True

For explanation I would say: Wind is the RELATIVE velocity between the air mass and the GROUND. USUALLY wind is assumed to have only a horizontal component but there exists VERTICAL component which may affect the AIRCRAFT at considerable height.

The CORRECT choice is (C) minimum drag

The explanation: The best economic FLIGHT is achieved by minimum drag speed and take minimum time to climb. It is must that the flight path is OBSTRUCTION free and need to continue flight WITHOUT any stop before take-off.

Right choice is (a) The rate at which the flight take-off the ground

The best I can EXPLAIN: The climb rate is the rate at which the flight take-off the ground. The airspeed best for rate of climb to CLOSE to that of their minimum DRAG SPEED. The operating height is achieved by maximum rate climb. The SAFE height of an aircraft for take-off above the airfield is 15000m.

Right choice is (b) Direct operating costs

For explanation: The ANALYSIS of the economic benefits of a minimum climb will depend on the direct operating costs of the AIRCRAFT. Direct operating costs include the costs that are directly RELATED to the operation of the flight during travel such as crew salary, seat COST, fuel costs etc.

Correct option is (a) Varying airspeed

Easiest explanation: The drag that is produced in the aircraft is increased by aerodynamic means or by varying the airspeed. The aircraft has a very WIDE range of DESCENT path for the AIRCRAFTS. The aircraft will DESCEND if the propulsive thrust is less than the airframe drag.

The CORRECT choice is (B) False

To elaborate: There are THREE phases of descend. They are:

En-route

Terminal area

airfield

The CORRECT answer is (a) L=Wcosγ2

Explanation: The correct lift BALANCING equation for THRUST PRODUCING engines is given by the equation L=Wcosγ2 where L is lift, W is weight and γ is the angle at which the FORCE is acting on the aircraft.

The correct option is (B) It occurs when difference in THRUST power and DRAG power is maximum

For explanation: The maximum rate of CLIMB occurs when the difference in thrust power and drag power is maximum. The rate of climb is a function of excess thrust power.The BEST rate of climb occurs when the excess thrust power (FNV) is maximum than the drag power (DV).

Correct option is (a) Thrust POWER increases with increase in true airspeed

Explanation: Thrust power increases with increase in true airspeed.The best rate of CLIMB occurs when the excess thrust power (FNV) is maximum than the DRAG power (DV). As the ideal power increases linearly with true airspeed the best gradient of climb is predicted to be at an airspeed greater than minimum drag speed.

Correct answer is (a) It occurs at minimum power SPEED

The EXPLANATION: The minimum sink rate occur in a gliding flight at minimum power speed. The rate of climb is given by λ-\(\frac{1}{2}\)[u^3+u^-1]=Emaxv where λ is dimensionless power, u is AIRSPEED and Emax is endurance.

Right OPTION is (a) It is used for calculating climb rate

To ELABORATE: The use of partial climbs is to calculate rate of climb. The other method for calculating climb rate is level accelerated method. Level accelerated is best suitable for THRUST producing engines. Partial climb is best suitable for power-producing engines.

The correct answer is (a) True

The explanation: In the case of gliders, the DESCENT is calculated by lift-drag RATIO. In this case the minimum RATE of descent occurs at the minimum power SPEED and minimum gradient occurs at the minimum drag speed.

Right choice is (a) By FLYING the aircraft at AIRSPEED less than minimum DRAG speed

Easy explanation: The descent of an aircraft can happen at flying the aircraft at airspeed less than minimum drag speed. In this CASE the minimum rate of descent OCCURS at the minimum power speed and minimum gradient occurs at the minimum drag speed.

The correct answer is (a) True

To explain: In the calculation of the specific climb the specific fuel consumption is KEPT constant and the thrust or power is kept at maximum position.The formula for specific climb is given by SC=\(\frac{dH/dt}{Q_f}\) where SC is specific thrust, \(\frac{dH}{dt}\) is RATE CHANGE of height with respect to TIME and Qf is fuel flow. The specific climb is calculated in the units of ft/kg.

Right choice is (a) Low airspeed which is closer to critical OPERATING airspeed

To EXPLAIN I would say: The best gradient of climb is ACHIEVED when the airspeed is low and closer to that of the critical operating speed. The SAFEGUARD path has a gradient of 2%. The safeguard distance extends from the take-off distance available on the runway to a distance of 15000m.

The CORRECT answer is (a) THRUST is less than airframe drag

The best explanation: The aircraft will descend if the propulsive thrust is less than the airframe drag. The descend flight path can be VARIED from a shallow DESCENT to a very steep descent either by REDUCING the engine thrust or by increasing airframe drag.

Correct answer is (a) True

To explain: In practical the specific fuel consumption is effected by the airspeed, mach number and temperature. These factors will optimize the minimum fuel climb of the aircraft. The analysis of the ECONOMIC BENEFITS of a minimum climb will DEPEND on the DIRECT operating costs of the aircraft.

Right choice is (b) FN-D=Wsinγ2+mV̇

Explanation: The correct thrust BALANCING EQUATION for thrust producing engines is given by the equation FN-D=Wsinγ2+mV̇ where FN is the NORMAL FORCE acting on the aircraft, γ is the angle at which the force is acting on the aircraft, D is DRAG produced, W is weight of the aircraft, m is mass and V̇ is velocity.

Correct answer is (a) True

The BEST EXPLANATION: The best rate of climb OCCURS when the excess thrust power (FNV) is maximum than the drag power (DV). As the ideal power increases linearly with true airspeed the best gradient of climb is PREDICTED to be at an airspeed greater than minimum drag speed.

Correct option is (a) True

For explanation: The relative airspeed for maximum gradient of climb is a FUNCTION of engine power. The BEST climb gradient is given by U^4+λu-1=0 where u is airspeed and λ is DIMENSIONLESS power.

Right option is (a) True

For EXPLANATION I would say: At the back side of the DRAG CURVE the rate of change of drag is negative. During this case the flight path gradient cannot be alone handled by the elevator control. If the aircraft is flying with airspeed greater than minimum drag speed then the flight path gradient of descent can be increased by increasing airspeed.

Correct choice is (a) 1/\(\SQRT[4]{3}\)

The explanation is: At the maximum rate i.e. dv/du=0 the value of u=1/\(\sqrt[4]{3}\) which is the minimum power SPEED. The rate of CLIMB is given by λ-\(\frac{1}{2}\)[u^3+u^-1]=Emaxv where λ is dimensionless power, u is airspeed and Emax is ENDURANCE.

The correct choice is (d) τu-\(\frac{1}{2}\)[u^3+u^-1]=Emaxv

Best explanation: The correct rate of CLIMB equation is given by τu-\(\frac{1}{2}\)[u^3+u^-1]=Emaxv where τ is dimensionless thrust, u is airspeed, Emax is ENDURANCE and γ is the ANGLE at which the force is acting on the AIRCRAFT.

Correct OPTION is (a) 2%

Explanation: The gradient of climb for one ENGINE inoperative is 2%. This is same in all CONDITIONS and also during the emergency conditions of one engine inoperative. In this situation the engine works at maximum AIRSPEED required for BEST gradient of climb.

Right choice is (d) DECREASE in CLIMB gradient

Explanation: The THRUST reduction during climb leads to reduction in RATE and climb gradient and increases the time taken for the take-off of the aircraft. The minimum fuel climb occurs at the airspeed for BEST rate of climb.

Right answer is (a) Emaxsinγ2=\(\Big[\frac{\lambda}{u}+\tau \Big]-\frac{1}{2}[u^2+u^{-2}]\)

Explanation: The correct performance equation for MIXED power-plants is GIVEN by the formula Emaxsinγ2=\(\Big[\frac{\lambda}{u}+\tau \Big]-\frac{1}{2}[u^2+u^{-2}]\) where u is AIRSPEED, γ is the angle at which the force is acting on the aircraft and τ is dimensionless thrust and Emax is endurance.

Right choice is (a) True

To ELABORATE: The USE of partial climbs is to calculate RATE of CLIMB. The other method for calculating climb rate is level accelerated method. Level accelerated is BEST suitable for thrust producing engines. Partial climb is best suitable for power-producing engines.

The correct choice is (a) True

For explanation: Thrust is directly proportional to airspeed. This MEANS the thrust increases with increase in airspeed and decrease with decrease in decrease in airspeed. The airspeed best for rate of CLIMB to close to that of their minimum DRAG speed.

Right answer is (b) size of the airfield

For EXPLANATION I WOULD say: The DEPARTURE path is defined as the path which is free from obstacles and the landing or take-off is guaranteed a safe and clear flight path. It is COMPLEX and depends on the size of the size of airfield and aircraft operations that are PERFORMED in that airport.

The correct choice is (a) A PATH of no obstructions

For explanation: The departure path is defined as the path which is free from obstacles and the LANDING or take-off is GUARANTEED a safe and clear flight path. It is complex and depends on the SIZE of the size of AIRFIELD and aircraft operations that are performed in that airport.

Correct answer is (d) 15000m

The explanation: The SAFEGUARD distance of the runway is 1500m. The safeguard path has a GRADIENT of 2%. The safeguard distance extends from the take-off distance AVAILABLE on the runway to a distance of 15000m.

The CORRECT choice is (a) True

To EXPLAIN: The operating HEIGHT is achieved by MAXIMUM rate climb. The safe height of an aircraft for take-off above the airfield is 15000m. The use of safe height is to avoid ground based obstacles and make a clear flight path to avoid aircraft collision and LOSS of life and property.

Correct option is (a) True

Best EXPLANATION: Engines with low noise signature comply with the noise regulations and provide a BETTER performance in the CLIMB. This is because they can operate at higher thrust levels WITHOUT EXCEEDING the noise limits.

The correct choice is (c) 0.3

To explain I WOULD say: 0.3 MUST be the value of thrust- to-weight ratio for a normal take-off so that the acceleration ASSOCIATED with the rate of climb is neglected. This WAY the climb can be ASSUMED with constant airspeed and mach number.

The correct ANSWER is (a) elevator

Explanation: The flight path can be controlled by the use of elevator control only. If the aircraft is flying with airspeed greater than minimum drag SPEED then the flight path GRADIENT of descent can be increased by INCREASING airspeed.

The correct option is (a) True

The BEST explanation: Accurate MANUAL control of the flight PATH will be difficult to handle without the auto-throttle. The precise control of the flight path GRADIENT is achieved by the changes in the thrust of the flight along with the elevator control inputs given to the system during the gradient of descent of the aircraft.

Correct choice is (c) 5%

Easiest explanation: The gradient of descent is 3° or 5%. Smaller transport AIRCRAFT are capable of DESCENDING in a steeper way which are depending on the condition like restricted approach path and separation of airport traffic arrivals.

The CORRECT choice is (a) τ-\(\frac{1}{2}\)[u^2+u^-2]=Emaxsinγ2

The explanation is: The correct climb gradient equation is GIVEN by the equation τ-\(\frac{1}{2}\)[u^2+u^-2]=Emaxsinγ2 where τ is dimensionless thrust, u is airspeed, Emax is ENDURANCE and γ is the angle at which the force is acting on the aircraft.