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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 \(s\).

Short Answer

Expert verified
Darcy's Law for an inclined surface incorporates gravitational components resolved along the slope.

Step by step solution

01

Review Darcy's Law in Vertical Flow

Darcy's Law for vertical flow is originally expressed as \[ V = -\frac{k}{\mu} \frac{d p}{d y} \]where \(V\) is the Darcy velocity, \(k\) is the permeability, \(\mu\) is the dynamic viscosity, and \(\frac{d p}{d y}\) is the pressure gradient in the direction of increasing depth \(y\). When the fluid is at rest, the pressure change is solely due to the gravitational force: \( \rho g \). Thus, for fluid motion, the adjusted pressure gradient is \( \frac{d p}{d y} - \rho g \). The corrected vertical velocity equation is \[ V = -\frac{k}{\mu} \left( \frac{d p}{d y} - \rho g \right). \]
02

Setup the Inclined Flow Scenario

For an inclined porous medium, consider the plane to be at an angle \(\theta\) to the horizontal. Define \(s\) as the downslope distance along the plane. The pressure gradient component along this surface is \(\frac{d p}{d s}\).
03

Resolve Gravitational Component Along Incline

The gravitational force along the incline acts in the downslope direction. This component can be expressed as \( \rho g \sin \theta \) due to the inclusion of trigonometry to resolve the force along the incline.
04

Apply Darcy's Law to Inclined Flow

Substitute the partial gravitational influence along the incline into the Darcy's Law expression. This gives the inclined Darcy's law for the flow component as: \[ q = -\frac{k}{\mu} \left( \frac{d p}{d s} - \rho g \sin \theta \right), \]where \(q\) now represents the volumetric flow rate per unit area along the inclined surface.

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Key Concepts

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

Porous Medium
A porous medium refers to a material containing numerous small holes or pores through which fluid can flow. Examples include soil, rocks, and sponge-like materials. The flow in a porous medium is characterized by how the fluid moves through the interconnected spaces inside it. The size and connectivity of these pores greatly influence the ease with which fluids can move. This property is known as permeability. Permeability is a very important factor in defining how a fluid will behave when passing through a porous medium.
Understanding the porous structure helps in predicting flow patterns and how substances or contaminants might disperse within it. This is fundamental in fields like hydrology, petroleum engineering, and environmental science, as it affects water availability, oil extraction, and pollution control.
Pressure Gradient
The pressure gradient is essentially the change in pressure over a certain distance. In any fluid system, pressure naturally wants to move from high-pressure areas to low-pressure areas. This is the driving force that causes fluids to flow.
In the context of Darcy's law, the pressure gradient is an important factor that influences fluid velocity in a porous medium. Specifically, for a vertical flow, it is represented as \(\frac{d p}{d y}\), indicating how pressure changes with depth.
  • When the fluid is stationary, the pressure gradient is balanced by gravitational forces.
  • When moving, it is this gradient, minus the gravitational influence, that drives the fluid's flow.
Understanding this concept aids in grasping why and how pressure needs to escalate more rapidly to facilitate upward fluid movement.
Gravitational Force
Gravitational force is the force exerted by the Earth that pulls objects toward its center. It plays a significant role in how fluids behave in a porous medium, especially in vertical and inclined flows. For a vertical flow, gravity acts in opposition to the flow if considering upward motion, exerting a downward force quantified as \(\rho g\), with \(\rho\) as the fluid density and \(g\) as the acceleration due to gravity.
In inclined flows, like in the described scenario of a medium on an inclined plane, gravity also influences flow direction and velocity. The component of gravitational force along the incline is calculated using trigonometry as \(\rho g \sin \theta\). This adjustment allows for the calculation of pressure effects on fluid flow along sloped surfaces, accounting for gravitational influence in a different orientation.
Dynamic Viscosity
Dynamic viscosity describes a fluid's resistance to flow. It is a measure of internal friction within the fluid. Thicker, more viscous fluids, like honey, flow more slowly compared to less viscous ones, like water.
In Darcy's Law, dynamic viscosity, denoted as \(\mu\), influences how easily a fluid can pass through a porous medium. It appears in the denominator of the Darcy's equations, meaning higher viscosity leads to lower fluid velocity through the medium. This indicates that a fluid's viscosity must be accounted for when determining flow characteristics in porous materials.
  • Viscosity plays an essential role in engineering applications where fluid flow needs to be predicted or controlled.
  • Understanding dynamic viscosity helps in managing industrial processes, environmental systems, and natural phenomena involving fluid movement through porous structures.

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Most popular questions from this chapter

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 \(\delta=1 \mathrm{~mm} ?\)

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 $$ \left(f \frac{d f}{d \xi}\right)_{\xi=0+}=-\frac{1}{2} $$

Consider an unconsolidated (uncemented) layer of soil completely saturated with groundwater; the water table is coincident with the surface. Show that 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.

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 \(1 \mathrm{~km} ?\) $$ \text { Take } R_{V}=0.462 \mathrm{~kJ} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}, L=2500 \mathrm{~kJ} \mathrm{~kg}^{-1}, T_{b 0}= $$ \(373 \mathrm{~K}, p_{0}=10^{5} \mathrm{~Pa}, \rho_{l}=1000 \mathrm{~kg} \mathrm{~m}^{-3}, g=10 \mathrm{~m} \mathrm{~s}^{-2}\)

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