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K L University 1 By G.SUNITA DEPARTMENT OF PHYSICS.

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Presentation on theme: "K L University 1 By G.SUNITA DEPARTMENT OF PHYSICS."— Presentation transcript:

1 K L University 1 By G.SUNITA DEPARTMENT OF PHYSICS

2 ULTRASONICS LECTURE 2

3 3 If mechanical pressure is applied to one pair of opposite faces of certain crystals like quartz, equal and opposite electrical charges appear across its other faces. This is called as piezo- electric effect. The converse of piezo electric effect is also true. If an electric field is applied to one pair of faces, the corresponding changes in the dimensions of the other pair of faces of the crystal are produced.This is known as inverse piezo electric effect or electrostriction. Principle : Inverse piezo electric effect

4 4 Quartz Crystal

5 5 The circuit diagram is shown in Figure Piezo electric oscillator

6 6 The quartz crystal is placed between two metal plates A and B. The plates are connected to the primary (L 3 ) of a transformer which is inductively coupled to the electronics oscillator. The electronic oscillator circuit is a base tuned oscillator circuit. The coils L 1 and L 2 of oscillator circuit are taken from the secondary of a transformer T. The collector coil L 2 is inductively coupled to base coil L 1. The coil L 1 and variable capacitor C 1 form the tank circuit of the oscillator.

7 7 When H.T. battery is switched on, the oscillator produces high frequency alternating voltages with a frequency. Due to the transformer action, an oscillatory e.m.f. is induced in the coil L 3. This high frequency alternating voltages are fed on the plates A and B. Inverse piezo-electric effect takes place and the crystal contracts and expands alternatively. The crystal is set into mechanical vibrations. The frequency of the vibration is given by n = where P = 1,2,3,4 … etc. for fundamental, first over tone, second over tone etc., Y = Young’s modulus of the crystal and ρ = density of the crystal. Working

8 8 Advantages Ultrasonic frequencies as high as 5 x 108Hz or 500 MHz can be obtained with this arrangement. The output of this oscillator is very high. It is not affected by temperature and humidity. Disadvantages The cost of piezo electric quartz is very high The cutting and shaping of quartz crystal are very complex.

9 9 a)Kundt's tube method b)Sensitive flame method c)Piezo-electric detector: a) Thermal detection method b)By acoustic grating method

10 10 a)Kundt's tube method Lycopodium Powder is used Heaps at the Nodes and blown off at Antinodes Measure the Wavelength and Velocity of ultrasonic sound

11 11 Wavelength: The average distance btw two successive nodes or heaps is taken as ‘d’. It should be equal to the half of the wavelength of ultrasonic waves. Velocity: Velocity of ultrasonic wave is V ϑ frequency of ultrasonic wave This method is suitable for measuring velocity of low frequency ultrasonic waves. It can not be used for high frequency ultrasonic waves.

12 12 b) Sensitive flame method When a narrow sensitive flame is moved in a medium of ultrasonic waves. Flame remains stationary at antinodes and flickers at nodes. c) Piezo-electric detector Quartz crystal – for detection of ultrasonic One pair of faces of quartz subjected to Ultrasonics. Varying electric charges are produced. These charges are very small and they can be amplified.

13 13 e) acoustic grating method Principle : Ultrasonic waves are propagating through a liquid medium (stationary waves formed) Density of liquid varies from layer to layer. Monochromatic light is passed – perpendicular direction. Liquid behaves as diffraction grating. Probe made of thin platinum wire. Temperature of the medium changes due to alternate compressions and rarefactions. Resistances of the platinum wire changes at node and remains constant at antinodes. Detected by sensitive resistance bridge. d) Thermal detection method

14 14 Reflected waves are called echos. Superposition of the direct and reflected waves - Longitudinal stationary waves produced. Nodes and Antinodes are formed – refractive index changed

15 15 λ is the wavelength of ultrasonic wave d = λ /2 Velocity of ultrasonic wave Working: Monochromatic light Diffraction pattern consists of central maxima, first order maxima d – distance btw two nodes or antinode planes. Wavelength of light is given by 2d sinθ = n λ

16 16 Applications of Ultrasonic Waves in Engineering (1)Detection of flaws in metals (Non Destructive Testing –NDT) Principle: Ultrasonic waves are used to detect the presence of flaws or defects in the form of cracks, blowholes porosity etc., in the internal structure of a material By sending out ultrasonic beam and by measuring the time interval of the reflected beam, flaws in the metal block can be determined.

17 17 Experimental setup 1. Non Destructive Testing –NDT Master timer Signal pulse generator Time base amplifier Echo signal amplifier C R O Transducer Metal under Test It consists of an ultrasonic frequency generator and a cathode ray oscilloscope(CRO), transmitting transducer(A), receiving transducer(B) and an amplifier.

18 18 NDT Working Process delamination plate IP F BE IP = Initial pulse F = Flaw BE = Backwall echo Probe Flaw Sound travel path Work piece s Pulse echo systems:

19 19 In flaws, there is a change of medium and this produces reflection of ultrasonic at the cavities or cracks. The reflected beam (echoes) is recorded by using cathode ray oscilloscope. The time interval between initial and flaw echoes depends on the range of flaw. By examining echoes on CRO, flaws can be detected and their sizes can be estimated. NDT Working Process

20 20 NDT Working Process Transmission testing systems: Through transmission signal T T R R Flaw

21 21 NDT Working Process Resonance Systems: Ultrasonic standing waves are setup with in the specimen causing the specimen to vibrate at greater amplitude. Resonance is then sensed by CRT (cathode ray tube), and that frequency is useful to detect the discontinuity of material.

22 22 This method is used to detect flaws in all common structural metals and other materials like rubber tires etc. The method is very cheap and of high speed of operation. It is more accurate than radiography. Features :

23 23 2. SONAR (Sound Navigation and Ranging ) High Frequency Ultrasonic Waves are Used To find : Distance and direction of submarines. Depth of sea Depth of rocks in the sea Crowd of fish in seas Sharp Ultrasonic beam is directed in various directions. The reflection of waves from any direction shows the presence of some reflecting body in the sea. Reflected waves collected by the receiver.

24 24 3. Depth of Sea: The ultrasonic transducer transmits the ultrasonic waves towards the bed of sea. The waves are reflected back from the bed and reflected signal (echo) collected by the receiver. Depth of sea ‘h’= v х t/2 Where ‘v’ velocity of sound through sea water ‘T’ is time interval between emitted signal and echo received.

25 25 3. Medical applications: To obtaining information about flow of blood through the heart and the about the condition of heart valves. Its used in blood less surgery Also used for detecting tumors and other defects in human body. To view the Fetus in its mother's womb, viewed in a sonogram.

26 26 Problems on Ultrasonic's: 1. For a quartz crystal of length 0.05cm, calculate the fundamental frequency of oscillation. In a piezoelectric oscillation oscillator if the velocity of longitudinal waves in the crystal is 5.5 x 10 3 m/sec. 2. A boat out ultrasonic pulse to determine the depth of the sea, if the echo is received after 80 msec. What id the depth of sea given that speed of sound in water is 1500m/sec. 3. A Quartz crystal of thickness 0.001m is vibrating at resonance. Calculate the fundamental frequency. Given (Y = 7.9 x N/m 2, ρ = 2.65 x 10 3 kg/m 3 ). 4. Design a piezoelectric oscillator which produces ultrasonic waves of frequency 10 6 Hz with an inductance of 1 Henry and what is a capacitance? 5. A particle is acted upon by a sound wave executing S.H.M is given by y=10sin(6t + π/3) in meter after 10 sec, find the displacement, velocity and time period of the particle.

27 27 6. A particle crystal in an ultrasonic interference produces stationary waves of frequency 1.5 MHz. If the distance between 6 consecutive nodes is 2.75mm. Find the velocity of ultrasonic waves. 7.A pulse of ultrasound is sent along the length of a piece of metal which is suspended to have an internal crack. The pulse is reflected from any cracks, and from the end of the metal bar and the reflected pulse is picked up by a detector beside the transmitter. The diagram below shows the trace on a CRO which picks up the echo of the sound. The “tick” mark on the time axis are 0.2 ms apart. Find the position of the crack and length of the metal rod. Remember the speed of sound in typical metal is 5000 m/s. 8. A piezo-electric crystal has thickness m. If the velocity of sound wave in crystal is 5750 m/s. Calculate the fundamental frequency of the crystal.

28 28 9. Design a magneto- strictive oscillator which produces ultrasonic wave of frequency 10 6 Hz with an inductance of 1Henrry. Then find the capacitance. 10. Find the fundamental frequency of a quartz crystal plate of thickness is √30 mm. (Given E= 8x 10 9 pascal, density of material = 2.7x10 3 kg/m 3 ).


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