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  - defined so that
/x=velocity in the x-direction
/x=velocity in the y-direction
(2) The stream function  defined so that no fluid
parcel crosses a streamline
 Apply the continuity and momentum concepts
 Apply the principle of irrotationality
 Find distributions of  and  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.
p+(dp/dr)dr
(r+dr)d
dr
e
p+dp/dr(r/2)
rd
p
r
V
Volume Element
for a Concentric
Circular Flow
d
There is a small fluid element a distance r from the
center, a length dr, unit thickness perpendicular to the
flow, a width rd at the inner face, and (r +dr) d 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:
(rd +(r+dr)d)dr/2(V2/r)= rddrV2/r
Where, the higher order term has been dropped
The strength of the circulatory flow is called the
circulation  defined as the line integral of Vcos
around a closed path in the fluid.
 V cos dl
 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,  is 0 and cos
is 1. 
   V (1)rd   A / r (r )d   Ad
Integrating through 2 gives  = 2A or A= /2
V=/2r
dl
r
d
Circulation with
concentric circle flow
An important point is that outside of a vortex core the
circulation is constant. Since V = /(2r) and for a
potential vortex V = A/r, there is not radial dependence
of circulation ( = 2A).
The superposition of a vortex flow and a uniform linear
flow can produce lift
PP+
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 Face -prd
Outer Face (p+(dp/dr)dr)(r+dr)d
Side Faces -(p +dp/dr(dr/2))(r+dr)d-rd
Neglecting higher order terms the total of the pressure
forces is:
rddp
Equating the pressure and centrifugal forces:
rddp= rddrV2/r
dp = V2dr/r
Substituting for dp in the Euler equation dp = -VdV
gives: -VdV = V2dr/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:
-prd
Outer Face:
(p+dp)(r+dr)d
(pr+pdr+dpr+dpdr)d
prd+pdrd+dprd
Side Faces:
-(p+dp/2)(r+dr)d-rd
-p+dp/2)(rd+drd-rd)
-(p+dp/2)drd
-drpd-dp/2(drd)
= neglect product of differentials
Inner Face
-prd
Outer Face
Side Faces
prd+pdrd+dprd
The total of the forces is rddp
-drpd
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 V on the upper
surface of a lifting body with cord c and an equal
reduction V on the lower surface
pl = p0 + /2 V20 -(V20 + 2V0 V +V2)
pl= p0 + /22V0V +V2
pu = p0 - /2(V20 + 2V0 V +V2) -V20 
pu= p0 - /22V0V +V2
The lift per unit span is: pl - pu = 2V0Vc
The circulation is:
   dVdx  2Vc
 Lift per unit span is: V0
Kutta-Joukowski Law
L = Lift per unit span =  V 
 = L/(V) = Total Lift/ (SV)
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