AE 8330 Advanced Architectural Acoustics 2014 HW3_Solution.pdf - Advanced Architectural Acoustics Architectural Engineering Program Univ of Nebraska AE

AE 8330 Advanced Architectural Acoustics 2014 HW3_Solution.pdf

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Unformatted text preview: Advanced Architectural Acoustics Architectural Engineering Program, Univ. of Nebraska AE 8330, Fall 2014 Homework ##3 – Solution (+110 total possible points) due Wednesday, Oct. 1 at the beginning of class Only select problems will be graded for 70% of the homework grade. The other 30% will be awarded for completing each of the other problems. To receive full credit for each problem, please show your work, work simplify expressions to a reasonable degree, label units appropriately, box your answers,, and staple the problems in order. 1. (+10) Show whether the following functions are solutions to the one one-dimensional dimensional wave equation. State clearly whether or not you think they are solutions and justify your answer. a. sin( x + 2ct ) b. (ct − x ) c. (ct + x)(4ct ) Solution: f = sin(x + 22ct ) a. 5 ∂f = cos( x + 2ct ) ∂x ∂2 f = − sin ( x + 2ct ) ∂x 2 ∂f = 2c[cos(x + 2ct )] ∂t ∂2 f = 4c 2 [− sin ( x + 2ct )] 2 ∂t 1 2 2 4c [− sin ( x + 2ct )] ≠ − sin (x + 2ct ) c [ No, not a solution b. f = (ct − x ) 5 ∂f ∂f 4 4 = 5(ct − x ) (c ) = −5(ct − x ) ∂t ∂x 2 ∂ f ∂2 f 3 2 3 = 20 ( ct − x ) c = +20(ct − x ) 2 2 ∂t ∂x 1 3 3 2 2 20(ct − x ) (c ) = 20(ct − x ) c [ Yes, this is a solution c. f = (ct + x)(4ct ) = 4c 2 t 2 + 4ctx ∂f = 8c 2t + 4cx ∂t ∂2 f = 8c 2 2 ∂t 1 2 2 8c ≠ 0 c ∂f = 4ct ∂x ∂2 f =0 ∂x 2 No, not a solution Advanced Architectural Acoustics 2. (+12) Consider the three fundamental linearized equations developed in class: (1) the Equation of Continuity, (2) Euler’s Equation, & (3) Equation of State. Using the hints given in class, combine these three equations into the following form of the wave equation: 1 ∂2s = ∇2s 2 2 c ∂t Solution: ∂s v r + ∇⋅u = 0 ∂t v v ∂u = −∇p 2. Euler’s Eqn: ρ 0 ∂t 3. Eqn. of State: p = βs 1. Eqn. of Continuity: r a. 1st eliminate u : v ( ) Take ∇ ⋅ of Eqn. 2: v ∂uv v v ∇ ⋅ ρ 0 = ∇ ⋅ (− ∇p ) ∂t ∂ v r ρ0 (∇ ⋅ u ) = −∇ 2 p ∂t Eqn. 4 b. Use Eqn. 1 in Eqn. 4: ∂ ∂s ρ 0 − = −∇ 2 p ∂t ∂t ρ0 ∂2s = ∇2 p ∂t 2 Eqn. 5 c. Use Eqn. 3 in Eqn. 5: 2 ∂ s = ∇ 2 (β s ) 2 ∂t ∂2s ρ0 2 = β∇ 2 s ∂t ρ0 ∂ 2 s 2 = ∇ 2 s β ∂t ρ0 Let c = β ρ0 » 1 ∂2s = ∇2s 2 2 c ∂t Advanced Architectural Acoustics 3. (Not graded; +17 for completing) Using Equation of State relationships: a. Beginning with the expression for an adiabatic process:  & (  (& ' ) , derive the following equation expressing the adiabatic temperature rise (∆T) produced in a gas by an acoustic pressure pp: % # 1   ∆  %  b. What is the amplitude of the temperature fluctuation produced by sound of intensity 10 W/m2 in air at 20° C and standard atmospheric pressure ( 1.01310  , assuming a plane wave? Solution: a. For adiabatic process:           →            Next eliminate ,  :                                ∶     &    ∆:           ∆       ∆  1       ∆  1    #  Use binomial expansion to linearize:    1    % # 1  ⋯ %  % # 1  ∆  1   #  %  % # 1   ∆  %  $1 Advanced Architectural Acoustics b. + 10 ,/./ 0 / + → 0 32 2+ 2 2 0 3241.21 5⁄.6 4344 ./ 410 ,/./ 91.24 Pa % # 1    %  .: /;6.°=4;./: > ' )=0.0756°A @ ∆ .: . 6? > Advanced Architectural Acoustics 4. (Not graded; +17 for completing) If there are body forces present in the fluid, a body r r force per unit volume F ( r , t ) (with units of N/m3) must be included in Euler’s Equation. Examples of this kind of force are those produced by a source that moves through the fluid without any change of volume, such as the cone of an unbaffled loudspeaker. . a. Show that if body forces and gravity are taken into account (where g is the v acceleration due to gravity and e z is the unit vector in the vertical direction), Euler’s Equation becomes: r r r r r r r v ∂u ρ + (u ⋅ ∇)u = −∇p + F (r , t ) − gρ e z ∂t b. Show that the linearized version of (3a) for a homogeneous, quiescent fluid is: r r r ∂u r v ρ0 + ∇p = F (r , t ) − gρ e z ∂t c. Use (3b) together with the linearized Equation of Continuity and the linearized Equation of State to derive the linear Wave Equation with body forces and gravity present. Solution: a. With the additional body force, force is expressed as FDGBH  ID ∙ 4D, K BH # BHL BCD #∇ FFFDM Because BCD DB. O FD FDQ DB. N  P FD ∙ ∇ FDR 4BH OK Combining these two equations and canceling dV, we get O FD FDQ FDG  ID ∙ 4D, K # LFFFDM  P FD ∙ ∇ FD #∇ OK b. Because   4  1 ; linearize equation 2a) based on   FD ≪   FD, we get  O FD FDG  ID 4D, K #  L #∇ FFFDM OK c. Combine all three equations to eliminate s and  FD, we get  1 O/G FD ∙ TID 4D, K # LDM # ∇ FDGU 0 ∇ 2 / OK / 1 O/G FD ∙ ID 4D, K # ∇ FDLDM # ∇/ G 0 ∇ 2 / OK / The term LDM only valid on the z-direction, expanding it becomes FDLDM ∇ O O LDM  LDM OV OV Therefore, the linearized Wave Equation with body forces and gravity becomes 1 O/G O FD ∙ ID 4D, K #  LDM # ∇/ G 0 ∇ / / 2 OK OV Advanced Architectural Acoustics 5. (+10) A plane wave in air at 20°C has LI = 90 dB (re 10-12 W/m2) What is the energy density and peak pressure for this wave? Solution: From T2.2 pg 54,  1.21 5/.6 , 2 344 ./ WX 10YZ4X X [\] , where +^_` 10/ cd w m2 + 10 × +^_` 1.00 × 106 f + + 2.91 × 10g Pa 2 w m2 0 / = / →f 2 2 2 2 / 0 32f 2 / 0.91 Pa 6. (+10) A plane wave in water @ 15°C has LP = 97 dB (re 1 Pa).What is the energy density and the time averaged intensity for this wave? Solution: From T2.2 pg 54,  998 5/.6 , 2 1450 ./ n 0[lm n , 0[\] W 10YZ4 where G^_` =1 Pa for water cq G^op 10/ × G^_` 70795 Pa +̅ / G^op w 3463 2  2 m +̅ 2.34 Pa 2 7. (+10) A simple source emits 0.032 W in a room with a room constant of 650 m2 (at the source frequencies). The source lies directly above a hard floor. Estimate the sound pressure level of this source for an observer located 1 meter from the source. f Solution: u Wu 10YZ u [\] . 6/  10YZ ' vwn ) 105.1 dB (re 1 pW) W Wu  10YZ x y 4  {A / 2z   : 105.1  10YZ |/}4n  g ~  0.1 97.4 dB (re 20 µPa) Advanced Architectural Acoustics 8. (+24) Charlotte is analyzing the acoustics of a multipurpose room that is 18’ tall x 36’ wide x 90’ long. The room is covered in materials with α500 of 0.02 for the floor and 0.08 for the walls. a. How many reflections per second does an average ray make with the walls in this room? Solution: 2€ 53.3 reƒlections/sec 4H ‹ℎLL € 24W? W  W? WM  W WM  b. What is the mean free path of the room? Solution:  : 21.2 ft=6.45 m Λ ‘ c. What is the average time between reflections for this room? Solution: K : “‘  ” 0.019 sec/reflections d. If the average loss in sound level per reflection is determined to be 0.8 dB at the 500 Hz octave band, what are the following quantities? i. the average absorption coefficient for the surfaces in the room at that octave band Solution: For each reflection, f^_`•_“–_— 41 # ˜ f™”“™—_”– f^_`•_“–_—  10 log41 # ˜ #0.8 Bš 10 log f™”“™—_”– ˜ 1 # 10  .œ  0.17 ii. the absorption coefficient of the ceiling material (assume uniform coverage) Solution: For 500 Hz, the average absorption  is calculated from ˜ ∑ €™ ˜™ €ž>•• 40.08  €`•ŸŸ^ 40.02  €“_™•™” 4˜“_™•™” ∑ €™ €ž>••  €`ŸŸ^  €“_™•™” Solve for ˜“_™• 0.44 iii. the reverberation time of the room at the 500 Hz octave band (neglecting air absorption) Solution: For ˜ ≥ 0.2, use Norris-Eyring’s equation for RT.  0.05H 1.41 sec #€ ∙ Y41 # ˜ Advanced Architectural Acoustics iv. the average modal damping constant for the 500 Hz octave band Solution: 6.91  ¢ ¢ £¤” ¥ ¢£¤” ¢¥ 4.91 s e. What is the reverberation time for each octave band shown, if variance from the mean free path is considered? Solution:  . 05H %/ #€ ∙ Y41 # ˜ x1  2 ln41 # ˜ { For the dimension z: x: y = 1:2:5, % / 0.403 from T5.1 on Blackboard. Therefore,  1.46 sec. ...
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