Waves in Fluids

Waves in Fluids. WWM Chapter 6, etc.January 8, 2006

First a derivation of the acoustic wave equation based on bulk modulus, and stress and strain.

We first recall that the bulk modulus is the change in pressure over the fractional decrease in volume

B = P/(-V/V) . (An increase in pressure decreases the volume)

P = P-Po is the acoustic pressure p. Next from our study of 1-D waves on bars that strain = lim x0 (xs - x)/x) = /x, where  is the displacement in x. This could be further written as

strain = (Lx - Lxo)/Lxo = /x , or

Lx = Lxo (1+/x) .

(Fluids do not support shear stresses, so we do not have to worry about stresses in y and z causing further strain in x. )

Then by considering a displacement in y of , we would find

Ly = Lyo (1+/y) .

Doing the same thing for a displacement  in z, we would arrive at

LxLyLz = LxoLyoLzo (1 + /x)(1 +/y)(1+/z) , or (since the partials are small)

V  Vo (1 + /x+/y+/z), and from here that

V/Vo = (/x+/y+/z)

Combining this with the definition of p and B at the top of the page we find

(/x+/y+/z) = p/B.

Then we apply F = ma in x to arrive at -p/x = 2/t2 , and by taking the x-derivative

-2p/x2 = 2/t2 (/x) .

Combining this with y and z directions, and applying (/x+/y+/z) = p/B we get

2p = /B 2p/t2,the acoustic wave equation .

On the next page we will derive the acoustic wave equation in terms of fluid concepts involving flow, gradient, divergence, and so on. This is patterned after a treatment in Kinsler, Frey, Coppens, and Sanders.

 = displacement from equilibrium, u= velocity = /t,

s = 'condensation' = fractional change in density = (-o)/ o (s taken to be<1).

The bulk modulus B is defined as B = -(P-Po)/V/V . The acoustic pressure

p = P-Po , p is the 'acoustic pressure'

the excess of the pressure over atmospheric, and the mass density  = m/V. Considering the mass in a given region to be constant we have B = -p/[dV/V] or, using V = m/, that B = p/[d/] .

B= p/[( -o)/ o];

p= Bs, ( p = acoustic pressure )

The equation of continuity states that no matter is gained or lost:

the flux of mass into a region = mass/time = /t ( dV )

Mass flux is the surface integral of u: mass flux = -udA . The reason for the - sign is that dA points out of the volume, and a positive flux means mass is leaving the volume. The ratio of flux to volume as V0 is the divergence. So as the volume shrinks on the right,  becomes constant over the tiny volume Vtiny and we have

mass flux = -tiny surface udA = /t Vtiny

When we divide both sides by Vtiny we get

-div (u) = /tThe equation of continuity

This 'equation of continuity' is linearized by neglecting changes in  inside the divergence:

-o div(u) = /t .

[o is a large quantity but grad is a small quantity. Since u is already a small quantity, the product of grad and u would be too small to keep.] Since o is a constant, /t = (-o)/t, and

div u = ((-o)/o)/t = s/t .

div u = u = s/t

This says that when we have a 'divergence' of the particle velocity, the 'condensation' s will be changing with time.

Then combining p = Bs, and div u = s/t we have

-div u = 1/B p/t .

Now we will apply F = ma to a small volume of fluid V = x y z, and start by examining the x-direction only.

The sketch shows how forces act

on this volume in the x-directionp(x) y z p(x+x) y z

The net force in the x-direction is Fx = y z ( -p(x+x) + p(x)). This must equal max for the tiny volume:

 Fx = y z ( -p(x+x) + p(x)) =  [ (x y z) ux ]/t

This equation is 'linearized' by neglecting changes in , treating  as constant in time. [The idea is that o large but /t is a small quantity. Since u is already a small quantity, the product of /t and u would be too small to keep.] Then when we pass to the limit of x0, with constant  we get

-p/x = ux/t

The 3D version of F = ma acting on a particle is then understood to be

grad p = -ou/t .

When we put the last two equations together we obtain the acoustic wave equation

div (grad p ) = o/t ( div u ) = o/B 2p/t2 .

Div grad p is the laplacian of p or 2p . In rectangular coordinates this is

2p/x2 + 2p/y2 + 2p/z2 = o/B 2p/t2 . ( see Meeks text, Eq. 6.5, p. 172 )

From here on out  will be taken to be o unless otherwise stated.

In air, the oscillations are isentropic (adiabatic), which means that

pV = constant,constant entropy { no heat exchanged in the process }

where  = (heat capacity at constant pressure/heat capacity at constant volume) = cp/cv. In air,  = 1.4.

Exercise 1. Show that the isothermal bulk modulus is BT = p. { Start from pV = constant, at constant temperature}

pV = C; take ln of both sides : ln p+ lnV = ln C.

Then take derivative wrt V (1/p) dp/dV + 1/V = 0, and BT = -V (dp/dV)T = p.

Exercise 2. Show that the isentropic bulk modulus is BS =  p . { Start from pV = constant }

pV= C; take ln of both sides : ln p+ lnV = ln C.

Then take derivative wrt V (1/p) dp/dV + /V = 0, and BS = -V (dp/dV)T =  p.

PLANE WAVES

The form of a plane wave is

exp(i kr - it), where k is the r k

'propagation vector'. The sketch r

at the right shows the propagation

vector k, and a position vector

r made up of a part r|| parallel r||

to k, and a part r, perpendicular

to k.

For a given value of r||, the phase of

the 'plane' wave is kr||, and does not

depend on r . The locus of pointsphase of wave = kr - t

with the same r|| and any r is a

plane: all points in this plane haver = r|| + r

the same phase. The sketch shows

dotted lines parallel to each otherFor a given r||, points with any r lie in a plane

which could perhaps be the crestsand all these points of the wave have the same phase

of an advancing 'plane wave'.which is kr|| -t.

For travelling plane waves of sound in a fluid we have oscillations of both pressure and velocity

p = po exp (i kr - it) , plane waves of sound

u = uo exp (i kr - it) .

Now we will show that velocity and pressure in a travelling wave are 'in phase' by utilizing

-grad p = u/t .

For the kr part of a plane wave we can write k = i^ kx + j^ ky + k^ kz, and r = i^ x + j^ y + k^ z .

Then kr = kx x + ky y + kz z . The gradient operator in x,y,z is

grad = i^ /x + j^ /y + k^ /z

so when we take the x-component of -grad p = u/t for a plane wave we get

-/x [ po exp( i{ kx x + ky y + kz z} - it) ] =/t [uox exp( i{ kx x + ky y + kz z} - it) ],

-i kx po = -i  uox .

When we assemble the x, y, and z components of -grad p = u/t we find

k po =  uo.

This says that the velocity u and the propagation vector k in a travelling plane wave of sound are in the same direction. Since c = /k we have a relation between pressure and velocity amplitudes

uo = po/(c) .

Then the velocity u can be written for a travelling plane wave in terms of pressure

u = (po/[c])exp ( i k . r - it) [ velocity amplitude is pressure amplitude divided by c ]

Acoustic impedance. The mechanical impedance Z started out in chapter 1 as the ratio of force/velocity, and now we come to the specific acoustical impedance z, which is the ratio of pressure to velocity

z = specific acoustic impedance = pressure/velocity

For a plane wave we have

z = po/uo = c.specific acoustic impedance

The product of density and speed is a very important material property and is tabulated for many materials in units of 'rayls' (named after Lord Rayleigh, in a manner similar to the farad being named after Michael Faraday.)

Power and Intensity.

Power is work/time. Since dW = Fdr, power has to be dW/dt = P = Fdr/dt = Fv.

This means the work done by one element of a fluid on adjacent element per unit time is

P = power = p u Awork per unit time in a plane fluid wave

Intensity = power/area = pu intensity in a fluid plane wave

Next page: time-averaged intensity

Time-averaged intensity. This is the time-average of the product of pressure and velocity. For a plane wave we need time-average the quantity

= = < po exp (i kr - it) uo exp (i kr - it) >

The understanding with complex notation is that that at the end, we will take the real part.

Re(p) = po cos (kr - t) [po understood to be real] and likewise for u. There is in general a phase difference between p and u. In the equations above the phase difference is . To shorten the notation we will take 1 = kr and 2 = kr + .

The time average is

= pouo (1/T) 0T dt [( cos (1-t) cos(2-t) )]

= pouo (1/T) 0T dt [( cos 1 cos t + sin 1 sin t) (cos 2 cos t + sin 2 sin t) )]

The time average of sin t cos t vanishes, and the time averge is 1/2 of both cos2t and sin2t, so

= 1/2 pouo ( cos 1 cos 2 + sin 1 sin 2 ) = 1/2 pouo cos(2-1) = 1/2 pouo cos()

We have been aiming to show that the time-averaged intensity can just as well be written as

= 1/2 Re (pu*),

where u* is the complex conjugate of u. In this form both p and u are written in complex notation:

= 1/2 Re (po exp(i(kr-t)) uo* exp(-i(kr-t + )) .

This reduces to the previous equation. Finally we can substitute for u in terms of p:

= 1/2 |po|2/(c) .time-average intensity, plane wave

Energy density (energy/volume) {Section 13.10, p. 420}.

In travelling waves, sound waves, or light waves or

others, we can think of energy being transported through

space at the speed of propagation of the wave. We could

imagine plane waves moving into a region without waves. A

Then a cylindrical region of volume dV = A (c dt) would

move a distance cdt into the 'unoccupied region', carrying

energy across a boundary at a rate P = energy/time. cdt

We'll call  the energy per unit volume so the energy within the volume is dE =  A cdt,. The energy per unit area per unit time is the intensity so we find I = 1/A dE/dt = c . This tells us that the energy density is

 = I/c Energy density = Intensity/velocity.