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Chapter 9: Problem 3

Assume that a porous medium can be modeled as a cubic matrix with a dimension \(b\); the walls of each cube are channels of thickness \(\delta\). (a) Determine expressions for the porosity and permeability in terms of \(b\) and \(\delta .\) (b) What is the permeability if \(b=0.1 \mathrm{~m}\) and \(\delta=1 \mathrm{~mm} ?\)

Short Answer

Expert verified
Porosity: \( \phi = \frac{(b-2\delta)^3}{b^3} \). Permeability: \( k = 2.80 \times 10^{-7} \) m² for the given values.

Step by step solution

01

Understand Porosity Definition

Porosity is the ratio of the volume of empty space (voids) to the total volume of the cube. In this cubic model, the voids are the channels between the solid walls, and the total volume is given by the cube's dimension, including both the walls and voids.
02

Calculate Porosity Expression

The volume of a single channel is based on the dimensions of the channel within the cubic matrix. For each side of the cube of dimension \(b\), the channel has a width \(\delta\), leading to the void volume. The total cube volume is \(b^3\), with wall thickness \(\delta\). The void within the cube is \((b-2\delta)^3\). The porosity \(\phi\) is given by:\[ \phi = \frac{(b-2\delta)^3}{b^3} \]
03

Understand Permeability Definition

Permeability is a measure of how easily fluid can flow through a porous material. It is a function of both the size and connectivity of the void spaces within the solid material. For this cubic model, it depends on the dimension \(b\), the channel thickness \(\delta\), and also the fluid properties.
04

Calculate Permeability Expression

Using the geometric model, and approximating using the capillary flow analogy (Hagen–Poiseuille equation), we approximate permeability \(k\) for channel at small \(\delta\) compared to \(b\). The permeability is given by: \[ k = \frac{(b-2\delta)^3}{32} \]
05

Insert Given Values

For part (b), substitute the given values: \(b = 0.1\) m and \(\delta = 1\) mm or 0.001 m, into the permeability expression:\[ k = \frac{(0.1 - 2 \times 0.001)^3}{32} \]
06

Simplify the Expression

Calculate the permeability by simplifying:\[ k = \frac{(0.098)^3}{32} \]Finally, evaluate the numeric expression, which calculates to approximately \( k = 2.80 \times 10^{-7} \) m^2.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Porosity
Porosity is a key concept in understanding how materials can hold and transmit fluids. Think of it as the capability of a material to have "holes" or "pores." In our cubic matrix model, these pores are the spaces between the solid walls, represented by channels of a certain thickness.
The porosity (\(\phi\)) is the fraction of these voids compared to the entire volume of the cube. This is important because it tells us how much of the cube's space can be filled with a fluid, like water or oil.
  • The total volume of the cube: \(b^3\)
  • The volume of the voids: \((b-2\delta)^3\)
  • Porosity formula: \(\phi = \frac{(b-2\delta)^3}{b^3}\)
In simpler terms, porosity measures how "holey" the material is. It's essential for determining the capacity of a material to store fluids.
Permeability
Permeability measures how easily fluids can move through a porous material. Imagine water pouring through a sponge. That's influenced by both the size of the pores and how they're connected. In our cubic model, permeability depends on the dimensions \(b\) and \(\delta\), which define the structure's "channels."
In practical terms, permeability (\(k\)) refers to the ease with which liquids can navigate through these channels. This is essential in fields like hydrology, petroleum engineering, and soil science.
For small channel dimensions, we approximate permeability using fluid dynamics principles similar to the Hagen-Poiseuille equation:
  • Permeability formula: \(k = \frac{(b-2\delta)^3}{32}\)
A higher permeability means fluids can travel through faster, which is critical in applications like oil extraction or groundwater flow.
Cubic Matrix Model
The cubic matrix model is a simplified representation to help us understand the flow of fluids in porous materials. Imagine a cube, where each side has a dimension \(b\), and the walls are channels with thickness \(\delta\).
This model allows us to create mathematical expressions that describe the behavior of fluids in porous media.
  • The solid parts of the cube represent the material structure.
  • The channels signify the voids or pores where fluids reside or travel through.
By using this model, we can precisely express and calculate both porosity and permeability, which are fundamental in assessing a material's suitability for applications involving fluid movement.
This approach is valuable for simplifying complex real-world materials into more manageable mathematical models, enabling us to make predictions and optimizations in practical scenarios.

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Most popular questions from this chapter

Consider an unconsolidated (uncemented) layer of soil completely saturated with groundwater; the water table is coincident with the surface. Show that of d the upward Darcy velocity \(|v|\) required to fluidize the bed is $$ |v|=\frac{(1-\phi) k g\left(\rho_{s}-\rho_{w}\right)}{\mu} $$ where \(\phi\) is the porosity, \(\rho_{s}\) is the density of the soil particles, and \(\rho_{w}\) is the water density. The condition of a fluidized bed occurs when the pressure at depth in the soil is sufficient to completely support the weight of the overburden. If the pressure exceeds this critical value, the flow can lift the soil layer.

To derive an upward flow in a porous medium, it is clear that pressure must increase more rapidly with depth \(y\) than it does when the fluid is motionless. Use this idea to justify writing Darcy's law for vertical flow in a porous medium in the form $$ v=-\frac{k}{\mu}\left(\frac{d p}{d y}-\rho g\right) $$ where \(v\) is the vertical Darcy velocity (positive in the direction of increasing depth), \(\rho\) is the fluid density, and \(g\) is the acceleration of gravity. Consider a porous medium lying on an impermeable surface inclined at an angle \(\theta\) to the horizontal. Show that Darcy's law for the downslope volumetric flow rate per unit area \(q\) is $$ q=-\frac{k}{\mu}\left(\frac{d p}{d s}-\rho g \sin \theta\right) $$ where \(s\) is the downslope distance and \(q\) is positive in the direction of \(s\).

Consider a porous layer saturated with water that is at the boiling temperature at all depths. Show that the temperature-depth profile is given by $$ \frac{1}{T_{b 0}}-\frac{1}{T}=\frac{R_{v}}{L} \ln \left(1+\frac{\rho_{l} g y}{p_{0}}\right) $$ where \(T_{b 0}\) is the boiling temperature of water at atmospheric pressure \(p_{0}, \rho_{l}\) is the density of liquid water which is assumed constant, and \(R_{r}\) is the gas constant for water vapor. Start with the hydrostatic equation for the pressure and derive an equation for \(d T / d y\) by using the formula for the slope of the Clapeyron curve between water and steam $$ \frac{d p}{d T}=\frac{L \rho_{l} \rho_{v}}{T\left(\rho_{l}-\rho_{v}\right)} \approx \frac{L \rho_{v}}{T} $$ where \(\rho_{v}\) is the density of water vapor. Assume that steam is a perfect gas so that $$ \rho_{v}=\frac{p}{R_{r} T} $$ Finally, note that \(p=p_{0}+\rho_{l} g y\) if \(\rho_{l}\) is assumed constant. What is the temperature at a depth of \(1 \mathrm{~km}\) ? Take \(R_{v}=0.462 \mathrm{~kJ} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}, L=2500 \mathrm{~kJ} \mathrm{~kg}^{-1}\) $$ T_{b 0}=373 \quad \mathrm{~K}, \quad p_{0}=10^{5} \quad \mathrm{~Pa}, \quad \rho_{l}=1000 \mathrm{~kg} \mathrm{~m}^{-3} $$ \(g=10 \mathrm{~m} \mathrm{~s}^{-2}\).

If fluid is injected along a plane at \(x=0\) at a rate of \(0.1 \mathrm{~m}^{2} \mathrm{~s}^{-1}\), how high is the phreatic surface at the point of injection and how far has the fluid migrated if \(\mu=10^{-3} \mathrm{~Pa} \mathrm{~s}, \phi=0.1, k=10^{-11} \mathrm{~m}^{2}\) \(\rho=1000 \mathrm{~kg} \mathrm{~m}^{-3},\) and \(t=10^{5} \mathrm{~s}\) ? A MATLAB solu- tion to this problem is provided in Appendix \(D\).

Consider one-dimensional flow through a confined porous aquifer of total thickness \(b\) and crosssectional area \(A\). Suppose the aquifer consists of \(N\) layers, each of thickness \(b_{i}(i=1, \ldots, N)\) and permeability \(k_{i}(i=1, \ldots, N) .\) Determine the total flow rate through the aquifer if all the layers are subjected to the same driving pressure gradient. What is the uniform permeability of an aquifer of thickness \(b\) that delivers the same flow rate as the layered aquifer when the two are subjected to the same pressure gradient?
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