Aerodynamics Flashcards
C_p equation (in terms of both pressure and velocity)
C_p= (p- p_∞)/(1/2 ρU_∞^2 )
Using Bernoulli’s: p_∞ + 1/2 ρU_∞^2 = p + 1/2 ρV^2
therefore C_p = 1 - V^2/U_∞^2
Boundary conditions
- No flow crossing the streamlines representing the body
- Flow returns to free stream, far away from the body
Semi-infinite bullet ultimate width (hence finding source/sink strength Λ)
Ultimate width = Λ/U_∞
Three related problems for any simple body and its reverse
- Full body
- Half body (same as full body, but only look at top/bottom half of streamlines etc)
- Near surface (mirror the elementary solution, so that it gives the effect of a floor)
- Equivalent problems in other direction, using a sink instead of source (hence -ve) or doublet going other way etc
Elementary soltuions for semi-infinite body
Uniform flow + source (going from LE to TE)
Uniform flow + sink (starting at TE going down to LE)
Elementary solutions for rankine-oval
Uniform flow + point source + point sink
Origin at midpoint between source and sink?
Elementary solution for cylinder
Uniform flow + doublet
Origin at doublet
Modification of elementary equations if a point is at a fixed distance from the origin
E.g. for the oval
now instead of x and y in the equations, replace them with (x - x’) and (y - y’), where x and y are your normal x and y (i.e. the point of interest), and x’ and y’ are the coordinates of the elementary solution (ie if placed at origin, x’ and y’ are 0 giving you the normal eqn)
Doublet strength
Κ=2πU_∞ R^2
R = radius of doublet
Rotating (lifting) cylinder elementary solutions
Uniform flow + doublet + vortex
C_p for lifting cylinder and its components
C_p= 1-4〖sin〗^2 (θ)- (2Γ sin(θ))/(U_∞ πR)-〖(Γ/(2U_∞ πR))〗^2
Base cylinder flow: 1-4〖sin〗^2 (θ)
- lift and drag contributions are 0 (as net lift and drag are 0)
Constant swirl term:〖(Γ/(2U_∞ πR))〗^2
- lift and drag contributions are also 0.
- i.e. no lift is created from just a cylinder or just a vortex
asymmetric swirl: (2Γ sin(θ))/(U_∞ πR)
- positive lift from upper and lower surface, but drag force is still zero due to pressure being horizontally symmetric
- combination of basic cylinder and vortex required for lift
Stagnation points on lifting cylinder
sin(θ) = -Γ/(4U_∞ πR)
- if Γ/(4U_∞ πR) < 1 (θ between 0 and -90 on right hand side of cylinder), there are two values of θ, which are symmetric about the y axis
- if Γ/(4U_∞ πR) = 1, sin(θ) = -1 so θ = -90, there is a single stagnation point at the vertical bottom
- if Γ/(4U_∞ πR) > 1, the stagnation travels away from the cylinder along the y axis
Kutta-Joukouwski lifting equation
L= ρU_∞ Γ (per unit span of cylinder, since it is 2D - N/m)
hence as Γ increases, so does the lift
- valid for any 2D body with a total ‘bound’ circulation Γ
- this year think of Γ as the vorticity
