Sensitivity

Theoretical considerations are presented here to estimate the sensitivity of piezo ceramic hydrophones, focusing on cylindrical and spherical designs. The provided equations predict the voltage generated as a function of external sound pressure, serving as an estimate for expected free-field sensitivities. However, realistic effective sensitivities, which account for the influence of attached electronics and nearby materials, can only be accurately determined through meticulous in-situ calibration.

References

https://engineeringlibrary.org/reference/thick-pressure-vessels-air-force-stress-manual

https://apps.dtic.mil/sti/pdfs/AD0759199.pdf

Cylinder

Lamé equations

For a Cylinder we have the following stress equations

radial tension

\begin{equation}T_3(r)=A-\frac{B}{r^2}\end{equation}

circumferential tension

\begin{equation}T_1(r)=A+\frac{B}{r^2}\end{equation}

axial tension (end-capped cylinder)

\begin{equation}T_2(r)=A\end{equation}

whereby

\begin{equation}A=\frac{P_aa^2-P_bb^2}{b^2-a^2}\end{equation}

\begin{equation}B=\frac{a^2b^2(P_a-P_b)}{b^2-a^2}\end{equation}

Assume for hydrophones internal pressure \(P_a=0\), then the two constants \(A\) and \(B\) become

\begin{equation}A=-P_b\frac{b^2}{b^2-a^2}\end{equation}

\begin{equation}B=-P_b\frac{a^2b^2}{b^2-a^2} = Aa^2\end{equation}

and the three stress equations are then

\begin{equation}T_3(r)=A(1-\frac{a^2}{r^2})\end{equation}

\begin{equation}T_1(r)=A(1+\frac{a^2}{r^2})\end{equation}

\begin{equation}T_2(r)=A\end{equation}

with \(A\) as defined above

Electric field

The electric field in radial direction \(E_3(r)\) due to external pressure becomes with \(g_{32}=g_{31}\)

\begin{equation}E_3(r)=-g_{31} (T_1(r)+T_2(r))-g_{33} T_3(r)\end{equation}

and the measured voltage \(V\)

\begin{equation}V = \int_a^b{E_3(r)dr}\end{equation}

Note

\begin{equation}\int_a^b{dr} = b-a\end{equation}

\begin{equation}\int_a^b{\frac{1}{r^2}dr} = -(\frac{1}{b}-\frac{1}{a})=\frac{b-a}{ab}\end{equation}

that is

\begin{equation}\int_a^b(1\pm {\frac{a^2}{r^2}})dr =(b-a)\pm\frac{a(b-a)}{b}=(b-a)(1\pm\frac{a}{b})\end{equation}

Generated Voltage

\begin{equation}V =-\int_a^b{(g_{31} (T_1(r)+T_2(r))+g_{33} T_3(r))dr}\end{equation}

\begin{equation}V =-A(b-a)\big(g_{31}(2+\frac{a}{b})+g_{33}(1-\frac{a}{b})\big)\end{equation}

\begin{equation}V =-P_b b \frac{b}{b+a}\big(g_{31}(2+\frac{a}{b})+g_{33}(1-\frac{a}{b})\big)\end{equation}

let \(t=\frac{a}{b}\) then

\begin{equation}V =-P_b b \frac{1}{1+t}\big(g_{31}(2+t)+g_{33}(1-t)\big)\end{equation}

The sensitivity \(M\) is then

\begin{equation}M=\frac{V}{P_b}=- b \frac{1}{1+t}\big(g_{31}(2+t)+g_{33}(1-t)\big)\end{equation}

Sphere

For a Sphere we have the following stress equations

\begin{equation}T_3(r)=P_b\frac{b^3}{a^3-b^3}(1-\frac{a^3}{r^3})\end{equation}

\begin{equation}T_1(r)=T_2(r)=P_b\frac{b^3}{a^3-b^3}(1+\frac{a^3}{2r^3})\end{equation}

With

\begin{equation}\int_a^b{\frac{1}{r^3}dr} = -\frac{1}{2}(\frac{1}{b^2}-\frac{1}{a^2})=\frac{1}{2}\frac{b^2-a^2}{a^2b^2}=\frac{1}{2}(\frac{b+a}{ab})(\frac{b-a}{ab})\end{equation}

and therefore

\begin{equation}\int_a^b(1- \frac{a^3}{r^3})dr =(b-a)-\frac{a^3}{2}\frac{b+a}{ab}\frac{b-a}{ab}=(b-a)(1-\frac{a}{2b}\frac{b+a}{b})=\frac{(b-a)}{2}(2-\frac{a}{b}-\frac{a^2}{b^2})\end{equation}

\begin{equation}\int_a^b(1+ \frac{1}{2}\frac{a^3}{r^3})dr =(b-a)+\frac{1}{2}\frac{a^3}{2}\frac{b+a}{ab}\frac{b-a}{ab}=(b-a)(1+\frac{1}{2}\frac{a}{2b}\frac{b+a}{b})=\frac{(b-a)}{4}(4+\frac{a}{b}+\frac{a^2}{b^2})\end{equation}

one gets

\begin{equation}V = -P_b\frac{b^3}{b^3-a^3}\frac{(b-a)}{2}\big(g_{31}(4+\frac{a}{b}+\frac{a^2}{b^2})+g_{33}(2-\frac{a}{b}-\frac{a^2}{b^2})\big)\end{equation}

and finally

\begin{equation}V = -P_b\frac{b}{2}\frac{b^2}{b^2+ab+a^2}\big(g_{31}(4+\frac{a}{b}+\frac{a^2}{b^2})+g_{33}(2-\frac{a}{b}-\frac{a^2}{b^2})\big)\end{equation}

which again is equivalent to

\begin{equation}V = -P_b\frac{b}{2}\frac{1}{1+t+t^2}\big(g_{31}(4+t+t^2)+g_{33}(2-t-t^2)\big)\end{equation}