• Author / Uploaded
• Marlene S. Beth

Citation preview

PROBLEM 2.1 An aircraft has mass of 50 000 kg and a wing area of 210 m2. With the engines giving a thrust of 23.4 kN it has a speed of 100 m s-1 at an altitude where the relative density is 0.75. Find the lift and drag coefficients. If the maximum lift coefficient is 1.42, find the stalling speed at this height. NOMENCLATURE 𝐶𝐿 :Lift coefficient

DATA 𝐶𝐿 = 1.42 𝑚 = 50000 𝑘𝑔 𝑆 = 210𝑚2

𝐶𝐷 :Drag coefficient 𝐷: Drag

𝑇 = 23.4 𝑘𝑁 𝑚 𝑉 = 100 𝑠 𝜎 = 0.75

𝐿: Lift 𝑚:Mass 𝑆: Wing area 𝑇: Thrust 𝑉: speed 𝜌:Density 𝜌0 :Density at sea level 𝜎:Relative Density

The first step is obtain the density from the formula: 𝜎=

𝜌 𝜌0

Clear to density 𝜌 𝜌 = 𝜎𝜌0 Substituting the values: 𝜌 = (0.75) (1.225

𝑘𝑔 𝑘𝑔 ) = 0.9187 3 3 𝑚 𝑚

The lift in cruise flight can also be considered as: 𝐿 = 𝑚𝑔 Using the formula of lift we can clear to the lift coefficient. 𝐿=

1 2 𝜌𝜈 𝑆𝐶𝐿 2

𝐶𝐿 = Substituting the values:

2𝑚𝑔 𝜌𝜈 2 𝑆

FORMULAE 𝜌 𝜎= 𝜌0 𝐿 = 𝑚𝑔 1 𝐿 = 𝜌𝜈 2 𝑆𝐶𝐿 2 𝐷=𝑇 1 𝐷 = 𝜌𝜈 2 𝑆𝐶𝐷 2

2(50000 𝑘𝑔) (9.81 𝐶𝐿 = (0.9187

𝑚 ) 𝑠2

𝑘𝑔 𝑚 2 ) (100 ) (210𝑚2 ) 3 𝑠 𝑚

= 0.5084

The drag in cruise flight can also be considered as: 𝐷=𝑇 Using the formula of drag: 𝐷=

1 2 𝜌𝜈 𝑆𝐶𝐷 2

We can clear to drag coefficient. 𝐶𝐷 =

2𝑇 𝜌𝜈 2 𝑆

Substituting the values: 𝐶𝐷 =

2(23.4 𝑥103 𝑁) = 0.0245 𝑘𝑔 𝑚 2 (0.9187 3 ) (100 ) (210𝑚2 ) 𝑠 𝑚

Using the formula of lift we can clear to the speed. 𝐿=

1 2 𝜌𝜈 𝑆𝐶𝐿 2

2𝑚𝑔 𝑉=√ 𝜌𝑆𝐶𝐿 The problem say that we should use the 𝐶𝐿 of 1.42 2(50000 𝑘𝑔) (9.81 𝑉=√

So, the lift coefficient at a speed of 100

𝑚 𝑠

𝑚 ) 𝑠2

𝑘𝑔 (0.9187 3 ) (210𝑚2 )(1.42) 𝑚

= 59.84

𝑚 𝑠

is 0.5084 and the drag coefficient is 0.0245. To 𝐶𝐿 of 1.42 we get a speed of 59.84

𝑚 𝑠

We can conclude that from the formulas of lift and drag we can get different parameters as lift and drag coefficient, the speed and other parameters.