# Tube-wave relationships

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
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 ${\displaystyle 2.15\times 10^{9}}$ Pa and a density of ${\displaystyle 1.20g/\mathrm {cm} ^{3}}$. The wall rock has ${\displaystyle \mathrm {\sigma } =0.250}$ and density ${\displaystyle 2.50g/\mathrm {cm} ^{3}}$. Calculate ${\displaystyle \mathrm {\mu } }$ and ${\displaystyle \mathrm {\alpha } }$ 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 ${\displaystyle {V_{T}}}$ is [see Sheriff and Geldart, 1995, equation (2.97)]

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

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

### Solution

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

{\displaystyle {\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

{\displaystyle {\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 ${\displaystyle \alpha _{r}}$ , we have

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

### Solution

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

{\displaystyle {\begin{aligned}\mu _{r}=8.80\times 10^{9}\ \mathrm {Pa} =\lambda _{r};\alpha _{r}=3.25\ \mathrm {km/s} .\end{aligned}}}

When ${\displaystyle V_{T}=1.30km/s}$,

{\displaystyle {\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 ${\displaystyle \mu _{r}}$ versus ${\displaystyle V_{T}}$, we get the following:

${\displaystyle V_{T}}$ ${\displaystyle \mu _{r}}$ ${\displaystyle \Delta V_{T}}$ ${\displaystyle \Delta \mu _{r}}$
1.05 3.44 ${\displaystyle \times 10^{9}}$
1.20 6.80 ${\displaystyle \times 10^{9}}$ +14${\displaystyle \%}$ +156${\displaystyle \%}$
1.30 35.7 ${\displaystyle \times 10^{9}}$ +24${\displaystyle \%}$ +938${\displaystyle \%}$

Since the relative change in ${\displaystyle \mu _{r}}$ is very much larger than the relative change in ${\displaystyle V_{T}}$, the method is very sensitive to changes in ${\displaystyle V_{T}}$, hence the accuracy is very poor.