A 0.8-m-diameter cylinder is placed in a 20-m/s atmospheric airflow. At what rotational speed should the cylinder rotate so that only one stagnation point exists on its surface? Calculate the minimum pressure acting on the cylinder if p = 0 at x=-∞

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[latex]v_θ=-2×20\sin θ - \frac{Γ}{2π× 0.4 } [/latex]. For one stag. pt.: [latex]v_θ = 0 at θ = 270^\circ [/latex]:

[latex]0=-2×20\sin 270^\circ - \frac{Γ}{2π× 0.4 } . ∴Γ =2 imes 20 imes 2\pi imes 0.4= 100.5 m^2/s[/latex].

[latex]Γ = 2\pi r_c^2\omega . ∴ \omega =\frac{Γ }{2\pi r_c^2}=\frac{100.5}{2\pi imes 0.4^2} =\underline{100 rad/s}[/latex] . (See Example 8.12.)

Min. pressure occurs where [latex] \left|v_θ ight| [/latex] is max, i.e., θ = π / 2. There

[latex]v_θ =-2 imes 20 imes 1-\frac{100.5}{2\pi imes 0.4} =80 m/s [/latex].

∴ [latex]p_{min}= p_\infty + \frac{V_∞^2}{2}ρ- \frac{v_θ^2}{2}ρ =0+\frac{20^2}{2} imes 1.22-\frac{80^2}{2} imes 1.22= \underline{3660 Pa}[/latex] .

Question 8.92: A 0.8-m-diameter cylinder is placed in a 20-m/s atmospheric ...

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