# Kutta–Joukowski theorem

Contents 1 Lift force formula 2 Circulation and the Kutta condition 3 Derivation 3.1 Heuristic argument 3.2 Formal derivation 4 Lift forces for more complex situations 5 See also 6 References 7 Bibliography Lift force formula The theorem applies to two-dimensional flow around a fixed airfoil (or any shape of infinite span). The lift per unit span {displaystyle L',} of the airfoil is given by[4] {displaystyle L^{prime }=rho _{infty }V_{infty }Gamma ,,}         (1) where {displaystyle rho _{infty },} and {displaystyle V_{infty },} are the fluid density and the fluid velocity far upstream of the airfoil, and {displaystyle Gamma ,} is the circulation defined as the line integral {displaystyle Gamma =oint _{C}Vcdot dmathbf {s} =oint _{C}Vcos theta ;ds,} around a closed contour {displaystyle C} enclosing the airfoil and followed in the negative (clockwise) direction. As explained below, this path must be in a region of potential flow and not in the boundary layer of the cylinder. The integrand {displaystyle Vcos theta ,} is the component of the local fluid velocity in the direction tangent to the curve {displaystyle C,} and {displaystyle ds,} is an infinitesimal length on the curve, {displaystyle C,} . Equation (1) is a form of the Kutta–Joukowski theorem.

Kuethe and Schetzer state the Kutta–Joukowski theorem as follows:[5] The force per unit length acting on a right cylinder of any cross section whatsoever is equal to {displaystyle rho _{infty }V_{infty }Gamma } and is perpendicular to the direction of {displaystyle V_{infty }.} Circulation and the Kutta condition Main article: Kutta condition A lift-producing airfoil either has camber or operates at a positive angle of attack, the angle between the chord line and the fluid flow far upstream of the airfoil. Moreover, the airfoil must have a sharp trailing edge.

Any real fluid is viscous, which implies that the fluid velocity vanishes on the airfoil. Prandtl showed that for large Reynolds number, defined as {displaystyle {mathord {text{Re}}}={frac {rho V_{infty }c_{A}}{mu }},} , and small angle of attack, the flow around a thin airfoil is composed of a narrow viscous region called the boundary layer near the body and an inviscid flow region outside. In applying the Kutta-Joukowski theorem, the loop must be chosen outside this boundary layer. (For example, the circulation calculated using the loop corresponding to the surface of the airfoil would be zero for a viscous fluid.) The sharp trailing edge requirement corresponds physically to a flow in which the fluid moving along the lower and upper surfaces of the airfoil meet smoothly, with no fluid moving around the trailing edge of the airfoil. This is known as the Kutta condition.

Kutta and Joukowski showed that for computing the pressure and lift of a thin airfoil for flow at large Reynolds number and small angle of attack, the flow can be assumed inviscid in the entire region outside the airfoil provided the Kutta condition is imposed. This is known as the potential flow theory and works remarkably well in practice.

Derivation Two derivations are presented below. The first is a heuristic argument, based on physical insight. The second is a formal and technical one, requiring basic vector analysis and complex analysis.

Heuristic argument For a heuristic argument, consider a thin airfoil of chord {displaystyle c} and infinite span, moving through air of density {displaystyle rho } . Let the airfoil be inclined to the oncoming flow to produce an air speed {displaystyle V} on one side of the airfoil, and an air speed {displaystyle V+v} on the other side. The circulation is then {displaystyle Gamma =Vc-(V+v)c=-vc.,} The difference in pressure {displaystyle Delta P} between the two sides of the airfoil can be found by applying Bernoulli's equation: {displaystyle {begin{aligned}{frac {rho }{2}}(V)^{2}+(P+Delta P)&={frac {rho }{2}}(V+v)^{2}+P,,\{frac {rho }{2}}(V)^{2}+Delta P&={frac {rho }{2}}(V^{2}+2Vv+v^{2}),,\Delta P&=rho Vvqquad {text{(ignoring }}{frac {rho }{2}}v^{2}),,end{aligned}}} so the downward force on the air, per unit span, is {displaystyle L'=cDelta P=rho Vvc=-rho VGamma ,} and the upward force (lift) on the airfoil is {displaystyle rho VGamma .,} A differential version of this theorem applies on each element of the plate and is the basis of thin-airfoil theory.

Formal derivation Formal derivation of Kutta–Joukowski theorem First of all, the force exerted on each unit length of a cylinder of arbitrary cross section is calculated.[6] Let this force per unit length (from now on referred to simply as force) be {displaystyle mathbf {F} } . So then the total force is: {displaystyle mathbf {F} =-oint _{C}pmathbf {n} ,ds,} where C denotes the borderline of the cylinder, {displaystyle p} is the static pressure of the fluid, {displaystyle mathbf {n} ,} is the unit vector normal to the cylinder, and ds is the arc element of the borderline of the cross section. Now let {displaystyle phi } be the angle between the normal vector and the vertical. Then the components of the above force are: {displaystyle F_{x}=-oint _{C}psin phi ,ds,,qquad F_{y}=oint _{C}pcos phi ,ds.} Now comes a crucial step: consider the used two-dimensional space as a complex plane. So every vector can be represented as a complex number, with its first component equal to the real part and its second component equal to the imaginary part of the complex number. Then, the force can be represented as: {displaystyle F=F_{x}+iF_{y}=-oint _{C}p(sin phi -icos phi ),ds.} The next step is to take the complex conjugate of the force {displaystyle F} and do some manipulation: {displaystyle {bar {F}}=-oint _{C}p(sin phi +icos phi ),ds=-ioint _{C}p(cos phi -isin phi ),ds=-ioint _{C}pe^{-iphi },ds.} Surface segments ds are related to changes dz along them by: {displaystyle {begin{aligned}dz&=dx+idy=ds(cos phi +isin phi )=ds,e^{iphi }\{}Rightarrow d{bar {z}}&=e^{-iphi }ds.end{aligned}}} Plugging this back into the integral, the result is: {displaystyle {bar {F}}=-ioint _{C}p,d{bar {z}}.} Now the Bernoulli equation is used, in order to remove the pressure from the integral. Throughout the analysis it is assumed that there is no outer force field present. The mass density of the flow is {displaystyle rho .} Then pressure {displaystyle p} is related to velocity {displaystyle v=v_{x}+iv_{y}} by: {displaystyle p=p_{0}-{frac {rho |v|^{2}}{2}}.} With this the force {displaystyle F} becomes: {displaystyle {bar {F}}=-ip_{0}oint _{C}d{bar {z}}+i{frac {rho }{2}}oint _{C}|v|^{2},d{bar {z}}={frac {irho }{2}}oint _{C}|v|^{2},d{bar {z}}.} Only one step is left to do: introduce {displaystyle w=f(z),} the complex potential of the flow. This is related to the velocity components as {displaystyle w'=v_{x}-iv_{y}={bar {v}},} where the apostrophe denotes differentiation with respect to the complex variable z. The velocity is tangent to the borderline C, so this means that {displaystyle v=pm |v|e^{iphi }.} Therefore, {displaystyle v^{2}d{bar {z}}=|v|^{2}dz,} and the desired expression for the force is obtained: {displaystyle {bar {F}}={frac {irho }{2}}oint _{C}w'^{2},dz,} which is called the Blasius theorem.

To arrive at the Joukowski formula, this integral has to be evaluated. From complex analysis it is known that a holomorphic function can be presented as a Laurent series. From the physics of the problem it is deduced that the derivative of the complex potential {displaystyle w} will look thus: {displaystyle w'(z)=a_{0}+{frac {a_{1}}{z}}+{frac {a_{2}}{z^{2}}}+cdots .} The function does not contain higher order terms, since the velocity stays finite at infinity. So {displaystyle a_{0},} represents the derivative the complex potential at infinity: {displaystyle a_{0}=v_{xinfty }-iv_{yinfty },} . The next task is to find out the meaning of {displaystyle a_{1},} . Using the residue theorem on the above series: {displaystyle a_{1}={frac {1}{2pi i}}oint _{C}w',dz.} Now perform the above integration: {displaystyle {begin{aligned}oint _{C}w'(z),dz&=oint _{C}(v_{x}-iv_{y})(dx+idy)\&=oint _{C}(v_{x},dx+v_{y},dy)+ioint _{C}(v_{x},dy-v_{y},dx)\&=oint _{C}mathbf {v} ,{ds}+ioint _{C}(v_{x},dy-v_{y},dx).end{aligned}}} The first integral is recognized as the circulation denoted by {displaystyle Gamma .} The second integral can be evaluated after some manipulation: {displaystyle oint _{C}(v_{x},dy-v_{y},dx)=oint _{C}left({frac {partial psi }{partial y}}dy+{frac {partial psi }{partial x}}dxright)=oint _{C}dpsi =0.} Here {displaystyle psi ,} is the stream function. Since the C border of the cylinder is a streamline itself, the stream function does not change on it, and {displaystyle dpsi =0,} . Hence the above integral is zero. As a result: {displaystyle a_{1}={frac {Gamma }{2pi i}}.} Take the square of the series: {displaystyle w'^{2}(z)=a_{0}^{2}+{frac {a_{0}Gamma }{pi iz}}+cdots .} Plugging this back into the Blasius–Chaplygin formula, and performing the integration using the residue theorem: {displaystyle {bar {F}}={frac {irho }{2}}left[2pi i{frac {a_{0}Gamma }{pi i}}right]=irho a_{0}Gamma =irho Gamma (v_{xinfty }-iv_{yinfty })=rho Gamma v_{yinfty }+irho Gamma v_{xinfty }=F_{x}-iF_{y}.} And so the Kutta–Joukowski formula is: {displaystyle {begin{aligned}F_{x}&=rho Gamma v_{yinfty },,&F_{y}&=-rho Gamma v_{xinfty }.end{aligned}}} Lift forces for more complex situations The lift predicted by the Kutta-Joukowski theorem within the framework of inviscid potential flow theory is quite accurate, even for real viscous flow, provided the flow is steady and unseparated.[7] In deriving the Kutta–Joukowski theorem, the assumption of irrotational flow was used. When there are free vortices outside of the body, as may be the case for a large number of unsteady flows, the flow is rotational. When the flow is rotational, more complicated theories should be used to derive the lift forces. Below are several important examples.