[유체] Chap 9. Force & Energy on a body

Chap 9. Force and Energy on a body

9.1 Force on a circular cylinder in a uniform flow

\[\vec{F}=\int p\vec{n}\,ds\]

$p$ : pressure

\[\vec{M}=\int \rho(\vec{r}\times\vec{n})\,ds\]

(Momentum)

From Bernoulli equation,

\[\frac{\partial\phi}{\partial t}+\frac{1}{2}|\nabla\phi|^2+gz+\frac{P}{\rho}=C(t)\] \[\vec{F}=-\rho\int\left\{\frac{\partial\phi}{\partial t}+\frac{1}{2}|\nabla\phi|^2+gz+C(t)\right\}\vec{n}\,ds\]

Ignore hydrostatic pressure $-\int \rho gz\vec{n}\,ds$

Ignore constant term $-\int C(t)\vec{n}\,ds=0$ for closed body (by divergence theorem)

Steady flow $\frac{\partial\phi}{\partial t}=0$

\[\therefore\vec{F}=-\int_{S_B}\frac{\rho}{2}|\nabla\phi|^2\vec{n}\,ds\]

9.2 Potential flow around the circular cylinder

\[\phi=U\cos\theta\left(r+\frac{a^2}{r}\right)=Ux+U\frac{a^2x}{r^2}\]

$a$ : radius of cylinder

\[\nabla\phi=\left\{\frac{\partial\phi}{\partial r},\frac{1}{r}\frac{\partial\phi}{\partial\theta}\right\}=\{0,-2U\sin\theta\}\] \[\vec{F}=\int-\rho\left(\frac{1}{2}|\nabla\phi|^2\right)\vec{n}\,ds\] \[=-\rho\int_0^{2\pi}\frac{1}{2}(2U\sin\theta)^2a\,d\theta\,\vec{n}\] \[=-\rho\int_0^{2\pi}2U^2\sin^2\theta\,a(-\cos\theta,-\sin\theta)\,d\theta=-(0,0)=\vec{0}\]

⇒ D’Alembert’s Paradox

Uniform + 2D dipole + 2D vortex

\[\phi=U\cos\theta\left(r+\frac{R^2}{r}\right)+\frac{\Gamma}{2\pi}\theta\] \[\nabla\phi\big|_{r=a}=\left(0,-2U\sin\theta+\frac{\Gamma}{2\pi a}\right)\] \[\vec{F}=\int_0^{2\pi}-\frac{\rho}{2}\left(4U^2\sin^2\theta-\frac{2U\Gamma}{\pi a}\sin\theta+\frac{\Gamma^2}{4\pi^2a^2}\right)(-\cos\theta,-\sin\theta)a\,d\theta\]

$F_x = 0$,

\[F_y=\int_0^{2\pi}-\frac{\rho}{2}\frac{2U\Gamma}{\pi a}\sin^2\theta\,a\,d\theta = -\rho U \Gamma \text{ (Lift) }\]

⇒ Kutta- Joukowski theorem

\[\text{2D:}\quad F_y=-\rho U\Gamma\] \[\text{3D:}\quad \vec{F}_y=\rho\vec{U}\times\vec{\Gamma}\]

In general form,

\[\vec{F}=\rho\vec{U}\times\sum\vec{\Gamma}\]

9.3 Translation of a sphere in infinite flow

• Called BVP (Boundary value problem)
  1. $\nabla^2\phi=0$ (Laplace Eqn)
  2. $\phi=0\quad\text{as}\quad r\to\infty$ for moving body
  3. $\vec{u}\cdot\vec{n}=U\cos\theta$ : Body force (no penetration)

Instantaneous Potential

\[\phi=-\frac{UR^3}{2r^2}\cos\theta\]

Kinetic Energy

\[\text{K.E.}=\frac{1}{2}\rho\int_{S_B} \phi \frac{\partial \phi}{\partial n}ds\] \[\begin{aligned}\text{K.E.} &= \frac{1}{2}\rho \iiint_{C.V.} |\nabla\phi|^2\,d\forall \\ &= \frac{1}{2}\rho \iiint \nabla\cdot(\phi\nabla\phi) - \phi\nabla^2\phi\,dV \\ &= \frac{1}{2}\rho \iint \phi\nabla\phi\cdot\vec{n}\,ds \\ &= \frac{1}{2}\rho \iint \phi\frac{\partial\phi}{\partial n}\,ds\end{aligned}\]

Recall : $\phi=-\frac{UR^3}{2r^2}\cos\theta$

\[\begin{aligned}\text{K.E.} &= \frac{1}{2}\rho \int \left( -\frac{UR^3}{2r^2}\cos\theta \right) \left( \frac{UR^3}{r^3} \right) (-\cos\theta)\,ds \\ &= \rho\int \left( \frac{UR}{2r} \right) \left( \frac{UR^3}{r^3} \right) \cos^2\theta\,ds \bigg|_{r=R}\\ &= \frac{\rho}{4}U^2R \int\cos^2\theta\,ds\end{aligned}\] \[\begin{aligned}\text{K.E.} &= \frac{\rho U^2R}{2} \int \cos^2\theta\cdot2\pi R^2\sin\theta\,d\theta \\ &= \frac{\rho U^2R^3\pi}{2} \int_0^\pi\cos^2\theta\sin\theta\,d\theta \\ &= \frac{1}{2}U^2 \left( \rho\cdot \frac{1}{3}\cos^3\theta\bigg|_0^\pi \right) R^3\pi \\ &= \frac{1}{2}U^2\rho \left( \frac{2}{3}\pi R^3 \right)\end{aligned}\] \[V=\frac{4}{3}\pi R^3, m=\frac{4}{3}\pi R^3\rho\] \[\text{K.E.} = \frac{1}{2}U^2\left(\frac{1}{2}m\right) = \frac{1}{2} m_a U^2\]

Total Energy

\[\mathrm{Total\ E} = \frac{1}{2}(M+m_a)U^2\]

in 3D : $m_a = \frac{1}{2}m$, 2D : $M$

Power to move the body

• Another perspective to approach on D’Alembert’s Paradox
• Power : Rate of Kinematic Energy increase

\[P=\frac{d}{dt}\left[K.E._{\mathrm{total}}\right]=(M+m_a)U\frac{dU}{dt}+\frac{1}{2}U^2\frac{d}{dt}(M+m_a)\]

Constant speed : $\frac{dU}{dt}=0$

Infinite fluid : $\frac{d}{dt}(M+m_a)=0$

$P=0, FU=0, U \neq 0 \rightarrow F=0$

to move an object at constant velocity

⇒ D’Alembert’s paradox

\[(M+m_a)U\frac{dU}{dt}=FU\] \[F=(M+m_a)\frac{dU}{dt}\]