Energy from an air-gun array

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 7 221 - 252 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

Problem 7.7

How much energy is released (approximately) by the air-gun array in Figure 7.7a when the input pressure is 2000 psi (14 MPa)? Assume that the change is adiabatic, that is, ${\displaystyle {\mathcal {PV}}^{1.4}=}$ constant, where ${\displaystyle {\mathcal {P}}}$, ${\displaystyle {\mathcal {V}}}$ are pressure and volume, that the final pressure is 2 atmospheres, and that the guns are far enough apart that they do not interact.

Background

An air gun consists of two chambers, both filled with air at high pressure. The two chambers are connected by a shuttle that is held in a closed position. When the restraining force is suddenly diminished the shuttle moves, allowing the air to vent into the water, creating the effect of an explosion. The energy release depends upon the change in air pressure and the volume of the chambers that discharge air into the water; the latter is usually given in cubic inches.

Solution

Energy released ${\displaystyle =}$ work done by expanding gas ${\displaystyle =W=\int _{{\mathcal {V}}_{0}}^{{\mathcal {V}}_{1}}{\mathcal {P}}{\rm {d}}{\mathcal {V}}}$. For an adiabatic change, ${\displaystyle {\mathcal {PV}}^{1.4}={\mathcal {P}}_{0}{\mathcal {V}}_{0}^{1.4}}$ or ${\displaystyle {\mathcal {P}}={\mathcal {P}}_{0}{\mathcal {V}}_{0}^{1.4}/{\mathcal {V}}^{1.4}=\kappa /{\mathcal {V}}^{1.4}}$. Thus,

{\displaystyle {\begin{aligned}W=\kappa \int _{{\mathcal {V}}_{0}}^{{\mathcal {V}}_{1}}{\rm {d}}{\mathcal {V}}/{\mathcal {V}}^{1.4}=-\left(\kappa /0.4\right)\left(1/{\mathcal {V}}_{1}^{0.4}-1/{\mathcal {V}}_{0}^{0.4}\right)\\=2.5\left({\mathcal {P}}_{0}{\mathcal {V}}_{0}-{\mathcal {P}}_{1}{\mathcal {V}}_{1}\right)\quad \left({\rm {since}}\quad \kappa ={\mathcal {P}}_{0}{\mathcal {V}}_{0}^{1.4}={\mathcal {P}}_{1}{\mathcal {V}}_{1}^{1.4}\right).\end{aligned}}}

We have

{\displaystyle {\begin{aligned}{\mathcal {P}}_{0}=14\,{\rm {MPa}}=14\times 10^{6}\ \mathrm {N/m} ^{2},\\{\mathcal {P}}_{1}=2\,{\rm {atm.}}=2/\left(9.87\times 10^{-6}\right)\ {\rm {Pa}}=2.0\times 10^{5}\ {\rm {Pa}}=0.20\times 10^{6}\ {\rm {N/m}}^{2},\\{\mathcal {V}}_{0}=\left(305+200+125+95+75+60+50\right)\ {\rm {in}}^{3}=910\ {\rm {in}}^{3}\\=910\times (2.54\ {\rm {cm/in}})^{3}=1.49\times 10^{4}\ {\rm {cm}}^{3}=0.0149\ {\rm {m}}^{3}.\end{aligned}}}

Since ${\displaystyle {\mathcal {P}}_{1}{\mathcal {V}}_{1}^{1.4}={\mathcal {P}}_{0}{\mathcal {V}}_{0}^{1.4}}$, ${\displaystyle {\mathcal {V}}_{1}={\mathcal {V}}_{0}({\mathcal {P}}_{0}/{\mathcal {P}}_{1})^{1/1.4},}$

{\displaystyle {\begin{aligned}V_{1}=0.0149\times (14\times 10^{6}/0.20\times 10^{6})^{0.71}=0.0149\times 70^{0.71}=0.30\ {\rm {m}}^{3}.\end{aligned}}}

Finally, ${\displaystyle W=2.5\left(14\times 10^{6}\times 0.0149-0.20\times 10^{6}\times 0.30\right)=0.15\times 10^{6}}$ J.