Sound radiation and sound insulation


 Claude Wilkinson
 11 months ago
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1 11.1 Sound radiation and sound insulation We actually do not need this chapter You have learned everything you need to know: When waves propagating from one medium to the next it is the change of impedance which determines what happen. In our case this will be the radiation impedance. Even here Snell s law will be valid Along the plate there is a pattern given by the bending wavelength λ B. This pattern has to be identical with the projection of the wavelength of the radiated sound. Only if the wavelength on the plate is bigger then the wavelength in air, sound radiation is possible (this is only true for infinite plates as we see later). In the case where the wavelength on the plate is shorter than the wavelength in air you cannot find an angle where the projection fits. This a relatively simple derivation of what is following in detail in later in the text. Radiation from infinite plates Three areas:! B <! air! B =! air! B >! air No radiation Only a nearfield Sound radiation parallel to the plate surface Sound radiation under a certain angle
2 11. This would be a good qualitative description. The effect that there is no sound radiation for! B <! air could be explained by the shortcircuiting between positive and negative displacement on the plate. Instead of getting the air in a wave motion, it is just moved parallel to the surface. When! B =! air a sound wave propagates parallel to the plate surface. For frequencies with! B >! air sound is radiated under a certain angle. Radiation efficiency To characterise the efficiency of a radiator the radiated power is compared with the radiation of a plane wave with the same surface and the same averaged velocity. The radiation efficiency! = W radiated "cs v where S is the radiating surface and v is the averaged effective value of the velocity of the radiator (i.e. average over the surface). The radiation efficiency defined in this way can become bigger than unity, since it is not relating for instance mechanical power to radiated power, but compares the radiation from a structure with what would be radiated from a plate, vibrating in phase and with the same amplitude as the structure has in average. In our case the radiation efficiency has to look as this => For this analysis we can summarise that there is one frequency where c B, the speed of sound on the plate, is identical with c, speed of sound in air. This frequency is called critical frequency and can be calculated as c air = c B or c air = B 4! => m
3 f c = c air! m = c air B! " P h( 1 # $ ) h 3 E = c air 1 1! h " P ( 1 # $ ) E = 0.55 c air = 1 K c hc L h c L is the speed of sound for longitudinal waves in structures. The factor K c depends on the material only. Typical values are material K c [m/s] concrete 18 stone light concrete 36 gypsum 3 steel 1 aluminium 1 glass 14 Many of the values are in the same order. This means that it is mainly the thickness of the material, which will lead to differences in the critical frequency. Radiation from finite plates From our daily experience we know that all structures radiate independently where the critical frequency is situated. The results before are only valid for infinite plates. The main difference between infinite and finite plates become obvious when looking at a plate (e.g. a window) built into a bigger and more rigid structure (e.g. a wall). It is obvious that this socalled hydrodynamic shortcircuiting does not work perfectly at the edges of the plate. There is some air, which cannot move from areas of high pressure to areas of low pressure. Indeed, one can show by theory that the edges of the plate are responsible for the radiated sound below the critical frequency. $! " % 1 Uc # Sf c f f c 0.45 Uc ( for f << f c for f " f c ) CHALMERS UNIVERSITY f c OF TECHNOLOGY Division of Applied Acoustics 1 for f >> f c ' * For a slightly damped plate the radiation efficiency can be calculated as where U is the circumference of the plate and S is its surface.
4 11.4 An increase of the radiated sound power does although occur when discontinuities are present such as stiffeners, material changes or close to the forces acting on the structure. Examples for the radiation efficiency are shown here For a vibrating plate built into a rigid wall: with decreasing size, the radiation efficiency is increasing. It approaches the radiation efficiency of a point source in front of a rigid wall. From radiation to excitation The radiation efficiency of walls, floors, ceiling etc. is and important factor for radiation of vibrations transported through the structure of the building. socalled flanking transmission is strongly depended on the radiation efficiency of the structures. the The
5 11.5 The source in the sending room excites a part of the structure. The structure is vibrating. These vibrations are transported to the next room and radiated into the room. Both, excitation and radiation depend on the radiation efficiency of the structure, this means on the relation between the wavelength on the structure (bending waves) and the wavelength in air. What can we do to keep the radiation efficiency low? One idea would b to chose the critical frequency in such a way that it will be very high. How to do this? Looking at the equation for the critical frequency f c = c air m! B It is clear that we need something heavy with low bending stiffness. Some materials might deliver this, eg. Heavy rubber plates. Alternative would be to make the plate softer by cutting small grooves into it. This will not affect the mass but making the structure softer. It is also clear that stiffening a structure by rips or beams might have a negative effect since it increases the bending stiffness strongly but will increase the mass only slightly. Is it not possible to modify the radiating wall or ceiling, one might cover the it with a second wall freely suspended. This second wall should then have a low radiation efficiency TASK: After all what you have learned draw your conclusions concerning sound insulation
6 11.6 Mathematical derivation of what we have learned (just for fun ) Sound radiation from bending waves on infinite plates In the Figure below a situation is shown, where a plate is vibrating with the wave length! B, and the velocity v P y = v B e! jk B y from the plate a wave might be radiated under a certain angel. Only one dimension of the plate is considered to simplify the problem. The radiated wave can be described as p rad ( x,y,! ) = p Ae " jk x x e " jk y y At the surface of the plate the velocity in the normal direction to the surface of the plate has to be identical with the particle velocity in air. This means that the wave number of the plate k B and the wave number in the y direction k y have to be identical (see Snell's law). The velocity in the field directly above the plate is v rad,x ( x = 0,y,! ) = k x p Ae # jk x x e # jkb y = v B e # jkb y which is equivalent with "! v rad,x k cos " ( x = 0,y,! ) = #! $ jk cos (" ) p A e x e $ jk B y = v B e $ jk B y This velocity has to be identical with the velocity of the plate. This allos us to calculate the unknown amplitude p A! = "! k x v B. The wave number in the x direction however, is not independent from the wave number in ydirection. For both the relation k = k x + k y has to be valid as we learned when discussing the oblique incidence of waves on a surface. Replacing k y by k B we obtain k x = k! k B or k x = k! k B where k is the wave number in air. The radiated pressure can finally be written as p rad ( x,y,! ) = v B "! k # k B e # j k #k B x e # jk B y Together with the velocity on the surface of the plate the intensity in the normal direction of the plate surface can be calculated
7 11.7 { v * y P } = 1 I x = 1 Re p x = 0,y #%!" '% v B Re $ ( % )% k x Three different cases can be distinguished. Case 1 : k air > k B or! air <! B In this case the wavelength in air is smaller than the wavelength on the plate. The root k! k B is real and consequently also the wave number k x. The radiated power per unit area is given as I = 1!" v B = 1!" v B k x k cos # = 1 v B!c cos # This can be interpreted as the effective value of the velocity times the radiation impedance for a wave radiated under the angle!. The angle! is determined by Snell's law k cos! = k " k B which gives cos! # or for the angle!! = cos "1 % % $ 1 " k B k = 1 " k B # ( % = cos "1 % 1 " ( % ' $ k c c B ( ( which we already have obtained ( ' for Snell's law. If the speed of sound in air is much smaller than the speed of sound for the bending waves on the plate, the angle! will become close to 0 degrees, i.e. the wave is radiated as a plane wave in normal direction to the plate. The radiation efficiency is! = W radiated "cs v = 1 v B "# k x S "cs 1 v B 1 = cos $ which becomes in this case close to unity. Case : k = k B,! =! B In this case the wavelength on the plate equals the wavelength in air. This also means that the speed of sound on the plate is identical with the speed of sound in air. This case gives in some way strange results. First of all the amplitude of the pressure wave as well as the radiated power and the radiation efficiency becomes infinity, since the
8 root k! k B 11.8 becomes zero. The angle! is in this case 90 degrees, which means that the wave is radiated in parallel to the plate. The somewhat strange results concerning pressure amplitude and radiated power is a consequence of that we do not include radiation loading. Creating a high pressure in front of the plate certainly means, that the plate will experience a force, which will act against the vibration. In most of the cases the radiation loading is neglected. However, in the case of very light structures or plates in a heavy medium like water, radiation load has to be considered. Case 3 k < k B,! <! B In this case the wavelength on the plate is shorter than the wavelength in air. The speed of sound on the plate is smaller than the speed of sound in air. The root k! k B becomes imaginary and consequently also the wave number in xdirection. Instead of a radiated wave only a nearfield (i.e. an in xdirection decreasing exponential function) is created. Since the wave number in xdirection is imaginary the intensity yields zero. This means there is no power radiated from the surface of the plate. The radiation efficiency consequently equals zero. The air particles in front of the surface move on circles as can be seen when considering both the velocity in xdirection and the velocity in y direction. Since the velocity amplitude in xdirection becomes imaginary, while the velocity amplitude in ydirection is still real, there is a phase difference of 90 degrees between both.
9 Sound insulation 11.9 In order to have a quiet room one needs possibilities to hinder the propagation of sound power W in into this room. The sound insulation of the walls (or whatever is around the room) determines how much sound power W tr is coming into the room. Definitions: The transmission factor! is a measure for the capability to isolate against the transport of sound power is! = W tr W in. The reduction index R is 10log 1! = 10logW in W tr. The reduction index can be measured for rooms where the sound field has statistical properties by measuring the averaged sound pressure level L S in the sending room the averaged sound pressure level L R in the receiving room the equivalent absorption area A R in the receiving room the surface area S through which the sound power is expected to be transported. The reduction index is R = L S! L R! 10log A R S. The reduction index weighted (i.e. compared with the reference curve) is called R w. A reduction index measured in field is indicated by a prime ( R!). If a construction consists of several parts with different reduction index the resulting reduction index is R res = 10log S 1 + S +...S n S 1 10! R S n 10! R n 10 "S = 10log i "S i # where S1, S= the area of the different parts andr1, R = the reduction index of the different parts. TASK: IS THERE A RELATION BETWEEN SOUND INSULATION AND SOUND RADIATION? Have a look on the notes from the last lecture on sound radiation. We found for a very big (infinite) plate:
10 11.10 no sound radiation below the critical frequency fc very high radiation parallel to the surface of the plate at the critical frequency radiation at frequencies above the critical frequency HAS THIS ANY CONSEQUENCE FOR THE SOUND INSULATON? Derivation of the theory for sound insulation of plates Airborne sound incident at a panel, wall or some other type of partition mainly excites bending waves in the construction. The sound insulation of plates can be derived in several ways. One way follows the approach we used for Snell's law (see notes). One assumes an incident wave, a reflected wave and a transmitted wave. All three waves together have to fulfil the boundary conditions at the plate: continuity of particle velocity the differe nce betwee n the pressure on both sides of the plate excites the plate to vibrations. Since we assume the same medium on both sides of the plate, the wavenumbers are the same as well as the angle of incidence and the angle under which the wave is radiated into the second medium. p i p r p t ( x,y,! ) = p A,i! ( x,y,! ) = rp A,i! ( x,y,! ) = tp A,i! "jk cos # e + jk cos " e " jk cos # e x e " jk sin # y x e #jk sin " x e "jk sin # The sin and cosine functions come from the projection of the waves in the different directions. Each of the individual exponential functions gives the changes of the phase due to a change in the x or y coordinate. y y
11 11.11 The particle velocities of the waves (in the directions of the waves) are obtained from the pressure using the relation p/u =!c. The components normal to the plate are a factor cos ϑ smaller. u i u r u t (! ) p ( x,y,! ) = "r A,i cos ( # ) p ( x,y,! ) = A,i cos " #c $c p ( x,y,! ) = t A,i cos " #c e$jk cos " x e $ jk sin " y + jk cos (# )x "jk sin # (! ) e e y cos (" )x $ jk sin " (! ) e$jk e y When discussing Snell's law we had continuity of the pressure. However, this is not the case when a plate is separating both sides. There the pressure is different on both sides. Otherwise the plate would not reduce sound; there would be complete transmission through the plate. The vibration of the plate is described by the bending wave equation! 4 " x,t! " x,t B + m = p 1 x,t!x 4!t displacement on the plate has the form! x," # p ( x,t ) Assuming harmonic vibrations the =! B ( " ) e # jk By or u p ( x,! ) = u B (! ) e "jk By becomes after multiplication with j! : = j" p 1 ( x," ) $ p ( x," )! 4 j"# x," B $ " mj"# x,"!x 4 j!" is just the velocity u p of the plate.! 4 u P x," B # " mu P x,"!x 4 which is [ ] [ ] = j" p 1 ( x," ) # p ( x," ) Bk B 4 u B e!jk By! " mu B e! jk By = j" p 1 x," [! p ( x," )] Now we can replace the pressure by p 1 = p i + p r and p = p t. Since the velocity on the plate has to be identical with the velocity of the sound field on both sides for all y. This means that
12 !jk sin (" )y e 11.1 = e! jk By orkb = k sin (! ) has to be valid. Comment the impedance is Z = p, from the equation above we get the impedance of a u plate when exposed to a wave: [ p 1! p ] Z = p u p = u p = Bk 4 B! " m j" Using all equation for the pressures and velocities the two unknowns (reflection coefficient and transmission coefficient) can be determined. Task: try this by your own Transmission coefficient: t = cos (! ) [ j"#c $" m + Bk 4 sin 4! ] The transmission factor is the squared magnitude of the transmission coefficient:! = % cos " 1 + ( ' * ' #c$ * ' )* 1 [ +$ m + Bk 4 sin 4 " ] Physic behind the equation: The relation shows that the transmission through the plate depends on the angle of incidence. We see that there is a possibility for total transmission when the expression in the second parenthesis in the nominator becomes zero. Total transmission occurs for:!" m + Bk 4 sin 4 # = 0 or when k 4 sin 4! = " m Or for cos (! ) = 0. B
13 11.13 From this expression we can calculate the frequency at which total transmission occurs. This frequency is called the coincidence frequency and the phenomenon the coincidence phenomenon. The forced bending wave velocity coincides with the sound velocity in air. Or expressed in wave number, the projection of the wave number of the exciting wave in air onto the surface of the plate coincides with the wave number of the free bending wave in the plate, compare with what we have learned about critical frequency Reduction index as function of frequency and angel of incidence k sin! = " m 4 = k B. B For each angle of incidence we will find one coincidence frequency. The lowest coincidence frequency is obtained for a wave propagating parallel to the plate. This lowest frequency is the critical frequency) f coincidence = c! sin " The coincidence phenomenon is a special type of resonance. We are used to link the resonance concept to the case where a driving force or pressure has a frequency, which equals a natural frequency (eigenfrequency) of the driven system. Here, instead, it is a driving field the spatial distribution of which equals a (natural) bending wave field of the driven system. As we recall from physics (see Physics, V), it is the losses that limit the response of a system driven at resonance. So, at coincidence, the losses in the system get important. The reduction index does not drop to zero because there are always losses (damping) in real systems, which then determine the reduction index. m B. We see that for 90 degrees incidence (gracing incidence, parallel to the plate) the reduction is zero over the whole frequency range since cos ϑ = 0 and therefore τ = 1. However, this result is somewhat artificial since for a parallel wave there is no normal component to the plate and therefore there is no sound power transmitted through the plate.
14 11.14 When we are well below the coincidence frequency, Bk 4 sin 4! << " m. The wall or panel 1 has mass character. In this case the transmission factor becomes! = % cos # 1 + "m ( ' * ' $c * ' )* The sound insulation of the plate follows a mass law, which says that it increases with 6 db when doubling the mass per unit area. This does not mean that the plate as the whole is vibrating like one rigid mass, but each element vibrates more or less independently from its neighbour following the vibration pattern imposed by the sound field. Summary: For the sound reduction index we get quite different expressions below and above the critical frequency. We can divide the frequency range into three ranges; f <f c, f f c, and f > f c. We get f < fc The construction has mass character. The reduction index follows a mass law. We call this Non resonant transmission. f f c The construction gets sound transparent for gracing incidence if the losses can be neglected. Incident sound is totally transmitted. Resonant transmission. f > f c For a certain angle of incidence the sound transmission is total if the losses are zero. Also for constructions with losses the transmission becomes big. Resonant transmission. Diffuse sound field (Sound comes from all directions with the same probability). For f > fc we always have sound waves fulfilling the coincidence condition if the sound field is diffuse. This cannot happen for f < f c. We consider diffuse sound incidence at a wall, i.e. all angles of incidence ϑ are equally probable. We have to take the average over the transmitted sound power and the incident sound power for all angels! sin! d!
15 ! d = $ # / W in (" )!(" ) cos" sin"d" # / =!(" ) cos" sin"d" # / $ 0 W in (") cos" sin"d" $ 0 How to do : The total solid angle is 4π. The fraction of the sound field having an angle of incidence ϑ is then! sin"d" 4! = sin"d" Projected area against this incident sound is S cos ϑ. We then have to take the average of the transmission factor for diffuse sound incidence, τ d as! d = $ # / 0!(" ) cos" sin"d" # / =!(" ) cos" sin"d" # / $ 0 cos" sin"d" $ 0 Using this expression, τ d may be calculated for different cases. For f f c we have to introducee the losses of the plate or wall through a complex Youngs modulus E = E 0 (1 + jη) where η is the loss factor. This expression for τ (ϑ) is then inserted in the integral above. It is a more tricky derivation. However, in this case it is found that the total transmission is dominated by that part of the incident sound field that fulfils the coincidence condition. The integration only needs to be performed over this resonance peak. The derivation is found in Cremer s paper from 194. The results are:! d =!(0),3log 1!(0) or R d R(0) 10 log (0,3 R(0)) With τ(0) obtained from the formula for the transmission factor setting ϑ = 0 we obtain for the three frequency areas of interest: For f < fc R d mass law 0 log m + 0 log f 49 db For f f c η is the loss factor of the wall material. R d = R d,masslaw + 10log! + 8dB Finally for f > f c R! R(0) + 10log( f f c " 1) + 10log# " db
16 11.16 Comments: R(0) increases with 6 db/octave. We can then see that above the critical frequency Rd increases with 9 db/octave if we can assume the loss factor η to be frequency independent. For η = 0,01 the depth in the coincidence valley is approximately 1 db. The influence of the coincidence phenomenon may be observed in the reduction index already around fc/. The pure mass law is therefore valid to approximately 0,5fc. Thick brick and concrete walls have such a low critical frequency that the expression describes the reduction index in most of the frequency range from 100 to a couple of 1000 Hz. shows in summary the schematical reduction index for a single wall of limited size. Observe that the frequency scale is normalised with regard to the critical frequency. For thin constructions such as gypsum boards, thin steel plates and thin glass plates the sound reduction index is determined by the mass law.
17 Problems to section a) A vibrating machine is mounted on a quadratic steel plate of thickness 10 mm and surface area of 10m. The machine is vibrating with a frequency of 500 Hz and the rms value of the velocity has been measured to 103 m/s. a) What is the critical frequency for the plate? b) What is the radiation factor at this frequency? c) How large is the radiated sound power from the plate? b) The steel plate in 11.1 is changed to a 16 cm thick quadratic concrete plate with the same area as in Assume that the vibration velocity in the concrete plate is the same as for the steel plate. The concrete has the density 300 kg/m 3 and a Young s modulus of 6 GPa. The Poisson s number can be set to 0,3. a. What is the critical frequency for the concrete plate? b. What is the radiation factor at this frequency? d) How large is the radiated sound power from the concrete plate in this case? c) An infinite plate vibrates with a frequency that is higher than the critical frequency. a) Why are sound waves radiating from a certain angle from the plate? c. Why does the infinite plate not radiate anything at all for frequencies below the critical frequency? d) Plane sound waves with a frequency of 50 Hz hits a 1 cm thick concrete plate ( E!, 5" Pa, ρ=300 kg/m 3,! " 0.3) with an angle of 45. a) What is the wave length of the forced bending waves in the plate? b) What is the wave length of the free bending waves in the plate at 50 Hz? c) At what angle of incidence will the wave length of the forced bending waves be equal to the free bending waves in the plate? e) A large plate vibrates in phase over the entire surface. The radiation factor is 1 then. The rmsvalue for the vibrational velocity is then v eff = 3,5!10 "5 m/s and the plate area is 0m. How large is the radiated power? f) The mean level was measured to 58 db (re 5!10 "8 m/s) for a wall of area 10m. Note the reference value for the velocity level. The wall is the only surface radiating to a room with the reverberation time 1s and dimensions 4 x 5 x,5 m 3. The resulting sound pressure level in the room was measured to 55 db (!10 "5 Pa). Calculate the radiation factor for the wall.
18 11.18 g) A machine is mostly vibrating below 700 Hz. The machine is builtin a sealed box in order to prevent radiation of sound from the machine to the surrounding. You shall now choose the material, which gives lowest amount of radiated sound. The largest available thickness is 80 mm due to limited available space. The available material is steel plate and aluminium plate with the following data: Steel: ρ=7880 kg/m 3, E=10 GPa, ν=0.3, thickness=44 mm. Aluminium: ρ=700 kg/m 3, E=70 GPa, ν=0.3, thickness=15 mm. V,T S v The floor of a car (see figure above) works as a source to the compartment when the car is driven on a road. Calculate the total sound level inside the compartment caused by the floor vibrations! Calculate the Schröder frequency. The following information is given for the indicated octave bands. f center [Hz] radiationfactor,! 0,0 0,01 0,30 0,80 1,00 T 6 0 [s] 0,15 0,10 0,08 0,09 0,07!v " [(m/s) ] 3,9x106 5,x106 9,81x108 3,1x107 1,98x106 Floor surface, S=3,5 m Compartment volume, V=5 m 3
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