### Fluid Flow II

Streamlines for irrotational flow around a cylinder: plotted for various values of the swirl B=UaA

B=0 B=1

B=2 B=3

Streamlines around an aerofoil: if we start with the flow around a cylinder and use a technique called conformal mapping (involving various coordinate transformations), we can calculate the flow around other shapes. We’re not going to go into this in detail here- instead we’ll just look it schematically for an aerofoil.

Boundary layers: even when ν is small, the viscosity is still important in a thin ‘boundary layer’ near a solid surface.

$Re=\frac{UL}{/nu} \rightarrow \infty$

Thickness of boundary layer ~ $\frac{L}{\sqrt{Re}}$

Boundary layer separation: in practice, low viscosity flows around objects are not quite as smooth as those shown above. The boundary layer tends to break away from the object, giving rise to a turbulent wake, as shown schematically below.

Turbulent wake -> dissipation, drag, inefficiency…

Aerofoil shapes are designed to give separation at the trailing edge only, so that they are more efficient.

Circulation, Γ: circulation ΓC around a closed curve C lying within a fluid is defined as the line integral

$\Gamma_C=\oint \bold{u}.d\bold{l}$

$\Gamma_C=\int_S (\nabla \times \bold{u}).d\bold{S}$ Stokes’ theorem

$\Gamma_C=\int_S \bold{w}.\bold{S}$

S is a surface enclosed by C; it must lie wholly within the fluid (i.e. same conditions everywhere) for Stokes’ theorem to apply/

For an irrotational flow, ΓC=0.

If we have inviscid swirling vortex flow around a cylinder of radius a, $\bold{u}=\frac{B}{r}\bold{i}_{theta}$, then w=0, but

$\Gamma_C=\oint \bold{u}.d\bold{l}=\int_0^{2\pi}\frac{B}{r}rd\theta=2\pi B \neq 0$

This may seem contradictory, but it’s actually fine as the irrotational flow is not across the whole surface bounded by C (we have the cylinder in there too), so our Stokes’ theorem arguments no longer hold.

Lift force

Normal force on a unit length element of cylinder adθ $F=-pad\theta$

Lift force is in y-direction $dL=-pa\sin\theta d\theta$

Use the potential we used in the aerodynamic extension’ section:

$\phi =U(r+\frac{a^2}{r})\cos \theta-A\theta$ (the last term is clockwise swirl)

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

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

On cylinder

$\frac{p}{\rho}=const-2U^2\sin^2\theta-\frac{2UA}{a}\sin\theta$ from Bernoulli’s equation

Total lift force for a unit length of cylinder (at r=a)

$L=-\int_0^{2\pi}pa\sin \theta d \theta$

$L=-\int_0^{2\pi} ( const \times \rho a \sin \theta - 2U^2\rho a \sin^3 \theta - 2UA \rho \sin^2 \theta ) d\theta$ but of course the first two terms in the integral will just integrate to zero, making life much easier for us.

So $L=2UA\pi\rho$

Circulation $\Gamma_C=\oint_C \bold{u}.d\bold{l}$

$\Gamma_C=\int_0^{2\pi}u_{theta}a d\theta$ taking uθ at r=a

$\Gamma_C=\int_0^{2\pi} (-2Ua\sin \theta-A) d\theta$

$\Gamma_C=-2\pi A$

-> $L=-\rho U \Gamma_C$

Travelling waves on deep water (2D): still assuming inviscid, irrotational, incompressible flow

Once again, put $\bold{u}=\nabla \phi$

$latex \frac{\partial^2 \phi}{\partial x^2}+\frac{\partial^2 \phi}{\partial y^2}=0 We’ll also need Bernoulli’s equation $\frac{\partial \phi}{\partial t}+\frac{1}{2}|\bold{u}|^2+\frac{p}{\rho}+gy=0$ At the air-water interface, take p=p­0=const (1 atm) Suppose |u|, φ, η are all ‘small’ so that terms like |u|2 are negligible. $\frac{\partial \phi(\eta, x, t}{\partial t}+\frac{1}{2}|\bold{u}|^2+\frac{p_0}{\rho}+g \eta=0$ Remember, constants like $\frac{p_0}{\rho}$ can be gotten rid of by adding $\int_0^t const dt'$ to φ (we already discussed this earlier). Also, we can do a Taylor expansion on $(\frac{\partial \phi}{\partial t})_{y=\eta}$ $(\frac{\partial \phi}{\partial t})_{y=\eta}=(\frac{\partial \phi}{\partial t})_{y=0}+\ldots$ (all higher order terms are negligible). $\frac{\partial \phi}{\partial t}+g\eta=0$ at y=0 Also, at y=η $\frac{\partial y}{\partial t}=\frac{\partial \eta}{\partial t}$ but also $\frac{\partial y}{\partial t}=v=\frac{\partial \phi}{\partial y}$ Then, doing a Taylor expansion and saying that these variable terms are approximately the same at y=0 as they are at y=η, we get $\frac{\partial \eta}{\partial t}=\frac{\partial \phi}{\partial y}=v$ Now look for a travelling wave solution $\eta(x,t)=A\cos(kx-\bar{\omega}t)$ By convention, k>0; $\bar{\omega}$ is called ‘pomega’. Try $\phi(x,y,t)=f(y)\sin(kx-\bar{\omega}t)$ $\frac{\partial^2 \phi}{\partial x^2}=-\frac{\partial^2 \phi}{\partial y^2}$ $k^2f=f''$ $f(y)=Ce^{ky}+ De^{-ky}$ f is bounded as $y \rightarrow \infty$, so D=0 $\phi=Ce^{ky}\sin(kx-\bar{\omega}t)$ $\frac{\partial \phi}{\partial t}=-g\eta$ and $\frac{\partial \eta}{\partial t}=\frac{\partial \phi}{\partial y}$ at y=0 Which gives us $C\bar{\omega}=gA$ and $A\bar{\omega}=Ck$ respectively. -> $C=\frac{A\bar{\omega}}{k}$ $\bar{\omega}^2=gk$ dispersion relation Now we can calculate the path of fluid blobs. $\bold{u}=\nabla \phi=(\frac{\partial X}{\partial t}, \frac{\partial Y}{\partial t}, 0)$ Where X, Y, 0 are displacements from the mean positions. $\frac{\partial X}{\partial t}=\frac{\partial \phi}{\partial x}$ $\frac{\partial X}{\partial t}=\frac{A\bar{\omega}}{k}e^{ky}k\cos(kx-\bar{\omega}t)$ $X=Ae^{ky}\sin(kx-\bar{\omega}t)$ ignore the constant of integration because we only care about displacements from the mean and no absolute positions. Similarly $\frac{\partial Y}{\partial t}=\frac{\partial \phi}{\partial y}$ $\frac{\partial Y}{\partial t}=A\bar{\omega}e^{ky}\sin(kx-\bar{\omega}t)$ $Y=-Ae^{ky}\cos(kx-\bar{\omega}t)$ Circles, radii $r \propto e^{ky}$ Phase speed $c_p=\frac{\bar{\omega}}{k}=\sqrt{\frac{g}{k}}$ Group velocity $c_g=\frac{d\bar{\omega}}{dk}=\frac{d}{dk}(\sqrt{gk})\frac{1}{2}\sqrt{\frac{g}{k}}=\frac{1}{2}c_p$ Crests and troughs moving faster than packet, so they advance through the packet; if there is a mix of wavelengths, longer waves appear towards the outer edge. Very viscous (Stokes) flow: going back to the Navier-Stokes equation, neglect g, assume steady flow, take ρ, ν constant -> $\bold{u}.\nabla \bold{u}+\frac{1}{\rho}\nabla p=\nu \nabla^2 \bold{u}$ We can neglect the non-linear term $\bold{u}.\nabla \bold{u}$ compared with the viscous term when the Reynolds Number $Re=\frac{UL}{\nu}\ll 1$ i.e. when viscosity is ‘large’ ν>>UL, length scales are ‘small’ L<< ν/U, velocities are ‘small’ U<< ν/L, or any combination of the above. Neglecting the non-linear term gives $\nabla p=\mu \nabla^2 \bold{u}$ (1) And since the flow is incompressible $\nabla.\bold{u}=0$ These are the Stokes flow equation. Now take the div of (1) $\nabla^2 p=\mu\nabla^2(\nabla.\bold{u})=0$ where we have used the vector identity $\nabla.(\nabla^2 \bold{u})=\nabla^2(\nabla.\bold{u})$ Now we have Laplace’s equation for pressure, but actually this isn’t terribly useful since we don’t usually know the boundary conditions on p. Now let’s take the curl of (1) $\nabla \times (\nabla p)=\mu \nabla^2(\nabla \times \bold{u})$ $0=\mu \nabla^2 \bold{w}$ $\nabla^2 \bold{w}=0$ as μ≠0 Here we have Laplace’s equation for w; this is actually three equations- one for each component. Reversibility: the solutions of the Stokes flow equations are reversible- if u(r) is a possible flow, then so is u(-r), provided the boundary conditions allow it (p-> –p+const to reverse the pressure gradient). For example, if a fish tries to swim in treacle, then if a flap of the tail to the right drives it forward, a flap to the left drives it back again. Therefore, it’s just as well that fish aren’t usually found trying to swim in treacle. Stokes flow past a sphere: consider a very viscous fluid moving past a stationary sphere of radius a. A long way from the sphere, the flow velocity is equal to Uk. If $Re=\frac{Uu\rho}{\mu}\ll 1$, we expect the Stokes flow equations to hold. Boundary conditions: u=0 on the sphere u=Uk at large distances from the sphere $p \rightarrow p_{\infty}$ (const) far from the sphere To get a formula for the drag, we can use dimensional analysis. Assume dependence of drag, D, goes as $D=\kappa \mu^{\alpha}a^{\beta}U^{\gamma}$ (l does not appear in the Stokes flow equations). Dimensions (in terms of mass, length and time) $D=MLT^{-2}$ $\mu=ML^{-1}T^{-1}$ $a=L$ $U=LT^{-1}$ Equate dimensions $MLT^{-2}=( ML^{-1}T^{-1})^{\alpha}L^{\beta}(LT^{-1})^{\gamma}$ Mass: $1=\alpha$ Length: $1=-\alpha+\beta+\gamma$ Time: $-2=-\alpha=\gamma$ So α, β and γ are all equal to 1 and drag $D=\kappa \mu a U$ We can actually go further and find the constant κ, but it does take a fair amount of calculation. Here is an outline of the procedure. Working in spherical polars, assume axisymmetry- $\frac{\partial}{\partial \phi}=0$, latex u_{\phi}=0$

Boundary conditions

$u_r=u_{\theta}=0$ at r=0

$u_r=U\cos\theta$, $u_{\theta}=U\sin\theta$ as $r \rightarrow \infty$

Plug through the algebra, and we get

$u_r=U\cos\theta(1-\frac{3a}{2r}+\frac{a^3}{2r^3})$

$u_{\theta}=-U\sin\theta(1-\frac{3a}{4r}+\frac{a^3}{4r^3})$

$p-p_{\infty}=-\frac{3}{2}\mu Ua\frac{\cos\theta}{r^2}$

p is a ‘dipole’ solution of Laplace’s equation, but we can’t see this by inspection because we don’t know the boundary condition on p at r=a in advance.

For $\cos\theta<0$, $p>p_{\infty}$ (downstream)

For $\cos\theta>0$, $p (upstream)

Vorticity is in the +ve φ direction.

Viscous stresses on the sphere

Components of stress

$\tau_r=-p+2\mu\frac{\partial u_r}{\partial r}$

$\tau_{\theta}=\mu\frac{\partial}{\partial r}(\frac{u_{\theta}}{r})+\frac{\mu}{r}\frac{\partial u_r}{\partial \theta}$

$\tau_{\phi}=0$

At r=a

$\tau_r=-p_{\infty}+\frac{3}{2}\frac{\mu U}{a}\cos\theta$

$\tau_{\theta}=-\frac{3}{2}\frac{\mu U}{a}\sin\theta$

The problem is symmetric in all directions except the direction of U (taken to be the z-direction), so the only component which doesn’t integrate to zero is

$\tau_z=\tau_r\cos\theta-\tau_{\theta}\sin\theta=-p_{\infty}\cos\theta+\frac{3\mu U}{2a}$

Total drag on sphere

$D=\int\int\tau_z dS$

$D=\int_0^{2\pi}\int_0^{\pi}( - p_{\infty}\cos\theta+\frac{3\mu U}{2a})a^2\sin\theta d\theta d\phi$

$D=2\pi a^2 \int_0^{\pi}(-p_{\infty}\cos\theta\sin\theta+\frac{3\mu U}{2a}\sin\theta)d\theta$

$D=2\pi a^2[-\frac{1}{2}p_{\infty}\sin^2\theta-\frac{3\mu U}{2a}\cos\theta]_0^{\pi}$

$D=2\pi a^2(\frac{3\mu U}{a})$

$D=6\pi \mu a U$ Stokes Law

So the constant κ=6π.

Using Stokes Law: drop a ball-bearing of density ρb into treacle. At terminal velocity

$6\pi \mu a U=\frac{4}{3}\pi a^3(\rho_b-\rho)$

i.e. Drag=gravity-upthrust

Flow in a thin film of fluid e.g. lubricating oil between two solid surfaces.

In x, y plane, horizontal flow speed ~ U, horizontal length scale ~ L

u=(u, v, w)

If h<<L, we have a ‘thin film’

$\frac{\partial^2 u}{\partial z^2} \sim \frac{U}{h^2}$

$\frac{\partial^2 u}{\partial x^2} \sim \frac{U}{L^2} \ll \frac{\partial^2 u}{\partial z^2}$

$latex \nabla^2 u= \frac{\partial^2 u}{\partial x^2}+ \frac{\partial^2 u}{\partial y^2}+ \frac{\partial^2 u}{\partial z^2} \simeq \frac{\partial^2 u}{\partial z^2} \sim \frac{U}{h^2} The fluid is incompressible, so $\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z}=0$ -> $\frac{W}{h}\sim\frac{U}{L}$ and $W \sim \frac{Uh}{L} \ll U$ $(\bold{u}.\nabla)u=(u\frac{\partial}{\partial x}+ v\frac{\partial}{\partial y}+ w\frac{\partial}{\partial z})u \sim \frac{U^2}{L}+\frac{UW}{h} \sim \frac{U^2}{L}$ (ignoring numerical prefactors) Compare inertial and viscous terms $\frac{|(\bold{u}.\nabla)u|}{|\nu\nabla^2u|} \sim(\frac{UL}{\nu})(\frac{h}{L})^2=Re(\frac{h}{L})^2$ We can neglect non-linear terms if Re(\frac{h}{L})^2 \ll 1$; this is less stringent than $Re\ll 1$ since we’ve already assumed h<<L.

Similarly, $(\bold{u}.\nabla)v$ and $(\bold{u}.\nabla)w$ are small, so we can neglect $(\bold{u}.\nabla)\bold{u}$, $\frac{\partial^2 \bold{u}}{\partial x^2}$ and $\frac{\partial^2 \bold{u}}{\partial y^2}$

Our Stokes flow equation from earlier becomes $\nabla p=\mu$latex \frac{\partial^2 \bold{u}}{\partial z^2} \$ (1)

$\frac{\partial p}{\partial z} \sim \frac{\mu w}{h^2}$

$\frac{\partial p}{\partial x}, \frac{\partial p}{\partial y} \sim \frac{\mu U}{h^2} \gg \frac{\partial p}{\partial z}$

So we can neglect vertical pressure gradients and say $p\simeq p(x,y)$, i.e. pressure is a function of x and y only.

x-component of (1) gives $\mu \frac{\partial^2 u}{\partial z^2}=\frac{\partial p(x,y)}{\partial x}$

Integrate to get

$\frac{\partial u}{\partial z}=\frac{1}{\mu}\frac{\partial p}{\partial x}z+f(x,y)$

$u=\frac{1}{2\mu}\frac{\partial p}{\partial x}z^2+fz+g(x,y)$

f and g are arbitrary functions of x and y (functions of integration).

Similarly, the y-component of (1) gives $\mu \frac{\partial^2 v}{\partial z^2}=\frac{\partial p(x,y)}{\partial y}$

$v=\frac{1}{2\mu}\frac{\partial p}{\partial y}z^2+f'z+g'$

The slider bearing

Flat boundary at z=0 moves with speed U past a fixed block with a thin film between.

Take this to be a 2D problem: u=(u,0,w); p=p(x)

At z=0, u=U -> U=g (remember that’s the function of integration g, not acceleration due to gravity)

At z=h, u=0 -> $0=\frac{1}{2\mu}\frac{dp}{dx}h^2+fh+U$

$f=-\frac{h}{2\mu}\frac{dp}{dx}-\frac{U}{h}$

So $u(x,z)=[\frac{U}{h}-\frac{z}{2\mu}\frac{dp}{dx}](h-z)$

Volume flux $Q=\int_0^{h(x)}udz$

As the fluid is incompressible, this must be independent of z.

$Q=const=\int_0^{h(x)}U-\frac{Uz}{h}-\frac{zh}{2\mu}\frac{dp}{dx}+\frac{z^2}{2\mu}\frac{dp}{dx} dz$

$Q=Uh-\frac{Uh}{2}-\frac{h^3}{4\mu}\frac{dp}{dx}+\frac{h^3}{6\mu}\frac{dp}{dx}$

$Q=\frac{Uh}{2}-\frac{h^3}{12\mu}\frac{dp}{dx}$

Now let’s try to find an expression for the pressure, starting with

$\frac{dp}{dx}=\frac{12\mu}{h^3}(\frac{1}{2}Uh-Q)$

Integrate wrt x, using boundary conditions p=p0 at the edges of the block (x=0, x=L).

$p-p_0=U\int_0^x\frac{dx'}{h^2}-2Q\int_0^x\frac{dx'}{h^3}$

If x=L, p=p0 and

$0= U\int_0^L\frac{dx'}{h^2}-2Q\int_0^L\frac{dx'}{h^3}$

We’ll work this out using a simple example: $h(x)=h_1-\alpha x$, $\alpha=\frac{h_1-h_2}{L}$

To work out the integrals, make the substitution $y=h_1-\alpha x'$, $dy=-\alpha dx'$

$\int_0^L\frac{dx'}{(h_1-\alpha x')^2}= \int_{h_1}^{h_2}\frac{dy}{\alpha y^2}=-\frac{1}{\alpha}[\frac{1}{h_1}-\frac{1}{h_2}]$

Similarly

$\int_{h_1}^{h_2}\frac{dy}{\alpha y^3}=[\frac{-1}{2\alpha y^2}]_{h_1}^{h_2}=-\frac{1}{2\alpha}[\frac{1}{h_1^2}-\frac{1}{h_2^2}]$

So

$-\frac{U}{\alpha}(\frac{1}{h_1}-\frac{1}{h_2})=-\frac{Q}{\alpha}(\frac{1}{h_1^2}-\frac{1}{h_2^2})$

$U(\frac{h_2-h_1}{h_1h_2})=Q(\frac{h_2^2-h_1^2}{h_1h_2})$

$U(h_2-h_1)=Q\frac{(h_2-h_1)(h_2+h_1)}{h_1h_2}$

$Q=\frac{Uh_1h_2}{h_1+h_2}$

Also

$\frac{p-p_0}{6\mu}= - U\frac{1}{\alpha}[\frac{h-h_1}{hh_1}]+\frac{U}{\alpha}\frac{h_1h_2}{h_1+h_2}[\frac{h^2-h_1^2}{h^2h_1^2}]$

But remember we can substitute for α using $\alpha=\frac{h_1-h_2}{L}$

$\frac{p-p_0}{6\mu UL}=\frac{-(h-h_1)}{h_1-h_2}\{\frac{1}{hh_1}-\frac{h_2(h+h_1)}{h^2h_1(h_1+h_2)}\}$

Now put the contents of the squiggly bracket under a common denominator to get

$\frac{p-p_0}{6\mu UL}=\frac{-(h-h_1)}{h_1h^2(h_2^2-h_1^2)}(-hh_1-hh_2+h_2h+h_2h_1)$

Which, when cleaned up, gives

$\frac{p-p_0}{6\mu UL}=\frac{(h_1-h)(h_2-h)}{h^2(h_2^2-h_1^2)}$

As h lies between h1 and h2, if h2< h1 then p-p0>0, implying a net upward force on the block if the width of the film decreases in the direction of the flow.

Syrup on a rotating rod: a thin layer of viscous fluid, thickness h(θ) covers the outer surface of a rotating cylinder whose circumference rotates at a speed U under gravity.

User local Cartesian coordinates.

Going back to $\nabla p=\mu\frac{\partial^2 \bold{u}}{partial z^2}$ and this time including gravity:

x-component $\frac{\partial p}{\partial x}=\mu\frac{\partial^2 u}{\partial z^2}-\rho g \cos \theta$

z-component $\frac{\partial p}{\partial z}=\mu\frac{\partial^2 w}{\partial z^2}-\rho g \sin \theta$

For a thin film

$\frac{\partial p}{\partial x} \sim \frac{P}{L} \ll \frac{P}{h} \sim \frac{\partial p}{\partial z}$

$\frac{\partial^2 u}{\partial z^2} \sim \frac{U}{h^2} \gg \frac{w}{h^2} \sim \frac{\partial^2 w}{\partial z^2}$

So we can neglect $\frac{\partial p}{\partial x}$ and $\frac{\partial^2 w}{\partial z^2}$

-> $0=\mu\frac{\partial^2 u}{\partial z^2}-\rho g \cos \theta$

and $\frac{\partial p}{\partial z}=-\rho g \sin \theta$

$\nu\frac{\partial^2 u}{\partial z^2}=g\cos\theta$

Boundary conditions:

u=U at z=0 (no slip)

$\mu\frac{\partial u}{\partial z}=0$ at z=h (zero stress)

Integrate to get

$\frac{\partial u}{\partial z}=\frac{gz}{\nu}\cos\theta+A$

$A=\frac{-gh}{\nu}\cos\theta$

$\frac{\partial u}{\partial z}=\frac{gz}{\nu}\cos\theta(z-h)$

Now integrate again

$u=\frac{g}{\nu}\cos\theta(\frac{z^2}{2}-Zh)+B$

B=U

$u=\frac{g}{\nu}\cos\theta(\frac{z^2}{2}-Zh)+U$

Volume flux across any section θ =constant, Q

$Q(\theta)=\int_0^{h(\theta)}u dz$

$Q=[Uz+\frac{g}{\nu}\cos\theta(\frac{z^3}{6}-\frac{z^2h}{2}]_0^h$

$Q=Uh-\frac{gh^3}{3\nu}\cos\theta$

Put $H(\theta)=\frac{Uh(\theta)}{Q}$, $\alpha=\frac{gQ^2}{3\nu U^3}$

-> $\frac{1}{H^2}-\frac{1}{H^3}=\alpha\cos\theta$

As $\cos \theta \leq 1$, H(θ) can only exist for all values of θ if $\alpha \leq \frac{4}{27}$ (as $\frac{1}{H^2}-\frac{1}{H^3}$ must be $\leq \frac{4}{27}$

This means that the fluid can only be kept on the cylinder if $U \geq \frac{g\bar{h}^2}{\nu}\times \frac{27}{4\times 3 \times 1.06^2}$

where $\bar{h}$ is the mean thickness (not the Planck constant!).

$\bar{h}=1.06\frac{Q}{U}$ when $\alpha=\frac{4}{27}$, leading to the numbers given above.

Hele Shaw Cell: we use the same procedure as we did for the slider bearing, except now the flow is between two flat, rigid horizontal boundaries at z=0 and z=h, so we adjust our boundary conditions accordingly (u=0 at z=0 and z=h as per the good old ‘no slip’ condition). Armed with this information, I’m sure you can plug through the maths yourself.

Additional- ship waves: waves in the wake of a ship (or even a duck) lie within a wedge of arcsin 1/3 (19.5°). At each instant, the ship generates a spectrum of waves of different wavelengths (and hence phase and group velocities) which propagate in all directions. As the ship moves on, a constructive/destructive interference pattern is formed.