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Correct option is (a) The state of equilibrium where the forces and MOMENTS are balanced

The EXPLANATION: Trim is a state of equilibrium where the forces and moments are balanced. In this condition

FN=D

L = W

Where FN is normal FORCE acting on the aircraft, D is drag, L is LIFT and W is weight.

Correct CHOICE is (B) False

To explain I would say: In cruise performance the aircraft is considered to be steady, level, straight, symmetric FLIGHT with no acceleration or maneuver.

The CORRECT answer is (d) 46082.75 N

The EXPLANATION: Aircraft weight=mg=11000*9.81=107910 N.

Engine thrust: From equating the HORIZONTAL forces we GET the following equation: T cos α=D+Wsin γ

T=\(\frac{D+Wsin\gamma}{cos\alpha}\)=\(\frac{9000+107910*sin(20)}{cos(5)}\)

T=46082.75 N.

The correct choice is (b) 9220.07 N

Easy explanation: Aircraft weight=mg=11000*9.81=107910 N.

Engine thrust: From equating the forces we get the FOLLOWING equation: T COS α= D+W sin γ

T=\(\frac{D+Wsin\gamma}{cos\alpha}\)=\(\frac{9000+107910*sin(20)}{cos(5)}\)

T=46082.75 N.

From equating the VERTICAL forces we get the following equation: L+Tsinα=Wcosγ

L=107910*cos(20)–46082.75*sin(5)

L=97385.85 N.

Right option is (a) horizontal distance flown per unit of fuel consumed

The best EXPLANATION: Specific AIR range (SAR) is the horizontal distance flown per unit of fuel consumed. It can be expressed as SAR=\(\frac{V}{Q_f}\) where V is true airspeed and Qf is fuel mass FLOW. The units of SAR are length/mass i.e. nm/Kg.

The CORRECT option is (a) 400 m/kg

For EXPLANATION I would say: The answer is 1.33. Given Qf=0.75 kg/s. The SPECIFIC endurance is given by SAR=\(\FRAC{V}{Q_f}\)

On substituting we get SAR=\(\frac{300}{0.75}\)

SAR=400m/kg.

Correct choice is (c) 1.33 s/kg

Explanation: The ANSWER is 1.33. GIVEN Qf=0.75 kg/s. The specific endurance is given by SE=\(\frac{1}{Q_f}\)

On substituting we get SE=\(\frac{1}{0.75}\)

SE=1.33s/kg.

Correct option is (d) It is INSTANTANEOUS flight time per UNIT of fuel consumed

For EXPLANATION I WOULD say: Specific endurance (SE) is the instantaneous flight time per unit of fuel consumed. Specific endurance (SE) can be EXPRESSED as SE=\(\frac{1}{Q_f}\) where Qf is fuel mass flow. The units of SAR are time/mass i.e. hr/Kg.

Right choice is (b) It is the fuel flow rate per unit thrust

Easy EXPLANATION: The SPECIFIC thrust is defined as the fuel flow rate per unit thrust produced by the aircraft. The units of specific thrust are kg/N hr. The expression for specific fuel consumption is given by C=\(\FRAC{Q_f}{F_N}\) where Qf is fuel mass flow and FN is thrust produced by the aircraft.

The correct CHOICE is (d) SPECIFIC fuel CONSUMPTION is constant

The explanation: The significance of first fuel flow law is that the specific fuel consumption is constant. The significance of SECOND fuel flow law is that it considers temperature and effects of mach number. The significance of third fuel flow law is that assumptions are taken for EMPIRICAL fuel flow data.

The correct choice is (c) three

The BEST explanation: There are three fuel flow laws. They are:

i. C=C1, here SPECIFIC fuel CONSUMPTION is constant

ii. C=C2θ^1/2M^n , here it considers temperature and effects of mach number

iii. C=C3+C4M and C=C5+C6FN, here the assumptions are taken for empirical fuel flow data.

Right OPTION is (b) considers temperature and effects of mach number

The explanation: The significance of first fuel flow LAW is that the specific fuel consumption is constant. The significance of second fuel flow law is that it considers temperature and effects of mach number. The significance of third fuel flow law is that assumptions are TAKEN for empirical fuel flow DATA.

Right choice is (c) Assumptions are TAKEN for empirical fuel FLOW DATA

The best I can explain: The significance of first fuel flow law is that the specific fuel consumption is constant. The significance of SECOND fuel flow law is that it considers temperature and effects of mach NUMBER. The significance of third fuel flow law is that assumptions are taken for empirical fuel flow data.

The correct ANSWER is (a) True

The explanation is: The significance of first fuel flow law is that the specific fuel consumption is constant. The significance of second fuel flow law is that it considers temperature and EFFECTS of mach number. The significance of third fuel flow law is that assumptions are TAKEN for empirical fuel flow data. The second fuel flow law is used in the turbojet or turbofan. ALSO it takes in ACCOUNT the temperature and effects of mach number.

Correct answer is (c) RATIO of initial weight of aircraft to that of FINAL weight of aircraft

The explanation: The FUEL ratio is the ratio of initial weight of aircraft to that of final weight of aircraft. It is denoted by ω. The EQUATION is given by ω=\(\frac{W_i}{W_f}\) where Wi is initial weight of aircraft and Wf is final weight of aircraft.

Correct option is (c) SAR=\(\frac{V}{C}\frac{L}{D}\frac{1}{W}\)

Easy explanation: The equation for specific air range is GIVEN by SAR=\(\frac{V}{C}\frac{L}{D}\frac{1}{W}\) where V is true AIRSPEED, C is specific fuel consumption, L is lift, W is weight. The significance of first fuel FLOW law is that the specific fuel consumption is constant.

The CORRECT option is (a) \(\big[\frac{E_{max}}{C}\big]\)

The explanation: The endurance factor is given by \(\big[\frac{E_{max}}{C}\big]\) where Emax is endurance and C is SPECIFIC fuel consumption. The endurance equation is given by E=\(\big[\frac{E_{max}}{C}\big]\big\{\frac{2u^2}{U^4+1}\big\}\)lnωwhere 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 (a) cruise method 1

To explain I would say: Cruise climb technique is INVOLVED in cruise method 1. 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.

The CORRECT choice is (c) W/D RATIO is kept constant

Best 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.

Right CHOICE is (d) Aircraft WEIGHT is DECREASED

The best I can EXPLAIN: 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.

The correct answer is (c) 0.6

Explanation: The value of ‘N’ exponential in the fuel flow law for a high bypass RATIO TURBOFAN is 0.6. The value of ‘n’ exponential is directly proportional to the ENGINE bypass ratio i.e. it increases when the bypass ratio increases.

The CORRECT answer is (d) C=\(\FRAC{Q_f}{P}\)

The explanation is: The unit of specific fuel consumption is given by kg/kW-hr. The FORMULA for specific fuel consumption is C=\(\frac{Q_f}{P}\) where Qf is fuel flow rate, P is power and C is specific fuel consumption.

Right choice is (a) True

For explanation I would say: Relative airspeed acts as a shaping FACTOR. Relative airspeed is given by \(\Big\{\frac{2u^3}{u^4+1}\Big\}\) where u is relative airspeed. 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 ANSWER is (d) CONSTANT mach number and constant altitude number

Easiest explanation: Cruise METHOD 3 also known as constant altitude and constant mach number method. In cruise method 3 altitude, mach number and RELATIVE airspeed are kept constant i.e. M, u and h are constant throughout the method.

Correct answer is (a) True

For explanation I would say: The ENDURANCE of cruise method 2 is IDENTICAL to that of cruise method 1. E=\(\Big[\frac{E_{MAX}}{C}\Big]\Big\{\frac{2u^2}{u^4+1}\Big\}l_n\omega\) 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 choice is (b) False

The explanation is: The correct formula for fuel FLOW is C=C2\(\theta^{\frac{1}{2}}\)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.

Correct choice is (d) 494.37

To explain I would say: The answer is 494.37.

Given n=0.2, VMD=400m/s. From the FORMULA \(\big[\frac{3-n}{1+n}\big]^{\frac{1}{4}}\)Vmd. On SUBSTITUTING the values we GET \([\frac{3-0.2}{1+0.2}]^{\frac{1}{4}}\)*400=494.37.

The correct OPTION is (c) ηP=VD

To explain I would say: The correct EQUATION for PERFORMANCE equation is ηP=VD where ŋ is propeller efficiency, P is drag power, D is drag and V is velocity. In a power-producing engine the shaft power is PRODUCED as well as propulsive thrust produced by the propeller which is converted into power.

The correct answer is (d) 380.43

For explanation: The answer is 380.43.

Given n=0.2, Vmd=400m/s. From the formula \(\BIG[\FRAC{2-n}{2+n}\big]^{\frac{1}{4}}\)Vmd. On SUBSTITUTING the values we get \(\big[\frac{2-0.2}{2+0.2}]^{\frac{1}{4}}\)*400=380.43.

The correct ANSWER is (B) thrust

For explanation I WOULD say: The residual energy in the exhaust is converted into thrust by the exhaust NOZZLE. The shaft power is converted into thrust through the propeller. This way the cruise performance characteristics of an aircraft wiith MIXED power-plants lies between that of aircraft with pure thrust or pure power-producing plants.

Correct choice is (c) R=\(\Big[\frac{V_{mdi}}{C}E_{MAX}\Big]2u_i\Big\{tan^{-1}\Big[\frac{1}{u_{i}^2}\Big]-tan^{-1}\Big[\frac{1}{\omega u_{i}^2}\Big]\Big\}\)

To elaborate: The correct equation of cruise method 3 is R=\(\frac{V}{C}\int_{W_i}^{W_f}\frac{dW}{D}\) where R is RANGE, V is VELOCITY, C is SPECIFIC fuel consumption, D drag and Wf, Wiare final and initial weights.

The correct answer is (d) Relative air speed

To elaborate: The FACTOR that MUST be reduced in cruise method 3 (also known as CONSTANT angle of ATTACK 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.

Correct OPTION is (a) Increase in initial weight of aircraft implies increase in FUEL available for CRUISE

To elaborate: Increase in initial weight of aircraft implies increase in fuel available for cruise. The increase in initial weight increases the range of the aircraft which results in INCREASING the aircraft efficiency.

The CORRECT answer is (d) kg/kW-hr

For explanation: The unit of SPECIFIC fuel consumption is given by kg/kW-hr. The formula for specific fuel consumption is C=\(\frac{Q_f}{P}\) where Qf is fuel flow rate, P is POWER and C is specific fuel consumption.

The correct choice is (b) 57.6hr

To explain I would SAY: The answer is . Given \(\frac{L}{D}\)=10, W=50,000kg, η=85%, C=4.25×10^-5kg/kW-hr, V=69.44km/hr. From the EQUATION SE=\(\frac{\eta}{CV}\frac{L}{D}\frac{1}{W}\). On substituting the values we get SE= \(\frac{0.85}{69.44 \times 4.25 \times 10^{-8}}\)*10*\(\frac{1}{50000}\).

On solving above equation we get SE=57.6hr.

The correct ANSWER is (a) SAR=V*SE

The explanation: The formula for specific air range is given by: SAR=\(\FRAC{V}{CP}\) where v is velocity, C is specific FUEL CONSUMPTION and P is power. The formula for specific ENDURANCE is given by: SE=\(\frac{1}{CP}\) where C is specific fuel consumption and P is power.

Correct option is (d) R1 = \(\frac{V}{C}\frac{L}{D}\) ln ω

Easiest 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.

The CORRECT option is (b) False

The BEST EXPLANATION: Maximum range is given when the aircraft is fly’s at u=1.316. 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.