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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 is \( \frac{6\delta}{b} \); Permeability is \(1 \times 10^{-6} \text{ m}^2\) for given values.

Step by step solution

01

Define Porosity

Porosity is the ratio of the volume of void spaces (the channels in this case) to the total volume of the medium. In a cubic matrix with side length \(b\) and channel thickness \(\delta\), the volume of the voids is \(6b^2 \delta\) (considering there are six faces with voids). The total volume is \(b^3\). Therefore, porosity \( \phi \) is given by:\[\phi = \frac{6b^2 \delta}{b^3} = \frac{6\delta}{b}\].
02

Define Permeability

In a porous medium like this model, permeability \(k\) is related to the geometrical properties of the medium. The permeability is generally proportional to the square of the characteristic dimension of the passage through which fluid flows. For this problem, a reasonable approximation is that the permeability is proportional to \(\delta^2\), leading to:\[k \propto \left(\frac{b \delta}{b}\right)^2 = \delta^2\].
03

Determine Specific Expression for Permeability

To get a specific expression, consider the equation based on dimensional analysis and known permeability principles for cubic structures:\[k = C \delta^2\]where \(C\) is a constant depending on the structural geometry and tortuosity of the flow path. A simple assumption is that \(C\) can be determined experimentally or through more complex models, but here we assume:\[k = \left(\frac{\delta}{b}\right)^2 b^2\].
04

Calculate Permeability for Given Values

Given \(b = 0.1 \text{ m}\) and \(\delta = 1 \text{ mm} = 0.001 \text{ m}\), substitute these into the expression for permeability:\[\begin{align*}k &= \left(\frac{0.001}{0.1}\right)^2 (0.1)^2 \&= (0.01)^2 \times 0.01 \&= 0.000001 \end{align*}\]Thus, the permeability is \(1 \times 10^{-6} \text{ m}^2\).

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Key Concepts

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

Understanding Porosity in a Porous Medium
In the context of a porous medium, porosity is a measure of how much empty space there is within the material. This is crucial as it determines the capacity of a material to hold fluids. Specifically, porosity, typically denoted by the symbol \( \phi \), is the ratio of the volume of void spaces to the total volume. When looking at a cubic matrix model, one can think of these voids as the channels through which fluids move.

For a cubic model with a side length \(b\) and channel thickness \(\delta\), the voids are represented by these channels on each face of the cube. Given a cube has six faces, we calculate the volume of the voids as \(6b^2 \delta\). The entire cube's volume is simply \(b^3\).

Therefore, the formula for porosity becomes:
  • Porosity \( \phi = \frac{6b^2 \delta}{b^3} = \frac{6\delta}{b}\).
This formula allows us to understand that porosity is directly influenced by the thickness of the walls of the cube and the size of the cube itself.
Exploring Permeability in Porous Materials
Permeability is another vital characteristic of porous materials. It describes the material's ability to aid the flow of a fluid through its void spaces or pores. Just like porosity, permeability is intrinsically linked to the material's structure.

In the case of a cubic matrix model, permeability \(k\) becomes essential to measure, as it helps predict how easily fluids traverse these channels. Mathematically, permeability is often proportional to the square of the characteristic dimensions of the flow path. Here, it is proportional to the square of the channel thickness \(\delta\), reflected in the relation:
  • \(k \propto \delta^2\).
To hone this understanding, we derive the permeability from the geometric ratio, and it looks like:
  • Specific permeability formula: \(k = \left(\frac{\delta}{b}\right)^2 b^2\).
Utilizing this model, we can deduce the permeability for specific cases, such as when \(b = 0.1 \text{ m}\) and \(\delta = 0.001 \text{ m}\). The calculated permeability is \(1 \times 10^{-6} \text{ m}^2\).

This measurement demonstrates how permeability is influenced by the dimensions of the matrix.
The Cubic Matrix Model
The cubic matrix model is a simplistic but effective representation of a porous medium. It is widely used in mathematical modeling to study the flow of fluids through porous materials. This model treats the porous medium as a 3D lattice of cubes, accounting for channels or voids that control the fluid movements.

By using a cubic matrix, one accurately approximates the volume of voids and the connectivity between them. This is crucial, as it decides not only how much fluid can pass through, but also how quickly it can do so. The model presumes the voids occupy the edges of the cubes, emphasizing that fluid path is primarily linked to the channel thickness \(\delta\).

A primary advantage of the cubic matrix model is its simplicity and the ease with which it allows us to derive mathematical expressions for attributes like porosity and permeability. Even with these simplifications, it provides valuable insights into the macroscopic behavior of porous materials. Understanding the cubic matrix model lays the groundwork for studying more complex porous systems.
The Role of Dimensional Analysis in Porous Media
Dimensional analysis is a pivotal mathematical tool when examining porous media. It's particularly useful in deriving expressions for physical quantities by considering their dimensions. This process helps simplify the complex relationships between different physical variables by comparing them dimensionally.

In the context of our cubic matrix model, dimensional analysis is instrumental. It assists in ensuring that equations for porosity and permeability maintain correctness across varying units and dimensions. For example:
  • While deriving permeability \(k\), dimensional analysis ensures that the expression \( \left(\frac{\delta}{b}\right)^2 b^2 \) remains dimensionally consistent and leads to a correct physical interpretation.
Through this approach, dimensional analysis helps in validating theoretical models and making them applicable to real-world scenarios.

The beauty of dimensional analysis lies in its ability to provide insight into the significance of various physical quantities and to uncover scaling laws, which further aid in understanding the flow characteristics within porous media.

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Most popular questions from this chapter

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 \(R_{V}\) 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 $$ \frac{d p}{d T}=\frac{L \rho_{I} \rho_{v}}{T\left(\rho_{l}-\rho_{v}\right)} \approx \frac{L \rho_{v}}{T} $$ where \(\rho_{V}\) is the density of water vapor. Assume that steam is a perfect gas so that $$ \rho_{v}=\frac{p}{R_{v} T} $$ Finally, note that \(p=p_{0}+\rho_{l} g y\) if \(\rho_{l}\) is assumed constant. What is the temperature at a depth of \(1 \mathrm{~km} ?\) $$ \text { Take } R_{V}=0.462 \mathrm{~kJ} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}, L=2500 \mathrm{~kJ} \mathrm{~kg}^{-1}, T_{b 0}= $$ \(373 \mathrm{~K}, p_{0}=10^{5} \mathrm{~Pa}, \rho_{l}=1000 \mathrm{~kg} \mathrm{~m}^{-3}, g=10 \mathrm{~m} \mathrm{~s}^{-2}\)

Consider the upwelling of a mixture of water and steam in a porous medium. Because of the cold temperatures near the surface, the mixture will reach a level where all the steam must abruptly condense. There will be a phase charge interface with upwelling water just above the boundary and upwelling steam and water just below it. Show that the temperature gradient immediately above the interface \((d T / d y)_{2}\) is larger than the temperature gradient just below the interface \((d T / d y)_{1}\) by the amount \(-L \rho_{s} V_{s},\) where \(L\) is the latent heat of the steam-water phase change, \(\rho_{s}\) is the density of the steam, and \(-v_{s}\) is the upwelling Darcy velocity of the steam.

Let \(h_{0}\) be the height of the laterally spreading groundwater mound at \(x=0\) and \(t=t_{0}\). Let the half-width of the mound at its base be \(l_{0}\) at \(t=t_{0}\). Show that the height of the mound at \(x=0\) and \(t=\) \(t_{0}+t^{\prime}\) is given by $$ h_{0}\left(1+\frac{6 k \rho g h_{0} t^{\prime}}{\mu \phi l_{0}^{2}}\right)^{-1 / 3} $$ In addition, demonstrate that the half-width of the mound at its base at time \(t=t_{0}+t^{\prime}\) is $$ I_{0}\left(1+\frac{6 k \rho g h_{0} t^{\prime}}{\mu \phi l_{0}^{2}}\right)^{1 / 3} $$ We next determine the height of the phreatic surface \(h\) as a function of \(x\) and \(t\) when water is introduced at \(x=0\) at a constant volumetric rate \(Q_{1}\) per unit width. For \(t<0, h\) is zero; for \(t>0,\) there is a constant input of water at \(x=0 .\) Half of the fluid flows to the right into the region \(x>0\), and half flows to the left. From Equation \((9-24)\) we can write the flow rate to the right at \(x=0+\) as $$ \frac{-k \rho g}{\mu}\left(h \frac{\partial h}{\partial x}\right)_{x=0+}=\frac{1}{2} Q_{1} $$ The water table height \(h(x, t)\) is the solution of the Boussinesq equation \((9-43)\) that satisfies condition \((9-73)\) Once again we introduce similarity variables. The appropriate similarity variables for this problem are $$ \begin{array}{l} f=\left(\frac{k \rho g \phi}{Q_{1}^{2} \mu t}\right)^{1 / 3} h \\ \xi=\left(\frac{\phi^{2} \mu}{k \rho g Q_{1} t^{2}}\right)^{1 / 3} x \end{array} $$ Aside from numerical factors these variables are the same as the ones in Equations \((9-62)\) and \((9-63)\) if we replace \(V_{1} / t\) in those equations by \(Q_{1}\). The introduction of these similarity variables into the Boussinesq equation yields $$ f \frac{d^{2} f}{d \xi^{2}}+\left(\frac{d f}{d \xi}\right)^{2}+\frac{2}{3} \xi \frac{d f}{d \xi}-\frac{1}{3} f=0 $$ The boundary condition at \(x=0+\) given in Equation \((9-73)\) becomes $$ \left(f \frac{d f}{d \xi}\right)_{\xi=0+}=-\frac{1}{2} $$

Calculate pressure as a function of depth in a vapor-dominated geothermal system consisting of a near-surface liquid layer \(400 \mathrm{~m}\) thick overlying a wet steam reservoir in which the pressurecontrolling phase is vapor. Assume that the hydrostatic law is applicable and that the liquid layer is at the boiling temperature throughout. Assume also that the steam reservoir is isothermal.

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 \(s\).
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