Physics Made Easy

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


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

Rectilinear flow between parallel plates

flow-between-plates.jpg\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)

parabolic-flow.jpgWe 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

pipe-flow.jpgAssume 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.


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



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

flow-between-cylinders.jpg(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}



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


Try u_{\theta}=r^n



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


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}


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}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 vorticity-meter.jpg

|\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}


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


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


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


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).

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