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Consider the possibility of using two rotating cylinders to replace the conventional wings on an airplane for lift. Consider an airplane flying at 150 km/hr through the Standard Atmosphere at 2,000 m. Each "wing cylinder" has a 1.0-m radius. The surface velocity of each cylinder is 20 km/hr. Find the length ℓ of each wing to develop a total lift of 40 kN. Assume potential flow and neglect end effects. Hints: = 2 r V see example 6.6

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Answer:

27.35m

Explanation:

For the calculation of the Support Force we rely on the formula for obtaining the force in a cylinder of a certain length l,

[tex]F_y = - \rho Ul\Gamma[/tex]

Here each term is,

[tex]F_y[/tex]= Lift force

[tex]\rho[/tex]= density of air

[tex]\Gamma[/tex] = vortex strength

For this last equation, its mathematical representation is given by,

[tex]\Gamma = 2\pi av_{\theta}[/tex]

Here each term is,

a= 1m, radios of cylinder

[tex]v_{\theta}= 20 Km/hr=5.5m/s[/tex], the velocity of cylinder surface.

[tex]\Gamma = 2\pi (1)(5.5) = 34.90m^2/s[/tex]

In order to find the density of the area at 2000m we will refer to the table of Standard Atmosphere of the United States, that is [tex]1.007kg/m^3,[/tex]

[tex]U= 150Km/hr = 41.6m/s, F_y = 40000N, \Gamma = 34.90m^2/s[/tex]

Replacing the values,

[tex]40000 = -(1.007)(41.6)l(34.90)[/tex]

Clearing l and solving for it we have,

[tex]l=-27.35m[/tex]

In this way we can conclude that the length of the cylinder must be 27.35m

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