Physics Made Easy

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<p_{\infty} (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.

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