Consider the nonlifting flow over a circular cylinder. Derive an expression for the pressure coefficient at an arbitrary point (r, θ) in this flow, and show that it reduces to Equation Cp = 1 − 4 sin^{2} θ on the surface of the cylinder.

Consider the nonlifting flow over a circular cylinder. Derive an expression for the pressure coefficient at an arbitrary point (r, θ) in this flow, and
show that it reduces to Equation Cp = 1 − 4 sin^{2} θ on the surface of the cylinder.

techAir Asked on 4th November 2019 in Aerodynamics.
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    For the nonlifting flow over a circular cylinder stream function \psi=\left ( V_{\infty}r sin\theta \right )\left ( 1-\frac{R^{2}}{r^{2}} \right )

    V_{r}=\frac{1}{r}\frac{\partial \psi}{\partial \theta}=V_{\infty}cos\theta \left ( 1-\frac{R^{2}}{r^{2}} \right )

    V_{\theta}=-\frac{\partial \psi}{\partial r}=-\left ( 1+\frac{R^{2}}{r^{2}} \right )V_{\infty}sin\theta

    V^{2}=V_{r}^{2}+V_{\theta}^{2}=\left ( 1-\frac{R^{2}}{r^{2}} \right )V_{\infty}^{2}cos^{2}\theta+\left ( 1+\frac{R^{2}}{r^{2}} \right )^{2}V_{\infty}^{2}sin^{2}\theta

    C_{p}=1-\frac{V^{2}}{V_{\infty}^{2}}=1-\left ( 1-\frac{R^{2}}{r^{2}} \right )cos^{2}\theta-\left ( 1+\frac{R^{2}}{r^{2}} \right )^{2}sin^{2}\theta

    At the surface of the cylinder ; r=R

    Therefore C_{p}=1-4sin^{2}\theta

    Worldtech Answered on 4th November 2019.
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