Fuselage wake contraction effect

For a configuration with a fuselage of significant size, such as the one shown in Figure 5.5, the average sheet velocity Va can no longer be assumed to be parallel to the freestream, so the wake does not trail straight back from the wing trailing edge. The actual velocities and streamline trajectories can be determined from a panel or slender-body model of the fuselage (see Section 6.6). Nikolski [48] used instead a simple axisymmetric fuselage flow model shown in Figure 5.13, where Va is assumed to be parallel to axisymmetric streamtubes. Conservation of mass between the streamtube cross-section at the wing and in the Trefftz Plane gives

m(y) = pTOV* n (y2 – (d/2)2) = pTO V* ny2 (5.83)

y(y) = sjy2 + (d/2)2 (5.84)

which assumes that the mass flux pV magnitudes adjacent to the fuselage of maximum diameter d are nearly the same as in the freestream.

Fuselage wake contraction effect

Figure 5.13: Axisymmetric streamtube flow around fuselage determines wake contraction from wing to the Trefftz plane.

Equation (5.84) is the correspondence function which specifies the wing location y which is connected to wake location y by an average streamline. For a given wing circulation distribution Г(у), the potential jump in the wake is then given in terms of the correspondence function.

A p(y) = r(y(y)) (5.85)

As an example, consider the case of an elliptical spanwise loading in the wake.

A <p(y) = A P’0 J 1 – (2 у/b)2 = Ap0 sin d (5.86)

b2 = b2 – d2 (5.87)

Fuselage wake contraction effect Fuselage wake contraction effect
Подпись: 2
Подпись: 1
Подпись: ds
Подпись: MQ
Подпись: (5.92) (5.93)

Only the lift constraint will be assumed first to simplify the initial discussion of the concepts. Adding other constraints will then be considered.

which holds for any two fields f, g which satisfy V2/ = 0 and V2g = 0. For our case we choose / = p and g = 5p, in which case the identity shows that the two terms in the equation (5.96) integrand are actually equal. Hence, omitting the second term and doubling the first term will not change the result.

SDi = – p^J ASip^ ds (5.97)

If Di is to be a minimum it’s necessary that it be stationary, specifically that 5Di = 0, for any admissible A 5p(s) distribution along the sheet. This is satisfied by a normal velocity distribution which is everywhere proportional to the local cos в,

—^-(s) = Л cos 9(s) (5.98)

dn

where Л is some constant. This solution can be verified by putting it into the £Di expression (5.97), to give

5Di = —poo f А-Sip A cos 6 ds = —SL = 0 (5.99)

as required. The conclusion is that a normal velocity which is given by (5.98) results in the smallest possible induced drag for a given lift and a given wake shape. This result is exactly consistent with the result (5.76) obtained via the Fourier series approach for the flat wake case. The great advantage of (5.98) is that it applies to a wake of any shape.