Relationship between acoustic power and frequency

[lupok2001]

Hi everybody,
I am working with ultrasound-induced cavitation (i.e. what is used in ultrasonic water-bath cleaners) and I can't make my mind around the concept. I recon cavitation happens when the acoustic pressure (p) induced by the sound waves is greater than the hydraulic pressure of the medium through which the waves are traveling.

The acoustic pressure is calculated as
p = √(P Z A^-1)

Where P is the acoustic power, Z is the impedance of the medium and A is the surface area of the wave-front.

By looking at just the physics it seems that the acoustic pressure does not depend on the frequency of the sound wave. However, only ultrasounds (> 20 kHz) are used for cavitation. What am I missing? Is the acoustic power related to the frequency?

Thanks for the help
Fabio

 

[eljose79]

Hello Fabio,

Thank you for your question. You are correct in your understanding that cavitation occurs when the acoustic pressure is greater than the hydraulic pressure of the medium. However, the acoustic pressure is not the only factor that determines cavitation. The frequency of the sound wave also plays a crucial role.

The acoustic power, P, is indeed related to the frequency of the sound wave. It is calculated as P = ω ²ρcA²V, where ω is the angular frequency, ρ is the density of the medium, c is the speed of sound in the medium, A is the surface area of the wave-front, and V is the velocity of the medium. As you can see, the frequency is included in the equation through the angular frequency, ω.

In ultrasonic water-bath cleaners, high frequency sound waves are used to induce cavitation because they have a higher acoustic power compared to lower frequency sound waves. This means that they are able to create a greater acoustic pressure, which is necessary for cavitation to occur.

I hope this helps clarify your confusion. Please let me know if you have any further questions. Best of luck with your research on ultrasound-induced cavitation.