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HW12_solution473

# HW12_solution473 - ECE 473/TAM 413 Homework Assignment#12...

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Homework Assignment #12 Solutions - 1 ECE 473/TAM 413 Homework Assignment #12 Solutions Final Exam : Monday, May 5, 2008; 1:30 - 4:30 pm, 163 EL 1. A ventilation duct propagates a strong and very annoying signal at 250 Hz. It is desired to use a Helmholtz resonator to eliminate that signal. One engineer has already drilled a hole of diameter 1 cm in the thin walled duct. What volume must you make for the resonator in order to eliminate the annoying signal? L' = 1.6a = 1.6 0.005m ( ) = 0.008m , S = ! a 2 = 0.00007854m 2 , and c = 343 m/s V = c 2 S ! o 2 L' = 468 cm 3 2. Pipe 1 branches into pipes 2 & 3 at x = 0, where each pipe has the cross sectional areas: S 1 = 0.04m 2 , S 2 = 0.02m 2 and S 3 = 0.01m 2 . Pipes 2 & 3 can then be considered to extend to infinity. If the incident wave has a pressure amplitude of 10 Pa, then (a) determine the pressure amplitude of the reflected wave and (b) determine the pressure amplitudes of the waves transmitted down pipes 2 & 3. Let Z o be the acoustic impedance that the incident wave sees at x = 0. Then a) Z o = Z 2 Z 3 Z 2 + Z 3 = ! o c S 2 ! o c S 3 ! o c S 2 + ! o c S 3 = ! o c S 2 + S 3 R = p r p i = Z o ! " o c S 1 Z o + " o c S 1 = " o c S 2 + S 3 ! " o c S 1 " o c S 2 + S 3 + " o c S 1 = S 1 ! S 2 ! S 3 S 1 + S 2 + S 3 = 0.143 Therefore, p r = Rp i = 1.43Pa b) p 2 = p 3 = p i + p r = 10 + 1.43 = 11.43Pa 3. Problem 10.8.1 in Kinsler et al. Use flanged approach. For part (d), it may be useful to use the arguments on pp 284-285 of the text for estimating effective mass and effective stiffness. a) V = 4 3 ! 0.1 m 2 " # \$ % & 3 = 0.0005236 m 3 L' = 1.7a , S = ! a 2 , and c = 343 m/s ! o = c S L' V = 343 m / s " a 2 1.7a ( ) 0.0005236 m 3 ( ) = 20,377 a ! o = 2 " f = 2 " 320 / s ( ) = 2,010.6 = 20,377 a # a = 0.00974 m Diameter of opening = 2a = 1.95 cm

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Homework Assignment #12 Solutions - 2 b) From Eqs. 10.8.12 and 10.8.13, Q = P c P = 2 ! V L' S " # \$ % & 3 Thus, P + 20 μ bar P = 2 ! 0.0005236 1.7a ! a 2 " # \$ % & 3 where a = 0.00974 m P + 20 μ bar P = 59.536 ! P = 0.342 μ bar c) S new = 2S part(a) ; Radius new = 2 Radius (from part (a)) !
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HW12_solution473 - ECE 473/TAM 413 Homework Assignment#12...

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