Engineering Questions with Answers - Multiple Choice Questions
MCQs on Flight Mechanics – Steady Climbing and Descending Flight
1 - Question
What do you mean by rate of climb?
a) Vertical velocity of an aircraft
b) Lift of an aircraft
c) Thrust required
d) Drag polar
View Answer
a
Explanation: Rate of climb is nothing but the vertical velocity of aircraft. It is the climbing rate of an aircraft. Lift is an aerodynamic force which holds the aircraft in the air. Weight is the force acting due to gravity. Thrust is propulsive force.
2 - Question
Ratio of vertical distance travelled to the horizontal distance travelled is known as __________
a) climb gradient
b) lift curve slope
c) power required
d) thrust loss
View Answer
a
Explanation: Ratio of vertical distance travelled to the horizontal distance travelled is known as Climb gradient. Lift curve slope is change lift coefficient to the change in angle of attack. Power required is product of Thrust required and velocity of the aircraft.
3 - Question
Which of the following is correct in terms of fate of climb (R/C)?
a) R/C = excess power/weight
b) R/C = lift/drag
c) R/C = excess lift/drag
d) R/C = excess power/Thrust required
View Answer
a
Explanation: Rate of climb is nothing but the vertical velocity of the aircraft. Rate of climb can be defined as the ratio of excess power of the aircraft to the aircraft weight. Hence, Correct relation between rate of climb and power is as follows: R/C = excess power/weight. Where, R/C = rate of climb.
4 - Question
Consider the vertical velocity of the aircraft is 10m/s and horizontal velocity is 12 m/s. Determine the value of climb gradient.
a) 0.833
b) 1.89
c) 8
d) 2.483
View Answer
a
Explanation: Climb gradient = vertical velocity/ horizontal velocity = 10/12 = 0.833.
5 - Question
Find lift to weight ratio if climb angle is 45°.
a) 0.707
b) 1
c) 1.34
d) 0.992
View Answer
a
Explanation: Lift to weight ratio = cos(climb angle) = cos(45°) = 0.707.
6 - Question
Determine the value of climb angle if, excess thrust is 40 unit and weight of the aircraft is 60 unit. Consider steady climb.
a) 41.8
b) 50
c) 78
d) 12
View Answer
a
Explanation: Climb angle = arcsine (excess Thrust/weight) = arcsine (40/60) = 41.8°.
7 - Question
Find the approximate value of climb angle if Thrust is 1500N, drag is 1000N and weight of the aircraft is 2500N.
a) 11.53 degree
b) 30 degree
c) 40 degree
d) 1 degree
View Answer
a
Explanation: Given, steady climb, Thrust=1500N, drag D=1000N, weight W=2500N.
Climb angle = arcsine [{T-D}/W] = arcsine [{1500-1000}/2500] = 11.53 degree
8 - Question
Determine the value of fuel burned during steady climb if SFC C is 0.000029 per second, thrust is 150KN and time required is 1500s.
a) 6.525 KN
b) 100N
c) 345
d) 87.87 N
View Answer
a
Explanation: Fuel burned = SFC*Thrust*Time
= 0.000029*150*1500*1000 = 6.525KN.
9 - Question
Find the approximate value of velocity of best climb. Consider Thrust loading as 1.08, wing loading as 377 unit, CD0 as 0.025, K as 0.0075 and sea level density as 1.225 unit.
a) 98.46
b) 37.45
c) 12.1
d) 189.982
View Answer
a
Explanation: Given, Thrust loading T = 1.08, wing loading w=377, CD0=0.025, K = 0.0075, density d=1.225 unit
Velocity of best climb = w∗[T+T2+12CD0∗K√]3∗d∗CD0−−−−−−−−−−−−−−−√
= 377∗[1.08+1.082+12∗0.025∗0.0075√]3∗1.225∗0.025−−−−−−−−−−−−−−−−−−−−−√
= 4488.09∗2.16−−−−−−−−−−−√ = 98.46 unit.
10 - Question
Which of the following is correct value for rate of climb if an aircraft has free stream velocity of 125 m/s? Consider steady climb at climb angle of 10°.
a) 21.70 m/s
b) 12 m
c) 35 m/min
d) 1.302 min/m
View Answer
a
Explanation: Rate of climb = free stream velocity*sine (climb angle) = 125*sin (10) = 21.70 m/s.
11 - Question
To operate at R/C of 7.08 m/s, determine the value of excess power. Consider weight of the aircraft as 13127.5 N.
a) 92.942 kW
b) 1000W
c) 13127 N
d) 1.312KN
View Answer
a
Explanation: Excess power = (R/C) * Weight = 7.08*13127.5 = 92.942 kW.
12 - Question
Consider steady climb from an altitude of 10km to 15km. If rate of climb is 20 m/s then, determine the time to climb.
a) 250s
b) 25 min
c) 4.16 hr
d) 0.07 min
View Answer
a
Explanation: Time to climb = change in altitude/rate of climb = (15-10)*1000/20 = 250s.