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Now consider a finite-span wing with the same profile, chord and attack angle flying at 5 km altitude with a speed of 50 m/s. The wing aspect ratio is 5, area is 5 m2 , and span efficiency factor is 0.9. Determine the wing lift.

Respuesta :

Chrisnando

Answer:

Wing lift, L = 211.57kg

Explanation:

We are given:

•Attitude= 5Km

•V = 50m/s => 164.042 ft/s

• Area= 5m² = 53.8196ft²

•Aspect ratio = 5

•span efficiency factor = 0.9

Table is attached

For density at 5km, we have: d = 16404ft

To find wing lift, we use the formula:

[tex] L = \frac{1}{2}dv^2*S*C_L*e [/tex]

Substituting figures, we have:

(1/2)*(0.0014314)*(164.042)²*(53.8196)*0.5*0.9

= 466.4382 lb

= 211.57

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