Magnitude of disturbance from a seismic source

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Problem 2.3a

Firing an air gun in water creates a pressure transient a small distance away from the air gun with peak pressure $ {\mathcal {P}} $ of 5 atmospheres ($ 5.0\times 10^{5} $ Pa). If the compressibility of water is $ 4.5\times 10^{-10} $/Pa, what is the peak energy density?

Background

Air guns (see problem 7.7) suddenly inject a bubble of high‐pressure air into the water to generate a seismic wave.

Stresses acting upon a medium cause energy to be stored as strain energy, because the stresses are present while the medium is being displaced. Strain energy density (energy/unit volume) $ E $ is equal to [see Sheriff and Geldart, 1995, equation (2.22)]


$ {\begin{aligned}{E={\frac {1}{2}}\sum \limits _{i}\sum \limits _{j}{\mathrm {\sigma } }_{ij}{\mathrm {\varepsilon } }_{ij}={\frac {1}{2}}{\mathrm {\lambda } }{\mathrm {\Delta } }^{2}+{\mathrm {\mu } }\left({\mathrm {\varepsilon } }_{xx}^{2}+{\mathrm {\varepsilon } }_{yy}^{2}+{\mathrm {\varepsilon } }_{zz}^{2}\right)}\\{{\qquad }{\qquad }{\qquad }{\qquad }+{\frac {1}{2}}{\mathrm {\mu } }\left({\mathrm {\varepsilon } }_{xy}^{2}+{\mathrm {\varepsilon } }_{yz}^{2}+{\mathrm {\varepsilon } }_{zx}^{2}\right).}\end{aligned}} $ (2.3a)

Solution

From problem 2.1c, we see that $ k=1/\left(4.5\times 10^{-10}\right)=2.2\times 10^{9} $ Pa. Also, $ {\mathrm {\mu } }=0 $ for water, so $ E={\frac {1}{2}}{\mathrm {\lambda } }{\mathrm {\Delta } }^{2} $. From equation (7,5) of Table 2.2a we find that $ {\mathrm {\lambda } }=k $ when $ {\mathrm {\mu } }=0 $. Also, $ k=-{\mathcal {P}}/{\mathrm {\Delta } } $ (see equation (2.1f)), so


$ {\begin{aligned}\Delta =\left|{\cal {P}}/k\right|=5.0\times 10^{5}/2.2\times 10^{9}=2.3\times 10^{-4}.\end{aligned}} $

Using equation (2.3a) we find that


$ {\begin{aligned}E={\frac {1}{2}}\left(2.2\times 10^{9}\times 2.3^{2}\times 10^{-8}\right)=58\,\,\,\mathrm {J/m} ^{3}.\end{aligned}} $

[The dimensions of $ E $ are the same as those of stress, since strains are dimensionless. Thus, stress units are $ {\rm {N/m}}^{2}={\rm {Nm/m}}^{3}={\rm {J/m}}^{3} $.]

Problem 2.3b

If the same wave is generated in rock with $ {\mathrm {\lambda } }={\mathrm {\mu } }=3.0\times 10^{10} $ Pa, what is the peak energy density? Assume a symmetrical $ P $-wave with $ {\mathrm {\varepsilon } }_{xx}={\mathrm {\varepsilon } }_{yy}={\mathrm {\varepsilon } }_{zz},{\mathrm {\varepsilon } }_{ij}=0 $ for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): i\ne j .

Solution

We have Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\mathrm{\Delta}} ={\mathrm{\varepsilon}} _{xx} +{\mathrm{\varepsilon}} _{yy} +{\mathrm{\varepsilon}} _{zz} =3{\mathrm{\varepsilon}} _{xx} , Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\mathrm{\varepsilon}} _{ij} =0, i\ne j , so equation (2.3a) becomes


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): \begin{align} E=\frac{1}{2} {\mathrm{\lambda}} {\mathrm{\Delta}} ^{2} +3{\mathrm{\mu}} {\mathrm{\varepsilon}} _{xx}^{2} ={\mathrm{\lambda}} {\mathrm{\Delta}} ^{2} \left(\frac{1}{2} +\frac{1}{3} \right)=\frac{5}{6} {\mathrm{\lambda}} ({\mathcal P}/k)^{2}. \end{align}

Equation (9,3) in Table 2.2a gives Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): k={\mathrm{\lambda}} +\frac{2}{3} {\mathrm{\mu}} =\frac{5}{3} {\mathrm{\lambda}} so that


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): \begin{align} E=\frac{5}{6} {\mathrm{\lambda}} {\mathcal P}^{2} \left/\right.\left(\frac{5}{3} {\mathrm{\lambda}} \right)^{2} =0.3\times (5.0\times 10^{5} )^{2} /3\times 10^{10} =2.5\ \mathrm{J/m}^{3}. \end{align}

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