[유체] Chap 8. Flows Around Singularities

Chap 8. Flows Around Singularities

8.1 Superposition

Solution that satisfies Laplace equation has superposition property

\[\nabla^2\phi_1=0,\quad \nabla^2\phi_2=0\]

then

\[\nabla^2(\phi_1+\phi_2)=0\]

8.2 Rankine Half Body

2D Rankine Half body

\[\phi=U_1x+\frac{m}{2\pi}\ln\sqrt{x^2+y^2}\] \[u=\frac{\partial\phi}{\partial x}=U_1+\frac{m}{2\pi}\frac{x}{x^2+y^2}\]

$x_s$ : stagnation point, point where $u=0$

\[U_1+\frac{m}{2\pi}\frac{x}{x^2+y^2}=0\quad (y=0)\] \[x_s=-\frac{m}{2\pi U_1}\]

By continuity,

\[m=D\times U_1\]

$m$ : Flux generated by source

\[\therefore D=\frac{m}{U_1}\]

Or, can be calculated using Stream function $\psi$

\[\psi=U_1r\sin\theta+\frac{m}{2\pi}\theta\] \[\left(r,\theta\right)=\left(-\frac{m}{2\pi U_1},\pi\right)\] \[\psi_{x_s}={m \over 2}\]

Streamline or body

\[U_1r\sin\theta+\frac{m}{2\pi}\theta=\frac{m}{2}\] \[\lim_{\substack{\theta\to0\\r\to\infty}}2r\sin\theta=\lim_{h\to\infty}\left[\lim_{\theta\to0}\left\{\frac{m}{U_1}\left(1-\frac{\theta}{\pi}\right)\right\}\right]=\frac{m}{U_1}\]

3D Rankine Half-body

\[\phi=Ux-\frac{m}{4\pi r}\] \[u=\frac{\partial\phi}{\partial x}=U-\frac{mx}{4\pi r^3}\]

Stagnation point $x_s$

\[u=0=U-\frac{mx_s}{4\pi x_s^3}\] \[x_s=-\sqrt{\frac{m}{4\pi U}}\quad (x_s<0)\]

By continuity,

\[\pi R^2U=m\] \[\therefore R=\sqrt{\frac{m}{\pi U}}\]

8.3 Rankine Oval Body

2D Rankine Oval body

\[\phi=Ux+\frac{m}{2\pi}\left(\ln\sqrt{(x+a)^2+y^2}-\ln\sqrt{(x-a)^2+y^2}\right)\] \[u=\frac{\partial\phi}{\partial x}= U+\frac{m}{2\pi} \left[ \frac{x+a}{\sqrt{(x+a)^2+y^2}} - \frac{x-a}{\sqrt{(x-a)^2+y^2}} \right]\]

Applying $y=0,\quad x=x_s,\quad \phi>0$

\[-\frac{U\cdot 2\pi}{m}=\frac{1}{x_s+a}-\frac{1}{x_s-a}\] \[x_s=\pm\sqrt{a^2+\frac{ma}{\pi U}}\]

Applying $x=0$, $u=u(y)$

\[u(y)=U+\frac{m}{2\pi}\left[\frac{a}{\sqrt{y^2+a^2}}-\frac{-a}{\sqrt{y^2+a^2}}\right]\] \[=U+\frac{2am}{2\pi}\frac{1}{\sqrt{y^2+a^2}}\] \[2\int_0^h u(y)\,dy=m\] \[2Uh+\frac{2am}{\pi} \left[ \tan^{-1}\left(\frac{h}{a}\right) \right] = m\]

Solution is not closed form

3D Rankine Oval (Ovoid)

\[\phi=Ux-\frac{m}{4\pi}\left\{\left((x+a)^2+y^2+z^2\right)^{-1/2}-\left((x-a)^2+y^2+z^2\right)^{-1/2}\right\}\] \[u=\frac{\partial\phi}{\partial x}=U-\frac{m}{4\pi}\frac{-\frac{1}{2}\cdot2(x+a)}{\left((x+a)^2+y^2+z^2\right)^{3/2}}+\frac{m}{4\pi}\frac{-\frac{1}{2}\cdot2(x-a)}{\left((x-a)^2+y^2+z^2\right)^{3/2}}\]

Applying $y=z=0$

\[0=U+\frac{m}{4\pi}\left(\frac{x+a}{(x+a)^3}-\frac{x-a}{(x-a)^3}\right)\] \[-\frac{4\pi}{mU}=\left(-\frac{1}{(x+a)^2}-\frac{1}{(x-a)^2}\right)\] \[\left((x+a)^2(x-a)^2\right)\left(-\frac{4\pi}{mU}\right)=(x-a)^2-(x+a)^2 = -4ax\] \[(x_s^2-a^2)^2=\frac{maU}{\pi}x_s\]

Solution is not closed form

Let $r=\sqrt{y^2+z^2}$

\[u_R=\frac{\partial\phi}{\partial x}=U-\frac{m}{4\pi}\left[\frac{-(x+a)}{\left((x+a)^2+R^2\right)^{3/2}}+\frac{x-a}{\left((x-a)^2+R^2\right)^{3/2}}\right]\]

Applying $x=0$,

\[\begin{aligned}u(R)&=U-\frac{m}{4\pi}\left(\frac{-2a}{(a^2+R^2)^{3/2}}\right)\\ &= U+\frac{ma}{2\pi} \frac{1}{(a^2+R^2)^{3/2}}\end{aligned}\] \[\int_0^{R'}\left(U+\frac{ma}{2\pi}\frac{1}{(a^2+R^2)^{3/2}}\right)2\pi R\,dR\] \[=\pi R^2U-\frac{ma}{(a^2+R^2)^{1/2}}+m= m\] \[m=\pi UR^2\sqrt{1+\frac{R^2}{a^2}}\]

8.4 Rankine Doublet

2D Rankine Doublet

\[\phi=Ux+\frac{\mu x}{2\pi r^2}=\left(Ur+\frac{\mu}{2\pi r}\right)\cos\theta\] \[u_r=\frac{\partial\phi}{\partial r}=\left(U-\frac{\mu}{2\pi r^2}\right)\cos\theta\] \[u_\theta=\frac{1}{r}\frac{\partial\phi}{\partial\theta}=-\left(U+\frac{\mu}{2\pi r^2}\right)\sin\theta\]

Applying $u_R =0, \theta = 0$ to get Radius $R$

\[U=\frac{\mu}{2\pi R^2}, R=\sqrt{\frac{\mu}{2\pi U}}\] \[\therefore\phi=Ur\left(1+\frac{R^2}{r^2}\right)\cos\theta\] \[u_\theta=-\left[U\left(1+\frac{R^2}{r^2}\right)\right]\sin\theta\] \[|U_{\theta\max}|=2U\]

When $r=R, \theta = \frac{\pi}{2}, \frac{3\pi}{2}$

3D Rankine doublet

\[\phi=Ux+\frac{\mu x}{4\pi r^3}=\left(Ur+\frac{\mu}{4\pi r^2}\right)\cos\theta\] \[u_r=\frac{\partial\phi}{\partial r}=\left(U-\frac{\mu}{2\pi r^3}\right)\cos\theta\]

$u_r = 0, \theta = 0$ to get $R$

\[U=\frac{\mu}{2\pi R^3}, R=\sqrt[3]{\mu \over 2\pi U}\] \[u_\theta=-U\sin\theta\left(1+\frac{R^3}{2r^3}\right)\]

Therefore $U_{\max}=\frac{3}{2}U$

\[P=\frac{1}{2}\rho\left[U^2\sin^2\theta\cdot\frac{9}{4}+\frac{1}{2}U^2\right]\]

8.5 Methods of Image

• Note : Must repeat source with y-axis infinitely