# Tube-wave relationships

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 2 7 - 46 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem 2.16a

A tube wave has a velocity of 1050 m/s. The fluid in the borehole has a bulk modulus of $2.15\times 10^{9}$ Pa and a density of $1.20g/\mathrm {cm} ^{3}$ . The wall rock has $\mathrm {\sigma } =0.250$ and density $2.50g/\mathrm {cm} ^{3}$ . Calculate $\mathrm {\mu }$ and $\mathrm {\alpha }$ for the wall rock. Figure 2.15a  Response of vertical and horizontal geophones to different waves. (a) Direct P-wave, (b) reflected P-wave, (c) converted S-wave, (d) Rayleigh wave, (e) Love wave. U, D = up, down; A, T = away (from), toward (source); L, R = left, right.

### Background

Several types of tube waves exist (Sheriff and Geldart, 1995, Section 2.5.5). The classical type consists of a P-wave traveling in a fluid within a tubular cavity (such as a borehole) in a solid medium, the wall of the tube expanding and contracting as the pressure wave passes. Because the wall material interacts with the fluid, the tube-wave velocity depends upon the properties of both the wall material and the fluid. The formula for the tube-wave velocity ${V_{T}}$ is [see Sheriff and Geldart, 1995, equation (2.97)]

 {\begin{aligned}V_{T}^{2}={\frac {1}{p}}\left({\frac {1}{k}}+{\frac {1}{\mu }}\right)^{-1},\end{aligned}} (2.16a)

$\rho$ being the density and $k$ the bulk modulus of the fluid while $\mu$ is the rigidity modulus of the wall material.

### Solution

Assuming the wall material to be rock, we write $\rho _{r}$ , $\mu _{r}$ , $\lambda _{r}$ , $\sigma _{r}$ , and $\alpha _{r}$ for the rock, $\rho _{f}$ and $k_{f}$ for the fluid. We solve equation (2.16a) for $\mu _{r}$ but first we note that $k$ and $\mu$ are in units of $N/m^{2}$ , so we express $\rho _{f}$ as $1200\;\mathrm {kg} /m^{3}$ . Then,

{\begin{aligned}\mu _{r}=\left({\frac {1}{p_{f}V_{T}^{2}}}-{\frac {1}{k_{f}}}\right)^{-1}=\left({\frac {1}{1200\times 1050^{2}}}-{\frac {10^{-9}}{2.15}}\right)^{-1}=3.44\times 10^{9}\ \mathrm {Pa} .\end{aligned}} Using equation (5,5) of Table 2.2a, we have

{\begin{aligned}\lambda _{r}=\mu _{r}\left[2\sigma _{r}/\left(1-2\sigma _{r}\right)\right]=\mu _{r}=3.44\times 10^{9}\ \mathrm {Pa} .\end{aligned}} To get $\alpha _{r}$ , we have

{\begin{aligned}\alpha _{r}&=[\left(\lambda _{r}+2\mu _{r}\right)/\rho _{r}]^{1/2}\\&=(3\times 3.44\times 10^{9}/2500)^{1/2}=2.03\ \mathrm {km/s} .\end{aligned}} ## Problem 2.16b

Repeat for $V_{T}=1200m/s$ and 1300 m/s. What do you conclude about the accuracy of this method for determining $\mu$ ?

### Solution

When $V_{T}=1.20km/s$ ,

{\begin{aligned}\mu _{r}=8.80\times 10^{9}\ \mathrm {Pa} =\lambda _{r};\alpha _{r}=3.25\ \mathrm {km/s} .\end{aligned}} When $V_{T}=1.30km/s$ ,

{\begin{aligned}\mu _{r}=35.7\times 10^{9}\ \mathrm {Pa} =\lambda _{r};\alpha _{r}=6.55\ \mathrm {km/s} .\end{aligned}} Summarizing the results for $\mu _{r}$ versus $V_{T}$ , we get the following:

$V_{T}$ $\mu _{r}$ $\Delta V_{T}$ $\Delta \mu _{r}$ 1.05 3.44 $\times 10^{9}$ 1.20 6.80 $\times 10^{9}$ +14$\%$ +156$\%$ 1.30 35.7 $\times 10^{9}$ +24$\%$ +938$\%$ Since the relative change in $\mu _{r}$ is very much larger than the relative change in $V_{T}$ , the method is very sensitive to changes in $V_{T}$ , hence the accuracy is very poor.