Basic Physics of Underwater AcousticsDefinitionsOne-dimensional Case cont.Slide Number 4In Three Dimensions: A CUBEParticle VelocitySlide Number 7Spreading in Three-SpaceDecibels (dB)Spreading Losses with RangeAttenuation Losses with RangeThe Piezo-Electric ActuatorThe Piezo-Electric SensorSlide Number 14Massachusetts Institute of Technology 2.017Basic Physics of Underwater AcousticsReference used in this lecture: Lurton, X. 2002. An introduction to underwater acoustics. New York: Springer.Massachusetts Institute of Technology 2.017Definitionsp: pressure, measured relative to hydrostatic, Pa: density, measured relative to hydrostatic, kg/m3E: bulk modulus of the fluid, Pa, p= E [ [u,v,w]: deflections in [x,y,z]-directions, relative to the hydrostatic condition, mThen in one dimension (pipe)p = E [ -u/ x]xuu + u+xundeformeddeformedMassachusetts Institute of Technology 2.017One-dimensional Case cont.Newton’s Law: p= - utt xORpx = - uttConstitutive Law:p = - E u/ xORp = - E uxxpp + pudiff wrt xdiff wrt ttpxx = [ / E ] ptta wave equation!Massachusetts Institute of Technology 2.017Let p(x,t) = Po sin(t – kx)Insert this in the wave equation:-Po k2 sin( ) = - [ / E ] Po 2 sin( ) [ / k ]2 = E / Wave speed c k = [ E This is sound speed in fluid, independent of pressure, or frequency.In water: ~ 1000 kg/m3, E ~ 2.3e9 N/m2 c ~ 1500 m/sWavelength k = cc/f1kHz : 1.5m in watertimespaceMassachusetts Institute of Technology 2.017In Three Dimensions: A CUBENewton’s Law:px = - utt pxx = - uttxpy = - vtt pyy = - vttypz = - wtt pzz = - wttzConstitutive Law:-E ux = p / 3 -E uttx = ptt / 3-E vy = p / 3 -E vtty = ptt / 3-E wz = p / 3 -E wttz = ptt / 3xyzpxx +pyy +pzz = ptt / c2or ∇2p = ptt / c2Lead to Helmholtz Equation:All directions deform uniformlydeformedundeformedwhere ∇2 is the LaPlacian operatorMassachusetts Institute of Technology 2.017Particle VelocityConsider one dimension again (Newton’s Law): px = - utt px = - (ut )tIf p(x,t) = Po sin(t - kx) and ut (x,t) = Uto sin(t - kx) -kPo cos( ) = - Uto cos( ) Uto = Po / c Note velocity is in phase with pressure!c]: characteristic impedance; water: c ~ 1.5e6 Rayleighs “hard”air: c ~ 500 Rayleighs “soft”In three dimensions:∇p= - Vt where ∇p= px i + py j + pz k andV = ut i + vt j + wt kMassachusetts Institute of Technology 2.017Note equivalence of the following: = c / f and / k = cThere is no dispersion relation here; this is the only relationship between and k!Consider Average Power through a 1D surface:P(x) = [ 1 / T ] ∫ T p(x) ut (x) d [ 1 / T ] ∫ T Po Uto sin2( -kx) d Po Uto Po c = Uto2 c/ 2 Acoustic Intensity in W/m2 If impedance c is high, then it takes little power to create a given pressure level; but it takes a lot of power to create a given velocity levelPower per unit area is pressure times velocityMassachusetts Institute of Technology 2.017Spreading in Three-SpaceAt time t1 , perturbation is at radius r1 ; at time t2 , radius r2 P(r1 ) = Po2(r1 ) / 2 cP(r2 ) = Po2(r2 ) / 2 c Assuming no losses in water; then P(r2 ) = P(r1 ) r12 / r22 = Po2 (r1 ) r12 / 2 c r22andPo (r2 ) = Po (r1 ) r1 / r2Let r1 = 1 meter (standard!) P(r) = Po2(1m) / 2 c r2Po (r) = Po (1m) / rUto (r) = Po (1m) / c rrr2r1Pressure level and particle velocitydecrease linearly with rangeMassachusetts Institute of Technology 2.017Decibels (dB)10 * log10 (ratio of two positive scalars):Example:x1 = 31.6 ; x2 = 1 1.5 orders of magnitude difference10*log10 (x1 /x2 ) = 15dB10*log10 (x2 /x1 ) = -15dBRECALL log(x12/x22) = log(x1 /x2 ) + log(x1 /x2 ) = 2 log(x1 /x2 )In acoustics, acoustic intensity (power) is referenced to 1 W/m2 ;pressure is referenced to 1 Pa10*log10 [ P(r) / 1 W/m2 ] = 10*log10 [ [ Po2(r) / 2 c] / 1 W/m2 ] = 20*log10 [ Po (r) ] – 10*log10 (2c)= 20*log10 [ Po (r) / 1Pa ] – 120 - 65Massachusetts Institute of Technology 2.017Spreading Losses with RangePressure level in dB at range r is 20 log10 [ Po (r) / 1Pa ] - 185 = 20 log10 [ Po (1m) / r / 1Pa ] - 185 = 20 log10 [ Po (1m) / 1Pa ] – 20 log10 [r] -185Example: At 100m range, we have lost 40dB or four orders of magnitude in sound intensity40dB or two orders of magnitude in pressure (and particle velocity)Massachusetts Institute of Technology 2.017Attenuation Losses with Range0.1 1 10 100 10000.0010.010.11101001000kHz, in dB/km (pressure)At 100 Hz, ~1dB/1000km:OK for thousands of km,ocean-scale seismics and communications At 10kHz, ~1dB/km:OK for ~1-10km, long-baseline acousticsAt 1MHz, 3dB/10m:OK for ~10-100m,imaging sonars, Doppler velocity loggersLinear approximation!Francois & Garrison (1982) modelTL = 20log10 r + rAcoustic power does have losses with transmission distance – primarily related to relaxation of boric acid and magnesium sulfate molecules in seawater. Also bubbles, etc.(pressure transmission loss)Massachusetts Institute of Technology 2.017The Piezo-Electric Actuatorstrain = constant X electric field = d X E ort/ t = d X ( V / t )where d = 40-750 x 10-12 m / V Drive at 100V, we get only 4-75 nm thickness change!ttEV+Series connection amplifies displacement**still capable of MHz performance**V-Massachusetts Institute of Technology 2.017The Piezo-Electric Sensorelectric field = constant X stressE = g X orV = t g where g = 15-30 x 10-3 Vm/NIdeal Actuator: Assume the water does not impede the driven motion of the materialIdeal Sensor: Assume the sensor does not deform in response to the water pressure wavesMassachusetts Institute of Technology 2.017Typical Transducer: 120 to 150 dB re 1Pa, 1m, 1V means106 –107.5 Pa at 1m for each Volt applied or1-30 Pa at 1m for each Volt appliedTypical Hydrophone:-220 to -190 dB re 1Pa, 1V means10-11 to 10-9.5 V for each Pa incident or10-5 to 10-3.5 V for each Pa incidentSo considering a transducer with 16Pa at 1m per Volt, and a hydrophone with 10-4 V per Pa:If V = 200V, we generate 3200Pa at 1m, or 3.2Pa at 1km,
View Full Document
Unlocking...