11-06-2016, 08:44 AM
Lift forces on Joukowski airfoils. [ PERJOW, JOW, AJOW ]
From the author’s Engineering Collection, included in the ETSII4 module (ETI4 on the CL Library)
These programs calculate the lift forces on a Joukowski airfoil immersed in a uniform flow U with an angle of incidence (alpha), and the airfoil is characterized by its thickness (tau) and camber angle (2*beta)
The lift force L in the direction perpendicular to the oncoming stream and upwards is given by the product of the density ro, the velocity flow and the circulation around the airfoil G, that is:
L = - ro U G
According to the Kutta condition that requires the velocities be finite at the airfoil’s trailing edge, there’s just one possible value of the circulation and it’s expressed as follows, where “a” is half the distance to the edge of the foil:
G = -4pi U a tau sin(alpha+beta);
Therefore the lift force is expressed as: L = 4pi ro U^2 a tau sin(alpha+beta) ;
The main driver program prompts for all the required data automatically. The camber angle (beta) is expressed as a function of the maximum camber within the profile (at point of null x-coordinate) and X-coordinate of the tail point (cusp), using the formula:
2*beta = asin { 2 f xt / (f^2 + xt^2)
The airfoil is parameterized into its image cylinder using the Joukowski transform (with parameter a = xt/2) by the subroutine “JW”, and the results are output sequentially including the transformation parameters (circle origin coordinates and radius), the circulation and the xy components of the lift force.
Programs JOW and AJOW are also available to calculate the equivalent coordinates between the image plane (circles) and back to the original plane (Joukowski airfoils) – using the transform equations:
z = z' + a^2/z' ; and
z' = z/2 +- sqr[(z/2)^2 + a^2]
Examples.
Calculate the lift forces on a Joukowski airfoil immersed in a uniform flow of density 1.2 kg/m^3, and uniform velocity 150 km/h with an angle of attack of 45 deg. The maximum camber of the profile (where x=0) is f=0.214 m, and the coordinates of the head and tail points are xh = -0.904m, xt = 0.8m
The obtained results are as follows:
beta =14,9760 deg
R'=0,5499 m
X0'=-0,1312 m
Y0'=0,1421 m
CIR=-249,3013 N s m^2 /kg
FX=-8.814,1304 N
FY=8.814,1304 N
SUST=12.465,062 N
Note. The routines JW and AJW can be re-written much more efficiently using functions from the 41Z module, (pity it wasn’t around 30 years ago ;-) as shown below. The data is expected in the stack registers on entry as shown: R00: parameter a, Y: Imaginary part, X: Real part
PS. This link shows a graphical calculation of the general problem in a very didactic way:
http://s6.aeromech.usyd.edu.au/aerodynam...=Flowfield
From the author’s Engineering Collection, included in the ETSII4 module (ETI4 on the CL Library)
These programs calculate the lift forces on a Joukowski airfoil immersed in a uniform flow U with an angle of incidence (alpha), and the airfoil is characterized by its thickness (tau) and camber angle (2*beta)
The lift force L in the direction perpendicular to the oncoming stream and upwards is given by the product of the density ro, the velocity flow and the circulation around the airfoil G, that is:
L = - ro U G
According to the Kutta condition that requires the velocities be finite at the airfoil’s trailing edge, there’s just one possible value of the circulation and it’s expressed as follows, where “a” is half the distance to the edge of the foil:
G = -4pi U a tau sin(alpha+beta);
Therefore the lift force is expressed as: L = 4pi ro U^2 a tau sin(alpha+beta) ;
The main driver program prompts for all the required data automatically. The camber angle (beta) is expressed as a function of the maximum camber within the profile (at point of null x-coordinate) and X-coordinate of the tail point (cusp), using the formula:
2*beta = asin { 2 f xt / (f^2 + xt^2)
The airfoil is parameterized into its image cylinder using the Joukowski transform (with parameter a = xt/2) by the subroutine “JW”, and the results are output sequentially including the transformation parameters (circle origin coordinates and radius), the circulation and the xy components of the lift force.
Programs JOW and AJOW are also available to calculate the equivalent coordinates between the image plane (circles) and back to the original plane (Joukowski airfoils) – using the transform equations:
z = z' + a^2/z' ; and
z' = z/2 +- sqr[(z/2)^2 + a^2]
Examples.
Calculate the lift forces on a Joukowski airfoil immersed in a uniform flow of density 1.2 kg/m^3, and uniform velocity 150 km/h with an angle of attack of 45 deg. The maximum camber of the profile (where x=0) is f=0.214 m, and the coordinates of the head and tail points are xh = -0.904m, xt = 0.8m
The obtained results are as follows:
beta =14,9760 deg
R'=0,5499 m
X0'=-0,1312 m
Y0'=0,1421 m
CIR=-249,3013 N s m^2 /kg
FX=-8.814,1304 N
FY=8.814,1304 N
SUST=12.465,062 N
Note. The routines JW and AJW can be re-written much more efficiently using functions from the 41Z module, (pity it wasn’t around 30 years ago ;-) as shown below. The data is expected in the stack registers on entry as shown: R00: parameter a, Y: Imaginary part, X: Real part
Code:
01 LBL "JWZ"
02 Z^2
03 ZENTER^
04 ZENTER^
05 RCL 00
06 X^ 2
07 +
08 ZSQRT
09 Z+
10 Z<>W
11 LASTZ
12 Z-
13 END
01 LBL "AJWZ"
02 ZENTER^
03 1/Z
04 ZENTER^
05 0
06 RCL 00
07 X^2
08 Z*
09 Z+
10 ENDPS. This link shows a graphical calculation of the general problem in a very didactic way:
http://s6.aeromech.usyd.edu.au/aerodynam...=Flowfield