# Question 15.17: Consider fluid flows around a circular cylinder as shown in ......

Consider fluid flows around a circular cylinder as shown in Fig. 15.11. The flow separates at θ = 120°. Till the point of separation, the pressure distribution follows $p=p_{\infty}+\frac{1}{2} \rho U_{\infty}^2\left(1-4 \sin ^2 \theta\right)$. Beyond the separation point, the pressure in the wake region remains approximately constant. Neglecting the skin frictional drag on the cylinder, determine the drag coefficient. Fig. 15.11

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Given that:

$p=p_{\infty}+\frac{1}{2} \rho U_{\infty}^2\left(1-4 \sin ^2 \theta\right) \text { for } 0 \leq \theta \leq 120^{\circ}$                  $\text { for } 0 \leq \theta \leq 120^{\circ}$

At θ = 120°              $p=p_{\infty}+\frac{1}{2} \rho U_{\infty}^2\left(1-4 \sin ^2 120^{\circ}\right)=p_{\infty}-\rho U_{\infty}^2$

It is given that the pressure in the wake region remains approximately constant that means

$p=p_{\infty}-\rho U_{\infty}^2 \text { for } 120^{\circ} \leq \theta \leq 180^{\circ}$

Consider an elemental area subtending an angle dθ as shown in Fig. 15.12. Drag force acting on the elemental area due to pressure (neglecting the skin frictional drag) is

$d F_D=p d A \cos \theta=p \cos \theta R d \theta L$     (15.30)

The total drag force acting on the cylinder is obtained by integrating Eq. (15.30) over the entire surface of the cylinder as

$F_D=\int_A d F_D=2 \int_{\theta=0}^{\theta=120^{\circ}} p \cos \theta R L d \theta+2 \int_{\theta=120^{\circ}}^{\theta=180^{\circ}} p \cos \theta R L d \theta$

$=2 \int_{\theta=0}^{\theta=120^{\circ}}\left\{p_{\infty}+\frac{1}{2} \rho U_{\infty}^2\left(1-4 \sin ^2 \theta\right)\right\} \cos \theta R L d \theta$

$+2 \int_{\theta=120^{\circ}}^{\theta=180^{\circ}}\left(p_{\infty}-\rho U_{\infty}^2\right) \cos \theta R L d \theta$

$=2 R L\left(p_{\infty}+\frac{1}{2} \rho U_{\infty}^2\right) \int_{\theta=0}^{\theta=120} \cos \theta d \theta-4 \rho U_{\infty}^2 R L \int_{\theta=0}^{\theta=120} \sin ^2 \theta \cos \theta d \theta$

$+2\left(p_{\infty}-\rho U_{\infty}^2\right) R L \int_{\theta=120^{\circ}}^{\theta=180^{\circ}} \cos \theta d \theta$

$=\left.2 R L\left(p_{\infty}+\frac{1}{2} \rho U_{\infty}^2\right) \sin \theta\right|_0 ^{120^{\circ}}-\left.4 \rho U_{\infty}^2 R L \frac{\sin ^3 \theta}{3}\right|_0 ^{120^{\circ}}$

$+\left.2\left(p_{\infty}-\rho U_{\infty}^2\right) R L \sin \theta\right|_{120^{\circ}} ^{180^{\circ}}$

$=2 R L\left(p_{\infty}+\frac{1}{2} \rho U_{\infty}^2\right) \frac{\sqrt{3}}{2}-4 \rho U_{\infty}^2 R L \frac{3 \sqrt{3}}{8}-2\left(p_{\infty}-\rho U_{\infty}^2\right) R L \frac{\sqrt{3}}{2}$

$=2 R L \rho U_{\infty}^2 \frac{\sqrt{3}}{2}=\sqrt{3} R L \rho U_{\infty}^2$

The drag coefficient is found to be

$C_D=\frac{F_D}{\frac{1}{2} \rho U_{\infty}^2 A}$

$=\frac{\sqrt{3} R L \rho U_{\infty}^2}{\frac{1}{2} \rho U_{\infty}^2 2 R L}=\sqrt{3}$

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