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

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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
View Answer

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.
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2. 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

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. 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

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. 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

Answer: a
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
View Answer

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

Answer: a
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
View Answer

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. 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

Answer: a
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
View Answer

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 = \(\sqrt{\frac{w*[T+\sqrt{T^2+12CD0*K}]}{3*d*CD0}}\)
= \(\sqrt{\frac{377*[1.08+\sqrt{1.08^2+12*0.025*0.0075}]}{3*1.225*0.025}}\)
= \(\sqrt{4488.09*2.16}\) = 98.46 unit.
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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
View Answer

Answer: a
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
View Answer

Answer: a
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
View Answer

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

Sanfoundry Global Education & Learning Series – Aircraft Design.

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To practice all written questions on Aircraft Design, here is complete set of 1000+ Multiple Choice Questions and Answers.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn