Water Waves

$\underline{Assumptions:}$

• $v=0:\frac{\partial}{\partial{y}}=0$
• inviscid
• $\text{incompressible}\implies\nabla^2\Phi=0$
• $\text{irrotational}\implies\nabla^2\Phi=0$
• g is only body ‘force’
• no surface tension
• displacements, velocities small – can ignore products
• water at free surface stays there
• starts from rest

$\underline{Waves:Bernoulli}$

Bernoulli’s Equation

$\frac{p}{\rho}+\frac{1}{2}\mid\underline{u}\mid^2+gz+\frac{\partial{\Phi}}{\partial{t}}=F(t)$

top surface – far from distance

$\frac{p}{\rho}+\frac{1}{2}\mid\underline{u}\mid^2+gz+\frac{\partial{\Phi}}{\partial{t}}=\frac{p_0}{\rho}$

$t\text{ anything, top surface }p=p_0$

$\frac{1}{2}\mid\underline{u}\mid^2+gz+\frac{\partial{\Phi}}{\partial{t}}=0$

$\text{Linearity -ignore the}\mid\underline{u}\mid^{2}\text{ term}$

$gz+\frac{\partial{\Phi}}{\partial{t}}|_{t=0}=0$

$\Phi(x,h,t)=\Phi(x,0,t)+h\frac{\partial{\Phi}}{\partial{z}}(x,0,t)+...$

$\text{Becomes }gh+\frac{\partial{\Phi}}{\partial{t}}|_{z=0}=0$

Potential Flow

$\underline{Velocity Potential:}$

Incompressible flow

• Divergence zero
• $\nabla\cdot\underline{u}=0$

Irrotational flow

• curl zero
• $\nabla\times\underline{u}=\underline{0}$
• Implies existence of velocity potential
• $\nabla\Phi=\underline{u}$

Ideal flow: incompressible and irrotational

• $\text{Irrotational: }\nabla\Phi=\underline{u}$
• $\text{Incompressible: }\nabla\cdot\underline{u}$
• $\nabla\cdot\underline{u}=\nabla\cdot\nabla\Phi=\nabla^{2}\Phi=0$
• Laplace’s equation

$\underline{Cartesian}$

Stream function

• $\frac{\partial{\psi}}{\partial{x}}=-v$
• $\frac{\partial{\psi}}{\partial{y}}=u$

Velocity Potential

• $\frac{\partial{\Phi}}{\partial{x}}=u$
• $\frac{\partial{\Phi}}{\partial{y}}=v$

$\frac{\partial{\Phi}}{\partial{x}}=u=\frac{\partial{\psi}}{\partial{y}}$

$\frac{\partial{\Phi}}{\partial{y}}=v=-\frac{\partial{\psi}}{\partial{x}}$

– Cauchy-Riemann equations

$\omega(x+iy)=\Phi(x,y)+i\psi(x,y)$

$\implies\omega(x+iy)\text{ is differentiable complex function}$

Velocity potential example:

$\text{Uniform stream parallel to x-axis (speed }U_{0})$

$\psi=U_{0}y$

$(u,v)=(\frac{\partial{\psi}}{\partial{y}},\frac{-\partial{\psi}}{\partial{x}})=(U_{0},0)$

Velocity potential:

$\frac{\partial{\Phi}}{\partial{x}}=u=U_{0}$

$\frac{\partial{\Phi}}{\partial{y}}=v=0$

$\implies\Phi=U_{0}x+\not{c}$

$\underline{Polar}$

Streamfunction

• $\frac{\partial{\psi}}{\partial{r}}=-u_{\theta}$
• $\frac{1}{r}\frac{\partial{\psi}}{\partial{\theta}}=u_{r}$

Velocity Potential

• $\frac{\partial{\Phi}}{\partial{r}}=u_{r}$
• $\frac{1}{r}\frac{\partial{\Phi}}{\partial{\theta}}=u_{\theta}$

Source – Streamfunction

$\psi=m\theta$

$\text{Velocity }\underline{u}=u_{r}\underline{e}_r+u_{\theta}\underline{e}_{\theta}$

$u_{r}=\frac{1}{r}\frac{\partial{\psi}}{\partial{\theta}}=\frac{m}{r}$

$u_{\theta}=-\frac{\partial{\psi}}{\partial{r}}=0$

Velocity potential:

$\frac{\partial{\Phi}}{\partial{r}}=u_{r}=\frac{m}{r}$

$\frac{1}{r}\frac{\partial{\Phi}}{\partial{\theta}}=u_{\theta}=0$

$\implies\Phi=m\ln(r)+\not{c}$

Flow around a cylinder

We know that the flow velocity must be tangential to the surface of the cylinder, since no fluid can penetrate through the solid surface of the cylinder. As this is the only surface condition for an inviscid flow we can assume that for any inviscid flow, streamlines can be thought of as solid surfaces and vice versa.

Let’s look at an example:

$\psi=U_{0}r\sin{\theta}(1-\frac{a^2}{r^2})$

$\text{(1) }\frac{\partial{\phi}}{\partial{r}}=U_{r}=\frac{1}{r}\frac{\partial{\psi}}{\partial{\theta}}=U_{0}\cos{\theta}(1-\frac{a^2}{r^2})$

$\text{(2) }\frac{1}{r}\frac{\partial{\phi}}{\partial{\theta}}=U_{\theta}=-\frac{\partial{\psi}}{\partial{r}}=-U_{0}\sin{\theta}(1+\frac{a^2}{r^2})$

$\text{(1) }\implies\psi=U_{0}\cos{\theta}(1+\frac{a^2}{r^2})+f(\theta)$

$\frac{1}{r}\frac{\partial{\phi}}{\partial{\theta}}=-U_{0}\frac{r\sin{\theta}}{r}(1+\frac{a^2}{r^2})+f'(\theta)\rightarrow\text{(3)}$

$\text{(2), (3) }\implies{f}'(\theta)=0\implies{f}(\theta)=c$

$\psi=U_{0}r\cos{\theta}(1+\frac{a^2}{r^2})+\not{c}$

$\text{In the following graph, we can see that the streamlines and isopots always intersect at right angles.}$

$\underline{Sources:}$

Examples of Bernoulli’s Principle

$\text{Simple Bernoulli example: Flow from a reservoir}$

$\text{Bernoulli: } \frac{1}{2}u^2+\frac{p}{\rho}+\Phi=const.$

$\text{Now }-\nabla\Phi{=}\underline{F}=\text{ body force/unit mass }=-g\underline{k}$

$\implies\Phi{=}gz+\not{c}\text{ (w.l.o.g.)}$

$\text{Bernoulli becomes: }\frac{1}{2}u^2+\frac{p}{\rho}+gz=const.$

$\text{So, }\frac{1}{2}0^{2}+\frac{p_0}{\rho}+gz_{1}=\frac{1}{2}u_{2}^{2}+\frac{p_0}{\rho}+gz_{2}$

RHS = top surface (1), LHS=tap (2)

$\implies{u_{2}}=\sqrt{2g(z_{2}-z_{1})}$

$\underline{\text{Bernoulli's equation in Potential Flow - Derivation}}$

$\text{From Euler's Equation: }\frac{D\underline{u}}{Dt}=-\frac{1}{\rho}\nabla\text{+}\underline{F}$

$\text{From Conservative Body Force: } \underline{F}=-\nabla\Phi$

$\text{From Irrotational Flow: }\nabla\times\underline{u}=\underline{0},\underline{u}=\nabla\Phi$

$\text{Euler becomes: }\frac{D(\nabla\Phi)}{Dt}=-\frac{1}{\rho}\nabla\text{p - }\nabla\Phi$

$\text{i.e. }\frac{\partial(\nabla\Phi)}{\partial\text{t}}+\nabla(\frac{1}{2}\mid{u}\mid^2)-\underline{u}\times\text{}curl(\underline{u})=-\nabla(\frac{p}{\rho})-\nabla\Phi$

$\text{Thus, }\nabla(\frac{\partial{\Phi}}{\partial{t}})+\nabla(\frac{1}{2}\mid{u}\mid^2)+\nabla(\frac{p}{\rho})+\nabla\Phi=\underline{0}$

$\nabla(\frac{\partial{\Phi}}{\partial{t}}+\frac{1}{2}\mid{u}\mid^{2}+\frac{p}{\rho}+\Phi)=\underline{0}$

$\text{So, }\frac{\partial{\Phi}}{\partial{t}}+\frac{1}{2}\mid{u}\mid^{2}+\frac{p}{\rho}+\Phi=f(t)$

This is Bernoulli’s potential flow equation.

$\underline{Note:}$

Bernoulli’s Principle is not always valid. For example, if we consider the Rankine Vortex Tornado Model, there is low pressure at the origin, where the speed is also lowest. This is because the speed is constant along the streamlines for this flow.

Bernoulli’s Principle

Bernoulli’s Principle states, that for an inviscid flow, if the speed of the fluid is increased, then the pressure or potential energy of the fluid must decrease.

$\text{High speed }\longrightarrow\text{Low pressure}$

$\text{Low speed }\longrightarrow\text{High pressure}$

$\underline{Derivation:}$

$\text{From Euler's Equation: }\frac{D\underline{u}}{Dt}=-\frac{1}{\rho}\nabla\text{+}\underline{F}$

$\text{From Conservative Body Force: } \underline{F}=-\nabla\Phi$

$\text{Euler becomes: }\frac{D\underline{u}}{Dt}=-\frac{1}{\rho}\text{p -}\nabla\Phi$

$\text{i.e. }\frac{\partial\underline{u}}{\partial\text{t}}+\nabla(\frac{1}{2}u^2)-\underline{u}\text{ x curl}(\underline{u})=-\frac{1}{\rho}\nabla\text{p}-\nabla\Phi$

$\text{For steady flow, }\nabla(\frac{1}{2}u^2)-\underline{u}\text{ x curl}(\underline{u})=-\frac{1}{\rho}\nabla\text{p}-\nabla\Phi$

$\text{Combining terms: }\nabla(\frac{1}{2}u^2+\frac{p}{\rho}+\Phi)+\underline{u}\text{ x curl}(\underline{u})$

$\text{Consider unit tangent vector }\underline{s}\text{ along streamline, then take the dot product with equation: }$

$\underline{s}\cdot\nabla(\frac{1}{2}u^2+\frac{p}{\rho}+\Phi)=\underline(s)\cdot\underline{u}\text{ x curl}(\underline{u})=0$

$\text{i.e. }\frac{d}{ds}(\frac{1}{2}u^2+\frac{p}{\rho}+\Phi)=0$

$\text{Thus, the quantity in brackets is constant along a streamline: }$

$\frac{1}{2}u^2+\frac{p}{\rho}+\Phi=\text{const.}$

$\text{This is the streamline form of Bernoulli's Equation. It can also be written as:}$

$\frac{1}{2}\rho\text{}u^2+p+\rho\text{gz = const.}$

I found that this clip helped explain Bernoulli’s Principle better, and has 6 more examples to look at:

Euler’s Equations

$\underline{\text{Cartesian Coordinates}}$

$\text{x: }\rho(\frac{\partial{u}}{\partial{t}}+u\frac{\partial{u}}{\partial{x}}+v\frac{\partial{u}}{\partial{y}}+w\frac{\partial{u}}{\partial{z}})=-\frac{\partial{p}}{\partial{x}}+\rho\text{}F_{Bx}$

$\text{y: }\rho(\frac{\partial{v}}{\partial{t}}+u\frac{\partial{v}}{\partial{x}}+v\frac{\partial{v}}{\partial{y}}+w\frac{\partial{v}}{\partial{z}})=-\frac{\partial{p}}{\partial{y}}+\rho\text{}F_{By}$

$\text{z: }\rho(\frac{\partial{w}}{\partial{t}}+u\frac{\partial{w}}{\partial{x}}+v\frac{\partial{w}}{\partial{y}}+w\frac{\partial{w}}{\partial{z}})=-\frac{\partial{p}}{\partial{z}}+\rho\text{}F_{Bz}$

$\underline{\text{Polar Coordinates}}$

$\text{r: }\rho(\frac{\partial{v_r}}{\partial{t}}+v_r\frac{\partial{v_r}}{\partial{r}}+\frac{v_{\theta}}{r}\frac{\partial{v_r}}{\partial{\theta}}-\frac{v_{\theta}^2}{r}+v_z\frac{\partial{v_r}}{\partial{z}})=-\frac{\partial{p}}{\partial{r}}+\rho\text{}F_{r}$

$\text{r: }\rho(\frac{\partial{v_{\theta}}}{\partial{t}}+v_r\frac{\partial{v_{\theta}}}{\partial{r}}+\frac{v_{\theta}}{r}\frac{\partial{v_{\theta}}}{\partial{\theta}}-\frac{v_{r}v_{\theta}}{r}+v_z\frac{\partial{v_{\theta}}}{\partial{z}})=-\frac{1}{r}\frac{\partial{p}}{\partial{\theta}}+\rho\text{}F_{\theta}$

Lets look at an example of an unsteady 2-d flow:

$\underline{u}=(2yt,-2x)\text{ ; }\underline{F}=(2y,0)$

Find the pressure p(x,y,t), using cartesian Euler’s equation:

$\text{x: }\rho(\frac{\partial{u}}{\partial{t}}+u\frac{\partial{u}}{\partial{x}}+v\frac{\partial{u}}{\partial{y}})=-\frac{\partial{p}}{\partial{x}}+\rho\text{}F_{Bx}$

$\rho(2y+2yt*0+(-2x)*2t)=-\frac{\partial{p}}{\partial{x}}+\rho*2y$

$\text{This gives us 1:}\frac{\partial{p}}{\partial{x}}=\rho\text{4xt}$

$\text{y: }\rho(\frac{\partial{v}}{\partial{t}}+u\frac{\partial{v}}{\partial{x}}+v\frac{\partial{v}}{\partial{y}})=-\frac{\partial{p}}{\partial{y}}+\rho\text{}F_{By}$

$\rho(0+2yt*(-2)+(-2x)*0)=-\frac{\partial{p}}{\partial{y}}+0$

$\text{This gives us 2: }\frac{\partial{p}}{\partial{y}}=\rho\text{4yt}$

$\text{From 1 we get 3: }p=2\rho\text{}x^{2}t+f(y,t)$

$\text{We can now say 4: }\frac{\partial{p}}{\partial{y}}=\frac{\partial{f}}{\partial{y}}$

$\text{From 2 and 4: }\frac{\partial{f}}{\partial{y}}=4\rho\text{}yt$

$\text{So we get 5: }f(y,t)=2\rho\text{}y^{2}t+g(t)$

$\text{From 3 and 5: }p(x,y,t)=2\rho\text{}t(x^2+y^2)+g(t)$

$\text{To fix g(t) we need more information. If p=0 at (x,y)=(0,0) }\forall\text{t, then g(t)=0}$

$\text{Finally : p(x,y,t)=2}\rho\text{}t(x^2+y^2)$

$\text{We recognise that this will create circular lines of constant pressure, with centre (0,0) and radius }\sqrt{\frac{p_0}{2t\rho}}$

Dimensional Analysis

Dimensional Analysis is used to identify the physical dimensions of variables in an equation. It determines the units and analyses the relationship between the quantities.

We will be looking mainly at the fundamental units Mass (M in kilogrammes), Length (L in meters) and Time (T in seconds).

Dimensional Homogeneity is where both sides of the equation must have the same dimensions.

$\text{Force, }\underline{F}\text{ depends on:}$

• $\text{Diameter, d}$
• $\text{Forward velocity of the propeller, u}$
• $\text{Fluid density, r}$
• $\text{Rotation rate in revolutions per second, N}$
• $\text{Fluid viscosity, m}$

$\text{This gives us the equation: F=}\phi\text{(d, u, r, N, m)}$

$\underline{\text{Buckingham's p theorems}}$

$\text{1st Theorem:}$

A relationship between m physical variables can be expressed as a relationship between m-n non-dimensional groups of variables (called p groups), where n is the number of fundamental dimensions required to express the variables.

$\text{2nd Theorem:}$

Each p group is a function of n governing or repeating variables plus one of the remaining variables.

Sources: