### Fluid Flow I

The Navier Stokes equation: apply Newton’s Second Law to a small ‘blob’ of fluid moving with the flow (imagine we can watch it as it goes).

$\rho \delta V \bold{a}=\delta\bold{F}$

$\rho \delta x\delta y\delta z \bold{a}=\delta\bold{F}$

Normal pressure forces in x-direction

Net pressure force in x-direction $-\delta x\frac{\partial p}{\partial x}\delta A=-\delta V\frac{\partial p}{\partial x}$

Vector pressure force $\delta F_p=-(\delta V)\nabla p$

Viscous forces: take bulk fluid component u in the x-direction- assume it is independent of x and y.

Tangential viscous stress $\tau=\mu\frac{\partial u}{\partial z}$

μ is the dynamic viscosity (also known as η in kinetic theory, just to stop you getting complacent).

Net contribution to viscous force on blob

$\delta F_V= \frac{\delta\tau}{\delta z}\delta V=\mu \frac{d^2u}{dz^2}\delta V$

Including x and y variation

$\delta F_V=\delta V \mu (\nabla^2\bold{u}+\frac{1}{3}\nabla(nabla.\bold{u}))$

$\delta F_V=\delta V \mu\nabla^2\bold{u}$ when $latex(nabla.\bold{u})=0$ (incompressible)

Gravity force: mg downwards $\delta F_g=-\rho \delta Vg\bold{k}$

Putting it all together-

acceleration $\bold{a}=-\frac{1}{\rho}\nabla p+\nu\nabla^2u-g\bold{k}$

where $\nu=\frac{\mu}{\rho}$ is the kinematic viscosity.

Now to get the acceleration in terms of u:

The Lagrangian view is to follow the blob as it moves with the flow. The blob’s position is $\bold{r}(t)$, velocity $\frac{d\bold{r}}{dt}$, acceleration $\frac{d^2\bold{r}}{dt^2}$.

Of course, it’s hardly easy in practice to sit and watch a blob, so it’s also useful to have another way of thinking- the Eulerian picture.

Consider a time-varying velocity field referred to fixed points in space $\bold{u}(x,y,z,t)$

Langrangian velocity $\frac{d\bold{r}}{dt}=\bold{u}$ Eulerian velocity

Differentiate wrt t

$\bold{a}=\frac{d^2\bold{r}}{dt^2}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial t}+\frac{\partial u}{\partial z}\frac{\partial z}{\partial t}+\frac{\partial u}{\partial t}$

$\bold {a}=u\frac{\partial u}{\partial x}+ v\frac{\partial u}{\partial y}+ w\frac{\partial u}{\partial z}+\frac{\partial u}{\partial t}$

$\bold {a}=\frac{\partial u}{\partial t}+\bold{u}.\nabla u$

$\bold {a}=\frac{D\bold{u}}{Dt}$

where $\frac{D}{Dt}=\frac{\partial}{\partial t}+(\bold{u}.\nabla)$ is the material or advective derivative (see below).

Putting it all together, we finally get the Navier-Stokes equation for viscous fluid flow under gravity-

$\frac{D\bold{u}}{Dt}+\frac{1}{\rho}\nabla p+ g\bold{k}=\nu\nabla^2u$

Material derivative: the time rate of change following the motion of a fluid blob, rather than the rate of change at a fixed point ($\frac{\partial}{\partial t}$). Since $(\bold{u}.\nabla)$ is a non-linear term, this makes the maths harder, and in three dimensions, the non-linearity allows for chaos (the mathematical kind, rather than the realisation that you can’t work out the equations).

Mass conservation continuity equation:

Net mass inflow per unit time = rate of increase of mass in container

Net mass inflow in x-direction per unit time

= $(\rho (x)u(x)-\rho (x+ \Delta x)u(x+ \Delta x)) \Delta y \Delta z$

~ $-\frac{\partial}{\partial x}(\rho u)\Delta x\Delta y\Delta z$

= $-\frac{\partial}{\partial x}(\rho u)\Delta V$ Taylor expansion

Including y and z inflow -> total inflow = $-\nabla.(\rho \bold{u})\Delta V$

Rate of increase of mass in container = $\frac{\partial}{\partial t}(\rho\Delta V)$

For a fixed container, ΔV=constant.

$\therefore \frac{\partial\rho}{\partial t}+\nabla.(\rho\bold{u})=0$

$\frac{\partial\rho}{\partial t}+\bold{u}.\nabla\rho+\rho\nabla.\bold{u}=0$ using a vector identity

$\frac{D\rho}{Dt}+\rho\nabla.\bold{u}=0$

Following a blob of incompressible density, $\nabla.\bold{u}=0$. Incompressibility is a good approximation for liquids and gases where $\bold{u}<, the speed of sound. Under incompressible conditions, sound waves can be neglected.

Rectilinear flow between parallel plates

$\bold{u}=(u,0,0)$ in x-direction

$\nabla.\bold{u}=\frac{\partial u}{\partial x}=0 \Rightarrow u=u(y,z,t)$

Now use the Navier-Stokes equation, but neglect gravity.

$\frac{\partial \bold{u}}{\partial t}=(\bold{u}.\nabla)\bold{u}=-\frac{1}{\rho}\nabla\rho+\frac{\mu}{\rho}\nabla^2\bold{u}$

x-component: $\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}+w\frac{\partial u}{\partial z}=-\frac{1}{\rho}\frac{\partial p}{\partial x}+\frac{\mu}{\rho}\nabla^2u$

-> $\frac{\partial u}{\partial t}=-\frac{1}{\rho}\frac{\partial p}{\partial x}+\frac{\mu}{\rho}\nabla^2u$ as the other terms are zero.

If we take the x-derivative of the above, all terms involving u must vanish, because $\frac{\partial u}{\partial x}=0$

So $\frac{-1}{\rho}\frac{\partial^2p}{\partial x^2}=0$ and $\frac{\partial p}{\partial x}$ is independent of x.

y and z components: $\frac{\partial p}{\partial y}=\frac{\partial p}{\partial z}=0 \Rightarrow p=p(x,t)$

Now make the further assumption that the flow is steady and y-independent.

$\frac{\partial u}{\partial t}=0$ (steady) and $\frac{\partial u}{\partial y}=0 \Rightarrow u=u(z)$

For steady flow, p is also independent of t, so $\frac{dp}{dx}$ is a constant.

Define $G=-\frac{dp}{dx}$, the pressure gradient.

Putting this all into the original x-component equation, we end up with

$-G=\mu\frac{\partial^2u}{\partial z^2}$

Boundary conditions: for viscous flow, the no slip condition means that the fluid comes to rest at the walls, i.e. u=0 when z=±h.

$\Rightarrow u=(\frac{G}{2\mu})(h^2-z^2)$

We get a parabolic flow profile- an example of Poiseulle flow.

Volume flux per unit y distance for the above flow is $\int_{-h}^h udz=\frac{2}{3}\frac{Gh^3}{\mu}$

Flow in a circular pipe

Assume flow is independent of θ and t -> u=u(r)

$\mu\frac{1}{r}\frac{d}{dr}(r\frac{du}{dr}=-G$ taking the radial component of del-squared.

Boundary conditions: u=0 at r=a (pipe wall- no slip condition again) and no singularity at r=0.

$\frac{d}{dr}(r\frac{du}{dr})=\frac{-Gr}{\mu}$

$r\frac{du}{dr}=\frac{-Gr^2}{2\mu}+const$ putting r=0 -> const=0

$\frac{du}{dr}=\frac{-Gr}{2\mu}$

$u=\frac{-Gr^2}{4\mu}+const$

As u=0 at r=a, const=$\frac{ga^2}{4\mu}$

$\Rightarrow u=\frac{G}{4\mu}(a^2-r^2)$

Volume flux=$\int_0^a u 2\pi rdr=\frac{\pi Ga^4}{8\mu}$

Average speed=$\frac{vol flux}{\pi a^2}=\frac{Ga^2}{8\mu}=\bar{u}$

Reynolds number, Re, is a dimensionless number used to quantify the importance of viscosity. We do this by considering the ratio of inertial forces to viscous forces, and then simplifying with scale analysis.

$\frac{inertial}{viscous}=\frac{|(\bold{u}.\nabla)\bold{u}|}{|\nu\nabla^2\bold{u}|} \sim \frac{U\times U/L}{\nu U/L^2}=\frac{UL}{\nu}=Re$

where U=typical velocity scale, L=typical length scale of variation

If Re>>1, viscosity is usually unimportant

If Re<<1, viscosity is usually important

For Re>Recrit~1000, flow becomes turbulent (rectilinear flow is unstable). Below this value, the flow remains rectilinear.

Flow between rotating cylinders

(ir, iθ and iz are unit vectors in the respective directions r, θ and z.)

Inner cylinder is fixed.

Outer cylinder rotates with angular velocity Ω.

We want steady, incompressible flow with circular symmetry $\bold{u}=u_{\theta}(r)\bold{i}_{\theta}$

Neglect end effects and say there is no variation in z-direction.

Take the Navier-Stokes equation in cylindrical polar coordinates, neglecting gravity.

$(\bold{u}.\nabla)\bold{u}+\frac{1}{\rho}\nabla p=\nu \nabla^2 \bold{u}$

->

$\frac{-u_{\theta}^2}{r}\bold{i}_r+\frac{1}{\rho}\frac{dp}{dr}\bold{i}_r=\nu(\frac{d^2u_{\theta}}{dr^2}+\frac{1}{r}\frac{du_{\theta}}{dr}-\frac{u_{\theta}}{r^2})\bold{i}_\theta$

The two terms on the LHS are centrifugal and pressure gradient terms respectively, whilst the RHS terms are viscous terms.

ir and terms vanish separately.

For ν≠0

$\frac{d^2u_{\theta}}{dr^2}+\frac{1}{r}\frac{du_{\theta}}{dr}-\frac{u_{\theta}}{r^2}=0$

Try $u_{\theta}=r^n$

$\frac{du_{\theta}}{dr}=nr^{n-1}$

$\frac{d^2u_{\theta}}{dr^2}=n(n-1)r^{n-2}$

-> $n(n-1)r^{n-2}+nr^{n-2}-r^{n-2}=0$

$n^2=1$

$n=\pm 1$

-> $Ar+\frac{B}{r}$

Boundary conditions: uθ=0 at r=a and uθ=Ωb at r=b

At r=a: $0=Aa+\frac{B}{a}$

$A=\frac{-B}{a^2}$

At r=b: $\Omega b=Ab+\frac{B}{b}$

$\Omega b=\frac{-Bb}{a^2}+\frac{B}{b}$

$\Omega a^2b^2=B(a^2-b^2)$

$B=\frac{\Omega a^2b^2}{a^2-b^2}$

$A=\frac{-\Omega b^2}{a^2-b^2}$

We can now work out the torque needed to keep the inner cylinder at rest.

Tangential viscous stress on inner cylinder:

$\tau=\mu[r\frac{d}{dr}(\frac{u_{\theta}}{r})]_{r=a}$ per unit area of cylinder

Torque per unit z-distance on inner cylinder:

$T=\tau \times 2\pi a \times 1 \times a =\frac{4\pi a^2 b^2 \Omega \mu}{b^2-a^2}$

This can be used as a way of measuring μ (viscometer)

As Ω increases, laminar flow breaks down and we see waves and turbulence.

Finally, to find the pressure distribution, use $\frac{1}{\rho}\frac{dp}{dr}=\frac{u_{\theta}^2}{r}$

$\frac{1}{\rho}\frac{dp}{dr}=(Ar+\frac{B}{r})^2)\frac{1}{r}$

$\frac{1}{\rho}\frac{dp}{dr}=A^2r+\frac{2AB}{r}+\frac{B^2}{r^3}$

$\frac{1}{\rho}p(r)=\frac{A^2r^2}{2}+2AB\ln r-\frac{B^2}{2r^2}+const$

Vorticity, w, is the curl of the velocity field

$\bold{w}(\bold{r},t)=\nabla \times \bold{u}(\bold{r},t)$

Vorticity is the measure of the local ‘spin’ or ‘rotation’ of the fluid, but it is not in general a measure of large scale rotation.

Note: for 2D flow in the xy plane, w is of course in the z-direction.

a) Solid body’ rotation

$\bold{u}=\Omega r \bold{i}_{\theta}=(-\Omega y, \Omega x, 0)$

$\bold{w}=(0, 0, 2\Omega)$

Imagine dropping a ‘vorticity meter’ into the flow

$|\bold{u}| \propto r$ flow gets faster as r increases, meter spins.

b) Line vortex’

$\bold{u}=\frac{A}{r}\bold{i}_{\theta}=(\frac{-Ay}{x^2+y^2},\frac{Ax}{x^2+y^2}, 0)$

$\bold{w}=0$ for r≠0 -> flow is irrotational.

$|\bold{u}| \propto r^{-1}$ flow gets slower as r increases, in just such a way that the meter doesn’t spin.

c) Linear shear flow

$\bold{u}=(\beta y, 0, 0)$

No large scale rotation, but $\bold{w}=-\beta$ so there is local ‘spin’.

The vorticity equation: start with the Navier-Stokes equation, taking an incompressible fluid with ρ and ν constant.

$\frac{\partial \bold{u}}{\partial t}+(\bold{u}.\nabla) \bold{u}=-\frac{1}{\rho}\nabla \rho-g\bold{k}+\nu \nabla^2 \bold{u}$

$\frac{\partial \bold{u}}{\partial t}+(\bold{u}.\nabla) \bold{u}=-\nabla(\frac{p}{\rho}+gz)+ \nu \nabla^2 \bold{u}$

$\nabla.\bold{u}=0$

Vector identity $(\bold{u}.\nabla) \bold{u}=\frac{1}{2}\nabla|\bold{u}|^2-\bold{u}\times(\nabla\times \bold{u})$

$\frac{\partial \bold{u}}{\partial t}+\nabla(\frac{1}{2}|\bold{u}|^2+\frac{p}{\rho}+gz)-\bold{u}\times \bold{w}=\nu \nabla^2 \bold{u}$

Now take the curl of this equation, remembering that $\nabla \times (\nabla \phi)=0$ always and $\nabla \times (\nabla^2 \bold{u})=\nabla^2(\nabla \times \bold{u})$

$\frac{\partial}{\partial t}(\nabla \times \bold{u})-\nabla \times (\bold{u} \times \bold{w})=\nu\nabla^2(\nabla \times \bold{u})$

Vector identity $\nabla \times (\bold{u} \times \bold{w}=(\bold{w}.\nabla)\bold{u}-(\bold{u}.\nabla)\bold{w}-\bold{w}(\nabla.\bold{u})+\bold{u}(\nabla.\bold{w})$

The last two terms must vanish because we already established that $\nabla.\bold{u}=0$ and of course $\nabla.\bold{w}=\nabla.(\nabla \times \bold{u})=0$

With all this in mind, we get

$\frac{\partial \bold{w}}{\partial t}+(\bold{u}.\nabla)\bold{w}-(\bold{w}.\nabla)\bold{u}=\nu\nabla^2\bold{w}$

$\frac{D\bold{w}}{Dt}-(\bold{w}.\nabla)\bold{u}=\nu\nabla^2\bold{w}$ the vorticity equation.

Conservation of vorticity in 2D inviscid flow: for inviscid flow, ν=0 (Re=∞).

Considering 2D flow, u=(u, v, 0), w=(0, 0, w)

$\nabla^2 \bold{w}=0$

$(\bold{w}.nabla)\bold{u}=0$

So $\frac{Dbold{w}}{Dt}=0$

So, following the motion of a fluid blob, the vorticity of each blob is conserved. In particular, if the fluid is irrotational everywhere at t=0, then it remains irrotational for t>0$2D inviscid irrotational flow $\bold{u}=(\begin{array}{c} u(x,y,t)//v(x,y,t)// 0 \end{array})$ $\bold{w}=(\begin{array}{c} 0 // 0 // w \end{array})$ Irrotational, so $w=\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}=0$ Because $\nabla \times \bold{u}=0$, we can say that u is the gradient of some velocity potential φ $\bold{u}=\nabla \phi$; $u=\frac{\partial \phi}{\partial x}$; $u=\frac{\partial \phi}{\partial y}$ Incompressible, so $\nabla.\bold{u}=\nabla^2\phi=0$ Laplace’s equation. Bernoulli’s equation in 3D (take y-axis as ‘up’) Go back to $\frac{\partial \bold{u}}{\partial t}+\nabla(\frac{1}{2}|\bold{u}|^2+\frac{p}{\rho}+gy)-\bold{u}\times(\nabla \times \bold{u})=\nu \nabla^2 \bold{u}$ Put ν=0, w=0 (inviscid, irrotational flow), so that we can say $\bold{u}=\nabla \phi$$\latex \frac{\partial}{\partial t}(\nabla \phi)+\nabla(\frac{1}{2}|\bold{u}|^2+\frac{p}{\rho}+gy)=0\$

$\nabla(\frac{\partial \phi}{\partial t}+\frac{1}{2}|\bold{u}|^2+\frac{p}{\rho}+gy)=0$

i.e. the contents of the bracket on the LHS must be independent of x, y, z, so

$\frac{\partial \phi}{\partial t}+\frac{1}{2}|\bold{u}|^2+\frac{p}{\rho}+gy=F(t)$

But we can absorb F(t) into φ (remember, you can arbitrarily add constants to a potential without changing the gradient- all we care about is that $\bold{u}=\nabla \phi$

So put $\phi \rightarrow \phi -\int_0^t F(t')dt$

So now $\frac{\partial \phi}{\partial t}+\frac{1}{2}|\bold{u}|^2+\frac{p}{\rho}+gy=0$ Bernoulli’s equation.

We can also set the RHS equal to a constant other than 0.

2D inviscid, irrotational flow around a cylinder

Put a cylinder into a uniform flow (U, 0, 0); how does the flow go around the cylinder?

$\bold{u}=u_r\bold{i}_r+u_\theta\bold{i}_{\theta}=\nabla \phi$

$\nabla^2 \phi =0$

In cylindrical polar coordinates, Laplace’s equation becomes

$\frac{1}{r}\frac{\partial}{\partial r}r\frac{\partial \phi}{\partial r}+\frac{1}{r^2}\frac{\partial^2 \phi}{\partial \theta^2}=0$

Boundary conditions ur=0 on r=a (normal velocity=0) and uθ≠0 at r=a (because there is no viscosity, the fluid can slip on the surface).

As $r \rightarrow \infty$, $\phi \rightarrow Ux=Ur\cos \theta$ and $\nabla \phi \rightarrow U+\bold{i}$

All solutions must have the same θ dependence, so the solutions of Laplace’s equation that we need are

$\phi=U(r+\frac{a^2}{r})cos \theta$ for r≥a

So

$u_r=\frac{\partial \phi}{\partial r}=U(1-\frac{a^2}{r^2})\cos \theta$

$u_{\theta}=\frac{1}{r}\frac{\partial \phi}{\partial \theta}=-U(1+\frac{a^2}{r^2})\sin \theta$

Using Bernoulli’s equation, with $\frac{\partial \phi}{\partial t}=0$ and g=0

On cylinder $\frac{p}{\rho}=const-2U^2\sin^2\theta$

The constant is usually set so that as $r \rightarrow \infty$, $p \rightarrow p_{\infty}$

The pressure on the cylinder has ‘left-right’ and ‘up-down’ symmetry (i.e. we can replace θ with π-θ or –θ and nothing changes), so because the flow is inviscid, there is no pressure induced drag on the cylinder.

Aerodynamic extension: now add an asymmetric ‘swirl’ term to the potential we found earlier.

$\phi=U(r+\frac{a^2}{r})\cos \theta-A\theta$ A>0

Now $u_{\theta}=- U(1+\frac{a^2}{r^2})\sin \theta-\frac{A}{r}$

Where the last term is like a line vortex, leading to additional clockwise swirl but zero vorticity for r≠0

Now, on the cylinder

$\frac{p}{\rho}=const-2U^2\sin^2\theta-\frac{2UA}{a}\sin\theta$

There is still ‘left-right’ symmetry, but not ‘up-down’ symmetry, so we have a lift force (pressure is greater below the x-axis than above).

The streamfunction: because of incompressibility, $\nabla.\bold{u}=0$, i.e. $\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0$ in 2D.

We can write $u=\frac{\partial \psi}{\partial x}$ and $v=-\frac{\partial \psi}{\partial y}$

$\bold{u}=\nabla \times (\psi\bold{k})$

$\bold{u}=\nabla\psi \times \bold{k}$

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

$w=-(\frac{\partial^2 \psi}{\partial x^2}+\frac{\partial^2 \psi}{\partial y^2})$

For an irrotational flow, we get Laplace’s equation for ψ as well as φ.

$\frac{\partial^2 \psi}{\partial x^2}+\frac{\partial^2 \psi}{\partial y^2}=0$

For our standard 2D flow example, $\bold{u}.\nabla\psi=0$, so the velocity is parallel to the lines of constant ψ (and perpendicular to the lines of constant φ).