13.021 - Marine Hydrodynamics, Fall 2003

Lecture 9

Copyright © 2003 MIT - Department of Ocean Engineering, All rights reserved.

13.021 - Marine Hydrodynamics
Lecture 9

Vorticity Equation

Return to viscous incompressible flow.

N-S equation:

$$\frac{\partial \vec{v} }{\partial t} + \left(\vec{v} \cdot \nabl... ...{v} = - \nabla \left( {\frac{p}{\rho } + gy} \right) + \nu \nabla ^2\vec{v}$$

Then,

$$\nabla \times \left(\mbox{N-S equation}\right)\buildrel \over \lo... ...a \times \left(\vec{v} \cdot \nabla\right)\vec{v} = \nu \nabla^2\vec{\omega }$$

since $\nabla \times \nabla \phi = 0$ for any $\phi$ (conservative forces)

Now consider the vector identities:

$$(\vec{v} \cdot \nabla)\vec{v} = \frac{1}{2}\nabla \left( \vec{v} \cdot \vec{v} \right) - \vec{v} \times \left( {\nabla \times \vec{v} } \right)$$

$$= \nabla \left( {\frac{v^2}{2}} \right) -\vec{v} \times \vec{\omega } v^2 \equiv \left\vert\vec{v} \right\vert^2 = \vec{v} \cdot \vec{v}$$

$$\nabla \times \left({ \vec{v}} \cdot \nabla \right) \vec{v} = \nabla \times \nabla \left( {\frac{v^2}{2}} \right) - \nabla \t... ...vec{\omega } \right) = \nabla \times \left(\vec{\omega } \times \vec{v} \right)$$

$$= \left( \vec {v} \cdot \nabla \right)\vec{\omega} - \left( { \v... ...d{array}}{\underbrace{ \vec {v} \left( {\nabla \cdot \vec{\omega }} \right)}}$$

Therefore,

$$\frac{\partial \vec{\omega } }{\partial t} + \left(\vec{v} \cdot \nabla \right) \vec{\omega } = \left(\vec{\omega } \cdot \nabla \right) \vec{v} +\nu\nabla^{2} \vec{\omega }$$ or

$$\frac{D \vec{\omega } }{Dt} = \left(\vec{\omega } \cdot \nabla \right) \vec{v} + \underbrace{\nu \nabla ^2 \vec{\omega} }_{\mbox{diffusion}}$$

• Kelvin's Theorem revisited.
  

  If $\nu \equiv 0,$ then $\frac{D \vec{\omega } }{Dt} = \left(\vec{\omega } \cdot \nabla \right) \vec{v}$, so if $\vec{\omega } \equiv 0$ everywhere at one time, $\vec{\omega } \equiv 0$ always.

• $\nu$ can be thought of as diffusivity of (momentum) and vorticity, i.e., $\vec{\omega }$ once generated (on boundaries only) will spread/diffuse in space if $\nu$ is present.
  

• Diffusion of vorticity is analogous to the heat equation: $\frac{\partial T}{\partial t} = K\nabla ^2T$, where K is the heat diffusivity
  

  Also since $\nu \sim$ 1 or 2 mm$^{2}$/s, in 1 second, diffusion distance $\sim O\left( {\sqrt {\nu t} } \right)\sim O\left( {mm} \right)$, whereas diffusion time $\sim O\left( {{L^2} \mathord{\left/ {\vphantom {{L^2} \nu }} \right. \kern-\nulldelimiterspace} \nu } \right)$. So for a diffusion distance of L = 1cm, the necessary diffusion time needed is O(10)sec.

• For 2D, $\vec{v} = \left( {u,v,0} \right)$ and $\frac{\partial }{\partial z} \equiv 0$. So, $\vec{\omega } = \nabla \times \vec{v}$ is $\bot$ to $\vec{v}$ (parallel to z-axis).

  Then,

  $$\left( \vec{\omega } \cdot \nabla \right) \vec{v} = \left( {\unde... ...a _z \underbrace {\frac{\partial }{\partial z}}_0} \right) \vec{v} \equiv 0,$$

  so in 2D we have

  $$\frac{D \vec{\omega } }{Dt} = \nu \nabla^2 \vec{\omega }.$$

  
  

  If $\nu$ = 0, $\frac{D \vec{\omega } }{Dt} = 0$, i.e. in 2D, following a particle, the angular velocity is conserved. Reason: in 2D, the length of a vortex tube cannot change due to continuity.

• For 3D,

  $$\frac{D\omega _i }{Dt} = \underbrace{\omega _j \frac{\partial v_i... ...ace {\nu \frac{\partial ^2\omega _i }{\partial x_j \partial x_j }}_{diffusion}$$

  
  

  For example:

  $$\frac{D\omega _2 }{Dt} = \underbrace {\omega _1 \frac{\partial u_... ...ac{\partial u_2 }{\partial x_3 }}_{\mbox{vortex turning}} + \mbox{diffusion}$$
  

Example: Pile on a River

What really happens as the length of the vortex tube L increases?

IFCF (Ideal fluid under the influence of conservative forces) is no longer a valid assumption.

Why?

Ideal flow assumption implies that the inertia forces are much larger than the viscous effects (Reynolds number).

$$R_e \sim \frac{UL}{\nu }$$    Length increases $$\Rightarrow$$    diameter becomes really small $$\Rightarrow \quad R_{e}$$    is not that big after all.$$$$

Therefore IFCF is no longer valid.

3.3 Potential Flow - ideal (inviscid and incompressible) and irrotational flow

If $\vec{\omega } \equiv 0$ at some time $t$, then $\vec{\omega } \equiv 0$ always for ideal flow under conservative body forces by Kelvin's theorem. Given a vector field $\vec{v}$ for which $\vec{\omega } = \nabla \times \vec{v} \equiv 0$, then there exists a potential function (scalar) - the velocity potential - denoted as $\phi$, for which

$$\vec{v} = \nabla \phi$$

Note that $\vec{\omega } = \nabla \times \vec{v} = \nabla \times \nabla \phi \equiv 0$ for any $\phi$, so irrotational flow guaranteed automatically. At a point $\vec{x}$ and time $t$, the velocity vector $\vec{v}(\vec{x},t)$ in cartesian coordinates in terms of the potential function $\phi(\vec{x},t)$ is given by

$$\vec{v} \left(\vec{x} ,t \right) = \nabla \phi \left(\vec{x} ,t \... ...x},\frac{\partial \phi }{\partial y},\frac{\partial \phi}{\partial z} \right)$$

The velocity vector $\vec{v}$ is the gradient of the potential function $\phi$, so it always points towards higher values of the potential function.

Governing Equations:

Continuity:

$$\nabla \cdot \vec{v} = 0 = \nabla \cdot \nabla \phi \Rightarrow \nabla ^2\phi = 0$$

Number of unknowns $\to \quad \phi$

Number of equations $\to \quad \nabla ^2\phi = 0$

Therefore the problem is closed. $\phi$ and $p$ (pressure) are decoupled. $\phi$ can be solved independently first, and after it is obtained, the pressure $p$ is evaluated.

$$p = f\left( \vec{v} \right) = f\left( \nabla \phi \right) \quad \to \phi ,$$

3.4 Bernoulli equation for potential flow (steady or unsteady)

Euler eq:

$$\frac{\partial \vec{v} }{\partial t} + \nabla \left( \frac{v^2}{2... ...) - \vec{v} \times \vec{\omega } = - \nabla \left( \frac{p}{\rho } + gy \right)$$

Substitute $\vec{v} = \nabla \phi$ into the Euler's equation above, which gives:

$$\nabla \left( {\frac{\partial \phi }{\partial t}} \right) + \nabl... ...\phi } \right\vert^2} \right) = - \nabla \left( {\frac{p}{\rho } + gy} \right)$$

or

$$\nabla \left\{ {\frac{\partial \phi }{\partial t} + \frac{1}{2}\left\vert {\nabla \phi } \right\vert^2 + \frac{p}{\rho } + gy} \right\} = 0,$$

which implies that

$$\frac{\partial \phi }{\partial t} + \frac{1}{2}\left\vert {\nabla \phi } \right\vert^2 + \frac{p}{\rho } + gy = f(t)$$

everywhere in the fluid for unsteady, potential flow. The equation above can be written as

$$p = - \rho \left[ {\frac{\partial \phi }{\partial t} + \frac{1}{2}\left\vert {\nabla \phi } \right\vert^2 + gy} \right] + F(t)$$

which is the Bernoulli equation for unsteady or steady potential flow.

Summary: Bernoulli equation for ideal flow.

• Steady rotational or irrotational flow along streamline.

  $$p = - \rho \left( {\frac{1}{2}v^2 + gy} \right) + C(\psi )$$

  
  

• Unsteady or steady irrotational flow everywhere in the fluid.

  $$p = - \rho \left( {\frac{\partial \phi }{\partial t} + \frac{1}{2}\left\vert {\nabla \phi } \right\vert^2 + gy} \right) + F(t)$$

  
  

• For hydrostatics, $\vec{v} \equiv 0,\frac{\partial }{\partial t} = 0$.

  $$p = - \rho gy + c \leftarrow$$

  
  

• Steady and no gravity effect ( $\frac{\partial }{\partial t} = 0,g \equiv 0$)

  $$p = - \frac{\rho v^2}{2} + c = - \frac{\rho }{2}\left\vert {\nabla \phi } \right\vert^2 + c \leftarrow$$

  
  

• Inertial, acceleration effect

  $$\begin{array}{c} p\sim - \rho \frac{\partial \phi }{\partia... ...\frac{\partial }{\partial t} \vec{v} + \cdots \\ \end{array}$$

  
  
  

3.5 - Boundary Conditions

• KBC on an impervious boundary

  $$\underbrace { \vec{v} \cdot \hat {n}}_{\hat {n} \cdot \nabla \ph... ... boundary\ } \Rightarrow \frac{\partial \phi }{\partial n} = U_n \mbox{ given}$$

  
  

• DBC: specify pressure at the boundary, i.e.,
  $$- \rho \left( {\frac{\partial \phi }{\partial t} + \frac{1}{2}\left\vert {\nabla \phi } \right\vert^2 + gy} \right) = \mbox{given}$$

  Note: On a free-surface $p = p_{atm}$.

3.6 - Stream function

• continuity: $\nabla \cdot\vec{v} = 0$; irrotationality: $\nabla \times \vec{v} = \vec{\omega } = 0$

• velocity potential: $\vec{v} = \nabla \phi$, then $\nabla \times \vec{v} = \nabla \times \left( {\nabla \phi } \right) \equiv 0$ for any $\phi,$ i.e. irrotationality is satisfied automatically. Required for continuity:

  $$\nabla \cdot \vec{v} = \nabla ^2\phi = 0$$

  
  

• Stream function $\vec{\psi }$ defined by

  $$\vec{v} = \nabla \times \vec{\psi }$$

  
  

  Then $\nabla \cdot \vec{v} = \nabla \cdot \left( {\nabla \times \vec{\psi } } \right) \equiv 0$ for any $\vec{\psi },$ i.e. satisfies continuity automatically.

  Required for irrotationality:

  $$\nabla \times \vec{v} = 0 \Rightarrow \nabla \times \left( \nabla... ...race{\nabla \left( \nabla \cdot \vec {\psi } \right) - \nabla^2\vec{\psi}}} = 0$$ (1)

  
  

• For 2D and axisymmetric flows, $\vec{\psi }$ is a scalar $\psi$ (so stream functions are more useful for 2D and axisymmetric flows).
  

  For 2D flow: $\vec{v} = \left( {u,v,0} \right)$ and $\frac{\partial }{\partial z} \equiv 0$.

  $$\vec{v} = \nabla \times \vec{\psi } = \left\vert {{\begin{array}{... ...}{\partial x}\psi _y - \frac{\partial }{\partial y}\psi _x } \right)\hat {k}$$

  Set $\psi _x = \psi _y \equiv 0$ and $\psi _z = \psi$, then $u = \frac{\partial \psi }{\partial y}; v = - \frac{\partial \psi }{\partial x}$
  

  So, for 2D:

  $$\nabla \cdot \vec{\psi } = \frac{\partial }{\partial x}\psi _x + \frac{\partial}{\partial y}\psi _y + \frac{\partial }{\partial z}\psi _z \equiv 0$$

  
  

  Then, from the irrotationality (see (1)) $\Rightarrow \quad \nabla ^2\psi = 0$ and $\psi$ satisfies Laplace's equation.

• 2D polar coordinates: $\vec{v} = \left( {v_r ,v_{_\theta }} \right)$ and $\frac{\partial }{\partial z} \equiv 0$.
  
  $$\vec{v} = \nabla \times \vec{\psi } = \frac{1}{r}\left\vert {{\b... ...heta - \frac{\partial }{\partial \theta }\psi _r } \right)}^{v_z }\hat {e}_z$$

  Again let $\psi _r = \psi _\theta \equiv 0$ and $\psi _z = \psi$, then $v_r = \frac{1}{r}\frac{\partial \psi }{\partial \theta }$ and $v_\theta = - \frac{\partial \psi }{\partial r}$.

• For 3D but axisymmetric flows, $\vec{\psi }$ also reduces to $\psi$ (read JNN 4.6 for details).

• Physical Meaning of $\psi$.

  In 2D: $u = \frac{\partial \psi }{\partial y}$ and $v = - \frac{\partial \psi }{\partial x}$.
  

  We define

  $$\psi(\vec{x},t) = \psi(\vec{x}_{0},t)+\underset{\begin{array}{c} ... ...v}\cdot\hat{n}dl}} = \psi(\vec{x}_{0},t)+\int_{\vec{x}_{0}}^{\vec{x}}(udy-vdx)$$

  
  
  

  For $\psi$ to be single-valued, $\int$ must be path independent.

  $$\int\limits_C { = \int\limits_{C'} }$$    or $$\int\limits_C { - \int\limits_{C'} { = 0} } \buildrel \over \lon... ...nt\!\!\!\int\limits_S \underbrace{ \nabla \cdot \vec{v} }_{continuity}ds = 0$$

  Therefore, $\psi$ is unique because of continuity.
  

  Let $\vec{x} _1, \vec{x} _2$ be two points on a given streamline ( $\vec{v} \cdot \hat {n} = 0$ on streamline)

  

  $$\underbrace {\psi \left( { \vec{x} _2 } \right)}_{\psi _{_2 } } ... ...\begin{array}{c} = 0 \\ \mbox{ along } \\ \mbox{ streamline} \end{array} }dl }$$

  
  

  Therefore, $\psi _{1} = \psi_{2}$,i.e., $\psi$ is a constant along any streamline. For example, on an impervious stationary body $\vec{v} \cdot \hat {n} = 0$, so $\psi$ = constant on the body is the appropriate boundary condition. If the body is moving $\vec{v} \cdot \hat {n} = U_n$

  $$\psi = \psi _0 + \int {\underbrace {U_n }_{given}dl }$$

  
  
  

  Flux $\Delta \psi = - v\Delta x = u\Delta y$. Therefore, $u = \frac{\partial \psi }{\partial y}$ and $v = - \frac{\partial \psi }{\partial x}$

  

Summary: Potential formulation vs. Stream-function formulation for ideal flows

table

	potential	stream-function
definition	$\vec{v} = \nabla \phi$	$\vec{v} = \nabla \times \vec {\psi }$
continuity $\nabla \cdot\vec{v} = 0$	$\nabla ^2\phi = 0$	automatically satisfied
irrotationality $\nabla \times \vec{v} = 0$	automatically satisfied	$\nabla \times \left( {\nabla \times \vec {\psi } } \right) = \nabla \left( {\nabla \cdot \vec {\psi } } \right) - \nabla ^2 \vec {\psi } = 0$
in 2D $w = 0,\frac{\partial }{\partial z} = 0$	$\nabla ^2\phi = 0$	$\psi \equiv \psi _z : \nabla ^2 y\vec {\psi } = 0$
Cartesian (x, y)	$$\begin{array}{l} u = \mbox{ }\frac{\partial \phi }{\partial... ...tial y} = - \frac{\partial \psi }{\partial x} \\ \end{array}$$	Cauchy-Riemann equations for ($\phi$, $\psi )$ = (real, imaginary) part of an analytic complex function of z = x +iy
Polar (r,$\theta )$	$$\begin{array}{l} v_r = \mbox{ }\frac{\partial \phi }{\parti... ...theta } = - \frac{\partial \psi }{\partial r} \\ \end{array}$$

For irrotational flow	use	$\phi$
For incompressible flow	use	$\psi$
For both flows	use	$\phi$ or $\psi$

Given $\phi$ or $\psi$ for 2D flow, use Cauchy-Riemann equations to find the other:

For example: $\phi$ = xy $\psi$ = ?

$$\left. {\begin{array}{l} \frac{\partial \phi }{\partial \mbox{x}... ...2 + f_2 (y) \\ \end{array}} \right\} \psi = \frac{1}{2}(y^2 - x^2) + const$$