D. W. Herrin, Ph.D., P.E. Department of Mechanical Engineering
Types of Mufflers 1. Dissipative (absorptive) silencer: Duct or pipe Sound absorbing material (e.g., duct liner) Sound is attenuated due to absorption (conversion to heat) 2
2. Reactive muffler: Types of Mufflers Sound is attenuated by reflection and cancellation of sound waves Compressor discharge details 40 mm 3
Types of Mufflers 3. Combination reactive and dissipative muffler: Sound absorbing material Perforated tubes Sound is attenuated by reflection and cancellation of sound waves + absorption of sound 4
Performance Measures Transmission Loss W i Wr Muffler W t Anechoic Termination Transmission loss (TL) of the muffler: ( ) db =10 log TL 10 W W i t 5
Performance Measures Insertion Loss SPL 1 Muffler SPL 2 IL (db) = SPL 1 SPL 2 Insertion loss depends on : TL of muffler Lengths of pipes Termination (baffled vs. unbaffled) Source impedance Note: TL is a property of the muffler; IL is a system performance measure. 6
Example TL and IL Expansion Chamber Muffler Source Inlet Pipe 24 2 6 12 Outlet Pipe 12 20 10 TL and IL (db) 0-10 0 200 400 600 800 1000-20 -30-40 Insertion Loss Transmission Loss -50 Frequency (Hz) Pipe resonances 7
Acoustic System Components Source Su Su Any acoustic system P (sound pressure reaction) Z t Input or load impedance z = P Su = r + jx Termination impedance z t = P t Su t = r t + jx t 8
Summary 1 Dissipative mufflers attenuate sound by converting sound energy to heat via viscosity and flow resistance this process is called sound absorption. Common sound absorbing mechanisms used in dissipative mufflers are porous or fibrous materials or perforated tubes. Reactive mufflers attenuate sound by reflecting a portion of the incident sound waves back toward the source. This process is frequency selective and may result in unwanted resonances. Impedance concepts may be used to interpret reactive muffler behavior. 9
The Helmholtz Resonator Named for: Hermann von Helmholtz, 1821-1894, German physicist, physician, anatomist, and physiologist. Major work: Book, On the Sensations of Tone as a Physiological Basis for the Theory of Music, 1862. von Helmholtz, 1848 10
Helmholtz Resonator Model V L x K = M 2 ρoc S V 2 B = ρ o S Lʹ B L is the equivalent length of the neck (some air on either end also moves). S B F = PS B M x + Kx = PS B x = jωu B x = u B jω " j ωm K % $ 'u B = PS B ω z B = P " 1 % = j$ 2 S B u B S B ' " ωm K % $ ω ' z B 0 when ω = Damping due to viscosity in the neck are neglected K M = c SB Lʹ V (resonance frequency of the Helmholtz resonator) 11
Helmholtz Resonator Example A 12-oz (355 ml) bottle has a 2 cm diameter neck that is 8 cm long. What is the resonance frequency? f f n n c 2π SB Lʹ V = 182 Hz 343 = 2π ( 0.02) = π 4 ( )( 6 0.08 355 10 ) 2 12
Helmholtz Resonator as a Side Branch TL ( db) = 10 log 10 1 + ωlʹ c S B 2S c 2 ωv 2 Anechoic termination V = 0.001 m 3 L = 25 mm S B = 2 x 10-4 m 2 S = 8 x 10-4 m 2 f n = 154 Hz TL (db) 20 15 10 5 0 0 50 100 150 200 250 300 Frequency (Hz) 35 Hz 13
Network Interpretation (any system) z B P z z A V z B z A z = z z B z A + z B A Can we make Z B zero? z z B = P! 1 $ = j 2 S B u B " S B %! ωm K $ " ω % z B 0 when ω = K M = c SB Lʹ V (Produces a short circuit and P is theoretically zero.) 14
A Tuned Dynamic Absorber M 2 F M 1 x Original System F K 2 M 1 x Tuned Dynamic Absorber K 1 K 1 x/f tune K 2 M 2 = K 1 M 1 Original system M 2 /M 1 =0.5 Tuned dynamic absorber ω/ω 1 15
Resonances in an Open Pipe P = 1 Pa source L p = 1 m First mode λ 1 = 2L p = c f 1 = 343 =171.5 Hz f 1 2( 1) Second Mode λ 2 = L p = c f 2 = 343 = 343 Hz f 2 1( 1) etc. 16
SPL at Pipe Opening No Resonator 17
Example HR Used as a Side Branch* TL ( db) = 10 log 10 1 + ωlʹ c S B 2S c 2 ωv 2 Anechoic termination V = 750 cm 3 L = 2.5 cm (L = 6.75 cm) D B = 5 cm (S B = 19.6 cm 2 ) D = 10 cm (S = 78.5 cm 2 ) f n = 340 Hz * e.g., engine intake systems 18
SPL at Pipe Opening with Resonator 19
The Quarter Wave Resonator The Quarter-Wave Resonator has an effect similar to the Helmholtz Resonator: S B z B L S TL 2 ( kl) 4( S S ) ( ) B 4 S SB 2 tan + = 10 log10 2 z B = jρ o c S B ω n = nπc 2L f n = nc 4L or cot( ωl c) = 0 when ωl c = nπ 2 n =1,3, 5... L = nc 4 f = n " λ % $ ' 4 20
Summary 2 The side-branch resonator is analogous to the tuned dynamic absorber. Resonators used as side branches attenuate sound in the main duct or pipe. The transmission loss is confined over a relatively narrow band of frequencies centered at the natural frequency of the resonator. 21
The Simple Expansion Chamber 18 2 6 2 30 2 1 2 1 2 TL = 10 log ( ) ( ) 10 4cos kl + m + sin kl 4 m where m is the expansion ratio (chamber area/pipe area) = 9 in this example and L is the length of the chamber. 25 TL (db) 20 15 10 5 0 0 100 200 300 400 500 600 700 800 Frequency (Hz) 22
Quarter Wave Tube + Expansion Chamber 9 18 2 2 6 2 30 25 TL (db) 20 15 10 5 0 0 100 200 300 400 500 600 700 800 Frequency (Hz) 23
Extended Inlet Muffler 18 2 9 6 2 TL (db) 30 25 20 15 10 5 (same for extended outlet) 0 0 100 200 300 400 500 600 700 800 Frequency (Hz) 24
Two-Chamber Muffler 9 9 4 6 50 40 TL (db) 30 20 10 0 0 100 200 300 400 500 600 700 800 Frequency (Hz) 25
Complex System Modeling We would like to predict the sound pressure level at the termination. Quarter-wave resonator Source Engine Pump Compressor (intake or exhaust) Area change Expansion chamber Helmholtz Resonator termination 26
The Basic Idea The sound pressure p and the particle velocity v are the acoustic state variables 1 For any passive, linear component: p 1 = Ap 2 + BS 2 u 2 p 1, u 1 any acoustic component 2! " $ p 1 S 1 u 1 S 1 u 1 = Cp 2 + DS 2 u 2 or % ( A B +! = * -" ' ) C D, $ p 2 S 2 u 2 % ' p 2, u 2 Transfer, transmission, or four-pole matrix (A, B, C, and D depend on the component) 27
The Straight Tube A B L p 1, u 1 p 2,u 2 (x = 0) (x = L) must have plane waves S Solve for A, B in terms of p 1, u 1 then put into equations for p 2, u 2. ( ) = Ae jkx + Be + jkx u( x) = 1 p x p( 0) = p 1 = A + B u( 0) = u 1 = A B ρ o c p( L) = p 2 = Ae jkl + Be + jkl u( L) = u 2 = Ae jkl Be + jkl ρ o c jkρ o c p 1 = p 2 cos( kl) + u 2 ( jρ o c)sin( kl) u 1 = p 2 ( j ρ o c)sin( kl) + u 2 cos( kl) ) jρ " $ p cos( kl) o c, + sin( kl). 1 $ S " + 2. $ ' = %$ S 1 u + 1 ($ js 1 S. + sin( kl) 1 cos( kl).%$ * + ρ o c S 2 -. (note that the determinant A 1 D 1 -B 1 C 1 = 1) dp dx p 2 S 2 u 2 $ ' ($ 28
Combining Component Transfer Matrices [ T ] i = Ai Ci Bi D i 2 2 Transfer matrix of i th component! " $ p 1 S 1 u 1 % ' = [ T 1 ][ T 2 ][ T 3 ] T n [ ]! " $ p 2 S 2 u 2 %! p = () T system * 2 + " ' $ S 2 u 2 % ' [ T ] system = A C system system B D system system 2 2 29
Straight Tube with Absorptive Material L k,z c (complex wave number and complex characteristic impedance)! " $ p 1 S 1 u 1 ( % * * = * ' * )* cos( k 'L) js 1 z c sin( k 'L) jz c sin( k 'L) S 2 S 1 cos( k 'L) S 2 + -! - -" -$,- p 2 S 2 u 2 % ' 30
Area Change p 1 = p 2 S 1 S 2 S 1 u 1 = S 2 u 2 1 2! " $ p 1 S 1 u 1 % ' ( = * 1 0 ) 0 1 +! -", $ p 2 S 2 u 2 % ' 31
Expansion Chamber Muffler L S S S straight tube area changes! [ T ] = 1 0 " 0 1! [ T ] = "! $ % " cos( kl) cos( kl) jρ o c sin( kl) S' js' sin( kl) cos( kl) ρ o c jρ o c sin( kl) S' js' sin( kl) cos( kl) ρ o c $ % $! " % 1 0 0 1 $ % 32
Expansion Chamber Muffler 18 2 6 2 S' S = 9 33
Transfer Matrix of a Side Branch S B p 1 = p 2 = p B S Su 1 = S B u B + Su 2 1 2 z B = p B S B u B = p 2 S B u B Su 1 = p 2 ( z B ) + Su 2! " $ p 1 Su 1 % ' ( = * )* 1 0 1 z B 1 +! -",- $ p 2 Su 2 % ' 34
Helmholtz Resonator Model V L x K = M 2 ρoc S V 2 B = ρ o S Lʹ B L is the equivalent length of the neck (some air on either end also moves). S B F = PS B M x + Kx = PS B x = jωu B x = u B jω " j ωm K % $ 'u B = PS B ω z B = P " 1 % = j$ 2 S B u B S B ' " ωm K % $ ω ' z B 0 when ω = Damping due to viscosity in the neck are neglected K M = c SB Lʹ V (resonance frequency of the Helmholtz resonator) 35
Performance Measures Transmission Loss W i Wr A C B D W t Anechoic Termination 1 2 Transmission loss (TL) of the muffler:! TL =10 log 10 " $ ( ) db =10 log TL 10 S in 4S out W W A + S outb ρc + ρcc S in i t + S out S in D 2 % ' 36
Performance Measures Insertion Loss [ ] = T 0! Design A 0 B $ of 0 Mufflers and Silencers " C 0 D 0 % SPL 1 Z S [ T ] =! " A C B D $ % Z T Z S Muffler SPL 2 Z T! IL = 20 log 10 " $ A Z S + B Z T Z S + C + D Z T A 0 Z S + B 0 Z T Z S + C 0 + D 0 Z T % ' 37
Sound Wave Reflections in Engines Waves leaving muffler Waves leaving engine Muffler Engine Reflected from engine Reflected from muffler Reflected from muffler Reflected from open end Resonances can form in the exhaust and tail pipes as well as within the muffler. 38
Acoustic Source Source Impedance Waves Leaving Source Attenuating Element (i.e. Load) Reflected from Source Source Load Reflected from Attenuating Element z s u L p s p L z L p s z s + z L = p L z L 39
Transfer Impedance Incident Wave p1 p2 Reflected Wave u 1 = u 2 Transmitted Wave z tr u p p1 2 z tr = p 1 p 2 Su 40
Source/Load Concept Source z s, p s Load z L, p L L 1 L 2 Muffler z t, p t p s z s p L z L ( ) ( ) IL = f TL, z s, z t p t = f TL, z s, z t, p s 41
Insertion Loss Prediction 60 50 40 30 IL (db) 20 10 0-10 -20-30 Actual source impedance Pressure source (Zs=0) Velocity source (Zs=infinite) Anechoic source (Zs=rho*c) 0 200 400 600 800 1000 Frequency (Hz) 42
Summary 3 The transfer matrix method is based on plane wave (1-D) acoustic behavior (at component junctions). The transfer matrix method can be used to determine the system behavior from component transfer matrices. Applicability is limited to cascaded (series) components and simple branch components (not applicable to successive branching and parallel systems). 43