- #1
rwooduk
- 762
- 59
... it's amplitude exerts.
During cavitation a sound wave is applied to a liquid and breaks it apart and gas pockets are formed. The frequency and the amplitude of the sound wave effect the bubble/s.
My question is, if I have the power output of the device then how would this relate to the "pressure amplitude" or "drive-pressure amplitude" exerted on the bubble/s?
The pressure amplitude as the name suggests is usually given in atmospheres, I'm struggling to somehow get Watts to atmospheres.
For example a paper might say "the frequency and the amplitude of the ultrasound are 20KHz and 1.4 bar" how would I relate the 1.4 bar to the power output of the sound wave?
edit
Thinking back, power is proportional to the square of the amplitude (I am unsure of what the constants would be to make it equate, especially if the medium is water), so if I square root the power, but then how would I get to pressure?
During cavitation a sound wave is applied to a liquid and breaks it apart and gas pockets are formed. The frequency and the amplitude of the sound wave effect the bubble/s.
My question is, if I have the power output of the device then how would this relate to the "pressure amplitude" or "drive-pressure amplitude" exerted on the bubble/s?
The pressure amplitude as the name suggests is usually given in atmospheres, I'm struggling to somehow get Watts to atmospheres.
For example a paper might say "the frequency and the amplitude of the ultrasound are 20KHz and 1.4 bar" how would I relate the 1.4 bar to the power output of the sound wave?
edit
Thinking back, power is proportional to the square of the amplitude (I am unsure of what the constants would be to make it equate, especially if the medium is water), so if I square root the power, but then how would I get to pressure?
Last edited: