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ASNT NDT LEVEL III

ULTRASONIC TESTING
MODEL QUESTION BANK
PROBLEMS AND SOLUTIONS

2. If a probe frequency with 2 MHz and 10 mm diameter is used, find the wave length in steel
for longitudinal wave testing. (VL, steel = 0.59 cm/ s )

v = 0.59 cm/ s f = 2 MHz =?

=v/f
= 0.59 / 2
= 0.3 cm = 3 mm

31. The reflected pulse reaching the immersion transducer from the back surface of a 3 inch
aluminum plate standing in a tank of water is equal to _____ of the energy pulse which was
transmitted from the transducer. (Zaluminum = 17 × 106 kg/cm2/s, Zwater = 1.5 × 106 kg/cm2/s)

Z1 = 17 × 106 kg/cm2/s Z2 = 1.5 × 106 kg/cm2/s

2
R = (Z1 - Z2) / (Z1 + Z2)2
2
= (17 1.5) / (17 + 1.5)2
= 0.702 = 70.2% 70%

T% = 100% R%
= 100 70
= 30%

Transmitted pulse (% ) from the front wall of Al = 30%

Reflected pulse (% ) from the back wall of Al = 0.3 × 0.7

= 0.21
= 0.21 × 100 = 21%

Reflected pulse (% ) reaching the immersion

transducer from the back wall of Al = 0.21 × 0.3
= 0.063
= 0.063 × 100 = 6.3%

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49. An ultrasonic longitudinal wave travels in aluminum with a velocity of 632 000 cm/s and has
a frequency of 1 MHz. The wavelength of this ultrasonic wave is

v = 632 000 cm/s f = 1 MHz =1 × 106 Hz =?

= v/ f
= 632 000 / 1 × 106
= 0.63 cm
= 6.3 mm

53. When inspecting aluminum block by the immersion method using water for a couplant, the
following information is known. Velocity of sound in water = 1.48 × 103 m/s, velocity of
longitudinal waves in aluminum = 6.32 × 103 m/s, and angle of incidence = 5°. The angle of
refraction for longitudinal wave is approximately

1 =5° 2= ?
3
VL, water = 1.49 × 10 m/s VT, aluminum = 6.32 × 103 m/s

sin 2 =. sin 1 × [ VT, aluminum / VL, water ]

= sin 5° × [ 6.32 / 1.48 ]

= 0.37
-1
2 = sin (0.37)

= 21.7°
22°

55. Since the velocity of sound in aluminum is approximately 245000 in./s, it takes how long for
sound to travel through 25 mm (1 in.) of aluminum?

v = 245000 in./s d = 1 in. t=?

v= d/t

t = d /v
= 1/ 245000
= 4 × 10-6 s
= 4 s

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59. The resonant frequency of a 2 cm (0.79 in.) thick plate of naval brass (v = 4.43 × 103 m/s) is

v = 4.43 × 103 m/s T = 2 cm f=?

= 4.43 × 103 ×100
= 4.43 × 105 cm/s

f = v / (2 T)
= 4.43 × 105 / (2 × 2)
= 1.111 × 105 Hz
= 0.111 × 106 Hz
= 0.111 MHz

69. An aluminum disc with 1000 mm thick is scanned by immersion testing and rotated at 15
RPM. What is the optimum PRR to be set so that the defect do not run past before it reaches the
probe? Minimum number of hits is 5. The effective diameter of probe including the prescribed
over lap is 5 mm.

Disc Thickness = D = 1000 mm Minimum number of hits = 5

Probe diameter = 5 mm RPM = 15 PRR =?

r = D/2
= 1000/2 = 500 mm

Speed of test part = 2 r (RPM / 60)

=2 × 500 (15 / 60)

= 785 mm/ s

PRR = (Speed of test part × Number of hits) / Effective diameter of probe

= (785 × 5) / 5

= 785 pulse/s

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73. What will be the probe separation distance for a tandem technique for the examination of 50
mm thick weld (double V joint) root using 45 probe?

T = 50 mm = 45

d = Root of double V joint = T/2

= 50/2

= 25 mm

Distance between T & R probes in Tandem technique = 2 tan (T d)

= 2 × tan 45 × (50 25)
= 50 mm

76. A 50 mm steel plate requires immersion examination with focusing at the center. 25 mm
minimum water path distance requires for the examination. What will be the radius of curvature
required on the Plexiglass shoe? (VL, plexiglass = 0.273 cm/ s VL, water = 0.148 cm/ s)

Water path = 25 mm R=?

T = 50 mm

Desired distance of focus in material = Center of steel plate

= T/2
= 50/2
= 25 mm

n = VL, plexiglass / VL, water

= 0.273/0. 148
= 1.84 1.8

VL, steel / VL, water = 0.59/ 0.148

=4

f = Water path + [Desired distance of focus in material × (VL, material / VL, water ) ]

= 25 + [ 25 × 4 ]

= 125 mm

R = f [(n - 1) / n]

= 125 [(1.8 - 1)/ 1.8]

= 55.5 mm

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77. When inspecting aluminum plate by using the 45° shear wave probe, if change in probe
angle is 2° and what would be change in angle of incidence of shear wave in aluminum? (VT, steel
= 0.323 cm/ s, V T, aluminum = 0.313 cm/ s).

1 =2° 2 = ?
V1 =0.323 cm/ s V2 = 0.313 cm/ s

sin 2 = sin 1 (V2/ V1 )

= sin 2 (0.313/ 0.323)

= 0.034

2 = sin-1 (0.034)

= 1.95°

79. In a water immersion test, what is off-set distance is needed to produce 45° shear wave for
testing of 200 mm diameter steel?(VT, steel = 0.323 cm/ s, VL, water = 0.148 cm/ s )

D = 200 mm = 45

x = (sin ) (D/ 2)

= (V L, water/ V T, material ) (sin ) (D/ 2)

= (0.148/ 0.323) (sin 45) (200/ 2)
= 32.4 mm

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80. When inspecting a 200 mm thick steel block by using 10 mm diameter 4 MHz normal probe,
what would be beam diameter at 100mm depth from surface of the block?(VL, steel = 0.59 cm/ s )

v = 0.59 cm/ s f = 4 MHz D = 10 mm

Beam diameter at 100 mm depth =?

= v/ f
= 0.59 / 4
= 0.15 cm
= 1.5 mm

sin /2 = 1.2 /D
= 1.2 × 1.5 / 10
= 0.18

/2 = sin-1 (0.18)
= 10.4°

Beam diameter at depth = 2 d tan ( /2)

= 2 × 100 × tan (10.4)

= 36.6 mm

83. The signal from 2 mm diameter FBH was kept at 80% of full screen height. The 1 mm
diameter FBH with the same equipment setting will produce signal with ______ % of full screen
height?

H 1 =? H2 = 80% FBH
D 1 = 1 mm D 2 = 2 mm

H1 / H2 = (D1)2 / (D2)2

H1 = H 2 (D 1 ) 2 / (D2)2

= 80 × 12 / 22

= 20%

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84. In an immersion examination of 15 mm titanium plate (VL = 0.610 cm/ s; VT = 0.312

cm/ s) with the probe tilt of 10 , what angle and mode of ultrasonic beam will produce in the
plate? (V L, water = 0.148 cm/ s)

1 =10 2 = ?
V1 = 0.148 cm/ s V2 = 0.610 cm/ s

sin 2 = sin 1 (V2/ V1 )

= sin 10 (0.61/ 0.148)

= 0.716

2 = sin-1 (0.716)

= 45.7°

45°

86. Calculate the sound path for pitch catch technique for the examination of 30 mm thick plate
with the region of interest is at 2/3 rd of plate thickness by using 60 probe.

T = 2/3 rd of plate thickness = 60

= 2/3 × 30 mm
= 20 mm

S = 2 T / cos
= 2 ×20 / cos 60]
= 40 / 0.5
= 80 mm

87. An immersion examination requires 10% overlap while using 10 mm diameter focused
probe with focusing size of 3 mm diameter. What shall be the maximum linear probe shift
required to satisfy the examination requirements?

D = 3 mm Overlap between scan = 10%

Effective dia. of probe = Dia. of probe 2 [ (Dia. of probe) × (Percent of overlap between scan / 100) ]

= 3 2 [ 3 × (10/ 100)]
= 2.4 mm

Maximum linear probe shift required = 2.4 mm

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89. A 4 MHz normal probe was used in a 10 mm thick titanium plate and if the first back wall
echo appeared at 80% of FSH, second back wall echo appeared at 50% of FSH. The attenuation
on the plate is

H 1 = 80% of FSH H 2 = 50% of FSH d = 10 mm

Attenuation = [1/ (2 d)] 20 log (H 1 / H 2 )

= [1/ (2 × 10)] 20 log (80 / 50)
= 0.2 dB/mm

90. In an immersion testing of steel plate the 5 MHz probe was tilted at 19 and the corner echo
was noted at 4.67 µs from the water-steel interface. What is the thickness of the plate?

t = 4.67 µs T=?

1 =19 2 = ?
V1 =0.148 cm/ s V2 = 0.323 cm/ s

sin 2 = sin 1 (V2/ V1 )

= sin 19 (0.323/ 0.148)

= 0.71

2 = sin-1 (0.71)

= 45.3° 45°

S=v×t/2

= 0.323 × 4.67 / 2

= 0.75 cm = 7.5 mm

Cos = T / S

T = S Cos

= 7.5 Cos 45
= 7.5 × 1.41
= 10.6 mm

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96. In a testing of 10 mm steel plate with a 20 MHz and 6 mm diameter normal probe, a signal
was noted at 1.67 µs from the test surface. What is the depth of the discontinuity?
(VL, steel = 0.59 cm/ s)

v = 0.59 cm/ s t = 1.67 µs. d =?

S = v × t/ 2

= 0.59 × 1.67/ 2

= 0.49 cm

= 4.9 mm

d=S
= 4.9 mm

97. An examination revealed an indication at 5.5 s, while testing with 45° longitudinal angle
beam probe on 50 mm thick steel specimen. What is the depth of the discontinuity? (VL, steel =
0.59 cm/ s )

= 45 t = 5.5 s VL, steel = 0.59 cm/ s d=?

S = v × t/ 2

= 0.59 × 5.5/ 2

= 1.62 cm

= 16.2 mm

d = S Cos
= 16.2 × Cos 45
= 11.4 mm
11 mm

98. In an immersion testing of a 25 mm steel plate the half beam divergence was noted in the
specimen as 5 . The water path was noted as 100 mm for the 15 MHz immersion probe. What is
the beam diameter?

/2 = 5 d = 100 mm Beam diameter at 100 mm depth =?

Beam diameter at depth = 2 d tan ( /2)

= 2 × 100 × tan 5

= 17.5 mm
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99. The back wall echo from 100 mm job is adjusted to 80% of FSH. The probe is placed on a
200 mm thick job and back wall was found to be 60% of FSH. The attenuation factor of the
material is ______ dB/m. The equipment is used in pulse echo A-scan mode.

H 1 = 80% of FSH H 2 = 60% of FSH

d1 = 100 mm d2 = 200 mm

d = d2 - d1
= 200-100
= 100 mm

A = [1/ (2 d)] 20 log (H 1 / H 2 )

= [1/ (2 × 100)] 20 log (80 / 60)

= 0.0125 dB/mm

= 0.0125 × 103

= 12.5 dB/m

12 dB/m

100. A FBH located from the transducer at a distance of 40 mm produced echo height 80% of
FSH. From the same size of FBH located at 80 mm produced some height in CRT and then it
was adjusted to 80% of FSH. How much dB should have been added to achieve this?

H1 = 80% of FSH H 2 (initial ht.) =? dB = ? (if H 2 =80% of FSH)

d1 = 40 mm d2 = 80 mm

H1 / H2 = (d 2 ) 2 / (d1)2

H 2 = H1 (d 1 ) 2 / (d2)2

= 80 × 40 2 / 802

= 20% of FSH.

dB = 20 log (H1 / H 2 )

= 20 log (80 / 20)

= 12
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102. The attenuation coefficient for structural steel at 2 MHz is 8 dB/m. For this condition a
sample of 100 mm thickness was tested using a pulse echo instrument. The first back wall echo
was adjusted to 80% of FSH. What would be the height of the second back wall echo?

d = 100 mm H1 = 80% of FSH. H2 = ?

Attenuation = A = 8 dB/m

= 8 dB/ 1000 mm

= 8 × 10-3 dB/ mm

A = [1/ (2 d)] 20 log (H 1 / H 2 )

log (H1 / H 2 ) = A 2 d / 20

= A d / 10

= 8 × 10-3 × 100 / 10

= 0.08

(H1 / H 2 ) = antilog (0.08)

= 1.2

H 2 = H1 / 1.2

= 80 /1.2

= 66.5% 66%

103. A 1000 mm diameter (OD) 150 mm thick forged block is to be tested using 45º angle beam
by moving the probe on the circumference for possible cracks propagating radially towards
centre from the rim. From the rim surface how far the inspection is valid?

OD = 1000 mm = 45

ID = OD sin

= 1000 × sin 45

= 700

The inspection is valid up to = (OD - ID)/ 2

= (1000 - 700)/ 2

= 150 mm

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104. A 2 mm diameter FBH showed 60 % of FSH located at 25 mm. A 4 mm diameter FBH

located at 50 mm would show an indication at ______ of FSH.

H1 = 60% of FSH. H2 = ?

d1 = 25 mm d2 = 50 mm

D1 = 2 mm D2 = 4 mm

H1 / H2 = [(d 2 ) 2 (D1)2 ] / [ (d1)2 (D2)2 ]

H 2 = [ H1 (d 1 ) 2 (D2)2 ] / [ (d2)2 (D1)2 ]

= [ 60 (25) 2 (4) 2 ]/ [ (50)2 (2) 2 ]

= 60% of FSH

107. By positioning the probe on 25 mm radius of V 2 block the fourth echo found to be on the
10th division on Cathode Ray Tube (CRT). The range of the equipment is

The range of the equipment up to = 25 + 75 + 75+ 75

= 250 mm

The range of the equipment = 0 250 mm

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108. A 100 mm OD and 15 mm thick steel tube is to be scanned by generating shear wave in the
tube, for its full volume by placing the job in immersion testing. The angle of incidence in water
is: (VL, water = 0.148 cm/ s and VT, steel = 0.323 cm/ s )

OD = 100 mm T = 15 mm

ID = OD (2 T)

= 100 (2 × 15)

= 70 mm

sin = ID / OD

= 70/ 100

= 0.70

= sin-1 (0.7)

= 44.4 45 = angle of diffracted shear wave in steel

1 =? 2 = 45°
V1 = 0.148 cm/ s V2 = 0.323 cm/ s

sin 1 = sin 2 (V1/ V2 )

= sin 45 (0.148/ 0.323)

= 0.324

1 = sin-1 (0.323)

=18.9°

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111. In a through transmission technique using 2 MHz probe the echo was set to 80% of FSH on
a material of 100 mm thick. The attenuation coefficient of the testing material consideration is
20 dB/m. If the material is tested with pulse echo technique with the same frequency and
diameter probe, what would be the height of the first back wall echo in terms of FSH?

d = 100 mm H1 = 80% of FSH. H2 = ?

A = 20 dB/m

= 20 dB/ 1000 mm

= 0.02 dB/ mm

= 0.02 × 100

= 2 dB/ 100 mm

dB = 20 log (H1 / H 2 )

log (H1 / H 2 ) = dB / 20

= 2/ 20

= 0.1

(H1 / H 2 ) = antilog (0.1)

= 1.26

H 2 = H1 / 1.26

= 80 /1.26

= 63.5% 60% of FSH

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112. A steel plate with thickness of 25 mm is immersed in water and the water path distance is
20 mm. What would be percentage of reflection observed on the CRT after the sound returns to
the transducer after hitting the back wall surface, assuming the incident acoustic intensity on
front wall of steel-water interface (first interface) is 100 % of FSH? (VL, steel = 0.59 cm/ s)

Incident sound intensity at water-steel interface is 100%,:

Amount of Reflection at water-steel or water-steel interface 88%
Amount of Transmission at water-steel or water-steel interface 12%

Transmitted pulse (% ) from the front wall of steel = 12%

Reflected pulse (% ) from the back wall of steel = 0.12 × 0.8
= 0.096
= 0.096 × 100 = 9.6%

Reflected pulse (%) from back wall reaching immersion probe = (0.096 × 0.12) × 100
= 1.15%

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113. A 100 mm solid shaft is to be scanned by immersing the shaft in water. It is desired to
cover at least 6.7 mm depth from the periphery. How much off-set from the centre of the shaft
for normal beam probe is required? (VL, water = 0.148 cm/ s and VT, steel = 0.323 cm/ s )

OD = 100 mm d = 6.7 mm

ID (to scan) = OD ( 2 × d)

= 100 (2 × 6.7)

= 86.6 mm

ID = OD sin

sin = ID / OD

= 86.6/ 100

= 0.866

= sin-1 (0.866)

= 60 = angle of diffracted shear wave in steel

x = (sin ) (D/ 2)

= (V L, water/ V T, material ) (sin ) (D/ 2)

= (0.148/ 0.323 ) (sin 60) (100/ 2)

= 19.8 mm
20 mm
= 1/5th of diameter

115. The pulser excited the crystal with 300 V to generate acoustic vibration of certain
frequency. After attenuation the crystal received back mechanical energy and the voltage
generated was 0.1 V. This loss is equal to ______dB.

H1 = 300 V H 2 = 0.1 V dB =?

dB = 20 log (H1 / H 2 )

= 20 log (300 / 0.1)

= 69.5

70

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116. The depth of a vertical crack in a double V joint is at 15 mm and the thickness of the
scanning surface plate is 30 mm. The probe used is 45 in tandem arrangement. The probes gap
from their exit point would be

T = 30 mm d = 15 mm = 45

Distance between T & R probes in Tandem technique = 2 tan (T d)

= 2 × tan 45 × (30 15)
= 30 mm

117. In an immersion examination of 10 mm steel plate (VL = 0.590 cm/ s; VT = 0.323 cm/ s)
with the probe tilt of 15 . Calculate the angle of refraction of transverse wave in steel? (V L, water
= 0.148 cm/ s)

1 =15 2 = ?
V1 = 0.148 cm/ s V2 = 0.323 cm/ s

sin 2 = sin 1 (V2/ V1 )

= sin 15 (0.323/ 0.148)

= 0.565

2 = sin-1 (0.565)

=34.4°

34°

119. A change in the beam angle is noticed while calibrating a 45 degree probe using IIW block.
The change was found to be + 2 . What would be change in the incident angle in Lucite
(perspex) wedge? (VL, perspex = 0.273 cm/ s and VT, steel = 0.323 cm/ s )

1 = ? 2 = 2
V1 =0.273 cm/ s V2 = 0.323 cm/ s

sin 1 = sin 2 (V1/ V2 )

= sin 2 (0.273/ 0.323)

= 0.0295

1 = sin-1 (0.0295)

= 1.69° 1.7°

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122. The attenuation coefficient for a particular UT system on a certain material is 20 dB/m. If
the echo from FBH is set to 75% of FSH, for a metal path of 50 mm in pulse echo system, what
would be echo height for 100 mm metal path in the same system of testing?

H1 = 75% of FSH. H2 = ?

dB = 20 dB/m

= 20 dB/ 1000 mm

= 0.02 dB/ mm

= 0.02 × 200

= 4 dB/ 200 mm

dB = 20 log (H1 / H 2 )

log (H1 / H 2 ) = dB / 20

= 4/ 20

= 0.2

(H1 / H 2 ) = antilog (0.2)

= 1.58

H 2 = H1 / 1.58

= 75 /1.58

= 47.5%

125. In an immersion testing of an aluminum disc with 100 mm thick is scanned by immersion
testing and rotated at 60 RPM. Minimum number of hits is 5. The effective diameter of probe
including the prescribed over lap is 10 mm. What is the Pulse Repetition Rate (PRR) to be set?

Thickness = Diameter = 100 mm Minimum number of hits = 5

r = D/2 = 100/2 = 50 mm RPM = 60
Effective diameter of probe = 10 mm PRR =?

Speed of test part (mm/ s) = 2 r (RPM / 60)

= 2 × 50 (60 / 60)
= 314

PRR = (Speed of test part × Number of hits) / Effective diameter of probe

= (314 × 5) / 10 = 157

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127. A 20 mm plate is scanned by selecting a probe of 70º so that the beam hits the fusion zone
perpendicular for maximum reflection from unfused surface. What is groove angle to meet this
condition?

= 70

Probe angle = 90 (Groove angle / 2)

Groove angle = 2 × (90 Probe angle)

= 2 × (90 70) = 40

128. In immersing testing a scanner is moving at 300 mm/s, and the effective diameter of a
focused beam on the area of interest is 10 mm and 10% overlap between the scan is to be
ensured. The number of hits inside the specimen shall be 4 for total decay of pulse. Find the
PRR

Minimum number of hits = 4 Effective diameter of probe = 10 mm

Scanner speed = 300 mm/s PRR =?

PRR = (Scanner speed × Number of hits) / Effective diameter of probe

= (300 × 4) / 10
= 120

129. An aluminum disc is scanned by immersion testing. The water path is 60 mm and a strong
reflection is found from a discontinuity, and arrived after 44.6 µs in pulse echo system of
scanning. Assume the velocity of sound in water is 0.148 cm/ s, and in aluminum is 0.632
cm/ s. What is the depth of the defect from water-aluminum interface from the entry surface?

Water path = 60 mm = 60 /10 = 6 cm tdiscontinuity= 89.21 µs

t1 = Time for acoustic wave to travel water path = Water path / VL, water
= 6 / 0.148
= 40.5 s

t2 = Time for acoustic wave to travel between water-aluminum interface and the defect
= tdiscontinuity - t1
= (44.6 - 40.5) = 4.1 µs

Depth of the defect = t2 × VL, aluminum

= 4.1 × 0.632
= 2.59 cm
= 25.9 mm
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130. In an immersion testing the effective diameter of a focused beam on the area of interest is 8
mm and PRR is 375 pulses per second. The number of hits inside the specimen shall be 4 for
total decay of pulse. Find the scanner speed.

Minimum number of hits = 4 Effective diameter of probe = 8 mm

PRR = 375 Scanner speed = ?

Scanner speed = (PRR / Number of hits) × Effective diameter of probe

= (375 / 4) × 8
= 750 mm/s

144. In a water immersion test, ultrasonic energy is transmitted into steel at an incident angle of
14°. What is the angle of the refracted shear wave within the material?
(VT, steel = 0.323 cm/ s, VL, water = 0.148 cm/ s)

1 = 14° 2 = ?
V1 = 0.148 cm/ s V2 = 0.323 cm/ s

sin 2 = sin 1 (V2/ V1 )

= sin 14° (0.323 /0.148)
= 0.52

2= sin-1 (0.52)
= 31°

145. If a probe frequency with 2 MHz and 8 mm diameter is used, what is the minimum length
of the defect would be detected in steel for longitudinal wave testing? (VL, steel = 0.59 cm/ s)

v = 0.59 cm/ s f = 2 MHz Sensitivity =?

= v/ f
= 0.59 / 2
= 0.295 cm
= 3 mm

Sensitivity= /2
= 3/ 2
= 1.5 mm

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149. A pipe with outside diameter 10.4 in. and thickness 1.5 in. is to be tested with angle probe
in circumferential direction. The approximate angle which will ensure complete coverage

OD = 10.4 in. T = 1.5 in. =?

ID = OD [ 2 T ]

= 10.4 [ 2 × 1.5]

= 7.4 in.

sin = ID / OD

= 7.4/ 10.4

= 0.711

= sin-1 (0.711)
= 45.4° 45°

150. What is the minimum water path required for a 6 in. block of steel to be tested by
immersion testing with 0.75 in. diameter probe? (VL, steel = 0.59 cm/ s, VL, water = 0.148 cm/ s)

T = 6 in. Min. water path = ?

Min. water path = 0.25 in. + T / (VL, material / VL, water )

= 0.25 + 6/ (0.59 / 0.148)
= 0.25 + 6/ 4
= 1.75 in.

151. In the far field of a probe a 2 mm diameter FBH gives a 25% of FSH echo. At the same
metal path a 4 mm diameter FBH (Flat Bottom Hole) in the same material will give an echo
whose height will be

H1 = 25% of FSH. H2 = ?

d1 = 2 mm d2 = 4 mm

H1 / H2 = (d1)2 / (d2)2

H 2 = H1 (d2)2 / (d1)2

= 25 × 4 2 22

= 100 % of FSH

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152. The beam width (diameter) at the depth of 75 mm in a thick aluminum disk (v = 0.632
cm/ s ) with a 10 mm diameter and 2 MHz probe is

v = 0.632 cm/ s f = 2 MHz D = 10 mm

d = 75 mm Beam diameter at depth of 75 mm in Al disk = ?

= v/ f
= 0.632 / 2
= 0.316 cm = 0.316 × 10 =3.16 mm

sin /2 = 1.2 ( / D )
= 1.2 (3.16 / 10)
= 0.38

/2 = sin-1 (0.38)
= 22.3°

Beam diameter at depth = 2 d tan ( /2)

= 2 × 75 × tan 22.3
= 61.5 mm

215. Calculate the frequency for a wave length of 1.5 mm. if v = 0.600 cm/ s.

v = 0.6 cm/ s = 1.5 mm f=?

= 0.6 × 10
= 6 mm/ s

f=v/
= 6 / 1.5
= 4 MHz

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227. A 4 MHz shear wave probe having 60° angle is used for testing steel plate (VT, copper = 0.226
cm/ s, VT, steel = 0.323 cm/ s ), the angle and the frequency of this probe in copper will be

1 =60 2 = ?
V1 =0.323 cm/ s V2 = 0.226 cm/ s

sin 2 = sin 1 (V2/ V1 )

= sin 60 (0.226/ 0.323)

= 0.566

2 = sin-1 (0.566)

= 38.3° 38°

229. The acoustic impedance for brass is (v = 4.43 × 105 cm / s, density = 8.42 g/ cm3)

v = 4.43 × 105 cm = 8.42 g/ cm3 Z=?

Z= v
= 8.42 × 4.43 ×.105
= 3.7 × 106 g/cm2/s

230. First critical angle for perspex and steel interface (VL, perspex = 0.273 cm/ s and VL, steel =
0.590 cm/ s) would be

VL,1 =0.273 cm/ s VL,2 = 0.590 cm/ s 1st, crit = ?

1st, crit = sin-1 (VL,1/ VL,2 )

= sin-1 (0.273/ 0.590)

= 27.5°

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246. The near field of a round 0.5 in. diameter contact L-wave transducer being used on a steel
test part (VL, steel = 0.590 cm/ s ) operating at 6 MHz is

v = 0.590 cm/ s f = 6 MHz N=?

D = 0.5 in. = 0.5× 2.54 = 1.3 cm

N = D2 f / (4 v)
= (1.3)2 × 6 / (4 × 0.59)
= 4.3 cm

248. A change in echo amplitude from 80 percent of full screen height (FSH) to 20 percent FSH
is a change of

H1 = 80% of FSH. H 2 = 20% of FSH dB = ?

dB = 20 log (H2 / H 1 )
= 20 log (20 / 80)
= -12

254. The full angle beam spread for 2.25 MHz and 13 mm (1/2 in.) diameter transducer
longitudinal wave probe in brass is: (VL, brass = 0.383 cm/ s)

v = 0.383 cm/ s f = 2.25 MHz =?

D = 13 mm

= v/ f
= 0.383 / 2.25
= 0.170 cm = 1.7 mm

sin /2 = 1.2 ( / D )
= 1.2 (1.7 / 13)
= 0.157

/2 = sin-1 (0.157)

= 9°

= /2× 2
=9×2
= 18°

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256. The percentage reflection of sound energy in the case of compressional wave (Zaluminum = 17
× 106 kg/cm2/s , Zwater = 1.5 × 106 kg/cm2/s) at water-aluminum interface is

Z1 = 17 × 106 kg/cm2/s Z2 = 1.5 × 106 kg/cm2/s R %=?

2
R = (Z1 - Z2) / (Z1 + Z2)2
2
= (17 1.5) / (17 + 1.5)2
= 0.702

R% = R ×100%
= 0.702 × 100
= 70.2% 70%

259. The incidence angle of 12º in water would produce refracted angle of longitudinal wave in
steel:

1 = 12 2 = ?
V1 = 0.148 cm/ s V2 = 0.590 cm/ s

sin 2 = sin 1 (V2/ V1 )

= sin 12 (0.590/ 0.148)

= 0.829

2 = sin-1 (0.829)

= 56º

273. In an immersion testing the effective diameter of a focused beam on the area of interest is 4
mm and the scanner speed is 600 mm/s. The number of hits inside the specimen shall be 5 for
total decay of pulse. Find the maximum PRR.

Minimum number of hits = 5 Effective diameter of probe = 4 mm

Scanner speed = 600 mm/s PRR =?

PRR = (Scanner speed × Number of hits) / Effective diameter of probe

= (600 × 5) / 4

= 750

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289. What would be the percentage of transmitted energy into steel if oil is introduced between
the probe and steel? (Z steel = 45 × 106 kg/cm2/s, Z oil = 1.50 × 106 kg/cm2/s)

Z1 = 45 × 106 kg/cm2/s Z2 = 1.5 × 106 kg/cm2/s T% = ?

2
R = (Z1 Z2) / (Z1 + Z2)2
2
= (45 1.5) / (45 + 1.5)2
= 0.875

R% = R ×100%

= 0.875 × 100

= 87.5%

T% = 100% R%
= (100 87.5)%
= 12.5%

329. If a shear wave probe having 33.4° angle while testing brass plate (VT, brass = 0.205 cm/ s,
VT, steel = 0.323 cm/ s ), the angle of this probe in steel will be

1 = 34.3 2 = ?
V1 = 0.205 cm/ s V2 = 0.323 cm/ s

sin 2 = sin 1 (V2/ V1 )

= sin 33.4 (0.323/ 0.205)

= 0.866

2 = sin-1 (0.866)

= 60°

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334. Second critical angle for perspex and steel interface would be (VL, perspex = 0.273 cm/ s and
VT, steel = 0.323 cm/ s)

VL,1 = 0.273 cm/ s T,2 = 0.323 cm/ s 2nd, crit =?

2nd, crit = sin-1 (VL,1/ VT,2 )

= sin-1 (0.273/ 0.323)

= 57.6°

335. If Plate thickness = 25.4 mm, pulse-echo, for back wall echo straight beam measured
elapsed time = 8.6 s . What is the most likely material? (VL, steel = 0.59 cm/ s)

S = 25.4 mm t = 8.6 s

S = v t /2

v = 2 S/ t
= 2 × 25 / 8.6
= 5.9 mm/ s
= 5.9 /10
= 0.59 cm/ s
= velocity of steel

336. First critical angle at water-aluminum interface will be (VL, Aluminum = 0.632 cm/ s, VL, water
= 0.148 cm/ s)

VL,1 = 0. 148 cm/ s VL,2 = 0.632 cm/ s 1st, crit =?

1st, crit = sin-1 (VL,1/ VL,2 )

= sin-1 (0.148/ 0.632)

= 13.54° 13.5°

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Prepared by:

PRAVEEN INSTITUTE OF RADIATION TECHNOLOGY

(NDT TRAINING DIVISION)
CHENNAI- 600 063
INDIA.

www.pirtchennai.com
Email: pirtndt@yahoo.in

13.4, 65q

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