Basics of Fluid Mechanics, 2014a

10.3. POTENTIAL FLOW FUNCTIONS INVENTORY 367
pressure needed to be evaluated. For this process the Bernoulli’s equation is utilized
and can be written as
P θ = P 0 − 1 2 ρ ( U r 2 + U θ
2 ) (10.189)
It can be noticed that the two cylindrical components were accounted for. The radial
component is zero (no flow cross the stream line) and hence the total velocity is the
tangential velocity (see equation (10.177) where r = a) which can be written as
U θ =2U 0 sin θ +
Γ
2 πa
Thus, the pressure on the cylinder can be written as
P = P 0 − 1 (
2 ρ 4 U 2 0 sin 2 θ + 2 U )
0 Γ sin θ
+ Γ2
πa 4 π 2 a 2
(10.190)
(10.191)
Equation (10.191) is a parabolic equation with respect to θ (sin θ). The symmetry
dictates that D’Alembert’s paradox is valid i.e that there is no resistance to the flow.
However, in this case there is no symmetry around x coordinate (see Figure 10.16). The
distortion of the symmetry around x coordinate contribute to lift and expected. The
lift can be calculated from the integral around the solid body (stream line) and taking
only the y component. The force elements is
dF = −j · P ndA (10.192)
where in this case j is the vertical unit vector in the downward direction, and the
infinitesimal area has direction which here is broken into in the value dA and the
standard direction n. To carry the integration the unit vector n is written as
n = i cos θ + j sin θ (10.193)
The reason for definition or split (10.193) to take into account only the the vertical
component. Using the above derivation leads to
j·n = sin θ (10.194)
The lift per unit length will be
∫ 2 π
L = −
[P 0 − 1 (
0 2 ρ 4 U 2 0 sin 2 θ + 2 U )]
0 Γ sin θ
eq.(10.194)
+ Γ2 {}}{
πa 4 π 2 a 2 sin θ adθ
(10.195)
Integration of the sin θ in power of odd number between 0 and 2 π is zero. Hence the
only term that left from the integration (10.195) is
L = − ρU 0 Γ
πa
∫ 2 π
0
sin 2 θdθ = U 0 ρ Γ (10.196)