1. 1. Preparatory Notes for MyASNT NDT Level III ExaminationUltrasonic Testing, UT Reading Two Part 3Some essential knowledgeMy pre-exam self study note - 2014
2. 2. At Works
3. 3. Pulse Echo Methods
4. 4. Delta Techniques:In some two-transducer pitch-catch techniques, both transducers are locatedon the same side of the material. The time between pulses corresponds to asingle traverse of the sound from the transmitter to the reflector and then tothe receiver. One approach uses a "tandem" pitch-catch arrangement, usuallyfor the central region of thick materials. In this technique, the transmittersends an angle beam to the midwall area of the material (often a double Vweld root) and deflections from vertical planar surfaces are received by one ormore transducers located behind the transmitter. Another pitch-catchtechnique, found in immersion testing, uses a focused receiver and a broad-beamtransmitter, arranged in the shape of a triangle (delta technique). Thistechnique relies on reradiated sound waves (mode conversion of shearenergy to longitudinal energy) from internal reflectors, with background noisereduction through use of the focused receiver.
5. 5. Delta TechniquesAngulation of Transmitter as requiredfor detection of predetermined flawhttp://www.ndt.net/article/ecndt2006/doc/We.2.1.1.pdfShear WaveLongitudinal Wave
6. 6. Calculation Practices
7. 7. Q1: A tubular product is tested by AUT. The tube is rotated at 500 RPM Ifbeam diameter is 10 mm and overlap between scan is 50%. Calculatemaximum length of the tube that can be tested.Overlap 50%Area scanned for eachDirection of travel RevolutionAnswer: The length travel for each revolution with specified overlap is 5mm,the rate of traverse is 500x 5= 2.5m/ minute
8. 8. Q2: A steel bar with 200 mm thick is scanned by UT. Minimum number of hitsrequired is 10. What is the maximum PRR to avoid the ghost echo. (VL, steel= 0.59 cm/ s )Answer:The total distance traverse in 10 hits = 200 x10x 2 (round trip) = 4mThe traverse time T for sound to cover 4m = 4/ 5900 sThe PRR= 1/T = 5900/4 = 1475 Hz.200 mm400 mm round trip4m for 10 hints
9. 9. Q3: Minimum number of hits required is 2. What is the maximum allowableaxial speed for a probe with effective diameter of 102 mm and PRR of 800 Hz.Answer:Axial speed max = PRR/2 x 102mm = 40.8m/s
10. 10. Q4: Assuming that Minimum 3 pulses are required to trigger the alarm in AUT.What would be scanning speed to detect 0.5 mm size defect while using 4MHz probe, 10 mm beam diameter with PRR of 0.5 kHz.Answer:Scanning speed = PRR/ hits x Effective CoverageScanning speed = 500/3 x 10mm = 1666.7mmHint: the defect size of 0.5mm is non-significant as it is smaller than thebeam diameter. Should the defect size id greater than the beam diameter theanswer is invalid.Speed VBeam diameter, 10mmPRR = 1/T = V/10V = 10 x PRR, for 1 hitV3 = 10 x PRR/3, for 3 hitsTime to cover traverse the beam coverage T= Beam diameter/ speed, 10/S
11. 11. Q5: What is the maximum PRR is needed for contact test of steel materialwith 100 mm thick using longitudinal wave. Minimum number of hits requiredis 10. (VL, steel = 0.59 cm/ s )Answer: Time traverse for1 hit (T) = 0.2 / 5900, PRR= 1/T = 5900/ 0.2 =29500 Hz.IIn order to record more than 1 hit, the PRR has to be reduced so that thereceiver is allocated more time in doing the receiving.PRR for 10 hits = 1/(hits x T)PRR for 10 hits = 1/T /hits = 5900/(0.2 x 10) = 2950Hz.Distance = .2m for round trip
12. 12. Q5: Answer: (put it in order way)Time traverse for1 hit, T = 0.2 / 5900,Time traverse for 10 hits, T10 = 0.2 x10 / 5900PRR for 10 hits = 1/T10 = 5900/(0.2 x 10) = 2950Hz.
13. 13. Q6: What is the maximum PRR is needed for immersion testing of aluminumwith 80 mm thick using longitudinal wave. The water path is 10 mm. Minimumnumber of hits required is 20. (VL, aluminum = 6.32 103 m/s, VL, water =1.48 103 m/s)Answer:Single Water path =10mm, Single Al path= 80mmLoop H2O = 0.02 m, Loop Al = 0.16 mTime for traversing single hit, T = [0.02/1480] + [0.16/6320] = 3.883 x 10-5 sTime for traversing 20 hits T = 20 x T = 3.883sPRR = 1/T = 1287HzStandard answer: 1920 HzShould some one else figured outthe correct answer is 1920Hz0.2m0.001mplease advice me, with thanks.0.16m 0.08m
14. 14. Keywords:Numbers of hits increase , the PRR decrease
15. 15. My repeated mistake!Have I double the distances to account for the round trip?0.001m0.2m0.16m 0.08m
16. 16. Q7: A steel plate size 6.2 m 1.8 m 0.1m is scanned using 25 mmdiameter normal probe and overlap between scan is 20%. Minimum numberof hits required is 15. Calculate the inspection time if scanning speed is 500mm/s.Answer:Area of plate A1= 6.2 x 1.8 m2Effective Area cover by the probe in 1 second = 0.5 x (0.025 x 0.8)Time to scan the area = A1/A3 = 1116 sHint:1/ 0.8 1.2 !Calculate without considering the Deff and multiple the answer with 1.2 toaccount for the 20% overlap gives incorrect answer.
17. 17. Q8: Assume that the minimum PRR is needed for contact test of a given steelplate is 240 Hz. How much volume would be covered under test if the PRR isset at 120 Hz.Answer: The inspection rate increase with PRR, lowering the PRR reducethe inspection volume.Volume cover PRR, New Volume cover = 120/240 = 0.5.
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35. 35. 9th September 2014Student log:Good Luck:Good Luck:Good Luck:Good Luck: Good Luck:Good Luck:Good Luck:Good Luck:Good Luck: Good Luck:Good Luck:Good Luck:Good Luck:Good Luck: Good Luck:Good Luck:Good Luck:Good Luck:Good Luck: Good Luck:Good Luck:Good Luck:Good Luck:Good Luck: Good Luck:Good Luck:Good Luck:Good Luck:Good Luck: Good Luck:Good Luck:Good Luck:Good Luck:Good Luck: