Kutta-Joukowski Theorem

M. Wilhelm Kutta was born in Pitschen, Germany, in 1867 and obatined a Ph.D. in mathematics from the University of Munich in 1902. He developed the idea that lift and circulation are related. Kutta wrote a paper entitled "Lift in Flowing Fluids", where he showed the connection between circulation and lift. However, Kutta did not give the precise quantitative relation between the circulation and the lift.

Nikolai Y. Joukowski was born in Orekhovo in central Russia, in 1847. He graduated with a Ph.D. in applied mathematics from Moscow University in 1882. Joukowski published a paper in 1906, wherein he gave, for the first time in history, the relation \(L' = \rho_\infty U_\infty \Gamma\) – the Kutta-Joukowski theorem.

They developed these notions independently (around the same time) and hence the theorem is named after them.

Statement

The Kutta-Joukowski theorem states that lift per unit span on a two-dimensional body is directly proportional to the circulation around the body for a 2D incompressible irrotational flow.

For a 2-D incompressible irrotational flow, it states that

\[L = \rho_\infty U_\infty \Gamma\]

where, \(\Gamma = \oint_{c} \vec{U}.d\vec{l}\) is the circulation. We note here that the clock wise circulation is taken as positive and lift in the Z direction (upwards) is taken as positive.

We present here two separate proofs for the theorem under different assumptions. The first one is the simplest and is valid for thin airfoils. The second proof is more general and is valid for any streamlined body.

Proof 1

(Source: MIT Course)

Since both the proofs use the thin airfoil assumption, we first discuss the thin airfoil assumption.

A thin airfoil is one where the thickness of the airfoil is much smaller than the chord length.

\[c >> t\]

This implies that the velocity difference between the upper and lower surfaces of the airfoil is much smaller than the free-stream velocity.

\[U_u - U_l << U_\infty,\]

where \(U_u\) and \(U_l\) are the velocities on the upper and lower surface of the airfoil respectively.

Secondly, a thin airfoil assumption also entails a small angle of attack.

\[\alpha << 1\]

If the angle of attack is large, then even a thin airfoil will appear to be a bluff body and the flow will be separated. The Kutta-Joukowski theorem is not valid in this case.

If the angle of attack is small, then the panels (small linear segments into which the airfoil is divided) are almost horizontal. This lets us approximate the length of a panel as \(dx\).

Under such (severe) assumptions, the lift is always perpendicular to the chord (as opposed to the flow direction). The differential contribution to lift from a panel is given by

\[dL = (p_l - p_u) dx\]

where \(dx\) is the length of the panel.

We further assume that the flow is steady, incompressible and inviscid. In such a case, Bernoulli’s equation can be used to relate the pressure difference across a panel to the velocity difference across the panel as follows:

\[p_l - p_u = \frac{1}{2} \rho_\infty (U_u^2 - U_l^2) = \frac{1}{2} \rho_\infty (U_u + U_l)(U_u - U_l)\]

For a thin airfoil, the variations of the velocity from the free-stream velocity will be small and we may approximate this by \(U_u + U_l = 2U_\infty\).

\[\therefore\ p_l - p_u \approx \rho_\infty U_\infty (U_u - U_l)\]

Then the total lift of the airfoil is given by

\[L = \int dL = \rho_\infty U_\infty \int_{0}^{c} (U_u - U_l) dx\]

where \(c\) is the length of the airfoil.

But,

\[\int_{0}^{c} (U_u - U_l) dx = \int_{0}^{c} U_u dx - \int_{0}^{c} U_l dx = \oint_{c} \vec{U}.d\vec{l} = \Gamma\]

From the above two equations, we can write

\[L = \rho_\infty U_\infty \Gamma\]

Proof 2:

This proof uses the application of momentum conservation equation under the assumptions of thin airfoil theory to derive the Kutta-Joukowski theorem.

We have to first define a control volume as shown in the figure. The far field boundary is denoted by \(\infty\) and the body boundary is denoted by \(b\). The far field boundary is taken to be at a distance where the flow is undisturbed by the body. It is of circular shape.

The free stream velocity is in the X direction. \(U_\infty = u_\infty \hat{i}\).

Since we are working under the thin airfoil assumptions, it is reasonable to assume that the velocity (\(\vec{U}\)) at any point in the domain is the sum of the free stream velocity and the perturbation velocity (\(\vec{U}'\)).

\[\begin{aligned} \vec{U} &= \vec{U}_\infty + \vec{U'} \\ &= (u_\infty + u')\hat{i}\ + (w')\hat{k} \end{aligned}\]

Also, since we are using the thin airfoil theory assumption, the lift is the net force in the Z direction (as opposed to the direction normal to \(\vec{U}_\infty\)). So let us write down the the momentum equation in Z direction.

\[ \frac{\dop}{\dop t} \iiint_{V(t)}(\rho w') dV = - \iint_{s} p\, d\vec{s}\cdot\hat{k}\]

We can convert this equation to the Eulerian form using RTT (Reynolds Transport Theorem) to get

\[\begin{aligned} \iint_{s}(\rho w')\vec{U}.d\vec{s} &= -\iint_{s}p\, d\vec{s}.\hat{k} \\ &= - \iint_{\infty} p \,d\vec{s}.\hat{k} - \iint_{b} p \,d\vec{s}.\hat{k} \end{aligned}\]

We are applying this equation to the entire control volume. Hence, the boundary terms on the right hand side can be split into two parts: the boundary at infinity and the body boundary. The second term is the force on the fluid in Z direction by the body which is nothing but \(-L\). Therefore,

\[\begin{aligned} L &= \iint_{s} (\rho w') \vec{U}.d\vec{s} + \iint_{\infty} p\,d\vec{s}.\hat{k} \\ &= \iint_{b} (\rho w') \vec{U}.d\vec{s} + \iint_{\infty} (\rho w') \vec{U}.d\vec{s} + \iint_{\infty} p \,d\vec{s}.\hat{k} \\ &= \iint_{\infty} (p \hat{k} + \rho w'\vec{U}).d\vec{s}\\ \end{aligned}\]

The term \(\iint_{b} (\rho w') \vec{U}.d\vec{s} = 0\) because \(\vec{U}.d\vec{s} = 0\).

Now, \(\vec{U} = (u_\infty + u') \hat{i} + w' \hat{k}\) where \(u'\) and \(w'\) are the perturbation velocities.

Also from Bernoulli’s equation,

\[ p + \frac{1}{2} \rho_\infty [{(u_\infty + u')}^2 + {w'}^2]= p_\infty + \frac{1}{2} \rho_\infty {U_\infty}^2 \]

Now, \(U_\infty = u_\infty \hat{i}\), so, \({U_\infty}^2 = {u_\infty}^2\).

Therefore,

\[ p = p_\infty - \frac{1}{2} \rho_\infty [2 u_\infty \tilde{u} + {\tilde{u}}^2 + {\tilde{w}}^2].\]

Hence,

\[\begin{aligned} L &= {\subset\!\supset} \llap{\iint}_{\infty} \left[ \rho \tilde{w} ((u_\infty + \tilde{u}) \hat{i} + \tilde{w} \hat{k})+ (p_\infty - \frac{1}{2} \rho_\infty [2 u_\infty \tilde{u} + {\tilde{u}}^2 + {\tilde{w}}^2]) \hat{k}\right].\vec{ds} \\\end{aligned}\]

Now it can be shown that

\[\begin{aligned} \lim_{r\to\infty} u_r \longrightarrow \frac{1}{r^2} \\ \lim_{r\to\infty} u_\theta \longrightarrow \frac{1}{r} \\\end{aligned}\] (Proof is given in Katz)

Then,

\[ \tilde{u} = u_r cos \theta + u_\theta sin\theta \lim_{r\to\infty} \tilde{u} \approx \frac{1}{r}\]

Similarly,

\[ \lim_{r\to\infty} \tilde{w} \approx \frac{1}{r}\]

So now, considering the boundary as a circle

\[ {\subset\!\supset} \llap{\iint}_{\infty} ({\tilde{u}})^2.\hat{k} (rd\theta\hat{r}) \approx {\subset\!\supset} \llap{\iint}_{\infty} \frac{1}{r^2} (r d\theta)\hat{k}.\hat{r} \approx 0\]

In this way \(({\tilde{u}})^2, ({\tilde{w}})^2, (\tilde{u} \tilde{w})\) terms go to zero.

Therefore,

\[ L = {\subset\!\supset} \llap{\iint}_{\infty} p_\infty\vec{ds} + {\subset\!\supset} \llap{\iint}_{\infty} [(\rho u_\infty \tilde{w})\hat{i} - (\rho u_\infty \tilde{u})\hat{k}].\vec{ds}\]

As \(P_\infty\) is constant around closed integral, \({\subset\!\supset} \llap{\iint}_{\infty} p_\infty\vec{ds} = 0\).

Therefore,

\[L = \rho u_\infty {\subset\!\supset} \llap{\iint}_{\infty} (\tilde{w} \hat{i} - \tilde{u} \hat{k}).\vec{ds}\]

Note here that \(\vec{ds}\) is nothing but a cylinder extending \([-\infty, \infty]\) in y direction. That is, \(\vec{ds} = (dl \times 1) \hat{n} = dl \hat{n}\).

\[\begin{aligned} L &= \rho u_\infty \oint_{c} [\tilde{w} (\hat{i}.\hat{n}]) - \tilde{u} (\hat{k}.\hat{n})] dl \\ &= \rho u_\infty \oint_{c} (\tilde{w} cos\theta - \tilde{u} sin\theta) dl \\ &= \rho u_\infty \oint_{c} (\tilde{w} (\hat{k}.\hat{t}) + \tilde{u} (\hat{i}.\hat{t})) dl \\ &= \rho u_\infty \oint_{c} (\tilde{u}\hat{i} +\tilde{w}\hat{k}) dl \\ &= \rho u_\infty \oint_{c} \vec{\tilde{u}}.\vec{dl} \end{aligned}\]

Now \(u_\infty = U_\infty\), also \(\oint_{c} \vec{U_\infty}.\vec{dl} =0\).

Therefore,

\[\begin{aligned} L &= \rho U_\infty [\oint_{c} (\vec{\tilde{u}} + \vec{u_\infty}).\vec{dl}] = \rho u_\infty \Gamma \\ {\bf L = \rho u_\infty \Gamma} \end{aligned}\]

Hence Proved. (Note: This is lift per unit span)

Now, we have a much simpler way of calculating lift. Instead of calculating the \(\oint_{B} p \vec{ds}.\hat{i}\), we can straightaway \(\Sigma \Gamma_{\phi_i}\) of the singularities and get the lift.

A similar analysis along x direction yields \(F_x = 0\) (D’Alembert’s Paradox). This means drag is zero. This happens only because of inviscid assumption. As we add viscosity, this paradox is resolved.

Vorticity is generated in the thin boundary layer. It is localised, not defused and it stays with the same elements. It is then pushed down towards the trailing edge, which leaves in the form of a vortex. \[\begin{aligned} \Gamma_{c3} &= 0 \\ \Gamma_{c1} - \Gamma_{c2} &= 0 \\ \Gamma_{c1} &= \Gamma_{c2} \\ L &= \Gamma_{c1} \neq 0 \\ \end{aligned}\]

This is how lift is generated.

Any flow that is irrotational has \[\begin{aligned} \nabla \times \vec{u} &= 0 \\ {\subset\!\supset} \llap{\iint}_{s} \nabla \times \vec{u} &= 0 \\ \oint_{c} \vec{u}.\vec{dl} &= 0 \hspace{2mm} [For \hspace{2mm} simply \hspace{2mm} connected \hspace{2mm} domain]\end{aligned}\]

For multiple connected domain, you cannot apply Stoke’s theorem. So one has to remove the points where \(\nabla \times \vec{u} \neq 0\). In vortices, the vortex point is such a point.

General proof uses complex algebra Wiki.