4: Specimen and Sound Beam

Question 1

The near zone is also called the what zone? And how does the sound act?

Answer 1

Also called Fresnel.

Sound intensity fluctuates.

Question 2

The far zone is also called the what? And how does the sound act?

Answer 2

Also called the Fraunhofer.

Sound diverges predictably.

Question 3

The length of the near field is directly proportional to what?

Answer 3

The diameter and frequency of the transducer.

Question 4

Why is it important to know how far the near zone extends into the part?

Answer 4

The signal amplitude (height) cannot be trusted to represent the size of the reflector.

Question 5

Where is the maximum sound intensity located?

Answer 5

End of the near field - beginning of far field

Question 6

As the sound transitions from the near zone to the far zone, the sound pressure builds and waves become uniform.. what are the waves now referred to?

Answer 6

Plane wave

Question 7

What is it referred to when the sound moves into the far zone and begins to spread in a cone shape?

Answer 7

Beam spread/beam divergence

Question 8

How can beam spread be altered?

Answer 8

By changing the diameter and/or frequency of the transducer.

Question 9

What happens to the beam spread when testing a narrow specimen?

Answer 9

The sound will hit the sides of the specimen and produce mode conversion.

Question 10

The angle of divergence is inversely proportional to what?

Answer 10

The diameter and frequency of the transducer.

An increase in diameter/frequency will produce a smaller angle of divergence.

Question 11

Where is the sound intensity greatest?

Answer 11

Along the centreline (100%) and weaker towards the edges

Question 12

When a defect is at its maximum amplitude, where does that mean it will be in relation to transducer?

Answer 12

On the centreline

Question 13

What formula is used to calculate beam spread at 0% intensity?

Answer 13

Sin theta = 1.22V/fD

Question 14

What formula is used to calculate intensity at 10% edge of beam?

Answer 14

Sin theta = 1.08wavelength/D

Question 15

What formula is used to calculate intensity at 50% edge of beam?

Answer 15

Sin theta = 0.56wavelength/D

Question 16

What is attenuation?

Answer 16

The loss of energy of the sound beam from the time it is sent by the crystal to the time it returns to the crystal.

Question 17

What are dislocations?

Answer 17

Crystalline defects

Question 18

When does scattering occur?

Answer 18

When the material contains reflectors that are large in relation to the half wavelength.

Question 19

How is the length of near field affected by an increase in frequency?

Answer 19

Length of near field is directly proportional to frequency.

As frequency increases = length of near field increases.

Question 20

How is the beam spread affected by an increase in probe diameter?

Answer 20

Beam spread decreases with an increase in diameter.

Question 21

What is the problem with inspecting in the near zone?

Answer 21

Reflectors can’t be easily evaluated due to pressure interference.

Question 22

What are the 3 factors that affect beam spread?

Answer 22

Frequency and diameter of transducer, velocity in material.

Question 23

What are the 2 main characteristics that will cause attenuation in the material?

Answer 23

Absorption and scattering.

Question 24

What could cause attenuation by absorption?

Answer 24

Absorption is caused by friction… which produce heat and reduce sound energy and crystalline dislocations… which reduce inelasticity of the material.

Question 25

What could cause attenuation by scatter?

Answer 25

Scatter is caused by porosity, large grain size, precipitates and phase changes.

Question 26

Besides the 2 main causes of attenuation, what are other causes?

Answer 26

Geometrical losses, couplings losses and beam divergence.

Question 27

How would a rough front surface be indicated on a CRT screen?

Answer 27

Increase the width of the initial pulse.

Question 28

Describe the problems that would be encountered when inspecting a specimen with a rough front surface.

Answer 28

Mainly related to reflection, refraction and scattering. And loss of near surface resolution.

Question 29

Describe the problems that could be encountered when working with a material with a coarse grain structure.

Answer 29

Main problem is scattering, which causes attenuation of sound and noise on screen.

Question 30

Would a higher or lower frequency transducer result in the greater ultrasonic attenuation loss?

Answer 30

Higher

Question 31

What does PADSCRAN stand for?

Answer 31

⬆️ FREQUENCY
⬇️ P - penetration 
⬆️ A - attenuation 
⬇️ D - divergence 
⬆️ S - sensitivity 
⬇️ C - crystal thickness
⬆️ R - resolution 
⬇️ A (lambda) - wavelength 
⬆️ N - near field