Your question doesn't have a simpe answer.
The power's density F=P/S space distribution depends on distance R, usound wavelength L, aperture D and medium shape. Here P - power, S - transducer square.
In case of infinite medium the diffraction assumption is used. For short ranges (RRb) F~1/R2. The boundary distance Rb depends on L and D - L/D=D/Rb. So Rb=D2/L
In case of finite medium You neds to take in account waves reflection and interference.
https://en.wikipedia.org/wiki/Diffraction
You could use this expression to calculate the power attenuation: I(x)=I0*exp(-2*a*x), where I0 is the initial power intensity, "a" is the atenuation constant and "x" is the distances.
The drawback is the difficulty of calculation the atenuation constant, which depends of the propagation medium. For water the ratio a/f^2 (f is the ultrasonic frequency) is equal to 21.5x10^-17 m-1 s2. Therefore you could determine the value of "a".
Mason, T. J., & Peters, D. (2004). Practical sonochemistry: Power ultrasound uses and applications: Horwood Publishing Limited.
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