Ultrasonic Solution

SOLUTION_QUESTION_BANK_ULTRASONIC_SOUMYA

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EI703 (a)

Soumyadev Adhikari 7 th Semester Instrumentation and Control Engineering Academy Of Technology Aedconagar-712121, Hooghly

DECEMBER, 2008 P.T.O.


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Part 1 Objective and MCQ

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Objective Type Questions (1): 1. What is ultrasonic wave? Sound wave with frequency greater than 20kHz (frequency of sound wave audible to human being is 20-20000Hz) is called ultrasonic wave. It obeys all the laws that the low frequency sound wave obeys. The ultrasonic frequency range used for non destructive testing and thickness gauging is about 100kHz to 100MHz 2. What is ultrasonic frequency range? Any sound wave having frequency above 20kHz is ultrasound. The frequency of ultrasound may be 200MHz also. 3. Mention the name of variable affecting the propagation of ultrasonic wave. Medium inertia and elasticity are these variables. 4. What do you mean by acoustic impedance offered by any medium? When wave travels through a medium, the medium offers an impedance to the wave to propagate. In case of longitudinal wave (sound) this impedance is called Acoustic impedance. From the force-voltage analogy, acoustic impedance = Acoustic pressure / particle velocity, velocity, where acoustic pressure is analogous to voltage and particle velocity is analogous to current. Acoustic pressure generates due to wave propagation. It can be shown that if acoustic impedance = Z, then E Z = ρ

Where, ρ = Density of the medium, E = Bulk modulus of the medium. 5. Why attenuation of sound occurs in absorbing medium? Attenuation is the exponential decrement of the amplitude of sound wave when it propagates through the absorbing medium. This is occurred because when sound wave propagates through an absorbing medium (e.g. sound wave in a stretched string immersed in a viscous liquid), the medium absorbs the energy of the sound wave as it propagates. 6. What is the relation between power level and total energy transmitted per unit length in case of transverse wave propagating through a stretched stribg? Power level = (Total energy per unit length) * (wave velocity) 7. What do you mean by ultrasonic wavefront? It is the plane perpendicular to the direction of propagation of ultrasonic wave, where the phase and the velocity of all waves are equal.

Figure 1: Ultrasonic wavefront

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8. Draw the graph that shows the decay of amplitude of the wave with distance in an absorbing medium.

Figure 2: Waves in absorbing medium

9. What do you mean by wave number of a wave? The wave number (n) is the number of complete waves per unit length. Therefore, n = 2 π / λ OR [ Wavenumber in most physical sciences is a wave property inversely related to −1 wavelength, having SI units of reciprocal meters (m ). Wavenumber is the spatial analog of frequency, that is, it is the measurement of the number of repeating units of a propagating wave (the number of times a wave has the same phase) per unit of space.] 10. Define phase velocity of the ultrasonic u ltrasonic wave.

Figure 3: Planes of equal phases

Wave velocity is called the phase velocity (V). It is the velocity with which planes of equal phases (for transverse wave, crest & trough and for longitudinal wave compression & rarefaction) travel through the medium. V = ν ∗ λ Where ν & λ = frequency & wavelength respectively of the wave

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11. What is the function of wear plate of the th e ultrasonic transducer?  Basic purpose is to protect the active element from the external testing environment. In case of contact transducers, the wear plate must be a corrosion resistant material (steel).  Also a wear plate having proper impedance (that depend upon the impedances of active element & test medium) and proper size ( λ / 4) can allow as much as energy possible to come out from the active element. 12. Name two materials that show piezoelectric effect. Two materials are barium titanate & quartz with 12*10^-3 Vm/N & 50*10^-3 Vm/N voltage sensitivity respectively. 13. Which parameters are measured in pulse echo method? A material in-homogeneity, when illuminated by a pulsed ultrasonic beam, reflects an echo which is measured by receiver probe. That is, the amplitude of the echo and the transit time of the pulse from the transmitter to the reflector and back are measured. 14. What is echo repetition frequency? 15. What is the test method by which US frequency can be measured? In transit time method, the transit time or the corresponding US (ultrasonic) frequency is measured. Also in frequency modulation method, US frequency is measured. 16. What do you mean by ring down effect? In the single element transducers, the element changes its status between transmitter mode and receiver mode alternatively. Now if the time taken by the element to change its status from one mode to another is anyhow less than the total transit time of the ultrasound, a problem occurs. For example, if the element takes 5mS to become a receiver from a transmitter and the total transit time (taken by the ultrasound to cover the path between TX and RX) is 2Ms, then in time of receiving the ultrasound, the element will can not be acting as a receiver, it will be taking 3 more mS to become a receiver. So, nothing can be received. This is called Ring Down Effect. It can be removed by use of dual element transducers, delay line transducers (where, an acoustic path provides a delay between the generation and arrival of the ultrasound). 17. Give two example of couplant? In contact ultrasonic testing the couplants are generally a thin film of oil, grease, glycerin and water used between transducer and test body. In immersion ultrasonic testing the transducer and the test body are immersed in the couplant which is generally water. 18. What do you mean by side lobe of ultrasonic transducer?

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19. What is necessary of suspended particles or bubbles in the liquid for ultrasonic flow measurement? Actually the suspended particles or bubbles act as the reflector of ultrasonic waves in the liquid e. g. in Doppler flow-meter, the particles act as moving source and moving observers. And the reflections of ultrasound in the particles cause the frequency shift. 20. What is reverse piezoelectric effect? If a varying potential is applied to the proper axis of a piezoelectric crystal then the dimension of the crystal will be changed thereby deformation in the crystal will be occurred. This is called reverse piezoelectric effect. 21. Which is better, magnetostrictive or piezoelectric transducer? Piezoelectric transducer is better because it has:i. Lower cost. ii. Smaller size. iii. Ease of fabrication. iv. Higher efficiency. 22. What is the major factor that severely limits the transmission transmission and detective devices used in Submarines Su bmarines using ultrasonic wave?

Figure 4: transmission of ultrasound from steel to water

Reflection coefficient of sound energy from steel to water is given by RC = ((Z1 – Z2) / (Z1+Z2)) Where, Z1, Z2 = acoustic impedance of steel & water respectively It is observed that the approximate value of RC in normal condition = 0.85 So, 85% of the energy transmitted by a transmitter from the inside of steel bodied submarine will be reflected from the interface between submarine wall (steel) and water and only 15% can be transmitted to destination. Therefore, the transmitted signal becomes very weak making the transmission doubtful.

Choose the correct alternative 23. Propagation of ultrasonic wave through the material medium can be treated as

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i. Isothermal ii. Adiabatic iii. Both (i) & (ii) iv. None of these 24. If a string of mass 1 gm/m is stretched with a force of 10N, the velocity of transverse wave on the string is i. 10m/s ii. 1000m/s iii. 200m/s iv. 100m/s 25. For a loss less medium, the impedance is i. Real ii. Complex iii. Both (i) & (ii) iv. None of these 26. In case of a wave propagating through an absorbing medium, the amplitude with distance i. Increases linearly ii. Decreases linearly iii. Falls exponentially iv. None of these 27. If E is the bulk modulus of a loss free gas and ρ is its density, the characteristic impedance offered by the gas to the sound wave traveling in it is given by i. Z = ρE 2 ii. Z=ρ E 0.5 iii. Z = (ρE) iv. none of these 28. During propagation of ultrasonic wave, if there is a rigid wall between two media, the incident wave is i. Completely transmitted ii. Completely reflected iii. Partly reflected and partly transmitted iv. None of these 29. The condition for which all the incident energy with the incident wave is transmitted with no reflection is that impedance of the th e coupling medium is i. Harmonic mean ii. Arithmetic mean iii. Product iv. Sum of the two impedances to be matched 30. The sum of reflection & transmission coefficient at junction between two media is i. Zero

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ii. One iii. Between zero and one iv. None of these 31. All the energy arriving at the boundary with the incident wave leaves the boundary with the i. Reflected wave ii. Transmitted Both (i) & (ii) iii. iv. None of these 32. If Z, ω, A are respectively impedance, angular frequency, amplitude the the rate at which energy is carried per unit length along with string is 2 i. Zω A 2 ii. ZωA 2 2 iii. 0.5 Zω A iv. None of these 33. The piezo-electric material which are used to design ultrasonic trans-receiver is i. barium chloride ii. potassium nitrate iii. ADP and barium chloride iv. None of these 34. Ultrasonic used for Doppler flow meter is of i. Low frequency ii. Medium frequency iii. Both (i) and (ii) None of these iv. 35. in case of ultrasonography, jelly used between probe and body surface for the purpose of i. pain relief ii. removal of etching iii. coupling iv. None of these 36. The mounting of ultrasonic sensors at the ________ of a tank for the measurement of level of fluid is advantageous i. Top ii. Side iii. Base Both (i) and (iii) iv.

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Part 2 5 Marks Questions

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Short answer type Questions (5): 1. What is ultrasound? Write down the advantages ad vantages of ultrasonic measurement. (2+3) Sound wave with frequency greater than 20kHz (frequency of sound wave audible to human being is 20-20000Hz) is called ultrasound. Actually pressure variations travel through a medium at a velocity of sound. The term ultrasonic refers to pressure disturbances (usually are short bursts of sine waves) whose frequency is above 20 kHz. It obeys all the laws that the low frequency sound wave obeys. The ultrasonic frequency range used for non destructive testing and thickness gauging is about 100 kHz to 100MHz.

Advantages of ultrasonic measurement  Higher frequency ultrasounds have shorter wave length. So, the diffraction or bending of the wave around the obstacle of given dimension is reduced. Therefore, it is easier to focus or to direct a beam of ultrasound.  Ultrasound can pass easily through the metal wall. That means that the entire measurement system can be mounted completely external to the fluid i.e. it is non invasive. It is important for hostile fluids such as corrosive acids, radioactive, explosives etc.  Ultrasound can propagate through biological tissues without causing any harm to them. So, it is applicable in medical field.  The silence of ultrasound makes it useful in military applications. 2. Distinguish between particle velocity & wave velocity and obtain the relation between them. (5) For, any types of waves (transverse or longitudinal), the particle velocity refers to the velocity of vibration of tiny particles of the medium (the vibration is responsible for the generation of the wave), while the wave velocity is the velocity at which the wave propagates through the medium as a result of the vibrations of tiny particles of the medium.

Figure 1: Transverse wave

We know the displacement of a general particle in a transverse wave (fig-1) is given by: y(x,t) = Asin ((2 π/λ)(Vt−x)) where, y=displacement of particle x=travel of wave in +ve X direction

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V=velocity of wave in +ve X direction A=amplitude of displacement curve Particle velocity = dy/dt = (2AV π/λ) cos ((2π/λ)(Vt−x)) = U (say) And, slope of the displacement curve = dy/dx = −(2Aπ/λ) cos ((2π/λ)(Vt−x)) = m (say) Therefore, U = −(2Aπ/λ) cos((2π/λ)(Vt−x)) * V = m*V Therefore, U = m*V 3. A simple harmonic wave of amplitude 8 units transverse a line of particles in the direction of +ve X axis. At any instant of time, at a distance 10cm 1 0cm from origin, the displacement is +6 units, and for a particle at a distance 25cm from the origin, the displacement is +4 units. Calculate the wavelength. (5)

Figure 2: Situation of question 3

We know, the displacement of particle for +X direction travel of the wave is given by: y(x,t) = Asin((2π/λ)(Vt−x)) -----------------(1) Given that, at a time instant, y = +6 units for x = 10cm and y = +4 units for x = 25cm Putting these values in (1) we get, 6 = 8sin ((2 π/λ)(Vt−10))--------------------------(2) and 4 = 8sin ((2π/λ)( Vt−25))--------------------------(3) From (2) we get, (6/8) = sin ((2 π/λ)( Vt−10)) or, 0.85 = (2π/λ)(Vt−10) or, 0.135 = Vt/ λ−10/λ λ−10/λ or, Vt/ λ = 0.135 + 10/λ ---------------------------(4) From (3) we get, (4/8) = sin ((2 π/λ)( Vt−25)) or, 0.52 = (2π/λ)(Vt−25) or, 0.08 = Vt/ λ−25/λ λ−25/λ or, Vt/ λ = 0.08 + 25/λ -----------------------------(5) From (4) & (5) we get 0.135 + 10/λ = 0.08 + 25/λ Therefore,

λ = 272.72cm

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4. Show that the characteristic impedance offered o ffered by a string to the th e traveling waves is given by Z = µ T ; where y is the linear density of the string and T is the tension with which it is stretched.

Figure 3: Situation of question 4

The periodic transverse force F = F oCosωt is applied on the string at x = 0 position. Now, particle displacement is given by: y(x,t) = Asin((2π/λ)(Vt−x)) --------------------(1) Now, dy/dx = −(2Aπ/λ) cos ((2π/λ)(Vt−x)), dy/dt = (2AVπ/λ)cos ((2π/λ)( Vt−x)) Therefore, (dy/dx) = −(1/V) (dy/dt) -----------------------------(2) Now, from fig-3, F = Focos ωt = −Tsinθ = −Ttanθ [Q for, small θ, sinθ = tanθ] = −T(dy/dx)x=0 [Q tanθ = Slope of curve at x = 0 = (dy/dx) x=0 ] = (T/V) (dy/dt) [from 2] = (T/V) (2AVπ/λ)cos ((2π/λ)(Vt−x)) x=0 = (T/V) (Vo) cos(2Vtπ/λ) (Vo = 2AVπ/λ) = (T/V) (Vo) cos( ωt) (ω = 2vπ/λ)

∴ Focosωt = (T/V) (Vo) cos(ωt) Now, characteristic impedance is given by, Z = (Transverse force amplitude) / (Transverse velocity amplitude) = Fo /Vo = T/V = µ T

[Q velocity of transverse wave on a string is V = T / µ ]

5. Define intensity of sound wave. Also prove that intensity of sound wave depends on amplitude, frequency and impedance. Intensity: - When wave travels through a medium, energy is transported from one part to another. The intensity of a sound wave is defined as the average energy crossing a unit cross sectional area perpendicular to the direction of propagation of the wave per unit time. Or, it may also be stated as the average power transmitted across the unit cross sectional area perpendicular to the direction of propagation of the wave.

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Consider an elemental volume ∆v containing n number of particles of a pipe containing a gas of density ρ through which sound wave is propagating. Energy of oscillation of each particle is given by: 2 2 E = mA ω / 2 Where, m = mass of particles, A= amplitude of their oscillation, ω = angular velocity of particle.

Figure 4a: Sound in a pipe

Figure 4b: Two consequent conse quent wave fronts

Now, E for n number of particles of that elemental volume ∆v is = nE 2 2 And, nE = (nm)A ω / 2 = (ρ∆v) A 2ω 2 / 2 [Q nm = ρ∆v] Now, energy per unit volume = U = nE / ∆v = ρ A 2ω 2 / 2 = ρ A 2(4 π 2 ν 2) / 2 [Q ω = 2πν] 2 2 2 = 2ρ A π ν Now, consider fig-4b. Here two consequent wave fronts are considered. The separation between them is given by ∆x = V∆t Where, V = velocity of sound, ∆t = time taken by the wave two form the second wave front after the first one Volume of the area between two wave fronts = S ∆x Where, S= wave front plane’s area Energy received and transmitted by this volume = U S ∆x Energy transfer per unit time per unit area i.e. the intensity is, therefore, given by I = U S∆x / S ∆t =UV [Q V= ∆x / ∆t] 2 2 2 = 2ρ A π ν V

I= 2ρ A 2 π 2 ν ν2 V 6. Define acoustic impedance. Prove that acoustic impedance offered by a loss E . Where, ρ is the free gas to the sound wave traveling in it is given by Z = ρ

density of the gas and E is the bulk modulus.

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Acoustic impedance: - When wave travels through a medium, the medium offers an impedance to the wave propagating. In case of longitudinal wave (sound) this impedance is called Acoustic impedance. From the force-voltage analogy, acoustic impedance = Acoustic pressure / particle velocity, where acoustic pressure is analogous to voltage and particle velocity is analogous to current. Acoustic pressure generates due to wave propagation.

Figure 5: Sound wave in pipes

Let us consider a pipe of uniform cross sectional area α which is filled with a lossfree gas of density ρ and through which sound wave is propagating along positive X direction (from left to right horizontally in fig-5). Consider two planes, separated by a distance ∆x, perpendicular to the axis of the pipe. Volume of this element is v = (α * ∆x )Due to the presence of a tuning fork at the left end of the pipe, the displacements of the particles in the planes A & B are respectively given by y(x1) & y(x2). The displacements might not be equal. At equilibrium, the particles of two planes experience the same pressure(P o), but due to propagation of wave, the change in pressure occurs. This is the acoustic pressure. Now, increment in length ∆x is = y(x2) − y(x1) Increment in volume V is ∆V = α * (y(x2) − y(x1)) = ∆v = α(δy/δ x)∆x [Q y(x2) = y(x1) + ((δy/δ x)x1)∆x +…… by Taylor’s series expansion with x2 = x1+∆x, and neglecting higher order terms]

∴ Volume strain = ∆v/v = (δy/δx) Now, bulk modulus E = −(change in pressure / volume strain) = − (∆p) / (∆v/v) ∴ ∆p = − E(∆v/v) = − E(δy/δx) -2 -1 Now, acoustic impedance = Z (rayls or kgm s ) -1 = Acoustic pressure (Pa) / particle velocity (ms ) = ∆p / (δy/δt) = (− E(δy/δ x)) / (δy/δt) = (− E )∗[−(2Aπ/λ) cos ((2π/λ)(Vt−x)] / [(2AVπ/λ) cos ((2π/λ)(Vt−x))] = E/V (V = velocity of sound) = ρ E

[Q we know V = E / ρ ]

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7. Derive the expression for intensity of sound in terms of maximum pressure.

Figure 6: Intensity of sound in a lossfree gas

Let us consider a pipe of uniform cross sectional area α which is filled with a lossfree gas of density ρ and through which sound wave is propagating along positive X direction (from left to right horizontally in fig-6). Consider two planes, separated by a distance ∆x, perpendicular to the axis of the pipe. Volume of this element is V = (α * ∆x )Due to the presence of a tuning fork at the left end of the pipe, the displacements of the particles in the planes A & B are respectively given by y(x1) & y(x2). The displacements might not be equal. At equilibrium, the particles of two planes experience the same pressure(P o), but due to propagation of wave, the change in pressure occurs. Now, increment in length ∆x is = y(x2) − y(x1) Increment in volume V is ∆V = α * (y(x2) − y(x1)) = ∆v = α(δy/δ x)∆x [Q y(x2) = y(x1) + ((δy/δ x)x1)∆x +…… by Taylor’s series expansion with x2 = x1+∆x, and neglecting higher order terms]

∴ Volume strain = ∆v/v = (δy/δx) Now, bulk modulus E = −(change in pressure / volume strain) = − (∆p) / (∆v/v) ∴ ∆p = − E(∆v/v) = − E(δy/δx) = − E(−(2Aπ/λ) cos ((2π/λ)(Vt−x))) [putting the value of δy/δx]

∴ Maximum excess pressure = P max = E A (2π/λ) = V 2ρA(2π/λ) [Q we know V = E / ρ ] = 2 πA ρνV [Q V/ λ = ν = frequency]

2 2 2 We know, Intensity of sound wave in in a gas is I = 2 π A ρν V 2 = (2 πA ρνV) / 2ρ V = (Pmax) / 2ρ V ∴I= (Pmax) 2 / 2ρ V

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8. (a) Define decibel (dB). (b) A sound is twice intense that another. a nother. What is their intensity ratio in dB? (c) What is attenuation? What is its unit? (a) Decibel is a unit to express the ratio of two values in the following manner: dB = 10log 10 ( value1 / value1 / value2 ) Generally it is a unit (dB) for measuring the relative strength of signal power. The number of decibels equals ten times the logarithm (base 10) of the ratio of the measured signal power to a reference power. Decibel is one tenth of a bell. Generally it is a logarithmic unit of sound intensity; 10 times the logarithm of the ratio of the sound intensity to some reference intensity.

(b) Let the intensities are I 1 = 2I & I2 = I. And the reference intensity is I o Intensity ratio in normal scale = I 1 : I2 Therefore, intensity ratio in dB is = 10log 10 ( I1 / I2) = 10log 10 ( 2I / 2I / I ) = 10log 10 ( 2 ) = 3.010299957 (c) When sound wave travels in an absorbing medium, the amplitude or intensity of the sound wave is reduced with the distance covered. This phenomena is known as attenuation. Units of attenuation are dB, neper. 9. Explain the terms “Attenuation constant” and “Attenuation length”. Give the relation between attenuation constant and energy flux.

Figure 7: Attenuation

Attenuation constant: - It is given by the decrement in amplitude, when wave travels in an absorbing medium, per unit amplitude per unit length. We know, the amplitude of wave that decreases with distance in absorbing medium is given by: A(x) = Ao exp(−αx) or dA(x)/dx = −αAo exp( −αx) = −αA(x) ∴ α = − (1/A(x)) (dA(x)/dx)

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Attenuation length: - It is the distance traveled by wave over which the amplitude is decreased by a factor “e = 2.718”, i.e. for x = 1/ α, A(x) = Ao exp(−1) = Ao /e. Now, We know, energy flux = intensity ∝ (Amplitude) 2 ∝ [A(x)] 2 = Κ (Ao) 2 exp (−2αx) [K = constant] 2 = Io exp (−2αx) [Io= ΚAo ] ∴ Energy flux =intensity = I = Io exp (−2αx) 10. Explain clearly what you mean by dispersive medium and dispersive relation. The displacement of particles of wave when traveling in an absorbing medium in positive X direction s given by: Y(x,t) = Ao exp(−αx) exp (i *(2 π/λ)∗(Vt – x)) = Ao exp[(2 πi/λ)(Vt – x)−αx] = Ao exp[(2Vπi/λ)(t – x/V)−αx] = Ao exp[2 νπi(t – x/V)−αx] [Q V/ λ = ν] = Ao exp[2 νπi(t – x/V−αx/2 νπi)] = Ao exp[2 νπi{ t – x(1/V+α /2 νπi)}] = Ao exp[2 νπi{t – x(1/V−αi/2 νπ)}] = Ao exp[2 νπi(t – x/V*)]

Where, 1/V* = 1/V −α i/2 νπ This V* is called the complex wave velocity applicable for the waves traveling through absorbing medium. In practice this method of introducing the complex wave finds application chiefly in optics in which we define the refractive index as n=c/V Where, c is the velocity of light in vacuum and v is the velocity of light at that medium. In absprbing medium the complex refractive index is given as: n* = c / V* = c / V − cαi/2 νπ = n − cαi/2 νπ This shows that the refractive index (and hence the wave velocity) depends on the frequency ν (or the wavelength λ ) of the wave in an absorbing medium. Such a medium is called dispersive medium and this relation between wave velocity and frequency (or wavelength) is called dispersive relation. 11. Define the terms Intensity Level (IL) and Sound Sou nd Pressure Level (SPL). Give the relation between them. Intensity Level: - It is defined as the ten times logarithmic (base 10) of the ratio of the sound intensity (I) to the reference intensity (I o). ∴ IL = 10log 10 (I/Io) [in dB] Sound Pressure Level: - It is defined as the twenty times logarithmic (base 10) of the ratio of the measured effective sound pressure to the reference pressure. ∴ SPL = 20log 10 (P r.ms /Po, r.ms)

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SOLUTION_QUESTION_BANK_ULTRASONIC_SOUMYA Now, IL = 10log 10 (I/Io) = 10log 10 ( P

2

ρ

17 Ι

r.ms / / oV ο)

[Q We know I = P 2max / 2ροV = ( 2 P r.ms) 2 / 2ροV = P r.ms 2 / ροV ]

∴ IL = 10log 10 (P 2 r.ms) −10log 10 (ροVΙο) = 20 log 10 (P r.ms) −10log 10 (ροVΙο)

∴ 20 log 10 (P r.ms) = IL + 10log 10 (ροVΙο)

--------------------------(1)

Now, SPL = 20log 10 (P r.ms /Po, r.ms) = 20log 10 (P r.ms)−20log 10 (Po, r.ms) = IL + 10log 10 (ροVΙο)−20log 10 (Po, r.ms) = IL + 10log 10 (ροVΙο)−10log 10 (P = IL − 10log 10 (P

2

o, r.ms)+

2

o, r.ms)

10log 10 (ροVΙο)

= IL − 10log 10 (P 2o, r.ms / ροVΙο)

∴ SPL =

IL − 10log 10 (P

2

o, r.ms

/ ροVΙο)

12. (a) The intensity of sound in a normal conversation at home is about 3*10^-6 2 W/m and the frequency of normal human voice is about 1000Hz. Find out the amplitude of waves, assuming that the air a ir is at standard condition. (Given; density of air = 1.29 kg/m^3 & velocity of sound =332m/s) 2 2 2 2 We know, the intensity of sound is I = 2 ρπ A ν V = 3*10^-6 W/m Where, ρ = density of traveling medium = 1.29 kg/m^3 A = Amplitude of the wave (to be calculated) ν = Frequency of the wave = 1000 Hz V = Velocity of the wave = 332m/s 2 2 ∴ Now, I = 2ρπ2A ν V gives: 2 2 2 3*10^-6 = 2* 1.29 * π ∗ A ∗ 1000 ∗ 332 or Α = 1.88 ∗ 10 ^-8 m = 18.8 nm 2 (b) A sound has an intensity of 1µW/cm . if the intensity of another sound is 14 dB higher, what is the intensity of the latter in dB? Let the two intensities in SI unit is I1 (1 µW/m 2) & I2. Io is the reference one. Now I1(dB) = 10log 10 (I1/I o), I2 (dB)= 10log 10 (I2/Io)

Given that, I2(dB) − I1(dB) = 14 dB or 10log 10 (I2/Io) −10log 10 (I1/Io) = 14 or 10log 10 (I2/I1) = 14 or log 10 (I2/I1) = 1.4 or I2/I1 = 10^1.4 or I2 = 10^1.4 * I1 = 10^1.4 * 1 (µW/cm 2) = 25.11 µW/cm 2

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= 25.11*10^-2 2 = 0.25 W/m (in SI unit) 13. (a) What is the need for matching layer in ultrasonic transmitter? The needs are: For impedance matching. The TX and the test body have different impedances. As a result, such a huge portion of transmitted wave may be reflected back to TX portion from the interface of TX and test body that the transmitted wave will be not enough to be detected by the receiver. The matching layer is used to transmit as much as transmission possible with nearly zero reflection from the interface.  Sometimes the matching layer provides isolation to the active element from harmful test bodies. (b) Derive the expression for characteristics cha racteristics impedance of matching layer for maximum over all transmission coefficients.

Figure 8: Situation in problem 13(b)

Transmission energy coefficients (from medium1 to 2 and 2 to 3) are given by:α12 = 4ZZ1 / (Z+Z1) 2 and α23 = 4ZZ2 / (Z+Z2) 2 respectively. But, transmission energy coefficient from medium 1 to 3 is given by:α13 = α12 * α23 = 16Z 2Z1Z2 / (Z+Z1) 2 (Z+Z2) 2 Now, for complete transmission from medium1 to 3, α13 should be maximum, i.e. d (α13) / dZ = 0 Or

2

2

2

2

2

[(Z+Z1) (Z+Z2) 32Z Z1Z2] - 16Z Z1Z2[(Z+Z1) 2(Z+Z2)+ (Z+Z2) 2(Z+Z1)] 2

=0

2 2

[(Z+Z1) (Z+Z2) ] Or

2

2

2

2

2

[(Z+Z1) (Z+Z2) 32Z Z1Z2] - 16Z Z1Z2[(Z+Z1) 2(Z+Z2)+ (Z+Z2) 2(Z+Z1)]

=0

32Z Z1Z2 (Z+Z1) (Z+Z2) [ (Z+Z1)(Z+Z2) −Z{(Z+Z1) + (Z+Z2)}] = 0 Or (Z+Z1)(Z+Z2) −Z(2Z+Z1+Z2) = 0

Or

Or Z 2 + ZZ2 + Z1Z + Z1Z2 = 2Z 2 + ZZ1 + ZZ2 Or Z 2 = Z1Z2 Or Z = Z 1 Z 2

∴ For complete transmission, imoedance of matching layer should be =

Z 1 Z 2

14. (a) Describe delay line transducer. (b) How dual element ultrasonic transducer works?

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(a) Delay line transducer: - As the name implies, the main feature of a delay line transducer is that it provides a delay between the generation and arrival of ultrasound from TX to RX. The delay is introduced by an acoustic path separated by the test body and the transmitter probe as shown in fig-9. This prevents the single element transducer from ring down effect. For example, say the sender/receiver alternating time is below the total transit time. In this case if a sufficient delay can be provided, then there will be no ring down effect. This enables the transducer to complete its sending function before it starts receiving.

Figure 9: Delay line transducer

They have variety of replaceable options. Removable delay line, surface conforming membrane and protective wear cap makes it useful in number of applications like: High precision thickness measurement of thin materials.  Delamination check in composite materials.  High temperature application as the delay element provides an isolation to the PZT from hot test body. (b) Dual element ultrasonic transducers: - It contains two independently operating crystal in a single housing. One of the crystals transmits and another one receives the ultrasound. Actually the TX and RX are positioned in such a way, so that the sound wave follows a crossed-beam sound path. They have no ring down effect as the TX and RX operates independently. As a result they are useful in:

Figure 10: Dual element transducer

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  

20

Measurement of very thin materials. Cases where the defects are at very distances from the surface of the test body. Inspection of course grain material.

15. (a) What is the reason of applying dc magnetic field for magnetostrictive transducer? Due to application of only an ac, the frequency of the US pulse becomes twice that of the input ac. For the two halves of one complete cycle of the input ac, two consequent pulses are generated (illustrated in fig-11(right)).

Figure 11a: Arrangements of ac and dc sources in series (left). Waveforms for only ac input (right)

The above happening is not a problem. But to obtain a US pulse whose frequency is same as that of the input ac we need to connect a dc source in series with the ac. After adding the dc source the resultant input waveform becomes as it is shown in fig-11b (Blue). In this case, the output vibration (below the blue input) frequency is as same as that of the input.

Figure 11b: Waveforms for resultant input

(b) What are the reasons of humming sound in transformer? Iron or ferrite core transformers if operate at frequency between 20- 20,000 Hz of the applied ac voltage (in India, it is 50Hz) the core material's length is extended & contracted in length by very small amount at that frequency known as magnetostriction process. Magnetostriction causes air being pushed back & forth

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21

at that frequency causing sound generation which is hummung sound. Transformers without core do not generate that sound. 16. An ultrasonic transmitter and receiver are attached to the same side of a steel block having a length of 75 cm. calculate the percentage of received energy in receiver. Neglect the noise due to cross talk. Given (αT)Q/S = 0.73, absorption coefficient co efficient of steel αW = 2.0 and (αT)S /air = 3.7*10^-5 Attenuation of ultrasound in a medium occurs exponentially. ∴ The ratio of received and transmitted energy is given by:air * exp(−αW * L )]*[( αT)S/Q] PRX / PTX = [(αT)Q/S * exp(−αW * L )]*[ ( αR)S/ air = [(0.73)*exp ( −2*0.75)]*[(1–3.7*10^–5)*exp( −2*0.75)]*[0.73]

= 0.0265 ∴ Required percentage = 2.65%

[Q (αR)S /air = 1 - (αT)S /air and (αT)Q/S = (αT)S/Q]

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Part 3 15 Marks Questions

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23

Long Answer Type Questions(15): 1. (a) Prove that the velocity of longitudinal waves through a medium V = E / ρ . Where, ρ is the medium density and E is the bulk modulus. (b) Also prove that velocity of longitudinal wave in a rod V = Y / ρ . Where, Y is the Young’s modulus of the rod and ρ is the density of the rod. (c) Calculate the velocity of sound in i. Water and ii. Steel. Given density of steel = 7800 kg/m^3. Young’s modulus of steel = 2* 10^10 2 2 N/m and bulk modulus of water = 0.20 * 10^10 N/m

(a)

Let us consider a pipe of uniform cross sectional area α which is filled with a lossfree gas of density ρ and through which sound wave is propagating along positive X direction (from left to right horizontally in fig-1). Consider two planes, separated by a distance ∆x, perpendicular to the axis of the pipe. Volume of this element is v = (α * ∆x )Due to the presence of a tuning fork at the left end of the pipe, the displacements of the particles in the planes A & B are respectively given by y(x1) & y(x2). The displacements might not be equal. At equilibrium, the particles of two planes experience the same pressure (P o), but due to propagation of wave, the change in pressure occurs.

Figure 1: Sound wave in a pipe

Now, increment in length ∆x is = y(x2) − y(x1) Increment in volume v is ∆v = α * (y(x2) − y(x1)) = ∆v = α(δy/δx)∆x [Q y(x2) = y(x1) + ((δy/δx)x1)∆x +…… by Taylor’s series expansion with x2 = x1+∆x, and neglecting higher order terms]

∴ Volume strain = ∆v/v = (δy/δx) Now, bulk modulus E = −(change in pressure / volume strain) = − (∆p) / (∆v/v) ∴ ∆p = − E(∆v/v) = − E(δy/δx) Now, change in pressure (decrement) = P(x2) − P(x1)

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= (δP(x)/δx)∆x = (δ(Po −∆ −∆p)/δx)∆x = − (δ∆p)/δx)∆x − E(∆v/v))/δx)∆x = − (δ(− = (δ(E(δy/δx))/δx)∆x [Q ∆v/v = (δy/δx)] = E∆x (δ 2y/δx 2 ) Net force = Cross sectional area of pipe * excess pressure 2 2 = αE∆x (δ y/δx ) = F1 (Say) Newtonian force = mass of the elemental area * acceleration = density * volume * acceleration 2 2 = (ρα∆x) (δ y/δt ) = F2 (Say) In dynamic equilibrium, F2 = F1 or, (ρα∆x) (δ 2y/δt 2 ) = αE∆x (δ 2y/δx 2 ) or, ρ(δ 2y/δt 2 ) = E(δ 2y/δx 2 ) ρ)(δ 2y/δx 2 ) or, ρ(δ 2y/δt 2 ) = (E/ ρ)(δ Comparing the above equation with the Differential equation of wave motion i.e.

(δ 2y/δt 2 ) = V 2(δ 2y/δx 2 ), we get, velocity of longitudinal wave V =

E / ρ

(b) Referring fig-7, the change in length is is =∆x = y(x2) − y(x1) = [(δy/δx)| ]∆x Longitudinal strain at x1 is given by ε = change in length / original length = (δy/δx)| Longitudinal stress at x1 is given by S1 = Y ε = Y(δy/δx)| Where, Y is the Young’s modulus of the rod material. Similarly, S2 = Y (δy/δx)| Now, S2 – S1 = Y(δy/δx)| – Y(δy/δx)| = Y{(δy/δx)| – Y(δy/δx)| } = Y (δ 2y/δx 2 )∆x 2 2 Net force = area * stress = αY (δ y/δx )∆x =F1 (say) Newtonian force = mass * acceleration = ( ρα∆x)* (δ 2y/δt 2 ) = F2 (say) In dynamic equilibrium, F2 = F1 Or (ρα∆x)* (δ 2y/δt 2 ) = αY (δ 2y/δx 2 )∆x (δ 2y/δt 2 ) = (Y/ ρ)(δ ρ)(δ 2y/δx 2 ) Comparing the above equation with the Differential equation of wave motion i.e. x1

x1

x1

x2

x1

x1

x2

x2

(δ 2y/δt 2 ) = V 2(δ 2y/δx 2 ), we get, velocity of longitudinal wave V =

Y / ρ

(c) i.

We know, velocity of longitudinal waves through a medium like water is given by V = E / ρ . Where, ρ is the medium density and E is the bulk modulus.

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2

Now, E = 0.20 * 10^10 N/m , ρ = 1000kg/m^3

∴V = =

E / ρ

(0.20 *10^10) /(1000)

= 1414.21 m/sec ii.

Velocity of longitudinal wave in a rod V = Y / ρ . Where, Y is the Young’s modulus of the rod material (steel) and ρ is the density of the rod. Now, Y = 2 * 10^10 N/m 2, ρ = 7800kg/m^3

∴V = =

E / ρ

(2 * 10^10) /(7800)

= 1601.28 m/sec 2. (a) Obtain an expression for total energy per wavelength in a string when transverse waves travel in it. (b) Show that the time averaged input power of the source generating waves in a string is equal to the total energy per unit length of string times the wave wav e velocity.

Figure 2: Situation of problem 2

(a) Consider a small element of length dx of the string that carries the transverse wave through it. Total energy of the small element in vibration = kinetic energy + potential energy 2 Now, kinetic energy of element = (1/2) (mass) (velocity) 2 Or dK = (1/2) (µdx) (dy/dt) −−−−−−−−−−−−−−−(1) (Where, µ = linear density of string and dy/dt = particle velocity) Now, potential energy of element is given by the work done by the tension T to change the length of the element from dx to ds during vibration. Therefore, potential energy of the element = dU = T (ds – dx) Now, from fig-2, dx/ds = cosθ 2 2 2 2 Or (ds/dx) = sec θ = 1 + tan θ = 1 + (dy/dx) 2 ds =sqrt[1 + (dy/dx) ]dx [Q tanθ = slope of the displacement curve = dy/dx] 2 = [1 + (1/2)(dy/dx) ]dx [neglecting higher order terms] ∴ ds –dx = [1 + (1/2)(dy/dx) 2]dx – dx = (1/2)(dy/dx) 2dx ∴ Potential energy = T (ds – dx) = dU = (1/2) T (dy/dx) 2dx ------------------(2)

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Now, from (1) and (2) the kinetic and potential energy density are given respectively 2 by: dK/dx = (1/2) (µ) (dy/dt) --------------------------------------------(3) dU/dx = (1/2) T (dy/dx) 2 --------------------------------------------(4) Now, displacement of particle a wave traveling in positive X direction is given by: y(x,t) = Asin ((2 π/λ)(Vt−x)) ∴ dy/dt = (2AVπ/λ)cos ((2π/λ)(Vt−x)) and dy/dx = −(2Aπ/λ)cos ((2π/λ)(Vt−x)) ∴ dK/dx = (1/2) ( µ) (dy/dt) 2 = (1/2) (µ)[(2AVπ/λ)cos ((2π/λ)(Vt−x))] 2 ∴ dK/dx|t=0 = (1/2) (µ)[(4A 2V 2π 2/λ 2)cos 2(2πx/λ)] ---------------------(5) And dU/dx = (1/2) T (dy/dx) 2 = (1/2) T[(2Aπ/λ)cos ((2π/λ)(Vt−x))] 2 ∴ dU/dx|t=0 = (1/2) T[(4A 2π 2/λ 2)cos 2 (2πx/λ)] -------------------------(6) Now, the total energy associated with one complete wavelength of the sinusoidal wave on the string is given by: E=

λ

λ

0

0

∫ (dU / dx)dx + ∫ (dK / dx)dx λ

2

2

∫

2

= (1/4) T (4A π /λ ) [1 + cos(4π x / λ )]dx

2

[ Q 2cos A = cos2A + 1]

0

λ

2

2

2

2

∫

+ (1/4) (µ) [(4A V π /λ ) [1 + cos(4π x / λ )]dx 0 2

2

2

2

2

=TA π /λ + µA V π /λ 2

2

2

2

2

2

= V µA π /λ + µA V π /λ 2

2

[Putting T = V 2µ in the first term Q V = T / µ ]

2

= 2µA V π /λ 2 2 2 = 2µA ν λπ

[

Q

V = νλ, ν being the frequency]

∴ E = 2µA 2π 2 νν 2λ (b) 3. (a) Obtain an expression of average energy density of ultrasonic wave when passing through acoustic medium. (b) Show that the intensity of plane sound wave in a gas is equal to the energy density times the wave velocity.

(a)Consider an elemental volume ∆v containing n number of particles of a pipe containing a gas of density ρ through which sound wave is propagating. Energy of oscillation of oscillation of each particle is given by: 2 2 E = mA ω / 2 Where, m = mass of particles, A= amplitude of their oscillation, ω = angular velocity of particle.

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Figure 3a: Sound in a pipe

27

Figure 3b: Two consequent conse quent wave fronts

Now, E for n number of particles of that elemental volume ∆v is = nE 2 2 And, nE = (nm)A ω / 2 2 2 = (ρ∆v) A ω / 2 [Q nm = ρ∆v] Energy density = energy per unit volume = U = nE / ∆v 2 2 = ρ A ω / 2 = ρ A 2(4 π 2 ν 2) / 2 [Q ω = 2πν] ∴ Energy density = U = 2ρ A 2 π 2 νν 2 (b) Now, consider fig-3b. Here two consequent wave fronts are considered. The separation between them is given by ∆x = V∆t Where, V = velocity of sound, ∆t = time taken by the wave two form the second wave front after the first one Volume of the area between two wave fronts = S ∆x Where, S= wave front plane’s area Energy received and transmitted by this volume = U S ∆x Energy transfer per unit time per unit area i.e. the intensity is, therefore, given by I = U S∆x / S ∆t =UV [Q V= ∆x / ∆t] 2 2 2 = 2ρ A π ν V

I= 2ρ A 2 π 2 ν ν2 V

∴ Intensity = energy density * wave velocity v elocity 4. (a) When ultrasonic wave is incident on a plane boundary between two media, then prove that the reflection and transmission amplitude coefficient depend on medium impedance. (b) Also prove that the total energy is conserved at the junction jun ction of two media.


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(a) Let, the ultrasonic wave is traveling from medium1 to medium2 and incident normally on the boundary of the two mediums. We know, V1 = E / ρ 1 , V2 = E / ρ 2 , Z1 = ρ1V1, Z2 = ρ2V2 Where, V = Velocity of wave, E = Bulk modulus. ρ = density of the medium Z = Acoustic impedance The displacement of particles in medium1 due to incident wave is given by: yi(x,t) = Ai sin( ωt-k1x) λ1 = 2πν /V1 where, k1 = 2π / λ1 The displacement of particles in medium1 due to reflected wave is given by: yr(x,t) = Ar sin( ωt+k1x) The displacement of particles in medium2 due to transmitted wave is given by: yt(x,t) = At sin( ωt-k2x) λ2 = 2πν /V2 where, k2 = 2π / λ2 Now, the boundary conditions gives: a) The displacement of particles is same to the left and right of the boundary at x=0, i.e. y(x,t) is continuous across the boundary at x=0, the velocity of the particle is also contiuous b) The excess pressure ( ∆p = -Eδy/δx) is continuous across the boundary at x = 0. Now, boundary condition a) gives: yi(x,t)[x = 0] + yr(x,t)[x = 0] = yt(x,t)[x = 0] or, Ai sin( ωt-k1x)[x = 0] + Ar sin( ωt+k1x)[x = 0] = At sin( ωt-k2x)[x = 0] or, Ai sin( ωt) + Ar sin( ωt) = At sin(ωt)

∴ Ai + Ar = At -------------------------(1) Now, boundary condition 2 gives: -Eδyi/δx [x = 0] - Eδyr/δx [x = 0] = -Eδyt/δx [x = 0] or, E Ai k1cos( ωt-k1x)[x = 0] - E Ar k1 cos( ωt+k1x) [x = 0] = E At k2 cos( ωt-k2x)[x = 0] or, E Ai k1 cos( ωt) - E Ar k1 cos( ωt) = E At k2 cos( ωt) or, E k1 Ai - E k1 Ar = E k2 At or, Ai 2πνZ1 – Ar 2πνZ1 = At 2πνZ2 [ Q Z = E / V ]

∴ Ai – Ar = At (Z2/Z1) ---------------------------(2) Now, adding (1) and (2) we get, Ai + Ar +Ai - Ar = At (Z1+Z2) / Z1 ∴ Ai = 0.5 At (Z1+Z2) / Z1 -----------------------------------------(3) Subtracting (2) from (1) we get, Ai + Ar - Ai + Ar = At (Z1-Z2) / Z1 ∴ Ar = 0.5 At (Z1-Z2) / Z1 -----------------------------------------(4)

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∴ Reflection amplitude coefficient coefficient = r12 = Ar/Ai Ar/Ai = (Z1-Z2) / (Z1+Z2) Now, from (1) we have Ai + Ar = At or, 1 + (Ar / Ai) = (At / Ai) [dividing both both sides by Ai] or, 1 + r12 = At / Ai or 1 + (Z1-Z2) / (Z1+Z2) = At / Ai

∴ Transmission amplitude coefficient = t12 = At/Ai = 2Z1 / (Z1+Z2) ( Z1+Z2) 2

2

(b) Incident energy = Pi = 0.5 Z1 Ai ω 2 2 Reflected energy = Pr = 0.5 Z1 Ar ω 2 2 Transmitted wave amplitude = Pt = 0.5 Z2 At ω 2 2 2 2 2 2 2 Now, we know, Ar = Ai ((Z1-Z2) / (Z1+Z2)) and At = 4 Ai Z1 / (Z1+Z2) 2 2 2 2 Now, Pr + Pt = 0.5 Z1 Ar ω + 0.5 Z2 At ω 2 2 2 2 2 2 2 = 0.5 Z1 ω Ai ((Z1-Z2) / (Z1+Z2)) + 0.5 Z2 ω 4 Ai Z1 / (Z1+Z2) 2 2 2 2 2 = 0.5 Z1 ω Ai ( Ζ1 −2Z1Z2 + Z2 + 4Z1Z2) / (Z1+Z2) = 0.5 Z1 ω 2 Ai 2 (Z1+Z2) 2 / (Z1+Z2) 2 = 0.5 Z1 ω 2 Ai 2 = Pi

∴ Pi = Pr + Pt (i.e. total energy is conserved at the boundary) 5. Deduce the expressions for the reflection and transmission amplitude and energy coefficient when a transverse wave traveling in the positive X direction in the string of impedance Z1 meets the junction of strings of impedance Z2.(15)

Let, the transverse wave is traveling from medium1 to medium2. Two mediums are represented by two different types of strings having different parameters. We know, V1 =

T / µ 1 , V2 =

T / µ µ 2 , Z1 = µ1V1, Z2 = µ2V2

Where, V = Velocity of wave, µ = Linear density of the medium T = Tension in string, Z = characteristic impedance


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The displacement of particles in medium1 due to incident wave is given by: yi(x,t) = Ai sin( ωt-k1x) λ1 = 2πν /V1 where, k1 = 2π / λ1 The displacement of particles in medium1 due to reflected wave is given by: yr(x,t) = Ar sin( ωt+k1x) The displacement of particles in medium2 due to transmitted wave is given by: yt(x,t) = At sin( ωt-k2x) λ2 = 2πν /V2 where, k2 = 2π / λ2 Now, the boundary conditions gives: c) The displacement of particles is same to the left and right of the boundary at x=0, i.e. y(x,t) is continuous across the boundary at x=0, the velocity of the particle is also contiuous d) The restoring force or the transverse component of tension (-Tsin θ = -Tδy/δx) is continuous across the boundary at x = 0. Now, boundary condition a) gives: yi(x,t)[x = 0] + yr(x,t)[x = 0] = yt(x,t)[x = 0] or, Ai sin( ωt-k1x)[x = 0] + Ar sin( ωt+k1x)[x = 0] = At sin( ωt-k2x)[x = 0] or, Ai sin( ωt) + Ar sin( ωt) = At sin(ωt)

∴ Ai + Ar = At -------------------------(1) Now, boundary condition 2 gives: -Tδyi/δx [x = 0] - Tδyr/δx [x = 0] = -Tδyt/δx [x = 0] or, T Ai k1cos( ωt-k1x)[x = 0] - T Ar k1 cos( ωt+k1x) [x = 0] = T At k2 cos( ωt-k2x)[x = 0] or, T Ai k1 cos( ωt) - T Ar k1 cos( ωt) = T At k2 cos( ωt) or, T k1 Ai - T k1 Ar = T k2 At or, Ai 2πνZ1 – Ar 2πνZ1 = At 2πνZ2 [ T k1 = T 2πν /V1 = 2πνZ1 Q Z1 = T / V1 and similarly, T k1 = 2 πνZ1]

∴ Ai – Ar = At (Z2/Z1) ---------------------------(2) Now, adding (1) and (2) we get, Ai + Ar +Ai - Ar = At (Z1+Z2) / Z1 ∴ Ai = 0.5 At (Z1+Z2) / Z1 -----------------------------------------(3) Subtracting (2) from (1) we get, Ai + Ar - Ai + Ar = At (Z1-Z2) / Z1 ∴ Ar = 0.5 At (Z1-Z2) / Z1 -----------------------------------------(4)

∴ Reflection amplitude coefficient coefficient = r12 = Ar/Ai Ar/Ai = (Z1-Z2) / (Z1+Z2) Now, from (1) we have Ai + Ar = At or, 1 + (Ar / Ai) = (At / Ai) or, 1 + r12 = At / Ai

[dividing both both sides by Ai]

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or 1 + (Z1-Z2) / (Z1+Z2) = At / Ai

∴ Transmission amplitude coefficient coefficient = t12 = At/Ai At/Ai = 2Z1 / (Z1+Z2) (Z1+Z2) 6. (a) Two strings of linear densities µ1 & µ2 are joined together and stretched with tension T. A transverse wave is incident on the boundary. Find (i) The fraction of incident amplitude reflected and transmitted at the boundary. (ii) The fraction of the incident energy reflected and transmitted at the boundary if µ2/µ1 = 4 & ¼ (b) A plane sound wave in air of density 1.29kg/m^3 falls on a water surface at normal incidence. The speed of sound in air is 334m/s and in water, the speed of sound is 1480m/s. Calculate (i) What is the ratio of the amplitude of sound wave enters water to that of incident wave? (ii) What fraction of the incident energy flux enters en ters the water?

(a) We know, V=

∴ Z1/Z2 =

T / µ and Z = T/V. Therefore, Z =

T µ

µ 1 / µ 2

(i)For, µ2/µ1 = 4, Z1/Z2 = 1/2, Z2 = 2Z1 Fraction of incident amplitude reflected = r12 = (Z1-Z2) / (Z1+Z2) = (Z1 – 2Z1) / (Z1+2Z1) = - 1/3 Fraction of incident amplitude transmitted = t12 = 2Z1 / (Z1+Z2) = 2Z1 / (Z1+2Z1) = 2/3 For, µ2/µ1 = 1/4, Z1/Z2 = 2, Z1 = 2Z2 Fraction of incident amplitude reflected = r12 = (Z1-Z2) / (Z1+Z2) = (2Z2 – Z2) / (3Z2) = 1/3 Fraction of incident amplitude transmitted = t12 = 2Z1 / (Z1+Z2) = 4Z2 / (3Z2) = 4/3 (ii) For, µ2/µ1 = 4, Z1/Z2 = 1/2, Z2 = 2Z1 Fraction of the incident energy reflected = Ir/Ii = ((Z1-Z2) / (Z1+Z2))

2

= 1/ 9 2

Fraction of incident amplitude transmitted = It/Ii = 4Z1Z2 / (Z1+Z2) =8/9 For, µ2/µ1 = 1/4, Z1/Z2 = 2, Z1 = 2Z2 Fraction of the incident energy reflected = Ir/Ii = ((Z1-Z2) / (Z1+Z2))

2

= 1/ 9 Fraction of incident amplitude transmitted = It/Ii = 4Z1Z2 / (Z1+Z2)

2

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32 =8/9

(b) Acoustic impedance of air = Z1 = ρair * Vair = 1.29 * 334 kg/m^2/s = 430.86 kg/m^2/s Acoustic impedance of water = Z2 = ρwater * Vwater = 1000 * 1480 = 1480000 kg/m^2/s Ratio of the amplitude enters water to that of incident wave = t12 = 2Z1 / (Z1+Z2) = 5.82*10^-4 Fraction of the incident energy flux enters the water = It/Ii 2 = 4Z2Z1 / (Z1+Z2) = 1.16*10^-3

7. (a) Describe ultrasonic cross-correlation flow meter. (b) Describe Doppler flow-meter with necessary diagram. Also draw and explain the signal processing unit for the same. (a) Cross correlation flow-meter: -

Figure 6: Cross correlation flow-meter flow-meter

In most flowing fluids there exist naturally occurring random fluctuations such as density, turbulence, and temperature which can be detected by suitably located transducers. If two such transducers are installed in a pipeline separated by a known distance (say L) , the upstream transducer will pick up a random fluctuation t seconds before the downstream transducer and the distance between the transducers divided by the transit time (say T) will yield flow velocity. In practice the random fluctuations will not be stable and are compared in a cross-correlator which has a peak response at transit time T, and correlation velocity V = UT, meters per second. This is effectively a non-intrusive measurement and could in principle be developed to measure flow of most fluids. Very few commercial cross-correlation systems are in use for flow measurement because of the slow response time of such systems. However, with the

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use of microprocessor techniques processing speed has been increased significantly, and several manufacturers are now producing commercial systems for industrial use. (b) Doppler flow-meter and signal conditioning circuit: -

Figure 7a: Doppler flowmeter

This type of flowmeter uses Doppler Effect as the principle. The transmitter TX sends an ultrasonic wave of frequency “f1” & velocity “c” at an angle “ θ ” relative to the direction of the flow. Bubbles, solid particles, eddies in the flow stream are considered as the moving observers (relative to the fixed transmitter i.e. source) with velocity V equal to that of flow velocity. Now, if the apparent frequency of the wave seen by the moving particle observers is “f2 ”, then the Doppler shift for fixed source & moving observer is given by: f2/ f1 = (velocity of waves relative to observer) / (normal wave velocity) = (c + Vcosθ ) / c -----------------------------------------------------------(1) The particles scattered the ultrasound in different directions. But a small amount of the scattered wave is received by the receiver RX. Thus, RX now acts as a fixed observer and particles act as moving source. Now, if the apparent wave length seen by the fixed observer RX is λ3 then the Doppler shift is given by: λ3 / λ2 = (velocity of wave relative to source) / (normal wave velocity) = (c − Vcosθ ) / c ---------------------------------------------------------(2) Q λ2 f2 = λ3 f3 = c, (2) can be written as f3 / f2 = c / (c − Vcosθ ) ------(3) From equation (1) and (3), elimination of f2 gives f3 = f1 (c + Vcosθ ) / (c − Vcosθ ) ∴ the frequency shift in Doppler flowmeter is = ∆f = f3 – f1 = f1 (c + Vcosθ ) / (c − Vcosθ ) – f1 = (2 f1 Vcosθ) / (c − Vcosθ ) Now, c − Vcosθ = c (1 – (V/c)cos θ) = c, for V<
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8. (a) Briefly discuss the working principle of pulse echo flaw detector. (b) Draw and describe the schematic block diagram of pulse echo method of flaw detection system. (c) What are the influences that affect the amplitude of the received echo? (d) What are the general specifications that are accessed in accordance with particular applications of pulse echo flaw detector? (e) What is the role of repetition frequency generator and base line voltage in pulse echo method of flaw detection? 3+4+2+3+3

(a) Working principle of pulse echo flaw detector:-

(a) Principle of pulse echo method (b) Display on the CR screen Figure 8: Pulse echo method of flaw detection

An ultrasonic pulsed wave, usually in the form of damped oscillation, is generated by transmitter (TX) and propagates through the specimen. Part of the wave will be reflected back to the receiver (RX) by an obstacle (if any) and the rest will also be reflected back to the receiver by the back wall of the specimen. One condition for the above to be true is the specimen should be small in size. The signal obtained from the receiver is displayed on a CR screen. The horizontal sweep is proportional to time, so that the transit time of the pulse to & from the reflector (t R) corresponds to the distance between initial peak to echo peak. And transit time of the pulse to & form the back wall (t B) corresponds to the distance between initial peak and back wall. To obtain standing image the pulses and the sweep of the CR tube are synchronized at repetition frequency. By calibrating the base line in in time per unit unit length the transit times t R & tB can be read from the CR screen and we obtain d = c*t / 2 Where, d = distance of reflector from front surface of the specimen c = velocity of ultrasound in the specimen tR = transit time for of pulse from reflector Usually the time scale is calibrated in terms of length if the thickness of the specimen is known. (b) Block diagram of pulse echo flaw detector:Pulse echo flaw detectors are basically oscilloscopes with some special features. It consists of a repetition frequency generator, sweep synchronization, pulse generator, transducer, amplifier and the CR tube. If only one probe is used as TX and RX (usual case) the pulse generator excites the transducer and the echoes received are fed to the CR tube after being amplified.

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Figure 9a: Block diagram of pulse echo e cho flaw detection system

The frequency generator triggers both the sweep generator and pulse generator. Now the pulse generator excites the transducer. Now the fact to be noted is that, the sweep generator is triggered a little before the pulse generator (as well as the transducer) is triggered (Tt>Ts). This is because we want to obtain the main echo rather than the side echoes for our measurement. Actually, during each ramp of the sweep generator output, the bright voltage of CRO exists and echoes are visible. If, during the main echo, ramp is not present then we will not be able to see that main echo.

Figure 9b: Waveforms of different blocks of pulse echo for Tt > Ts

(c) Influences that affect the amplitude of the received echo:i. Size of reflector. ii. Direction of the transmitter probe. iii. Power of transmitter pulse entering the sample. iv. Surface quality of the reflector. v. Position of reflector. vi. Size and direction of the receiver probe. vii. Losses at the receiver by reflection and coupling. viii. Attenuation of wave by absorption. ix. Shadow effect of any defect in front of the reflector.

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(d) General specifications:Ambient conditions A testing device must be fully operational even when subjected to adverse ambient conditions. According to DIN (a set of German standards), there must be a difference between the conditions for “storage & transport” and for “use”. Under the first, it is necessary for the device not to be damage when not in use. Under the second, it must be guaranteed that all equipments functions operate in accordance with performance specifications. The most important ambient conditions are: Range of temperature.  Humidity of the air and dew point.  Mechanical stress produced by shocks and vibrations.  Stability of the mains voltage.  Electromagnetic interference fields.  Rain, sprayed water, dust and dirt.  Ambient light intensity and its effect on the readability of the CRO screen and / or any digital display. Operation, reliability of inspection: Time needed for training of an operator.  Fail-safe properties.  Reproducibility of the results of any test.  Brightness of the screen, sharpness of focusing, size etc of the display unit.  Design, layout and number of control.  Calculating aids for sound beam calculation.  Simplicity of adjustments.  Potential for connection to computers. To these general specifications must be added the safety requirements of the VDE (Verband Deutscher Elektrotechniker) and the requirements for possible interference with communications by the German Post Office.. (e) Role of repetition frequency generator:To obtain stationary screen pictures, the repetition frequency generator triggers both the transmitter and the start of the sweep generator. Usually the transmitter pulses are triggered a little later than the sweep generator so that the main echo is visible on the screen. The problem that occurs when the TX pulse delay becomes less than the sweep generator delay is illustrated in fig-10. Here the TX pulse delay (Tt) is less than the sweep time delay (Ts). The bright portion of CRO voltage lasts during the each ramps. As a result, the side echos are only visible in the CRO screen rather than the main echo during Tp. So, we can’t get the actual result. The base line voltage (time base) was formerly given a fixed distance screen which is today built into the CR tube screen. It is calibrated in terms of length e.g. 50, 100 and 250mm etc by varying the sweep velocity.

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Figure 10: Waveforms of different blocks of pulse ec ho flaw detector forTt
9.

(a)???????????????????????? (b) Three strings are joined together and the composite string is stretched with a tension of 10N. The linear densities of strings 1 & 3 are 1*10^-3Kg/m 1*1 0^-3Kg/m & 4*10^3Kg/m respectively. A transverse wave of frequency 100Hz is produced in string 1. Calculate the linear density and length of intermediate string 2 so that the wave is completely transmitted through the composite string string without any loss due to reflection at the boundary. We know, Z = T / V = T / T / µ =

∴ Z1 = and Z3 =

T µ 1 = T µ 3 =

T µ 2

10 * 1 * 10^ −3 = 0.1kg / m / sec 2

10 * 4 * 10^ −3 = 0.2 kg / m / sec

For zero reflection, Z2 must be = Z 1 Z 3 = Again we know Z2 =

2

0.1 * 0.2 kg / m / sec

T µ 2

∴ µ2 = Z2 2 / T = 0.1*0.2 / 10 = 2 * 10^-3kg/m Now, velocity (V) = frequency (f) * wavelength ( λ) V2 = f * λ2 (Frequency of the wave remains unchanged in 3 mediums)

λ2 = V2 / f =

T / µ 2 / f =

10 /(2 * 10^ −3) / 100 =

5000 /100 = 0.707m

As per condition, the length of the 2 nd medium should be = L = λ2/4 = 0.176m 10. (a) Describe the integration method of thickness measurement. (b) With the help of neat diagram, briefly discuss the method of level measurement using ultrasonic level detectors. (c) A longitudinal wave traveling in a rod encounters discontinuity where the Young’s modulus suddenly doubles, the density remaining the same. Calculate the reflection and transmission amplitude coefficients. (d) Assuming the expression of reflection and transmission amplitude coefficient, prove that if a ultrasonic wave traveling in a medium of lower impedance meets the boundary of a medium of higher impedance, the wave reflected at the

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boundary undergoes a phase change of π.

(b) The ultrasonic level measurement is based on transit time of ultrasonic wave measuring (generally 30 to 300kHz is used) procedure. The ultrasound travels from the transmitter TX and a part of it is reflected back from the liquid-air interface and received by the receiver RX. When the TX & RX are placed at the bottom (bottom), the transit time is directly proportional to the level. But when TX & RX are placed at the (roof), the transit time is proportional to the gap (may be air gap) between roof and liquid level. In this case, one has to subtract the directly obtained gap length from the height of the tank to get the liquid level. In fig-12, the transit time measuring arrangement is shown. The pulse generator (PG) excites the Pulse oscillator (PO) at the same moment. Then and there the PO excites the TX crystal and US wave is sent to the tank. At that moment, the SR input of the SR F/F is S=1, R=0 and O/P=1. Now, after reflected back the US wave is received by RX and the RX generates a pulse which is detected by the amplifier-comparator and inputs of SR F/F becomes S = 1, R = 1 and output changes to 0. In fig-12, a single element TX/RX is shown. Now the time taken by the SR F/F O/P from 1 to 0 is nothing but the transit time and Fathometer is generally used to measure that time.

Figure 11: Ultrasonic level measurement. Unit at bottom (left) and unit at top (right)

Figure 12: Transit time measurement circuit

(c) Velocities of ultrasound in two mediums (1 & 2) = V1 and V2 Where, V1 =

Y 1 / ρ and V2 =

Y 2 / ρ

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Where, Y (1, 2) = Young’s modulus of medium (1, 2) ρ = density of mediums (same for both 1 & 2) Again, impedances of two mediums = Z1 and Z2 Where, Z1 = ρV1 and Z2 = ρV2 Now, Z1 / Z2 = V1 / V2 =

Y 1 / Y 2 = 1/ 2

∴ Z2 =

2 Z1 Now, reflection amplitude coefficient = r12 = (Z1-Z2) / (Z1+Z2) = (1- 2 ) / (1+ 2 ) = -0.17 Transmission amplitude coefficient = t12 = 2Z1 / (Z1+Z2) = 2 / (1+ 2 ) = 0.82

(d) We know, r12 = (Z1-Z2) / (Z1+Z2) and t12 = 2Z1 / (Z1+Z2) The condition given is Z2 > Z1 ∴ Always r12 < 0 But, r12 = Ar / Ai

Figure 13: Phase change of reflected wave

∴ Ar = −k Ai, where k is a constant depends on the values of Z1 & Z2 ∴ Ar is always negative with respect to Ai. So the reflected wave must undergo a phase change of π (as shown in fig-13). 11. (a) Discuss briefly how piezoelectric plate can be used as transmitter and receiver of ultrasound. (b) Briefly discuss the applications of ultrasonic waves in medical diagnosis.

(a) With a single crystal vertical probe (normal probe) the generated ultrasonic wave passes into the test body in a perpendicular direction. The probe is shown in fig-14 and is coupled to the test body by coupling layers. It mainly consists of active element, backing surface and wear plate. If necessary, there are also electrical matching elements and all are kept in a single housing.

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Figure 14: Ddesign characteristics of PZT transducer

Active element: - It is a piezoelectric element that converts electrical energy into mechanical wave. Quartz, Rochelle salt, lithium sulfate, ADP (ammonium dihydrogen phosphate) etc are used as active element. Thickness of the active element should be = (0.5*wavelength of ultrasound). Backing: - It is a highly attenuating high density material. It is used to absorb the waves (vibration) as much as possible from the back side of active element so that maximum possible waves are passed through the test body. Wear plate: - Its Basic purpose is to protect the active element from the external testing environment. In case of contact transducers, the wear plate must be a corrosion resistant material (steel). Also a wear plate having proper impedance (that depend upon the impedances of active element & test medium) and proper size (0.25*wavelength of ultrasound) can allow as much as energy possible to come out from the active element. (b) Applications of ultrasonic waves in medical diagnosis: In medical diagnosis, ultrasound of frequencies from 1.5 MHz to 20MHz is used. Since 1940, ultrasound has been applied for various medical purposes. However its acceptance as a powerful diagnostic tool was in the 1970s. Today, ultrasound is the second most utilized diagnostic imaging technology used in medicine, after X-Rays. Ultrasound scanning provides a less harmful but certainly effective method for “in vivo” diagnosis. Two common uses of ultrasound imaging are the scanning of the fetus during pregnancy and the scanning of the heart. Except these, ultrasound is used in the following areas:i. Brain ii. Pancreas iii. Peritoneal cavity iv. Blood flow measurement v. Ovary To keep up with the new developments in medical industry, and due to the necessity for better diagnostic methods, new ultrasound scanners provide 3D imaging capabilities. It should be noted that the ability of some tissues to absorb ultrasound restrict the use of ultrasound in the diagnosis of bones or lungs.

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12. Write short notes on any an y three of the following: a. Acoustic holography b. Ultrasonic transit time flow-meter c. A scan, B scan and C scan display d. Attenuation of ultrasonic waves in an absorbing a bsorbing medium e. Ultrasonography

b. Ultrasonic transit time flow-meter This device measures flow by measuring the time taken by an ultrasonic energy pulse to traverse a pipe section (of distance L), both with and against the flow of the fluid within the pipe (fig-15).

Figure 15: Ultrasonic transit type flow-meter

The time taken by the pulse to go from transducer A to B is (with the flow) tAB = L / (C + Vcosθ) −−−−−−−−−−−−−− (1) The time taken by the pulse to go from transducer B to A is (against the flow) tBA = L / (C − Vcosθ) −−−−−−−−−−−−−− (2) Where, V = fluid velocity C = velocity of ultrasonic pulse in the fluid L = acoustic path length (the distance the pulse traverse) in the fluid θ = the angle between the acoustic path and the pipe axis From (1) and (2) we get, t BA − tAB = L / (C − Vcosθ) − L / (C + Vcosθ) 2 2 2 = 2LVcosθ / (C − V cos θ) 2 = 2LVcosθ / C [ Q C>>V ] ∴ V = C 2 ∆t / 2Lcosθ -------------------- (3) Now let ∆t = tBA − tAB Now, tBA + tAB = L / (C − Vcosθ) + L / (C + Vcosθ) = 2LC / (C 2 − V 2cos 2θ) = 2L / C [ Q C>>V ] ∴ (tBA + tAB) / 2 = L / C = tA (say) = averaging transit time between the transducers ∴ C 2 = (L / tA) 2 2 Now, equation (3) gives V = C ∆t / 2Lcosθ = (L / t A) 2 ∆t / 2Lcosθ 2 = L ∆t / 2 t A cosθ 2 = K ∆t / tA (where, K = L / 2cosθ = constant) Now, if the cross sectional area (A) of the pipe is known, one can calculate the volumetric flow rate (Q) by the equation: Q = AV

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c. A-scan, B-scan and C-scan display Ultrasonic data can be controlled and displayed by number of formats. The 3 most common formats are known in NDT world as A-scan, B-scan and C-scan presentation. Each presentation mode provides a different way of looking at and evaluating the region of the material to be inspected. Modern computerized US scanning systems can display data in all 3 modes simultaneously. A-scan presentation It displays the amount of received US energy as a function of time. The relative amount of received energy is plotted along the Y-axis where the elapsed time (time taken by the US to travel within the material is plotted along X-axis. Most instruments with an A-scan display allow the received signal to be displayed in its natural radio frequency form and either rectified or non rectified form. The size of a defect is estimated by comparing the defect echo amplitude with a known defect-echo amplitude. Position of the defect is estimated by observing the position of the defect-echo signal on the horizontal sweep. The test body and the scanning direction are shown in fig-16(a). And the A-scan representation is shown in fig-16(b). When the transducer is in far left, only the IP and A (energy reflected back from the surface A) are visible. As the body is scanned from left to

(a)

(b) Figure 16: A-scan representation

right, the BW (back wall) signal will be displayed but at the right most end because the transit time, in this case, is maximum. Then signal B will appear and at last signal C will be appear but it is positioned before B as the transit time in case of C is less than that of B as the flaw C is nearer from the surface than flaw B is. B-scan presentation

Figure 17: B-scan representation

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It is a profile view of the test specimen. Here the time-of-flight of US wave is plotted along the Y-axis and the linear position of the transducer is plotted along the X-axis. Here, the depth of the reflector and the linear dimension of the reflector in the scanning direction are approximated. The B-scan is typically produced by establishing a trigger gate on the A-scan. When the signal intensity is great enough to trigger the gate, a point on the B-scan is produced. In the B-scan image of fig-17, line IP & A is produced as the transducer is scanned over surface A. then line BW, line B, line C will be gradually appear as the body is scanned from left to right. The limitation is that, reflectors may be masked by larger reflectors near the surface. C-scan presentation The C-scan representation produces a plane type view of the location and size of the test body features. The plane of the image is parallel to the scanning plane. C-scan representation is produced with an automated DAS. Typically, a data collection gate is established on the A-scan. Amplitude or the time of flight of the signal is recorded at regular interval as the test body is scanned. The signal amplitude or the time of flight is

Figure 18: C-scan representation

displayed as a shade of gray or a color for each of the positions where data was recorded. In this way, the C-scan produces an image of the reflectors within the test body (fig-18). d. Attenuation of ultrasonic waves in an absorbing medium In an absorbing medium (Sound waves in a viscous fluid), ultrasonic wave loses energy as it propagates through the medium. It has been experimentally found that the amplitude of the wave decays by a constant fraction of its value when the wave propagates through a certain distance. This means that the amplitude falls exponentially with distance and the amplitude (A(x)) and distance (x) relationship with respect to the origin at x = 0 is given as follows: A(x) = Ao exp(−αx) ------------------------ (1) Where, Ao is the amplitude at x = 0, the constant term “ α” is called the attenuation constant. From (1) we have: dA(x)/dx = −αAo exp(−αx) = −αA(x) ∴ α = − (1/Ao) (dA(x)/dx) Thus, attenuation constant (α) is given by the decrement in amplitude, when wave travels in an absorbing medium, per unit amplitude per unit length.

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Figure 19: Attenuation of ultrasonic waves in an absorbing medium

α). It is defined as the distance The inverse of “ α” is called the attenuation length (1/ α). traveled by wave over which the amplitude is decreased by a factor “e = 2.718”, i.e. for x = 1/ α, A(x) = Ao exp(−1) = Ao /e. Dispersive medium and dispersion relation:- The displacement of particles of wave when traveling in an absorbing medium in positive X direction s given by: Y(x,t) = Ao exp(−αx) exp (i *(2 π/λ)∗(Vt – x)) = Ao exp[(2 πi/λ)(Vt – x)−αx] = Ao exp[(2Vπi/λ)(t – x/V)−αx] = Ao exp[2 νπi(t – x/V)−αx] [Q V/ λ = ν] = Ao exp[2 νπi(t – x/V−αx/2 νπi)] = Ao exp[2 νπi{ t – x(1/V+α /2 νπi)}] = Ao exp[2 νπi{t – x(1/V−αi/2 νπ)}] = Ao exp[2 νπi(t – x/V*)] Where, 1/V* = 1/V −α i/2 νπ This V* is called the complex wave velocity applicable for the waves traveling through absorbing medium. In practice this method of introducing the complex wave finds application chiefly in optics in which we define the refractive index as n = c / V Where, c is the velocity of light in vacuum and v is the velocity of light at that medium. In absorbing medium the complex refractive index is given as: n* = c / V* = c / V − cαi/2 νπ = n − cαi/2 νπ This shows that the refractive index (and hence the wave velocity) depends on the frequency ν (or the wavelength λ ) of the wave in an absorbing medium. Such a medium is called dispersive medium and this relation between wave velocity and frequency (or wavelength) is called dispersive relation.

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Now, We know, energy flux = intensity ∝ (Amplitude) ∝ [A(x)] 2 = Κ (Ao) 2 exp (−2αx) (Where, K is a constant) ∴ Energy flux = Κ ( Ao) exp (−2αx) This implies that the attenuation length in case of intensity is half that of amplitude. This loss of intensity during the propagation of the ultrasonic wave is appears as heat in the system. 2

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Ultrasonic Solution

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