Solution to Exercise 5

William W. Woessner; Eileen P. Poeter

Solution to Exercise 5

5) A factory is disposing of hot waste water by injecting it into a 1000 m deep well that penetrates a confined aquifer. The sandstone aquifer contains water at 35 °C, a similar temperature as the waste water. A regulator wanted the company to model how far the contaminated water would travel in the aquifer over a 10-year period. Hydraulic conductivity values were sparse. They required the company to core part of the 50 m thick confined aquifer and determine a representative value of hydraulic conductivity. A portion of the core was placed in a constant head permeameter as illustrated in the accompanying figure.

5a) If an average of 34.0 ml/min was collected at the outlet, what is K in cm/s?

Hydraulic conductivity can be determined by rearranging Equation 15 of this book

$\displaystyle Q=-K\frac{\Delta h}{\Delta L}\ A$

where:

Q	=	volumetric flow rate (L³/T)
K	=	hydraulic conductivity, is the proportionality constant reflecting the ease with which water flows through a material (L/T)
Δh	=	difference in hydraulic head between two measuring points as defined for Equation 14 (L)
ΔL	=	length along the flow path between locations where hydraulic heads are measured (L)
$\displaystyle \frac{\Delta h}{\Delta L}$	=	gradient of hydraulic head (dimensionless)
A	=	cross-sectional area of flow perpendicular to the direction of flow (L²)

Rearranging:

$\displaystyle K=-\frac{Q\Delta L}{A\Delta h}$

$\displaystyle K=-\frac{\frac{34\ {\textup{cm}}^3}{\textup{min}}\ 30\ \textup{cm}\ \frac{1\ \textup{min}}{60\ \textup{s}}}{3.14\ \left(3\ \textup{cm}\right)^2(48.5\ \textup{cm}-60\ \textup{cm})}=\frac{\frac{17\ {\textup{cm}}^4}{\textup{s}}}{325\ {\textup{cm}}^{3\ }}=0.052\ \frac{\textup{cm}}{\textup{s}}$

5b) If the laboratory experiment was completed using 15 °C water, what is the intrinsic permeability of the sandstone in cm²?

If the test was conducted at 15 °C, the intrinsic permeability, k, of the sandstone in cm² can be calculated using Equation 31 of this book, using the relationships for the physical properties of water and temperature shown as Figure 28 of this book.

$\displaystyle K=\frac{Cd^{2}g\rho }{\mu }=\frac{kg\rho }{\mu }$

where

k	=	intrinsic permeability (L²)
ρ	=	fluid density (M/L³)
g	=	gravitational constant (acceleration of gravity) (L/T²)
μ	=	dynamic viscosity (M/LT)

rearranging:

$\displaystyle k=\frac{K\mu }{g\rho }=\frac{0.052\ \frac{\textup{cm}}{\textup{s}}\ 0.011\ \frac{\textup{gm}\cdot \textup{cm}}{\textup{s}}}{980\ \frac{\textup{cm}}{\textup{s}^{2}}\ 0.998\ \frac{\textup{g}}{\textup{cm}^{2}}}=5.8\times 10^{-7}\ \textup{cm}^{2}$

5c) What is the hydraulic conductivity of the sandstone if the water temperature is 35 °C (in cm/s)?

If the water temperature in the sandstone is 35 °C, Equation 31 of this book can be used to calculate, K, of the sandstone at that temperature in cm/s, using the relationships for the physical properties of water and temperature shown as Figure 28 of this book.

$\displaystyle K=\frac{5.8\times 10^{-7}\ {\textup{cm}}^{2}\ 980\ \frac{\textup{cm}}{\textup{s}^{2}}\ 0.994\ \frac{\textup{g}}{\textup{cm}^{3}}}{0.007\ \frac{\textup{g}\cdot \textup{cm}}{\textup{s}}}=\frac{0.08\ \textup{cm}}{\textup{s}}$

Return to Exercise 5

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