Ultrasonic pulse echo technique, a non-destructive ultrasonic testing, is employed for any possible flaw detection in a metallic bar of thikness 30 cm. If the arrival times of the ultrasonic pulses are 45 μs and 90 μs respectively, the distance of the flaw from one end of the steel bar at which the ultrasonic pulse initially enters the steel bar, will be:
15 cm
Ultrasonic Pulse Echo Technique Explained
The ultrasonic pulse echo technique is a widely used non-destructive testing (NDT) method for detecting internal flaws or discontinuities in materials like a metallic bar. In this technique, a short burst of high-frequency sound waves (ultrasonic pulses) is sent into the material. These pulses travel through the material, and when they encounter a different medium or a defect (flaw), a part of their energy is reflected back to the transducer, which also acts as a receiver. By measuring the time it takes for the pulse to travel to the flaw and return, and knowing the speed of sound in the material, the distance to the flaw can be accurately determined.
Flaw Detection Principle
When an ultrasonic pulse is introduced into a metallic bar, it travels through the material. If there's a flaw, a reflection will occur from that flaw. Another reflection will occur from the back wall of the bar, representing the full thickness. The question provides two arrival times, 45 μs and 90 μs. This implies:
- The first arrival time (45 μs) corresponds to the reflection from the flaw.
- The second arrival time (90 μs) corresponds to the reflection from the back wall, indicating the full thickness of the metallic bar.
Calculating Sound Speed in the Metallic Bar
To find the distance of the flaw, we first need to determine the speed of the ultrasonic pulse in the metallic bar. We can do this using the information provided about the full thickness of the bar and the time taken for the pulse to reflect from the back wall.
- Total thickness of the metallic bar = \(T = 30 \text{ cm}\)
- Time taken for the pulse to travel to the back wall and return = \(t_{\text{back}} = 90 \, \mu\text{s}\)
- The distance traveled by the pulse to the back wall and back is twice the thickness: \(2 \times T\).
The speed of sound \(v\) in the material is calculated as:
\[ v = \frac{\text{Distance traveled}}{\text{Time taken}} \] \[ v = \frac{2 \times T}{t_{\text{back}}} \]Substituting the given values:
\[ v = \frac{2 \times 30 \text{ cm}}{90 \, \mu\text{s}} \]Convert microseconds to seconds for consistency:
\[ 90 \, \mu\text{s} = 90 \times 10^{-6} \text{ s} \] \[ v = \frac{60 \text{ cm}}{90 \times 10^{-6} \text{ s}} \] \[ v = \frac{60}{90} \times 10^6 \text{ cm/s} \] \[ v = \frac{2}{3} \times 10^6 \text{ cm/s} \]Determining Flaw Distance
Now that we have the speed of sound in the metallic bar, we can calculate the distance to the flaw using the first arrival time.
- Time taken for the pulse to travel to the flaw and return = \(t_{\text{flaw}} = 45 \, \mu\text{s}\)
- Let the distance of the flaw from the end be \(d_{\text{flaw}}\).
- The distance traveled by the pulse to the flaw and back is \(2 \times d_{\text{flaw}}\).
Using the formula \( \text{Distance} = \text{Speed} \times \text{Time}\):
\[ 2 \times d_{\text{flaw}} = v \times t_{\text{flaw}} \] \[ d_{\text{flaw}} = \frac{v \times t_{\text{flaw}}}{2} \]Substitute the calculated speed \(v\) and the given flaw reflection time \(t_{\text{flaw}}\):
\[ d_{\text{flaw}} = \frac{\left(\frac{2}{3} \times 10^6 \text{ cm/s}\right) \times (45 \times 10^{-6} \text{ s})}{2} \]The \(10^6\) and \(10^{-6}\) terms cancel out:
\[ d_{\text{flaw}} = \frac{\frac{2}{3} \times 45 \text{ cm}}{2} \] \[ d_{\text{flaw}} = \frac{\frac{90}{3} \text{ cm}}{2} \] \[ d_{\text{flaw}} = \frac{30 \text{ cm}}{2} \] \[ d_{\text{flaw}} = 15 \text{ cm} \]Therefore, the distance of the flaw from the end where the ultrasonic pulse initially enters the steel bar is 15 cm.
| Parameter | Value |
|---|---|
| Thickness of metallic bar (\(T\)) | 30 cm |
| Time for back wall reflection (\(t_{\text{back}}\)) | 90 μs |
| Time for flaw reflection (\(t_{\text{flaw}}\)) | 45 μs |
| Calculated Speed of Sound (\(v\)) | \(\frac{2}{3} \times 10^6 \text{ cm/s}\) |
| Calculated Flaw Distance (\(d_{\text{flaw}}\)) | 15 cm |
The final answer is 15 cm.
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