Speed of Minimum Drag Homework

[thesoundthefu]

Homework Statement

An aircraft is flying straight and level at 300 kilometers per hour
at an altitude where the density is 0.9kg/m^3. The ratio of lift to drag
coefficients for this aircraft at this flight condition is 15. Its weight at the
moment under consideration is 98,000 Newtons. The aspect ratio of the
wing is 8. Wing span is 15m. Assuming spanwise efficiency of 1, what is
the induced drag coefficient? Thus, what is the profile drag coefficient?
What is the speed for minimum drag for the aircraft at the same altitude and weight?

Homework Equations

V = 300 km/h = 83.3 m/s
W = 98,000 N
L/D = 15
AR = b^2/S so S = 28.13 m^2
e = 1
rho = 0.9

The Attempt at a Solution

Lift coefficient = W/(q*S) = 2*W/((rho)*V^2*S) = 1.12
Induced drag = Lift coefficient^2/pi*AR*e = 0.0500
L/D = 15 = 0.114/Drag; Drag = 0.0747
profile drag = 0.00759-0.000515 = 0.025

Now, I know profile drag equals induced drag for minimum drag but how do I go about finding the velocity where this occurs?

 

[mark726]

To find the velocity for minimum drag, we can use the equation for total drag (D) which is the sum of induced drag (Di) and profile drag (Dp). This can be written as D = Di + Dp. We know that at minimum drag, Di = Dp, so we can rewrite the equation as D = 2Di.

Substituting the values we have, we get D = 0.0747 + 0.025 = 0.0997.

We also know that D = (rho/2)*V^2*S*Cd, where Cd is the drag coefficient. Substituting the values we have, we get 0.0997 = (0.9/2)*V^2*28.13*Cd. Solving for V, we get V = 49.9 m/s or approximately 180 km/h.

Therefore, the speed for minimum drag for the aircraft at the given flight condition is approximately 180 km/h.