ORE 654 Applications of Ocean Acoustics Lecture 3a Transmission and attenuation along ray paths
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Transcript of ORE 654 Applications of Ocean Acoustics Lecture 3a Transmission and attenuation along ray paths
ORE 654Applications of Ocean Acoustics
Lecture 3aTransmission and attenuation
along ray paths
Bruce HoweOcean and Resources Engineering
School of Ocean and Earth Science and TechnologyUniversity of Hawai’i at Manoa
Fall Semester 2011
04/20/23 1ORE 654 L3
Prologue
• Newton’s “corpuscular energy” moving along “rays” helped describe propagation, reflection and refraction.
• In WWII – Ewing and Worzel discovered the SOFAR channel – sound fixing and ranging
• Perth-Bermuda 1960 shots
04/20/23 ORE 654 L3 2Ewing and Worzel, 1948
Transmission and attenuation along ray paths
• Energy transmission in ocean acoustics• Ray paths and ray tubes• Ray paths in refractive medium• Attenuation• The “SONAR Equation”; source level, sound
pressure level, transmission loss
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Energy transmission in ocean acousticsImpulse sources
• Explosions• Sparkers and air guns• Implosions (light bulbs)• Often milliseconds• Finite, discrete delta
function• Continuous – Direc delta
function
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δ f (t) = 1 / Δt 0 ≤ t ≤ Δt
dt
Δt0
Δt
∫ = 1
g(t)δ (t − t1)dt−∞
∞
∫ = g(t1)
Light bulb impulse source
• “100 W” light bulb• At 18.3 m depth
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Heard et al., 1997
Umbrella
Underwater crater:92 m across6 m deep
Yield: 8 kilotonsLocation: EniwetokDate: 8June 1958
Depth: 48 m
Nuclear testing
04/20/23 6ORE 654 L1
1 kiloton TNT = 4 × 109 J
Pressure, particle velocity, and intensity in a pulse
• Far field, spreading spherically from point source
• p0 at R0
• Waveform shape of outward traveling p same as p0; u follows same form
• Just time delay and amplitude change
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p(R, t) =p0 (t−R / c)R0
R
uR(R,t) ≈p
ρAc=
p0 (t−R / c)ρc
R0
R
iR(R,t) =puR =p20 (t−R / c)
ρAcR2
0
R2
Pressure in a pulse - 2
• Typical pulse shape – satisfies two conditions
• Step/delta with exponential decay, finite for t ≥ R/c
• τs – time to decay to 1/e
• tg – duration of integral or “gate time”; 98% correct for tg > 4τs
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p0 (t) =0 t < 0
p0 (t) dt0
∞
∫ =finite
p(R,t) =p0R0
Rexp −
1τ s
t−Rc
⎛⎝⎜
⎞⎠⎟
⎡
⎣⎢
⎤
⎦⎥
p0R0
Rexp −
1τ s
t−Rc
⎛⎝⎜
⎞⎠⎟
⎡
⎣⎢
⎤
⎦⎥dt
Rc
Rc+tg
∫
=τ sp0R0
R1−exp −
tg
τ s
⎛
⎝⎜⎞
⎠⎟⎡
⎣⎢
⎤
⎦⎥
Particle velocity and intensity – transient signal
• Radial particle velocity and intensity in far field
• Message energy (Joules) pass through element ΔS at range R in gate open time tg, function of (t-R/c)
• Over full sphere ΔS = 4πR2
, total energy Em (J)• Energy flux at range R
transmitted during time interval pressure squared (tips) tg (J/m2)
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uR =p / (ρA / c) iR =puR
ΔEm =ΔS puR dt≈R2
0ΔSρAc0
∞
∫p0 (t−R / c)[ ]
2
R2 dtRc
Rc+tg
∫
ΔS=R2ΔΩ
Em ≈4πR2
0
ρAcp0 (t−R / c)[ ]
2dt
Rc
Rc+tg
∫
εg = puR dtRc
Rc+tg
∫ ≈tips[ ]R2
R20
ρAc
Power radiated by continuous wave signal
• Continuous wave sinusoid, peak value P0
• Source power over all angles is average energy over period T
• Same procedure for other signals
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p =P0ei2π ft
Π=4πR2
0
ρAcP0
2
2
P02
2=1T
P0 sin(2π ft) 2 dt0
T
∫ =Prms2
Ray paths and ray tubes
• Starting point -Spherical wave solution
• p0(t) is temporal function
• Time for message to arrive is R/c
• R0/R is spatial function• p(t,R) – output of
receiver at R
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p(t,R) =p0 (t−R / c)R0
R
Reflections along ray paths
• Reflected rays appear to come from image source• Pressure signal for bottom reflecting ray• Pressure signal for the bottom and surface reflected ray
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p(t) =p0 (t−Ra + Rb
c)
R0
Ra + Rb
R12
tpath =Ra +Rb
c
p(t) =p0 (t−Ra + Rb + Rc
c)
R0
Ra + Rb + Rc
R12R10
tpath =Ra + Rb + Rc
c
Multiple ray paths• Signal pressures
add vectorally and can constructively and destructively interfere at receiver
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Conservation of energy in ray tubes
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S1 =(ΔΩ)R12
S2 =(ΔΩ)R22
S1
ρA2c1p1(t−R1 / c1)[ ]
2dt
R1c1
R2
c1+tg
∫
=S2
ρA2c2p2 (t−R2 / c2 )[ ]
2dt
R2
c2
R2
c2+tg
∫
• Equality of energy• E1 = E2
Sound pressures in ray tubes• Sound pressures at 1 and 2
follow 1/R spherical divergence
• In homogenous media properties are constant, e.g., density and sound speed
• Change t-R/c to τ• Assume tg large enough to
include 1 and 2• Assume p1 proportional to p2,
so integrand must = 0• P changes as sqrt of area ratio,
or cross path ray tube diameter/scale
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τ1 = t − R1 / c; τ 2 = t − R2 / c
S1 p1(τ 1)[ ]2dτ
0
tg
∫ = S2 p2 (τ 2 )[ ]2dτ
0
tg
∫S1 p1(τ 1)[ ]
2− S2 p2 (τ 2 )[ ]
2⎡⎣
⎤⎦dτ0
tg
∫ = 0
p2 (τ 2 ) = +p1(τ 1)S1
S2
Ray paths in a refracting medium• Sounds speed / index of refraction usually function of space and time – can
be complicated• Use Snell’s Law to trace path of a small portion of a wave front –
determines direction• Ray is perpendicular to the local wave front• Gives direction and time• Valid for high frequencies:
– Changes in sound speed over scales large compared to wavelength– Water depth and range to receiver >> wavelength– Very good approximation when diffraction absent (wave effect)– Fails where rays cross (caustics), or shadow zones – use wave theory to patch
• Specular reflection at interfaces (mirror, flat)• Intensity losses along rays through geometric divergence (spherical
divergence modified by refraction), through absorption along paths, and reflections on interfaces
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Speed of sound - Seawater• From previous lecture• Sound speed (c or C m/s) is a complicated function of
temperature T °C, salinity S PSU, and pressure/depth z m
• Simple formula by Medwin (1975):
• Latest: http://www.teos-10.org/
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c = 1449.2 + 4.6T – 0.055T 2 + 0.00029T 3 + (1.34 – 0.010T )(S – 35) + 0.016z
Ocean stratification
• In general, sound speed = c(x,y,z,t)• Sound speed and other ocean properties such
as temperature and salinity (and pressure) have the largest variation in the vertical.
• Horizontal variations are typically smaller, over larger distances
• Assume c = c(z)
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Variation in sound speed – horizontal, vertical• Sargasso Sea at
750 m – main thermocline - +/- 5 m/s (1 °C)
• Perth to Bermuda – Blue 1,470 m/s, red 1,550 m/s
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Snell’s Law again
• Different layers• Define Ray parameter a – constant for one ray
no matter where along the ray you are• Start ray at z0 and c0. At depth z, angle is θz
and cz
04/20/23 ORE 654 L2 20
sinθ0
c0=
sinθ1
c1=
sinθz
cz
=a=constant
Formulating Ray integrals
• To determine ray paths (r,z) and travel time, must integrate, using Snell’s Law
• Dependent on z because c(z) only
• Differential distances dz, ds, dr and time dt
• Between “initial” and “final” positions
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ds =dz
cosθ
dt=ds
c(z)=
dzc(z)cosθ
dr =dztanθ
a=sinθ0
c0=constant
sinθ =ac(z)
cosθ = 1−a2c2 (z); ac(z) <1
tanθ =ac(z)
1−a2c2 (z)
tf −ti = dtzi
zf
∫ = dzzi
zf
∫1
c(z) 1−a2c2 (z)
rf −ri = drzi
zf
∫ = dzzi
zf
∫ac(z)
1−a2c2 (z)
dr
θds
dz
Ray tracing approximations
• Constant sound speed layers: – c(z) = cn for zn ≤ z ≤ zn+1
• Constant sound speed gradient layers, i.e., sound speed varies linearly in each layer:– c(z) = c(z1)+b(z-z1) for z1 ≤ z ≤ z2
– b = d(c(z))/dz
04/20/23 ORE 654 L3 22
Rays through constant sound speed layers• Real c(z) is continuous but
simplest solution - break into constant c layers
• Replace integrals with sums
Use Snell for next θ
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rf −ri = drzi
zf
∫ = dzzi
zf
∫ac(z)
1−a2c2 (z)
tf −ti = dtzi
zf
∫ = dzzi
zf
∫1
c(z) 1−a2c2 (z)sinθn =acn
cosθn = 1−(acn)2 ; acn <1
tanθn =acn
1−(acn)2
rN = (zn+1 −zn)tanθnn=0
N−1
∑
tN =(zn+1 −zn)cn cosθnn=0
N−1
∑
Rays through slowly changing sound speed layers
• Assume constant gradient in each layer
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c(z) =c(zn) +bn(z−zn) for zn ≤z≤zn+1
bn =c(zn+1)−c(zn)
zn+1 −zn
Rays through slowly changing sound speed layers
• Start with c(z) integrals
• discretize• change of
variables to simplify
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t f −ti = dtzi
zf
∫ = dzzi
zf
∫1
c(z) 1−a2c2 (z)
rf −ri = drzi
zf
∫ = dzzi
zf
∫ac(z)
1−a2c2 (z)
wn =z−zn +c(zn)bn
; dw=dz; c(w) =bnw
tn+1 −tn =dw
bnw 1−a2bn2w2( )
1/2wn
wn+1
∫
rn+1 −rn =abnwdw
1−a2bn2w2( )
1/2wn
wn+1
∫
Integral tables
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tn+1 −tn =dw
bnw 1−a2bn2w2( )
1/2wn
wn+1
∫
rn+1 −rn =abnwdw
1−a2bn2w2( )
1/2wn
wn+1
∫
dx
x (a2 −x2 )∫ =−1a
loga+ (a2 −x2 )
x
⎛
⎝⎜
⎞
⎠⎟
xdx
(a2 −x2 )∫ =− (a2 −x2 )
Rays through slowly changing sound speed layers
• Perform integrals• Sum over intervals to get total time and range
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tn+1 −tn =1bn
lnwn+1[1+ (1−a2bn
2wn2 )1/2 ]
wn[1+ (1−a2bn2wn+1
2 )1/2 ]=
1bn
lnwn+1(1+ cosθn)wn(1+ cosθn+1)
rn+1 −rn =1
abn
(1−a2bn2wn
2 )1/2 −(1−a2bn2wn+1
2 )1/2⎡⎣ ⎤⎦=1
abn
(cosθn −cosθn+1)
t= (tn+1 −tn)n=0
N−1
∑
r = (rn+1 −rn)n=0
N−1
∑
Rays through slowly changing sound speed layers• With linear sound
speed, ray paths are arcs of circles
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rn+1 −rn =1
abn
(cosθn −cosθn+1)
radius =1
abn
Rays through slowly changing sound speed layers
• Curvilinear path length• s = Rθ
04/20/23 ORE 654 L2 30
radius = Rn =1
abn
sn =Rn(θn+1 −θn)
Ray example - Arctic• Nearly isothermal• Linear sound
speed gradient = b ≈ 0.016 / s
04/20/23 ORE 654 L3 31
RayNorth Atlantic
• From Ewing and Worzel,1948.• Minimum at 1300 m
04/20/23 ORE 654 L3 32
RaysNorth Atlantic
• From Howe et al., 1987.• Minimum at 1300 m • 18°C mode water in upper layer• eigenrays
04/20/23 ORE 654 L3 33
RaysNorth Pacific
• From Dushaw et al., 1994.
• Minimum at 1000 m• Surface layer develops
as summer progresses• Transition from surface
reflecting to barely refracting
04/20/23 ORE 654 L3 34
Philippine Sea – 2009-2011
• ONR deep water acoustic propagation/tomography experiment
04/20/23 ORE 654 L3 35
Philippine Sea – profiles
04/20/23 ORE 654 L3 36
Philippine Sea - rays
04/20/23 ORE 654 L3 37
Philippine Sea – timefronts
04/20/23 ORE 654 L3 38
Attenuation• Seawater is a dissipative
propagation medium• Through viscosity or chemical
reactions to heat• Local amplitude decrease
proportional to the amplitude itself
• Acoustic pressure decreases exponentially with distance
• In terms of 1/e – neper/m• Then attenuation coefficient –
in terms of dB/km (relative power loss per km)
04/20/23 ORE 654 L3 39
dp =−αepdx
lnpp0
⎛
⎝⎜⎞
⎠⎟=αex
αe =1x
⎛⎝⎜
⎞⎠⎟ln
pp0
⎛
⎝⎜⎞
⎠⎟
p=p0e−αex
p(R,t) =p0 (t−R / c)R0
Re−αeR
α =1x
⎛⎝⎜
⎞⎠⎟10 log10
i1i2
⎛
⎝⎜⎞
⎠⎟⎡
⎣⎢
⎤
⎦⎥
α =1x
⎛⎝⎜
⎞⎠⎟
20 log10
p1
p2
⎛
⎝⎜⎞
⎠⎟⎡
⎣⎢
⎤
⎦⎥
1neper=8.68dB
8.68αe =α
Absorption losses in sea water: viscosities and molecular relaxation
• Coefficients of viscosity• μ dynamic or absolute
coefficient of shear viscosity – the viscosity =ratio shear stress to rate of strain
• Each component of stress is due to a shearing force parallel to area A, caused by velocity gradient (shear modulus)
• Bulk viscosity appears only in compressible media – i.e., acoustics – in compressible N-S, term proportional to rate of change of density – related to extensional modulus
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FxA
=μ∂u∂y
⎛⎝⎜
⎞⎠⎟
μ =dynamic viscosity
μb =bulk or volume viscosity
Absorption losses in sea water: Molecular relaxation
• Bulk viscosity – finite time for real fluid to respond to p change, or to relax back to a base state
• Ionic dissociation – activated, deactivated by condensation, rarefaction
• Magnesium sulfate and boric acid, even though minor parts of “salinity”
• Affects speed of propagation slightly (dispersive) – usually ignore this aspect
• Relaxation frequency• Damped oscillator
04/20/23 ORE 654 L3 41
c ≅cr
fr =1
2πτ r
αe =π fr / c( ) f 2
fr2 + f 2
α =Afr f 2
fr2 + f 2
α =Afr =constant for f ? fr
α =Afr
f2 : f 2 for f = fr
Molecular relaxation - 2
• Attenuation per wavelength
• For fresh water• Goes to zero very high
and very low frequency• Very high f – molecules
can’t respond• Very low – follow• Near fr – activated
molecules transfer energy from condensation to rarefaction – heat
04/20/23 ORE 654 L3 42
λ =c / f
αλ = Acfr( )f
fr2 + f 2
τ r =(4 / 3)μ + μ b
ρAc2
τ r = 2.1×10−12 s
f = 1 / (2π × 2.1×10−12 s) f = 1011Hz
Molecular relaxation - 2
• The fresh water curves• Relaxation frequency off
scale to right
04/20/23 ORE 654 L3 43
τ r = 2.1×10−12 s
f = 1 / (2π × 2.1×10−12 s) f = 1011Hz
Molecular relaxation in sea water
• During WWII, unexplained large attenuation at sonar frequencies – 20 kHz
• In 1950s – careful lab experiments by Leonard, Wilson and Bies
• Drive water filled sphere vs frequency - measure amplitudes of modes of oscillation and therefore damping, with various salts
• Determined magnesium sulfate, a relatively minor constituent by weight, was responsible
04/20/23 ORE 654 L3 44
Molecular relaxation in sea water
• Similar effects found by Mellen and Browning, 1970, for boric acid, moderated by pH
• Relaxation frequency ~ 1 kHz
04/20/23 ORE 654 L3 45
Molecular relaxation in sea water
• Final, including boric acid, magnesium sulfate and pure water
04/20/23 ORE 654 L3 46
α =A1P1 f1 f
2
f 2 + f12
+A2P2 f2 f
2
f 2 + f22
+ A3P3 f2
Molecular relaxation in sea water
• Boric acid• pH function of location• A1 in Pacific half of Atlantic
04/20/23 ORE 654 L3 47
α =A1P1 f1 f
2
f 2 + f12
+A2P2 f2 f
2
f 2 + f22
+ A3P3 f2
A1 =8.68
c10(0.78pH−5) dB km-1 kHz-1
P1 = 1
f1 = 2.8S
35⎛⎝⎜
⎞⎠⎟
0.5
10[(4 −1245 /(273+T )] kHz
Molecular relaxation in sea water
• Magnesium sulfate
04/20/23 ORE 654 L3 48
α =A1P1 f1 f
2
f 2 + f12
+A2P2 f2 f
2
f 2 + f22
+ A3P3 f2
A2 = 21.44S
c(1 + 0.025T ) dB km-1 kHz-1
P2 = 1 − 1.37 × 10−4 z + 6.2 × 10−9 z2
f2 =8.17 × 10[(8−1990 /(273+T )]
1 + 0.0018(S − 35)kHz
Molecular relaxation in sea water
• Pure water
04/20/23 ORE 654 L3 49
α =A1P1 f1 f
2
f 2 + f12
+A2P2 f2 f
2
f 2 + f22
+ A3P3 f2
A3 = 4.937 × 10−4 − 2.59 × 10−5T + 9.11 × 10−7T 2
− 1.50 × 10−8T 3 dB km-1 kHz2
P3 = 1 − 3.83 × 10−5 z + 4.9 × 10−10 z2
TL
04/20/23 ORE 654 L3 50
d
dR20 log10
R
R0
⎛
⎝⎜⎞
⎠⎟=
ddR
[α(R−R0 )]
Rt =8.68α
• Absorption dominant at long ranges, balance at R = Rt
Current interest in pH
• Change in sound absorption at 440 Hz due to CO2/pH
04/20/23 ORE 654 L3 51Hester et al., 2008
pH at ALOHA/HOT
• Variability? Time and space?
04/20/23 ORE 654 L3 52Duda, 2008
Borate absorption
• B(OH)3+ OH− ↔ B(OH)3 · OH− ↔ B(OH)4−
04/20/23 ORE 654 L3 53
Absorption - frequency
• Change in absorption with frequency vs frequency
• SNR(f,L) = SL − TLa(f,L) − TLg(L) − NL(f) − PG
• Big question is noise• Difficult!
04/20/23 ORE 654 L3 54
Scattering losses• Scattering of sound
out of the ray tube• “small” scale index of
refraction variation• Bubbles• Bodies – fish,
submarines, plankton, etc
• Address later
04/20/23 ORE 654 L3 55
The “SONAR Equation”: SL, SPL, and TL• Pressure can vary 10 orders
of magnitude• Use logarithmic scale –
decibel• Decibel = 10 times the log10
of the ratio of two powers• Ratio 10, 10 dB• Ratios – relative to 1 m and 1
μPa• Point source – spherical,
power Pac - Watts• At receiver, wave is plane,
intensity related to pressure• Transmitted acoustic
pressure brought back to R = 1 m
04/20/23 ORE 654 L3 56
I(R) =Pac
4πR2
I (R) =p2 (R)ρAc
p1m2 =
ρAc4π
Pac
p2 =p1m2 R0
2
R2 10−α(R−R0 )
energies/intensity/power dB = 10 log10
X1
X2
⎛
⎝⎜⎞
⎠⎟
pressure/velocity dB = 20 log10
x1x2
⎛
⎝⎜⎞
⎠⎟
re 1 μPa at 1 m
The “SONAR Equation”: SL, SPL, and TL• 1 bar = 1 atm = 105 Pa• 1 W = 170.8 db re 1 uPa at 1 m
04/20/23 ORE 654 L3 57
p1m2 =
ρAc4π
Pac
p2 =p1m2 R0
2
R2 10−α(R−R0 )
SPL(dB) =SL(dB)−TL(dB)
SL(dB) =10 log10 (Pac) +10 log10 (ρAc4π
) + 20 log10
106 μPa1Pa
⎛
⎝⎜⎞
⎠⎟
SL(dB) =10 log10 (Pac) + [(50.8 +120) =170.8]; re 1 μPa at 1 mTL(dB) =20 log10 (R / R0 ) +α(R−R0 )SPL(dB) =10 log10 (Pac) +170.8 −20 log10 (R / R0 )−α(R−R0 )
Re-cap: Transmission and attenuation along ray paths
• Energy transmission in ocean acoustics• Ray paths and ray tubes• Ray paths in refractive medium• Attenuation• The “SONAR Equation”; source level, sound
pressure level, transmission loss
04/20/23 ORE 654 L3 58