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Correct option is (d) RANGE of the aircraft is decreased

To ELABORATE: The following cases happen if the final fuel weight INCREASES in an aircraft:

The fuel ratio DECREASES as fuel ratio is the ratio of the initial fuel weight to final fuel weight

The range of the aircraft flying will be REDUCED.

Correct option is (a) Air TEMPERATURE increases with increase in the AIRCRAFT range

To EXPLAIN: The air temperature increases with increase in the aircraft range. This is because the true airspeed (TAS) increases and the aircraft FLIES further in the given time during which it burns same quantity of the FUEL.

The correct option is (a) TRUE

The explanation: The minimum DRAG speed in TERMS of equivalent airspeed is unaffected by the ALTITUDE. Also there will be an increase in the true airspeed, MACH number and minimum drag speed with decrease in altitude.

Right ANSWER is (a) True

To EXPLAIN: Specific air range is MAXIMUM when the aircraft is FLYING at minimum DRAG speed and specific endurance is maximum when the aircraft is flying at minimum power speed. This is the same relation with thrust producing engines.

Right option is (d) SE=\(\frac{\eta}{CV}\frac{L}{D}\frac{1}{W}\)

To explain: The correct formula for specific endurance is given by SE=\(\frac{\eta}{CV}\frac{L}{D}\frac{1}{W}\) where η is propeller EFFICIENCY, C is specific fuel consumption, L is lift, D is drag, V is velocity, W is weight. SE can also be written as SE=\(\frac{1}{CP}\) where C is specific fuel consumption and P is POWER.

Correct option is (d) R=\(\frac{1}{C}\Big(\frac{2}{\rho SC_L}\Big)^{\frac{1}{2}}\frac{L}{D}\int_{W_{i}}^{W_{f}}\frac{dW}{W^{\frac{1}{2}}}\)

The explanation: The correct RANGE equation of CRUISE method 2 is given by R=\(\frac{1}{C}\Big(\frac{2}{\rho SC_L}\Big)^{\frac{1}{2}}\frac{L}{D}\int_{W_{i}}^{W_{f}}\frac{dW}{W^{\frac{1}{2}}}\) where R is range of the cruise, C is specific fuel consumption, S is span, ρ is density, CL is coefficient of lift, L/D is lift to drag ratio, Wf and WI are FINAL and initial weights.

Correct choice is (a) True

The explanation is: The range factor is PROPORTIONAL to the square root of the initial cruising weight. This way we know that the range of the aircraft will increase with its weight. This IMPLIES that increase in weight increases the fuel availability for the CRUISE.

The correct answer is (b) False

Easiest EXPLANATION: In cruise method 1 the AIR pressure is DECREASED as WELL as the aircraft weight is decreased. In cruise method 1 the W/p ratio is maintained constant and ALSO L/D ratio is also kept constant.

Correct OPTION is (b) Angle of attack

Explanation: The factor that MUST be reduced in cruise method 3 (ALSO known as constant mach number and constant altitude method) is angle of attack. Also the RATIO of weight to lift coefficient and RELATIVE airspeed are kept constant i.e. W/CL ratio and u are constant throughout the method.

The correct option is (c) Increased time of flight

Best EXPLANATION: The disadvantage of cruise METHOD 2 is that to compensate the weight of aircraft the MACH number and relative airspeed are DECREASED and there is an increase in the time of flight thus does not MAKING any fuel saving.

The correct choice is (b) False

The best explanation: In cruise method 3 endurance is not much CONCENTRATED RATHER distance travelled is concentrated. This is the reason which makes this cruise method 3 best suitable for patrol and SURVEILLANCE purposes which mainly prefer more endurance and time of flight.

The correct choice is (c) mach number

Easiest EXPLANATION: CRUISE method 2 also known as constant angle of attack and constant altitude method. In cruise method 2 altitude, the ratio of lift to DRAG and RELATIVE airspeed are kept constant i.e. L/D ratio and u and h are constant throughout the method.

Correct choice is (c) constant angle of attack and constant altitude number

The best explanation: Cruise method 2 also known as constant angle of attack and constant altitude method. In cruise method 2 altitude, the ratio of lift to DRAG and relative AIRSPEED are kept constant i.e. L/D ratio and u and h are constant throughout the method.

The correct OPTION is (a) 1.316 Vmd

For explanation: Brequet RANGE EXPRESSION is given by R1 = \(\frac{V}{C}\frac{L}{D}\) ln ω where V is true airspeed, C is specific fuel CONSUMPTION, L is lift, D is drag, and ω is fuel ratio. This Brequet Range expression comes in the cruise method 1. Brequet Range expression value is maximum at 1.316 Vmd.

Right choice is (a) True

To explain: In CRUISE method 2 endurance is more CONCENTRATED than distance TRAVELLED. This is the reason which makes this cruise method 2 best suitable for patrol and surveillance purposes which mainly PREFER more endurance and time of FLIGHT.

The correct OPTION is (d) 7.43 kg/kW-hr

Best explanation: The answer is 7.43 kg/kW-hr. GIVEN Qf=42,800 kJ/kg and P=5760 kW-hr. From the formula C=\(\FRAC{Q_f}{P}\) we will find C. On substituting we get C=\(\frac{42800}{5760}\).

On solving above equation we get the value of C=7.43 kg/kW-hr.

The correct option is (a) True

Easiest EXPLANATION: Thrust PRODUCED is the functional form of ENGINE speed and flight Mach number. The temperature DECREASES as the altitude increases so that during the cruise-climb situation the rotational speed increases and mach number increases.

Correct choice is (d) CRUISE weight and air density

For explanation: Range factor consists of cruise weight and air density. Range factor in cruise method 1 is given by \(\left[\frac{V_{mdi}}{C} E_{MAX}\right]\). The range equation in cruise method 1 is given by R1=\(\left[\frac{V_{mdi}}{C} E_{max}\right]\Big\{\frac{2u^3}{U^4+1}\Big\}\) LN ω where V is true airspeed, C is SPECIFIC fuel consumption, L is lift, D is drag, Emax is endurance, u is relative airspeed and ω is fuel ratio.

The correct answer is (b) 4000 km

Easy explanation: The answer is 4000km. GIVEN \(\frac{L}{D}\)=10, W=50,000kg, η=85%, C=4.25×10^-5kg/kW-hr. From the equation SAR=\(\frac{\ETA}{C}\frac{L}{D}\frac{1}{W}\). On substituting the values we get SAR= \(\frac{0.85}{4.25 \times 10^{-8}}\)*10*\(\frac{1}{50000}\).

On SOLVING above equation we get SAR=4000 km.

Correct answer is (a) True

Easiest explanation: Step climb helps in saving fuel.Cruise method 1 is ideal for stratosphere and uses step climb which helps in saving fuel and is ECONOMICALLY efficient. The range equation in cruise method 1 is given by R1=\(\Big[\frac{V_{mdi}}{C}E_{max}\Big]\Big\{\frac{2u^3}{u^4+1}\Big\}\)lnω where V is true AIRSPEED, C is specific fuel consumption, L is lift, D is drag, Emax is endurance, u is RELATIVE airspeed and ω is fuel RATIO.

The correct ANSWER is (c) R=\(\frac{1}{C}\Big(\frac{2W_i}{S_{\rho}}\Big)^{\frac{1}{2}}\frac{C_{L}^{0.5}}{C_D}2\Big(1-\omega^{\frac{-1}{2}}\Big)\)

Explanation: The correct equation of CRUISE method 2 is R=\(\frac{1}{C}\Big(\frac{2W_i}{S_{\rho}}\Big)^{\frac{1}{2}}\frac{C_{L}^{0.5}}{C_D}2\Big(1-\omega^{\frac{-1}{2}}\Big)\) where R is range of the cruise, C is specific fuel CONSUMPTION, S is span, ρ is density, CL is coefficient of lift, Cl/Cd is ratio of coefficient of lift to drag, Wi is INITIAL weight and ω is fuel ratio.

The CORRECT answer is (d) C=C2θ^\(\frac{1}{2}\)M^n

To explain: The correct formula for fuel FLOW is C=C2θ^\(\frac{1}{2}\)M^n where θ is the temperature, M is MACH NUMBER and n varies from 0.2-0.6 for TURBOJET to turbofan. It depends on the bypass ratio of the combustion chamber.

Right ANSWER is (b) False

The best I can explain: The range of the aircraft is DECREASED if the aircraft is flown at a higher than optimum altitude. This is caused DUE to exceeded mach number and increase in drag. To avoid this condition an aircraft is ALWAYS PREFERRED to fly below optimum altitude.

The correct answer is (d) Patrol and surveillance

The BEST I can explain: Cruise method 3 is USED mainly for patrol and surveillance PURPOSES as there is more endurance and time of flight and short range which is best suitable for the surveillance PURPOSE. Rather than cruise method 3 we use cruise method 2.

The correct option is (c) \(\Big\{\frac{2u^3}{u^4+1}\Big\}\)lnω

For explanation I would say: The RANGE FUNCTION is given by \(\Big\{\frac{2u^3}{u^4+1}\Big\}\)lnω. The range EQUATION in cruise METHOD 1 is given by R1=\(\Big[\frac{V_{mdi}}{C}E_{max}\Big]\Big\{\frac{2u^3}{u^4+1}\Big\}\)lnω where V is true airspeed, C is specific fuel consumption, L is LIFT, D is drag, Emax is endurance, u is relative airspeed and ω is fuel ratio.

Right OPTION is (a) True

The EXPLANATION is: ESPH stands for Equivalent shaft horsepower. It is the COMBINATION of thrust output with the shaft POWER output to give the total output in power FORM as if the engine was a pure power producing engine.

Right ANSWER is (a) Mach NUMBER

The explanation: The factor that must be REDUCED in cruise method 2 (also known as constant angle of attack and constant ALTITUDE method) is mach number. Also the RATIO of lift to drag and relative airspeed are kept constant i.e. L/D ratio and uare constant throughout the method.

Right option is (d) R=\(\Big[\frac{V_{mdi}}{C}E_{max}\Big]2u_i\Big\{tan^{-1}\Big[\frac{1}{u_{i}^2}\Big]-tan^{-1}\Big[\frac{1}{\omega u_{i}^2}\Big]\Big\}\)

The explanation is: The CORRECT equation of CRUISE METHOD 3 is R=\(\Big[\frac{V_{mdi}}{C}E_{max}\Big]2u_i\Big\{tan^{-1}\Big[\frac{1}{u_{i}^2}\Big]-tan^{-1}\Big[\frac{1}{\omega u_{i}^2}\Big]\Big\}\) where R is range of the cruise, C is specific fuel consumption, EMAX is endurance, Vmdi is relative airspeed and ω is fuel RATIO.

Correct answer is (a) True

For explanation: Specific FUEL consumption effects the range of an aircraft. In general fuel flow laws the specific fuel consumption is ASSUMED to be constant but in reality it effects the range of an aircraft. The range is decreased further with DECREASE in TEMPERATURE and CONSIDERING specific fuel consumption.

Right answer is (a) True

Easiest EXPLANATION: Specific AIR range is MAXIMUM when the aircraft is flying at minimum drag SPEED and specific endurance is maximum when the aircraft is flying at minimum power speed. This is the same relation with thrust PRODUCING engines.

Correct answer is (a) SAR=\(\frac{\eta}{C}\frac{L}{D}\frac{1}{W}\)

The BEST explanation: The correct formula for specific air range is given by SAR=\(\frac{\eta}{C}\frac{L}{D}\frac{1}{W}\) where η is propeller efficiency, C is specific fuel consumption, L is LIFT, D is DRAG, W is weight. SAR can also be WRITTEN as SAR=\(\frac{V}{CP}\) where v is velocity, C is specific fuel consumption and P is power.

The correct OPTION is (b) net thrust

Easy explanation: The range and ENDURANCE are proportional only to net thrust. The fuel flow LAWS proved to be a constant idealized laws. Though fuel flow laws cannot DESCRIBE all the characteristics of the ENGINE they are proved to be a constant idealized laws.

Correct answer is (c) \(\big[\frac{3-n}{1+n}\big]^{\frac{1}{4}}\)Vmd

The BEST I can explain: The formula for velocity of maximum SPECIFIC AIR range is\(\big[\frac{3-n}{1+n}\big]^{\frac{1}{4}}\)Vmd where ‘n’ is the exponent and Vmd is maximum velocity. The VALUE of ‘n’ depends on the value of bypass RATIO of the engine. This implies that as the value of ‘n’ increases the bypass ratio increases.

Correct choice is (a) True

The best explanation: The expression of range and endurance is written in terms of endurance factor and relative SPEED FUNCTION. This helps in easy calculations and understanding of CRUISING performance. This also helps in analyzing the effects of drag characteristics and specific FUEL consumption.

Correct CHOICE is (C) Increased time of flight

To explain: The disadvantage of cruise METHOD 3 is that to compensate the weight of aircraft the mach NUMBER and relative airspeed are decreased and there is an increase in the time of flight thus does not making any FUEL saving.

The correct choice is (c) Fuel ratio for MEDIUM to long range aircraft is 1.3

For EXPLANATION: Fuel ratio for medium to long range aircraft is 1.3 and for SHORT range aircraft is 1.1 and very long range aircraft is 1.5. Fuel ratio is given by ω. Fuel ratio is given by ratio of initial WEIGHT of aircraft to that of final weight of aircraft.

The correct option is (a) True

To EXPLAIN I would SAY: The best RANGE is obtained at 1.316 Vmd. The optimum ALTITUDE will be attained by the aircraft at the optimum cruise airspeed and CRITICAL mach number so that the best range is flown at the highest airspeed.

Right option is (a) RP=\(\big[\frac{\eta}{C_P}E_{max}\big]\big\{\frac{2u^2}{u^4+1}\big\}\)lnω

The BEST explanation: The correct equation for range of thrust PRODUCING engine is given by the equation RP=\(\big[\frac{\eta}{C_P}E_{max}\big]\big\{\frac{2u^2}{u^4+1}\big\}\)lnω where V is true airspeed, C is SPECIFIC fuel CONSUMPTION, Emax is endurance, η is PROPELLER efficiency and ω is fuel ratio.