2. Carpenter2
Introduction
Joukowsky,aRussianmathematician,inventedamappingfunctionthatconvertsthe geometryandflow
physicsof the rotatingcylinderintothe geometryand flow physics of the Joukowskyairfoil, displayedin
Figure 1.
Figure 1
Figure 1 shows the geometry of the Joukowsky airfoil in the (x”, y”) coordinate system.
Whenworkingwithairfoilsthe goal isoftentoincrease the liftforce anddecrease the dragforce. The
liftforce isthe force perpendiculartothe incomingfreestreamvelocity andthe dragforce is the force
parallel tothe incomingfreestreamvelocity. Inorderto calculate the liftanddrag forcesit is necessary
to knowthe velocitydistribution aroundthe airfoil. The velocitydistributioncanthenbe usedto
calculate the pressure distribution,andthe pressure distributioncanbe integratedalongthe surface to
calculate the force. Whenworkingwithliftanddragit iscommon to speakof the coefficientof liftand
the coefficientof dragwhichare dimensionless numbers. Dimensionlessnumbersallow experimenters
to conductlessexperimentswhilestudyingthe effectsof anumberof variables. Theyare alsocompact,
easyto use,andcan be understoodinanycoordinate system. The thicknessandcamberof the airfoil
can be adjustedtomaximizethe coefficientof lift.
3. Carpenter3
The z’ Transformation and Joukowsky Transformation
Two transformationsare utilizedtoconvertacircle intothe Joukowskyairfoilshape. The firstisthe z’
transformation followedbythe Joukowsky transformation. The z’transformation (Equation 2) isapplied
to the equationof a circle centeredat (0, 0) inthe (x, y) plane witharadiusof 𝑎 = 1 (Equation 1).
𝑧 = 𝑎𝑒 𝑖𝜃 Equation 1
𝑧′ = 𝑧 + 𝑧 𝑐′, 𝑤ℎ𝑒𝑟𝑒 𝑧 𝑐′ = 𝑥 𝑐′+ 𝑖𝑦𝑐′ Equation 2
The z’ transformation convertsthe coordinatestothe new z’ coordinate system, changing the centerof
the circle as seenin Figure 2.
Figure 2
Figure 2 illustrates the conversion from the z coordinate system to the z’ coordinate system.
The blue circle representsthe original circle andthe redcircle isa resultof the z’ transformationwhere
𝑥 𝑐
′ = −0.1 and 𝑦𝑐
′ = 0.1. The second transformationisthe Joukowsky transformationwhichcanbe
expressedinthe followingway.
𝑧" = 𝑧′ +
𝑏2
𝑧′
Equation 3
The expressionforb inthe Joukowskytransformation isconveyedbelow.
𝑏 = 𝑥 𝑐
′ + √𝑎2 − 𝑦𝑐
′2 Equation 4
Thistransformationconvertsthe circle centeredat (xc’, yc’) inred intoanairfoil shape asshownin Figure
3.
4. Carpenter4
Figure 3
Figure 3 displays the Joukowsky transformation which converts a circle into an airfoil shape.
Withthe airfoil coordinatesdefineditisnow possibletocalculate the fluidvelocitiesatthe surface of
the airfoil. Thisisassuminginviscidflow.
Velocity Distribution
The velocitycalculationbeginswithastudyof the flow pasta circular cylinderwithcirculation. The
correspondingcomplex potential,w, expressionfollows.
𝑤 = 𝑧𝑈∗ +
𝑎2
𝑧
𝑈 +
𝑖Γ
2𝜋
𝑙𝑛
𝑧
𝑎
Equation 5
The complex potential forthe rotatingcircularcylinderincludesauniformstreamterm, adoubletterm,
and a vortex termfor the circulationwhichisnecessaryforlift. 𝑈∗ isthe complex conjugate of the
incomingfreestreamvelocity,while U issimplythe incomingfreestreamvelocity. Γisthe circulation.
Γ = 4πa| 𝑈| 𝑠𝑖𝑛𝛽, 𝑤ℎ𝑒𝑟𝑒 𝛽 = 𝛼 + 𝑠𝑖𝑛−1 𝑦𝑐
′
𝑎
Equation 6
| 𝑈| is the magnitude of the incomingfreestreamvelocity whichis1,and 𝛼 isthe angle of attack. Upon
differentiatingthe complex potentialthe x-velocity,vx, andy-velocity,vy,become apparent.
5. Carpenter5
𝑑𝑤
𝑑𝑧
= 𝑣𝑥 − 𝑖𝑣 𝑦 Equation 7
Differentiatingthe complexpotential functionyields Equation 8.
𝑑𝑤
𝑑𝑧
= 𝑈∗ −
𝑎2
𝑧2
𝑈 +
𝑖Γ
2𝜋𝑧
Equation 8
Thisis the velocityaroundthe rotatingcylinder. Tocalculate the velocityaroundthe Joukowskyairfoil,
the w mustbe differentiatedwithrespectthe airfoilcoordinates,z”.
𝑑𝑤
𝑑𝑧"
=
𝑑𝑤
𝑑𝑧
𝑧′2
𝑧′2−𝑏2
Equation 9
Figure 4 showsthe velocityvectorplotforthe Joukowskyairfoil foranangle of attack of zero.
Figure 4
Figure 4 representsthe velocity vectorsat each coordinateon the Joukowskyairfoil forangle of attack of
zero.
Pressure Distribution
Once the velocitydistributionhasbeencalculated,the coefficientof pressure, Cp,acrossthe airfoil can
be determinedfromthe followingequation,where v representsthe velocitydistribution.
𝐶 𝑝 = 1 − |
𝑣
𝑈
|
2
Equation 10
6. Carpenter6
The gage pressure, 𝑃𝑔, can be calculatedusingthe coefficientof pressure. The incomingfluiddensityis
𝜌.
𝑃𝑔 =
1
2
𝜌𝑈2 𝐶 𝑝 Equation 11
The pressure isdependentonthe coordinatesbecause the coefficientof pressure isdependenton
velocitywhichvariesacrossthe shape. Figure5 showsthe coefficientof pressure versus x”coordinates
for angle of attack of zero.
Figure 5
Figure 5 presentsthe coefficientof pressure at angle of attack of zero. The plot reveals a lower pressure
on the top of the airfoil and a higher pressure on the bottom of the airfoil.
Force Distribution
The pressure distributioncanbe utilizedtofindthe force distribution. The force isthe gage pressure
actingon the area, A,of the airfoil.
𝐹 = 𝐴𝑃𝑔 Equation 12
The differential force iscalculatedforelementsof the airfoil surfacebymultiplyingthe average pressure
appliedtothe elementbythe distance it’sappliedover,assumingthe spanwise thicknessinthe z-
directionisone.
𝑑𝐹𝑖𝑥 =
𝑃 𝑔𝑖+𝑃 𝑔(𝑖+1)
2
(𝑦𝑖
"
− 𝑦𝑖+1
"
) Equation 13
𝑑𝐹𝑖𝑦 =
𝑃 𝑔𝑖+𝑃 𝑔(𝑖+1)
2
(𝑥𝑖+1
"
− 𝑥 𝑖
"
) Equation 14
The total x-force, 𝐹𝑥,andy-force, 𝐹𝑦, are simplythe sumof the elementsof 𝑑𝐹𝑖𝑥 and 𝑑𝐹𝑖𝑦.
𝐹𝑥 = ∑ 𝑑𝐹𝑖𝑥 Equation 15
9. Carpenter9
Figure 6 presents the velocity distribution and the coefficient of pressure versus x” for various angles of
attack.
In the velocityplotforzeroangle of attack the velocityissplitaroundthe leadingedge at 𝑦” = 0. For
the two negative anglesof attack (-5⁰,-10⁰) the flow issplitslightlyabove 𝑦” = 0,whileforthe two
positive anglesof attack (5⁰, 10⁰) it issplitslightlybelow 𝑦” = 0. Afterthe splitthe fluidflowstangential
to the airfoil shape. Forthe largermagnitude of anglesof attackthere are largervectorsat the leading
edge. Inthe 10⁰ angle of attack vectorplotthe vectorsare verycloselyattachedtothe bottomsurface,
showingthe fluidpushingagainstthe bottomsurface. Onthe topsurface the fluidisgoingawayfrom
the surface on the leadingedge end. Thisindicatesareductionof pressure onthe topsurface. Withthe
vectorspushingagainstthe surface onthe bottomand movingawayfromthe surface onthe top,the lift
phenomenaispresent.
In the pressure plotsthere isalowpressure onthe top of the airfoil anda higherpressure onthe
bottomof the airfoil. Goingfrom 5⁰ to 10⁰ the pressure difference increases. The plotswith negative
angle of attacks have an interestingpatternwhere the topsurface andbottomsurface pressures
intersectsothat towardthe trailingedge the topsurface hasa higherpressure insteadof lower. At 0⁰
angle of attack the pressure difference isthe smallest.
Figure 7 givesthe coefficientof liftversusangleof attackwiththe followingparameters: 𝑎 = 1, 𝑥 𝑐
′ =
−0.1, 𝑦𝑐
′ = 0.1, | 𝑈| = 1. The coefficientof liftconvertsfromnegative topositive at around 𝛼 = −6⁰.
It linearlyincreasesfrom -10⁰to 10⁰. Thus,as the angle of attack increasesthe liftingforce increases.
10. Carpenter10
The maximumcoefficientof liftis 1.8763 at 10⁰. At0⁰ the value is 0.6917. The minimumvalue at -10⁰is
-0.5142.
Figure 7
Figure 7 expresses the coefficient of lift for a range of negative and positive angles of attack.
Accordingto Figure8 the coefficientof liftincreasesasthe camberor yc
’
increases. The thicknessorxc
’
doesn’tseemtohave aneffectonthe liftvalue. If thisproblemwasconcernedwithviscousflowthe
thicknesswouldhave aneffectondrag. A Joukowskyairfoil withthe followingparametershasahigher
liftcoefficientthanthe original plottedJoukowskyairfoil: 𝑎 = 1, 𝑥 𝑐
′ = −0.3, 𝑦𝑐
′ = 0.5,| 𝑈| = 1. This
airfoil isplottedin Figure9.
12. Carpenter12
Figure 8 shows plots of the coefficient of lift versus the angle of attack for various xc
’
and yc
’
values.
Figure 9
In Figure 9 the Joukowsky airfoil is optimized by changing the 𝑥 𝑐
′ from -0.1 to 0.3 and the 𝑦𝑐
′ from 0.1 to
0.5. This magnified thecamberto increase lift and enlarged the thicknessto increase structural support.
13. Carpenter13
Thisnewairfoil designhasanincreasedcamberandthicknesscomparedtothe Joukowskyairfoil in
Figure 3.4.3 of AdvanceFluid Mechanics by W.P.Graebel. The new airfoil providesmore liftasseenin
Figure 10.
Figure 10
Figure 10 shows the coefficient of lift for the original Joukowsky airfoil with xc
’
=-0.1 and yc
’
=0.1 and the
coefficient of lift for the optimized airfoil with xc
’
=-0.3 and yc
’
=0.5.
Conclusion
To conclude,the Joukowskyairfoil shapewasformedbycompletingthe z’ transformationandthe
Joukowskytransformation. The complex potential of the rotatingcylinderwasdifferentiatedwith
respectto the Joukowskycoordinatesinordertofindthe velocityaroundthe airfoil. Then,the equation
for the coefficientof pressure wasusedtoobtainthe correspondingpressures. Finally,the pressures
were integratedoverthe areaof the airfoil tofindthe forcesandultimatelythe liftforce. Withthe lift
force,density,freestreamvelocity,andchordlengththe liftcoefficientwascalculated.
The liftcoefficientlineincreasedwhenthe camberof the airfoil was increasedbyenlargingthe yc
’
value.
The thicknessterm, xc,’
didn’thave mucheffectonthe lift. Bychangingthe parametersfrom xc
’
=-0.1
and yc
’
=0.1 to xc
’
=-0.3 and yc
’
=0.5 the liftcoefficientof the Joukowskyairfoil wasincreased.
14. Carpenter14
MATLAB Code
Part1.
function [CL] = lift_coefficient2( xc1,yc1,alpha )
%UNTITLED2 Summary of this function goes here
% Detailed explanation goes here
a=1;
U=1;
rho=1000;
chord=3.6312;
b=xc1+sqrt(a^2-yc1^2);
Beta=alpha+asin(yc1/a);
Circulation=4*pi()*a*U*sin(Beta);
Uconj=U*exp(-i*alpha);
UU=U*exp(i*alpha);
zc1=xc1+yc1*i;
theta=linspace(0,2*pi(),100);
theta=theta(:);
z=a*exp(i*theta);
x=real(z);
y=imag(z);
figure();
plot(x,y);
xlabel('x');
ylabel('y');
title('Circle');
grid('on');
Circle=[x y];
xlswrite('Circle',Circle);
z1=z+zc1;
x1=real(z1);
y1=imag(z1);
figure();
plot(x1,y1);
xlabel('x1');
ylabel('y1');
title('Circle with Center (xc1,yc1)');
grid('on');
Circle2=[x1 y1];
