Two-Dimensional Flow: Lift and Drag
Questions
How can we determine the pressures at the surface of a body?
How does lift occur?
How to explain D’Alemberts paradox ?- That for a perfect inviscid fluid the drag force is zero.
Hydrodynamic Theory: The elements
· Establish two functions
(1)The Potential function f - defined so that ¶f/¶x=velocity in the x-direction
¶f/¶x=velocity in the y-direction
(2) The stream function y defined so that no fluid parcel crosses a streamline
· Apply the continuity and momentum concepts
· Apply the principle of irrotationality
· Find distributions of f and y that satisfy these principles throughout the fluid
· Apply the boundary conditions on the surface of the body
The Theory of Lift: Circulation
Consider a steady, two-dimensional flow in concentric circles in a perfect, incompressible fluid. When combined with a uniform linear flow such a combination will produce lift.
There is a small fluid element a distance r from the center, a length dr, unit thickness perpendicular to the flow, a width rdq at the inner face, and (r +dr) dq at the outer face.
The radially outward acting centrifugal forces on this element must be balanced by the pressure forces on the sides.
The centrifugal force is:
r(rdq +(r+dr)dq)dr/2(V2/r)= rrdqdrV2/r
Where, the higher order term has been dropped
The strength of the circulatory flow is called the circulation G defined as the line integral of Vcosq around a closed path in the fluid.
q is the angle between the velocity V and the path at a point on the contour
dl is an element of length along the contour
Now consider a circular path where V=A/r
Since the velocity is tangent to the path, q is 0 and cosq is 1. \
Integrating through 2p gives G = 2pA or A= G/2p
V=G/2pr
An important point is that outside of a vortex core the circulation is constant. Since V = G/(2pr) and for a potential vortex V = A/r, there is not radial dependence of circulation (G = 2pA).
The superposition of a vortex flow and a uniform linear flow can produce lift
Case of uniform flow with no circulation
Case of rotation with no uniform flow
Case of Uniform Flow with Rotation Producing Lift
The pressure forces positive in the radial inward direction are:
Inner FaceÞ -prdq
Outer FaceÞ (p+(dp/dr)dr)(r+dr)dq
Side FacesÞ -(p +dp/dr(dr/2))[(r+dr)dq-rdq]
Neglecting higher order terms the total of the pressure forces is: rdqdp
Equating the pressure and centrifugal forces:
rdqdp= rrdqdrV2/r
dp = rV2dr/r
Substituting for dp in the Euler equation dp = -rVdV gives: -rVdV = rV2dr/r or dV/V = -dr/r
Integrating we obtain:
Log V = -Log r + constant \ Log(Vr)=constant, and
Vr=constant
V = Constant/r = A/r
Such vortices are called potential vortices
Summary of Pressure Force Calculations
Inner Face: -prdq
Outer Face: (p+dp)(r+dr)dq
(pr+pdr+dpr+dpdr)dq
prdq+pdrdq+dprdq
Side Faces: -(p+dp/2)[(r+dr)dq-rdq]
-p+dp/2)(rdq+drdq-rdq)
-(p+dp/2)drdq
-drpdq-dp/2(drdq)
= neglect product of differentials
Inner Face Outer Face Side Faces
-prdq prdq+pdrdq+dprdq -drpdq
The total of the forces is rdqdp
In summary, for a cylinder in steady 2-d flow in a perfect fluid:
· The drag is zero
· If there is no circulation, the lift is zero
· If circulation exists around a body in uniform flow, lift is produced perpendicular to the freestream
The magnitude of the circulation can be calculated from the Kutta-Joukowski Law
Assume a constant increase of velocity DV on the upper surface of a lifting body with cord c and an equal reduction DV on the lower surface
pl = p0 + r/2[ V20 -(V20 + 2V0 DV +DV2)]
pl= p0 + r/2[2V0DV +DV2]
pu = p0 - r/2[(V20 + 2V0 DV +DV2) -V20 ]
pu= p0 - r/2[2V0DV +DV2]
The lift per unit span is: pl - pu = 2rV0DVc
The circulation is:
\ Lift per unit span is: rV0G
Kutta-Joukowski Law
L = Lift per unit span = r V G
\ G= L/(rV) = Total Lift/ (SrV)
Where S is the span
This indicates that heavy aircraft with smaller wing spans, moving slowly, will create the greatest circulation
This is why heavy aircraft during take off or landing present the greatest hazard to following aircraft