MIT OpenCourseWare http://ocw.mit.edu 6.013/ESD.013J Electromagnetics and Applications, Fall 2005 Please use the following citation format: Markus Zahn, 6.013/ESD.013J Electromagnetics and Applications, Fall 2005. (Massachusetts Institute of Technology: MIT OpenCourseWare). http://ocw.mit.edu (accessed MM DD, YYYY). License: Creative Commons Attribution-Noncommercial-Share Alike. Note: Please use the actual date you accessed this material in your citation. For more information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms� � � � � � � � � � � � 6.013 - Electromagnetics and Applications Fall 2005 Lecture 22 - Acoustic Waves Prof. Markus Zahn December 8, 2005 I. Useful identity (ζ - any quantity) � �� �� d 1 dt V dV ζ =Δt V (t+Δt) dV ζ(t + Δt) − V (t) dV ζ(t) V(t+∆t)da = ndaV(t+∆t)υ∆tV(t)___d 1 1 dt dV ζ =Δt dV [ζ(t + Δt) − ζ(t)] + Δt ¯dV ζ(t + Δt) V V (t) ΔV =da v¯Δt� � · ∂ζ ¯ = dV + da v ζ(t + Δt)¯∂t · V (t) S ∂ζ = dV ∂t + dV � · [ζ(t)¯v] V (t) V (t) ∂ζ = dV ∂t + � · (ζv¯) V (t) II. Conservation of Mass (ρ - mass density) d ∂ρ dt dV ρ = 0 = dV ∂t + � · (ρv¯) V V ∂ρ Since V is arbitrary: ∂t + � · (ρv¯) = 0 (Conservation of mass) 1 Image by MIT OpenCourseWare.yPPPPPPx∆x∆y∆zz� �� � � � � � � � III. Conservation of Momentum, ith component (i = x, y, or z) � �� �� d ∂ dV ρvi = dV (ρvi) + �· (ρviv¯) = dV FT idt V V ∂tV ���� Total force density ∂ (ρvi) + �· (ρviv¯) = FT i ∂t∂ρ ∂vi vi ∂t + ρ ∂t + vi� · (ρv¯) + ρ(¯v · � )vi = FT i � ∂ρ �� ∂vi � vi ∂t + � · (ρv¯) + ρ ∂t + (¯v · �)vi = FT i =0 (mass conservation) ρ ∂∂t v¯+ (¯v · �)¯v = F¯ T IV. Force density due to pressure (force/area) F¯ p = {− [p(x + Δx) − p(x)]¯ixΔyΔz − [p(y + Δy) − p(y)]¯iyΔxΔz − [p(z) − p(z − Δz)]¯izΔxΔy} ΔxΔ1 yΔz = −[p(x + ΔΔxx ) − p(x)]ix − [p(y + ΔΔyy ) − p(y)]iy − [p(z) − Δp(zz − Δz)]iz ¯ ¯ ¯∂p ∂p ∂p ¯ ¯ ¯ = − ∂x ix + ∂y iy + ∂z iz = −�p V. Governing Fluid Equations ∂ρ ∂t + � · (ρv¯) = 0 ∂v¯ρ ∂t + (¯v · �)¯v = −�p 2Image by MIT OpenCourseWare.VI. Small Perturbations About Equilibrium of Stationary Fluid ρ = ρ0 + ρ� (ρ� � ρ0) v¯ = 0 + ¯v� p = p0 + p� � ∂ρ� + ρ0� · v¯� � = 0 ∂ρ� + ρ0� · v¯� = 0∂t ∂t ∂v¯� + (¯⇒ ∂v¯�ρ0 ∂t � v� · ��� )¯v�� = −�p� ρ0 ∂t = −�p� =0 second order VII. Pressure / Density Constitutive Law A. Ideal Gas - p = ρRT , R is the Gas Constant = Rg/ molecular weight in grams Joules1. Isothermal (T constant) Rg = 8.31 × 103 kg (mole) K p = ρRT p� = RT ρ� where “mole” indicates the molecular weight in grams. 2. Adiabatic ∂p γp γp0 = p = constant ργ p� = ρ� ∂ρ ρ ⇒ ⇒ ρ0 cp 5 γ = = ratio of specific heats = (monatomic ideal gas) cv 3B. Liquid or Solid ∂p κ κ = p� = ρ�, where κ is the Bulk Modulus ∂ρ ρ ⇒ ρ0 VIII. Acoustic Wave Equation � ∂v¯� ∂ 2� · ρ0 ∂t = −�p� ⇒ ρ0 ∂t(� · v¯�) = −� · (�p�) = −� p� ∂ρ� 1 ∂ρ� ∂t + ρ0� · v¯� = 0 ⇒ � · v¯� = −ρ0 ∂t ∂2ρ�+�2 p� = + ∂t2 � �1/2 p� nt nt-m m2 cs = ρ� (Units: m2 kg = kg =s2 = (velocity)2 3m⎧ √RT Isothermal Ideal Gas ⎨� γp0 cs = ρ0 Adiabatic Ideal Gas ⎪⎪� ⎩ κ Liquid or Solidρ0 3 ⎪⎪� � � � In air: (Adiabatic, γ = 1.4), ρ0 = 1.29 kg/m3, p0 = 1.01×105 nt 2 (1 atmosphere), cs = γp0 mρ0 ≈ 330 m/s In water: cs ≈ 1500 m/s ρ�2 = pc�2 ⇒ �2 p� = c1 ∂∂t2p2 � (cs is the speed of sound) 2 2 s s ∂2p� ∂2p� ∂2p�2� p� = ∂x2 + ∂y2 + ∂z2 IX. Acoustic Waveguide A. Parallel plate waveguide p� = Re pˆ(x)ej(ωt−kz z) 1 ∂2p� d2pˆ ω2 �2 p� = c2 ∂t2 ⇒ dx2 − kz 2 pˆ = − c2 pˆd0XZYυx(x = 0) = 0υx(x = d) = 0s s d2pˆ+ � ω2 − k2 � pˆ = 0 dx2 c2 z � s �� � k2 x d2pˆ+ k2 pˆ = 0 pˆ(x) = A sin(kxx) + B cos (kxx)dx2 x ⇒ ∂v¯� dpˆρ0 ∂t = −�p� ⇒ ρ0jωvˆx = −dx = −kx [A cos(kxx) − B sin(kxx)] ρ0jωvˆz = jkzpˆvˆx(x = 0) = 0 A = 0 ⇒ vˆx = Bkx sin(kxx)ρ0jω vˆx(x = d) = 0 sin(kxd) = 0 kxd = mπ, m = 0, 1, 2, . . . ⇒ ⇒ pˆ(x) = B cos(kxx) v¯ = vx ¯ix + vz ¯iz = Bkx sin(kxx)¯ix + Bkz cos(kxx)¯izρ0jω ρ0ω ω2 �mπ �2 k2 + k2 + k2= = x z z c2 ds ω2 �mπ �2 kz = 2c− ds 4Image by MIT OpenCourseWare.� � � � � m = 0 - TEM mode v¯ = Bkz ¯iz = B ¯izρ0ω ρ0cs p = B ηs = p = ρ0cs is the Acoustic Impedance vz B. Rectangular Acoustic Waveguide y = 0y = bx = ax = 0YXZB.C. ux(x = 0) = ux(x = a) = 0 uy(y = 0) = uy(y = b) = 0 p� = Re pˆ(x, y)ej(ωt−kz z) pˆ(x, y) = [A sin(kxx) + B cos (kxx)] [C sin(kyy) + D cos(kyy)] kx 2 + ky 2 + kz 2 = ω2/c2 s 1 ∂pˆ vˆx = −ρ0jω ∂x 1 ∂pˆ vˆy = −ρ0jω ∂y vˆz = kz pˆρ0ω pˆ = A cos(kxx) cos(kyy) vˆ¯ = − A [− ¯ixkx sin(kxx) cos(kyy) − ¯iyky cos(kxx) sin(kyy) + kz ¯iz cos(kxx) cos(kyy)]ρ0jω mπ kx = , m = 0, 1, 2, . . . a nπ ky = , n = 0, 1, 2, . . . b ω2 �mπ �2 �nπ �2 =kz c2 − a − bs 5Image by MIT OpenCourseWare.� � � � � � � � � � � � � � X. Poynting Theorem v¯� ρ0 ∂v¯� ∂ 1 ρ0v¯� 2 = −(¯· ∂t = −�p� ⇒ ∂t 2| | v� · �)p� 1 ∂ρ� 1 1 ∂p� ∂ 2�p2 s p� � · v¯� p�(� · v¯�) = −= − = − ⇒ 2ρ0c2 ρ0 ∂t ρ0c ∂t ∂t s p�2∂ 1 1 ρ0|v¯�|2 + ∂t 2 = −� · (p�v¯�)2 2 ρ0cs 1 1 p�2 v¯�¯ d 2 +Integral form: p�v¯�da dV ρ0= − dt | | · 2 2 ρ0c2 sS V
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