HW5_solution473

# HW5_solution473 - ECE 473/TAM 413 Homework Assignment#5...

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Homework Assignment #5 Solutions - 1 ECE 473/TAM 413 Homework Assignment #5 Solutions Hour Exam 1: Wednesday, March 5, 2007, 7:00-8:30 pm Hour exams are open book (the text for the course) with one sheet of notes (you can use both sides). Your class notes and lecture notes are not permitted. You can use calculators. 1. What SPL (re: 20 μ Pa rms ) would be necessary in air (at 20˚C) to cause an rms temperature fluctuation of 1˚K? Provide your answer in dB. For your information, the answer to Kinsler et al’s problem 5.9.2a is ! T T o = " # 1 " \$ % & ( ) p P o \$ % & ( ) . p = P o ! T T o " # \$ % & ( ) 1 ( " # \$ % & = 101.3kPa 1˚K 293.16˚K " # \$ % & 1.402 ) 1 1.402 " # \$ % & = 1205 Pa (rms) SPL = 20 ! log 10 1205 Pa 20 μ Pa " # \$ % & = 155.6dB 2. Problem 5.12.8 in Kinsler et al. Use sea water at 13˚C. a) P o = 200 kPa I = p 2 2 ! o c o = 200,000 Pa ( ) 2 2 1026 kg / m 3 ( ) 1500 m / s ( ) = 12,995 W / m 2 b) ref:1 μ bar = 0.1Pa ! RMS value! SPL = 20 ! log 10 p p ref " # \$ % & = 20 ! log 10 200,000 Pa ! " # \$ % & = 123dB c) s o = k ! o = kU o " = P o # c 2 = 200,000 Pa 1026 kg / m 3 ( ) 1500 m / s ( ) 2 = 8.66 \$ 10 % 5 d) Recall the expression p = ! gh . In the case of water depth h, ! = 1026 kg / m 3 and g = 9.8m / s 2 . Also, recall that the pressure at the surface of the sea (sea level) is 1 atm = 101 kPa. Therefore, the additional pressure from water is 99 kPa (200-101 kPa). Thus, h = p ! g = 99 kPa 1026 kg / m 3 ( ) 9.8 m / s 2 ( ) = 9.8 m Thus, the water pressure of P o = 200 kPa is at a depth of 9.8 m 3. Calculate the missing values for a spherical outgoing wave propagating in the two liquids. distance from source (m) 0.1 0.1 frequency (kHz) 3 30

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Homework Assignment #5 Solutions - 2 angular frequency (kr/s) 18.850 188.496 propagation speed (m/s) 1500 1500 ambient temperature (˚C) 20 20 wave number (1/m) 12.57 125.66 wavelength (m) 0.500 0.050 equilibrium density (kg/m 3 ) 1000 1000 ratio of specific heats ( γ ) 1.01 1.01 equilibrium pressure (kPa) 100 100 isothermal bulk modulus (GPa) 2.228 2.228 adiabatic bulk modulus (GPa) 2.250 2.250 peak particle displacement amplitude (μm) 6.984 0.548 peak particle velocity amplitude (mm/s) 131.6 103.3 peak particle acceleration amplitude (m/s 2 ) 2481.4 19477.6 peak condensation amplitude (x10 -6 ) 87.76 68.89 peak acoustic pressure amplitude (kPa) 154.510 154.510 peak excess density amplitude (g/m 3 ) 87.8 68.9 SPL (re: 1 μPa rms ) 220.8 220.8 characteristic acoustic impedance (Mrayl) 1.500 1.500 average potential energy density (mJ/m 3 ) 2653 2653 average kinetic energy density (mJ/m 3 ) 4332 2669 average energy density (mJ/m 3 ) 6985 5322 average acoustic intensity (W/m 2 ) 7958 7958 total power of source (kW) 1.0 1.0 4. For a gas, develop an expression for the ratio of the discontinuity distance L d to the shock distance D, that is, L d /D. L d = 1 ! Mk = 2c o 2 " + 1 ( ) # U o and D = 2 !" B/ A P o p c + p r = ! c o /f ( ) ! # 1 P o p o L d D = ! " 1 ( ) ! + 1 ( ) # 5. a) A 20-kHz sonar source produces sound underwater in fresh water (assume B/A = 5 for water and 1.1 atmosphere ambient pressure at the depth of the source at 20˚C). If fully developed shock wave formation occurs (the discontinuity distance) at a distance of 1 m, then determine the SPL (re: 20 μPa rms) of the source. b) If the sonar source operates at 2 MHz (typically of diagnostic ultrasound frequencies) with the same SPL, then determine the distance that shock formation would occur.
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## This note was uploaded on 09/19/2009 for the course ECE 473 taught by Professor Obrian during the Fall '08 term at University of Illinois at Urbana–Champaign.

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HW5_solution473 - ECE 473/TAM 413 Homework Assignment#5...

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