Beer's Law ultrasound attenuation, account for spherical spreading?

Question

I've seen Beer's Law for narrow beam written as:

$$ I = I_0 \cdot e^{-\alpha x} $$

${\alpha }$ - attenuation coefficient (depends on absorption and scattering), $I_0$ - initial intensity, $I$ - intensity after distance x

I want to also account for the spreading that occours when the ultrasound is transmitted, say for instance the ultrasound spreads spherically. I initially tried to write the equation as:

$$ I = I_0 \cdot \left(\frac{e^{-\alpha x}}{A}\right) $$ where ${A=4\pi r^2}$ , the surface area of the sphere at the distance x

I noticed that this doesn't account for the size of the transducer that emits the signal, would it be correct to write the equation like this:

$$ I = I_0 \cdot \left(\frac{A_0}{A}\right) \cdot e^{-\alpha x} $$

where ${A_0}$ = the surface area of the initial beam? Or would I rather use the cross-sectional area of the beam at a certain distance? How would ${A_0}$ be calculated if the transducer is a cylinder shape?

Answer 1

Almost. Consider that you don't have a single beam, but an isotropic source that radiates waves in all directions. Then, change to spherical coordinates such that your coordinate becomes r (radial coordinate) instead of x. If the medium was ideal, the total power should be conserved between spherical layers. So, the intensity decreases as the distance (area) increases. We may write:

$$ I(r) = \frac{P_0}{4\pi r^2} $$

Where $P_0$ is the power of the source. Now, just add the scattering effect which produce losses:

$$ I(r) = \frac{P_0}{4\pi r^2} \exp(-\alpha r) $$

Done.