Expression for Lift Force Acting on Rotating Cylinder

Let us consider cylinder is rotating in a uniform flow field. Consider a small length of the an element on the surface of the cylinder.

Let $P_{s}=$ Pressure on the surface of the element on cylinder

$ds=$ Length of an element

$R=$ Radius of cylinder

$d \theta=$ Angle made by the length $ds$

$P=$ Pressure of the fluid far away from the cylinder

$v=$ Velocity of fluid far away from the cylinder

$u_{s}=$ Velocity of fluid on the surface of the cylinder.

Applying Bernoulli's equation, we have

$$\begin{aligned} \frac{P}{\rho g}+\frac{v^{2}}{2 g}&=\frac{P_{s}}{\rho g}+\frac{u_{s}^{2}}{2 g}\\ \therefore \quad \frac{P_{s}}{\rho g} &=\frac{P}{\rho g}+\frac{v^{2}}{2 g}-\frac{u_{s}^{2}}{2 g}\\ &=\frac{P}{\rho g}+\frac{v^{2}}{2 g}\left[1-\frac{u_{s}^{2}}{v^{2}}\right] \quad \ldots(i) \end{aligned}$$

The velocity on the surface of the cylinder is,

$$ u_{s}=u=2 v \sin \theta+\frac{\Gamma}{2 \pi R} $$

Substituting this value of $u_s$ in equation (i), we get

$$ \begin{aligned} \frac{P_{s}}{\rho g} &=\frac{P}{\rho g}+\frac{v^{2}}{2 g}\left[1-\frac{\left(2 v \sin \theta+\frac{\Gamma}{2 \pi R}\right)^{2}}{v^{2}}\right] \\ &=\frac{P}{\rho g}+\frac{v^{2}}{2 g}\left[1-\frac{\left(4 v^{2} \sin ^{2} \theta+\frac{\Gamma^{2}}{4 \pi^{2} R^{2}}+4 U \sin \theta \frac{\Gamma}{2 \pi R}\right)}{v^{2}}\right] \\ &=\frac{P}{\rho g}+\frac{v^{2}}{2 g}\left[1-\left(4 \sin ^{2} \theta+\frac{\Gamma^{2}}{4 \pi^{2} R^{2} v^{2}}+\frac{4 \sin \theta \Gamma}{v \times 2 \pi R}\right)\right] \\ P_{s} &=P+\frac{\rho g v^{2}}{2 g}\left[1-4 \sin ^{2} \theta-\frac{\Gamma^{2}}{4 \pi^{2} R^{2} v^{2}}-\frac{4 \sin \theta \Gamma}{v \times 2 \pi R}\right] ...(ii) \end{aligned} $$

The lift force acting on the small length $d s$ on the element, due to pressure $p_{s}$ as

$=$ Component of $P_{s}$ in the direction perpendicular to flow $\times$ Area of the element $=\left(-P_{s} \sin \theta\right) \times(d s \times L)$ Negative sign is taken, as the component of $P_{s}$ perpendicular to the flow is acting in the downward direction. Now $L=$ length of the cylinder $\therefore$ Lift force on the element $d s=R \times d \theta$ $(\because d s=R d \theta) $ The total force is obtained by integrating equation (iii) over the centre surface of the cylinder. $\therefore$ Total lift, $$\begin{aligned}F_{L} &=\int_{0}^{2 \pi}-P_{s} \sin \theta \times R d \theta \times L\ &=\int_{0}^{2 \pi}-P_{s} \times R \times L \times \sin \theta d \theta \end{aligned}$$

Substituting the value of $P_{s}$ from equation (ii), we get

$$ F_{L}=\int_{0}^{2 \pi}-\left[P+\frac{\rho g v^{2}}{2 g}\left(1-4 \sin ^{2} \theta-\frac{\Gamma}{4 \pi^{2} R^{2} v^{2}}-\frac{4 \sin \theta \Gamma}{v \times 2 \pi R}\right)\right] R L \times \sin \theta d \theta $$

$$ =-R L \int_{0}^{2 \pi}\left[p \sin \theta+\frac{\rho g v^{2}}{2 g}\left(\sin \theta-4 \sin ^{3} \theta-\frac{\Gamma \sin \theta}{4 \pi^{2} R^{2} v^{2}}-\frac{4 \sin ^{2} \theta \Gamma}{v \times 2 \pi R}\right)\right] d \theta $$

But $\quad \int_{0}^{2 \pi} \sin \theta d \theta=\int_{0}^{2 \pi} \sin ^{3} \theta d \theta=0$

$$ \begin{aligned} \therefore \quad F_{L} &=-R \times L \int_{0}^{2 \pi} \frac{\rho g v^{2}}{2 g}\left(-\frac{4 \sin ^{2} \theta \Gamma}{v \times 2 \pi R}\right) d \theta \\ &=R \times L \times \frac{\rho g v^{2}}{2 g} \times \frac{4 \Gamma}{v \times 2 \pi R} \int_{0}^{2 \pi} \sin ^{2} \theta d \theta \\ &=\frac{L}{g} \frac{\rho g v \Gamma}{\pi} \int_{0}^{2 \pi} \sin ^{2} \theta d \theta \end{aligned} $$

But $\quad \int_{0}^{2 \pi} \sin ^{2} \theta d \theta=\left[\frac{\theta}{2}-\frac{\sin 2 \theta}{4}\right]_{0}^{2 \pi}=\left(\frac{2 \pi}{2}-\frac{\sin 4 \pi}{4}\right)=\pi$

$$ \therefore \quad F_{L}=\frac{L}{g} \frac{\rho g}{\pi} v \Gamma \times \pi=\frac{L}{g} \rho g v \Gamma=\frac{\rho g}{g} L v \Gamma=\rho L v \Gamma ...(iii) $$

Equation (iii) is known as Kutta-Joukowski equation.