# Aircraft Design Questions and Answers – Flight Mechanics – Steady Climbing and Descending Flight

«
»

This set of Aircraft Design written test Questions & Answers focuses on “Flight Mechanics – Steady Climbing and Descending Flight”.

1. 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

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. Ratio of vertical distance travelled to the horizontal distance travelled is known as __________
b) lift curve slope
c) power required
d) thrust loss

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. 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

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. 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

Explanation: Climb gradient = vertical velocity/ horizontal velocity = 10/12 = 0.833.

5. Find lift to weight ratio if climb angle is 45°.
a) 0.707
b) 1
c) 1.34
d) 0.992

Explanation: Lift to weight ratio = cos(climb angle) = cos(45°) = 0.707.

6. 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

Explanation: Climb angle = arcsine (excess Thrust/weight) = arcsine (40/60) = 41.8°.

7. 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

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. 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

Explanation: Fuel burned = SFC*Thrust*Time
= 0.000029*150*1500*1000 = 6.525KN.

9. 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

Velocity of best climb = $$\sqrt{\frac{w*[T+\sqrt{T^2+12CD0*K}]}{3*d*CD0}}$$
= $$\sqrt{\frac{377*[1.08+\sqrt{1.08^2+12*0.025*0.0075}]}{3*1.225*0.025}}$$
= $$\sqrt{4488.09*2.16}$$ = 98.46 unit.

10. 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

Explanation: Rate of climb = free stream velocity*sine (climb angle) = 125*sin (10) = 21.70 m/s.

11. 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

Explanation: Excess power = (R/C) * Weight = 7.08*13127.5 = 92.942 kW.

12. 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

Explanation: Time to climb = change in altitude/rate of climb = (15-10)*1000/20 = 250s.

Sanfoundry Global Education & Learning Series – Aircraft Design. 