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Correct choice is (a) True

The explanation: THRUST to weight ratio is an important parameter. This ratio can affect the PERFORMANCE of an aircraft. HIGHER thrust to weight ratio represents more thrust for given weight. HENCE, more QUICK acceleration, more maximum speed etc. can be achieved through higher values of thrust to weight ratio.

The correct choice is (a) True

The best explanation: Wing loading will affect the DRAG of aircraft. Wing loading directly AFFECTS the wetted area and span of the wing. HENCE, based on adequate wing loading we can DETERMINE lift co-efficient as well.

Right ANSWER is (a) 0.48

To explain: Given, LOAD factor n=1.2, lift to drag L/D = 2.5

Now, n at CRUISE is given by, n = (T/W)*(L/D).

Hence, T/w is given as,

T/W = n / (L/D) = 1.2/2.5 = 0.48.

The correct ANSWER is (a) 700

To elaborate: Given, a prop Aircraft.

BHP/E = 0.08, DENSITY RATIO σ = 0.2, CL at take-off = 3.

Take-off parameter TOP = \(\frac{W/S}{σCL BHP/W}\) = 35 / (0.2*3*0.08) = 700.

The correct option is (a) WEIGHT and COST of aircraft

To explain: LOW wing LOADING shows larger wings are used. Larger wing will increase weight and as wing is large material requirement will increase as well. This will affect the cost PARAMETER.

Right option is (a) 48.225 mph

Easiest explanation: Given, Power LOADING P.L. = 12.

P.L. is given by,

W/hp =1 / (a*Vmax^c), for twin TURBO prop a=0.012, c=0.5.

Hence, MAXIMUM velocity Vmax is given by,

12 = 1 / (0.012*Vmax^0.5)

0.012*Vmax^0.5 = 1/12.

Vmax^0.5 = 1 / (0.012*12) = 6.944

Takin log at both sides,

0.5*LN (Vmax) = ln (6.944) = 1.937

Now, taking anti-log,

Vmax = e^(1.937/0.5) = e^3.875 = 48.225mph.

Right answer is (a) 5.832m/s

The BEST explanation: GIVEN WING loading W/S = 50, maximum CL = 2.4, density ρ = 1.225

Now, stall speed of aircraft is given by,

Vstall = \(\sqrt{2*(\frac{W}{S})(\rho*CL)} = \sqrt{2*(50)/(1.225*2.4)} = \sqrt{34.01}\) = 5.832m/s.

Right choice is (a) 0.096

To ELABORATE: Given, maximum lift to DRAG = 12.

THRUST loading at CRUISE T/W = 1 / (L/D at cruise)

For, jet aircraft cruise lift to drag = 86.6% of maximum lift to drag = 86.6% of 12 = 0.866*12 = 10.392

Now, thrust loading at cruise,

Thrust loading at cruise T/W = 1 / (L/D at cruise)

= 1/10.392 = 0.096.

Right CHOICE is (a) RATIO of THRUST produced by an engine to the weight of the AIRCRAFT

The explanation: Thrust loading is nothing but the thrust to weight ratio. It is defined as the ratio of thrust produced by the engine to the weight of the aircraft. Thrust loading will VARY from aircraft to aircraft. It can be between 0.2-0.6 typically.

The correct option is (a) True

For explanation I would SAY: For finite WING loading value of THRUST loading should be greater than the climbing gradient. If it is not greater than the climb gradient then, it will result in either negative or IMAGINARY values or both.

Correct option is (a) W/S = \(q*\sqrt{\pi eARCD0/3}\)

For explanation: From range formula of JET aircraft maximum range is POSSIBLE if aircraft is operating at wing loading W/S = \(q*\sqrt{\pi eARCD0/3}\). W/S = \(q*\sqrt{\pi eARCD0}\) is wing loading for maximum loiter performance.

The CORRECT choice is (a) 7.54

The explanation: Given, Wing loading for maximum loiter W/S = 32, CD0 = 0.06,

q = 20unit, Oswald efficiency E=0.6.

Wing loading at maximum loiter is,

W/S = q*\(\sqrt{3*ᴨ*AR*e*CD0}\)

32 = 20*\(\sqrt{3*ᴨ*AR*0.6*0.06}\)

1.6= \(\sqrt{3*ᴨ*AR*0.6*0.06}\)

Hence, by SQUARING the aspect ratio AR,

AR = 7.54

The correct answer is (a) weight of aircraft to the HORSEPOWER by prop engine

To explain I would say: POWER loading will be defined as the ratio of weight of the aircraft to the power; typically horsepower PRODUCED by prop engine. It will IMPACT on acceleration, climb, maximum speed etc. In general, it VARIES from 10-20.

Correct answer is (a) Comparison of the engine’s available thrust at cruise to the estimated DRAG of aircraft

For explanation I would say: Thrust matching is nothing but the comparison of an engine’s available thrust at cruise to the estimated drag of aircraft. It is USED for better initial estimation of the thrust to WEIGHT ratio.

The correct CHOICE is (a) True

Best explanation: POWER available is affected by the density at a GIVEN altitude. Based on density ratio power available decreases with INCREASING altitude given that the engine is not supercharged.

Correct option is (a) 10.80

For explanation I would say: Given, CD = 0.01+0.2CL^2, AR=8.5, e=0.7, DYNAMIC PRESSURE q=25Pa.

From CD = 0.01+0.2CL^2, CD0 = 0.01.

Wing loading for maximum LOITER is given by,

W/S = \(q*\sqrt{\pi eARCD0}\) = W/S = 25*\(\sqrt{\pi*0.7*8.5*0.01}\) = 10.80.

The correct choice is (a) Yes

Easiest EXPLANATION: Thrust loading and WING loading are one of the crucial parameter of an aircraft design. A designer can find both of them EITHER by estimating thrust loading and then EVALUATING the wing loading or vice-versa.

The correct option is (a) fuel burns with each mission phase

To explain: Thrust loading is GENERALLY measured BASED on take-off conditions. At each phase of mission PROFILE weight will vary. Fuel burns throughout the flight which varies aircraft weight and which changes thrust loading as WELL.

Correct choice is (a) 12

Easiest EXPLANATION: Given, WEIGHT at CRUISE W = 2275kg, POWER = 190hp.

Hence, power loading = W/hp = 2275/190 = 11.97 = 12.

Correct answer is (a) take-off distance ESTIMATION

Easy explanation: Above diagram is representing variation of take-off parameter with take-off distance. It is used for TAKEOFF distance estimation. Take-off parameter will vary with the WING loading and thrust loading. Power loading is ratio of power to WEIGHT for jet.

Correct option is (a) 0.57

The best EXPLANATION: Given, JET fighter with MAX Mach number M = 2.0

Now for jet fighter thrust loading is given by,

T/W0 = a*M^c

For a jet fighter a=0.514, c=0.141

Hence, T/W0 = 0.514*2.0^0.141 = 0.566 = 0.57.

Right option is (a) True

To explain I would say: Wing loading and thrust loading are very IMPORTANT parameter for designer. They will AFFECT the number of crucial performance parameter of an aircraft. If ADEQUATE estimation is not done prior to the initial sizing, then it can affect the actual layout and performance of an aircraft.

Right option is (a) T/W = (550*η / V) * (hp/W)

For explanation I would say: Power is defined as thrust into velocity. Propeller will be OPERATING with some FINITE value of propeller efficiency ƞ. 550 we MULTIPLY as to make units similar both the side.

In general, for PROP aircraft thrust loading and power loading is given as,

T/W = (550*η / V)*(hp/W)

The correct option is (a) static sea-level conditions, take-off requirement ETC

The explanation: When designer speaks for a thrust loading it GENERALLY refers to the value which has been calculated for STANDARD day conditions, static sea-level conditions, take-off performance etc. Different phases incorporate different values of thrust loading. HENCE, a designer typically converts them to take-off conditions.

Correct option is (a) 10.5m/s

The best I can explain: GIVEN, civil aircraft, MAXIMUM CL = 3, wing loading W/S = 120

Approach speed of civil aircraft is given by,

VA = 1.3*Vstall

Where, Vstall = \(\SQRT{2*(\frac{W}{S})/(ρ*CL)}

= \sqrt{2*(120)/(1.225*3)}\) = 8.08m/s

Now, Va = 1.3*Vstall = 1.3*8.08 = 10.5m/s.

The correct option is (a) T/W ≥ G + 2*\(\sqrt{\frac{CD0}{ᴨ*e*AR}}\)

Best explanation: Climb GRADIENT is defined as the ratio between VERTICAL distance and horizontal distance travelled during climb. For positive wing loading, thrust loading should FOLLOW following condition. T/W ≥ G + 2*\(\sqrt{\frac{CD0}{ᴨ*e*AR}}\).

Correct option is (a) higher maximum speed

The BEST EXPLANATION: Thrust loading is nothing but a ratio. It is the ratio of aircraft thrust to the WEIGHT of the aircraft. Higher thrust loading means higher thrust for GIVEN weight. This will improve maximum speed of an aircraft.

Correct answer is (a) drag to lift RATIO at CRUISE

Explanation: At IDLE cruise CONDITION, lift = weight and thrust = drag. Hence, thrust loading T/W = D/L = Drag to lift ratio at the cruise. Aerodynamic efficiency is the ratio of lift to drag. Thrust loading at cruise is inverse of aerodynamic efficiency.

Right choice is (a) INCREASE wing loading

For explanation I would say: Wing loading is directly RELATED to the weight of the aircraft. HIGH wing loading REFERS to the small wings and vice-versa. Hence, to decrease the empty weight, we can increase the wing loading by reducing REFERENCE area.

Correct option is (a) greater or equal to one

For explanation: Power LOADING for propeller driven aircraft is an important parameter REPRESENTING the relation between power and weight. Power loading will ALWAYS be greater than one and in much idle case it will be one. It cannot be zero. Lift and drag are forces.

The CORRECT answer is (a) 21.5

Easiest explanation: Given CD0=0.02, AR =8, e=0.6, q=39.20Pa.

For maximum RANGE wing LOADING is given by,

W/S = \(q*\sqrt{\pi eARCD0} = 39.2*\sqrt{\pi*0.6*8*0.02}\) = 21.5.

Right option is (a) 1.80m/s

The explanation: Given, lift coefficient = 2.8, q=10Pa, n=5

Since DENSITY is not given we will take as sea level.

Load factor n = q*CL/(W/S)

(W/S)/CL = q/n = 10/5 = 2.

Now, stall speed V is given by,

V = \(\sqrt{2*(\FRAC{W}{S})/(\rho*CL)}\)

= \(\sqrt{2*2/1.225}\) = 1.80m/s.

The correct answer is (a) TOP = \(\FRAC{W/S}{σCL \frac{T}{W}}\)

To explain: Take-off PARAMETER is used to determine take-off distance for the aircraft. TOP is affected by wing loading and THRUST loading both. Hence, proper estimation of such parameters are crucial. TOP for jet and prop DRIVEN will be different.

For jet, TOP = \(\frac{W/S}{σCL \frac{T}{W}}\).

Right OPTION is (a) 26.97 POUND of FORCE

Easy explanation: Thrust in pound of force = Thrust in NEWTON*0.225 = 120*0.225 = 26.97 LB of force.

The correct OPTION is (a) 0.941

Easy explanation: Given thrust at take-off = thrust at CRUISE = T

Cruise thrust LOADING is given by,

(T/W)cruise = (T/W)take-off * (Wtake-off / Wcruise).

For cruise Wcruise = 0.956 * Wtake-off

Hence,

(T/W)cruise = (T/W)take-off * (Wtake-off / 0.956*Wtake-off) = 0.9/0.956 = 0.941.

Correct choice is (a) True

The BEST I can explain: Different type of AIRCRAFTS will have different requirements. WEIGHT of different aircraft will be based on MISSION requirements. Thrust produced by engine will be different for all aircraft. Hence, thrust LOADING is not same for all the aircraft.

Right answer is (b) decrease

The BEST EXPLANATION: Thrust loading of the aircraft is defined as the ratio of the thrust of the aircraft to the weight of that aircraft. HENCE, thrust loading is INVERSELY proportional to the aircraft weight. Hence, if weight is increased then the corresponding value of thrust loading will decrease.

Correct ANSWER is (a) Lowest among all three

The best I can explain: In GENERAL, we chose the lowest among all the values. This is done if and only if the lowest VALUE can satisfy almost every mission phase requirement; from take-off to LAND. If it does not satisfy then, we should estimate another value.

Correct OPTION is (C) stall SPEED, climbing, ground roll etc

Best explanation: Ratio of the weight of an aircraft to the reference area of wing is termed as wing loading. Wing loading will show the RELATION between surface area and the weight of an aircraft as a WHOLE. Wing loading will have impact on stall speed, climbing, ground roll, take-off etc.

Correct option is (a) more SPACE for fuel STORAGE in wing

Best EXPLANATION: Wing loading will affect the SIZE of the wing, weight of aircraft etc. Wing loading is given by, Weight off aircraft divided by reference area. If wing loading is LOW then, the reference area is more. This shows that large wings are used which can store more fuel.

Right CHOICE is (a) 0.097

The explanation: Given, climb GRADIENT G = 1.5, DRAG D = 20N, Weight W=220kg.

Now thrust to weight ratio in terms of G is given by,

T/W = (G+D)/W = (1.5+20)/220 = 0.097.

The correct CHOICE is (a) 1.5

For EXPLANATION: Given, lift COEFFICIENT CL = 2.5, THRUST loading T/W = 0.6.

Now, thrust loading at CRUISE is given by,

T/W = CD / CL

0.6 = CD / 2.5

CD = 0.6*2.5 = 1.5.

Right option is (a) THRUST LAPSE at cruise

Explanation: Above diagram is SHOWING thrust lapse at cruise. It is the variation of MAXIMUM thrust at cruise and take-off with altitude. Temperature lapse RATE shows how temperature is changing with the altitude.

Correct CHOICE is (a) 3.5

The best explanation: Given, W/S = 12, lift co-efficient CL = 2.1 and q=20Pa

Now load factor N is given by, n = q*CL / (W/S) = 20*2.1/12 = 3.5.