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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
Gravity affects fluid flow downslope by adding \( \rho g \sin \theta \), adjusting Darcy's law accordingly.

Step by step solution

01

Understand Darcy's Law for Vertical Flow

Darcy's law for vertical flow in a porous medium is given as \( v = -\frac{k}{\mu}\left(\frac{d p}{d y} - \rho g\right) \). This equation accounts for pressure gradients and gravitational effects. Here, \( \frac{d p}{d y} \) represents the rate of pressure change with depth, \( \rho g \) is the gravitational pressure term assuming flow direction is aligned with gravity, while \( v \) is the Darcy velocity.
02

Establish the Condition for Downslope Flow

When considering a porous medium on an inclined surface, flow occurs downslope along the surface, not vertical. We define \( s \), the downslope distance, which aligns our coordinate system to the slope of inclination, allowing us to describe the flow along \( s \) rather than directly vertically.
03

Apply Trigonometric Inclination Corrections

Inclination impacts gravitational force along the downslope. The component of gravity acting along \( s \) is \( \rho g \sin \theta \), where \( \theta \) is the angle with the horizontal. This term captures how gravity influences the flow parallel to the slope rather than perpendicular.
04

Integrate Downslope Pressure Gradient

As with vertical flow, pressure changes along \( s \) affect flow. Thus, we need to consider \( \frac{d p}{d s} \), representing the pressure gradient along the downslope direction. This gradient is analogous to \( \frac{d p}{d y} \), but aligned to the slope.
05

Formulate Darcy's Law for Downslope Flow

By incorporating the pressure gradient along the slope \( \frac{d p}{d s} \) and the gravitational component \( \rho g \sin \theta \), Darcy's law for downslope flow becomes:\[ q = -\frac{k}{\mu}\left(\frac{d p}{d s} - \rho g \sin \theta\right) \]This final equation accounts for both the inclined gravitational effect and pressure gradient in the inclined direction, appropriately modeling the physical behavior of the flow.

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Key Concepts

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

Porous Medium Flow
The concept of porous medium flow is crucial for understanding how fluids move through materials filled with tiny holes or pores. This can be thought of as water moving through a sponge. In such cases, the fluid doesn't flow freely, like in an open river, but instead it navigates through an intricate network of narrow pathways.
Darcy's Law is a primary tool used to describe this kind of flow. It relates the flow velocity to how easily the fluid can move through the medium (permeability), to how much force is pushing it through (pressure gradient), and to the fluid's physical properties like viscosity. This relationship is akin to using a garden hose: squeeze it harder (pressure increases), and water will flow out faster.
Pressure Gradient
In fluid dynamics, a pressure gradient is essentially a difference in pressure that causes fluid to flow. Imagine a long hill where one end is higher than the other, creating a slant. In fluids, this slant is represented by pressure.
The formula \(\frac{d p}{d y}\) or \(\frac{d p}{d s}\) describes how pressure changes over a certain vertical or inclined distance. If pressure changes abruptly or steadily over distance, it means there is a strong or weak pressure gradient. This gradient acts as the force propelling the fluid through the porous medium, analogous to pushing a sled down a snowy hill, with the hill’s inclination determining the sled's speed.
Gravity in Fluid Dynamics
Gravity plays a significant role in how fluids behave, especially in porous media. It impacts the direction and strength of fluid flow. Picture gravity as a gentle but persistent force always pointing downwards due to Earth's pull.
In a vertical flow scenario, gravity helps the fluid move downwards. The term \(\rho g\) in Darcy's Law represents the gravitational force acting on the fluid. This term essentially counteracts upward flow, meaning there's a need for additional pressure to overcome gravity when pushing fluid upwards. This is why on an incline, the gravitational component changes, taking into account only part of gravity that actually drives the fluid along the slope.
Inclined Surfaces Impact on Flow
When considering a slope, like a hillside, water doesn't just fall straight down due to gravity; it also runs down along the surface. The angle of the slope, described by \(\theta\), directly influences this flow direction and rate.
Inclined surfaces introduce an additional factor into fluid dynamics – the gravitational force component parallel to the slope, which is \(\rho g \sin \theta\). This effect is critical for understanding why water flows faster down a steep hill than a gentle incline.
The modified Darcy's Law equation for inclined surfaces, \[ q = -\frac{k}{\mu}\left(\frac{d p}{d s} - \rho g \sin \theta\right) \], highlights how slightly tilting the surface changes the flow dynamics entirely, emphasizing the balance between pressure gradient and gravitational pull as they now operate along the slope rather than vertically.

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Most popular questions from this chapter

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 pressure-controlling 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.

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

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.

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

Determine the minimum critical Rayleigh number for the onset of convection in a layer of porous material heated from below with an isothermal and impermeable lower boundary and an isothermal constant pressure upper boundary. This boundary condition corresponds to a permeable boundary between a saturated porous layer and an overlying fluid. What is the horizontal wavelength that corresponds to the minimum value of \(\mathrm{Ra}_{\mathrm{cr}}\) ? Take the layer thickness to be \(b\), and let the upper boundary, \(y=0\), have temperature \(T=T_{0}\) and the lower boundary, \(y=b\), have temperature \(T=T_{1}\). Assume that at the onset of convection \(T^{\prime}\) has the form $$ T^{\prime}=T_{0}^{\prime} \sin \frac{2 \pi x}{\lambda} Y(y) $$ and show that \(Y(y)\) is a solution of $$ \frac{d^{4} Y}{d \bar{y}^{-4}}-2 a^{2} \frac{d^{2} Y}{d y^{-2}}+Y\left(a^{4}-a^{2} R a\right)=0 $$ where $$ a \equiv \frac{2 \pi b}{\lambda} \quad \bar{y} \equiv \frac{y}{b} $$ Show that the general solution of Equation (9.135) can be written as $$ Y=c_{1} e^{\gamma \bar{y}}+c_{2} e^{-\gamma \bar{y}}+c_{3} \sin \delta \bar{y}+c_{4} \cos \delta \bar{y} $$ where \(c_{1}, c_{2}, c_{3},\) and \(c_{4}\) are constants of integration and $$ \begin{array}{l} \gamma^{2}=a^{2}+a \sqrt{\mathrm{Ra}} \\ \delta^{2}=a \sqrt{\mathrm{Ra}}-a^{2} \end{array} $$ Show that the boundary conditions are $$ \begin{aligned} Y=0 & \text { on } \bar{y}=0 \text { and } 1 \\ \frac{d^{2} Y}{d y^{2}}=0 & \text { on } \bar{y}=1 \\ \frac{d}{d \bar{y}}\left(\frac{d^{2} Y}{d \bar{y}^{2}}-a^{2} Y\right)=0 & \text { on } \bar{y}=0 . \end{aligned} $$ Substitute Equation (9.137) into each of these boundary conditions to obtain four homogeneous equations for the four unknown constants \(c_{1}, c_{2},\) \(c_{3},\) and \(c_{4} .\) Show that a nontrivial solution of these equations requires $$ \gamma \tan \delta+\delta \tanh \gamma=0 $$ This transcendental equation is an eigenvalue equation that implicitly gives \(\mathrm{Ra}_{\mathrm{cr}}\) as a function of \(a\), since both \(\gamma\) and \(\delta\) are defined in terms of \(\mathrm{Ra}\) and \(a\) in Equations (9.138) and (9.139) . The critical Rayleigh number can be found by numerically solving Equations (9.138),(9.139) , and (9.143) . The value of min \(\left(\mathrm{Ra}_{\mathrm{cr}}\right)\) turns out to be \(27.1 .\) One way of proceeding is to choose a value of \(a\) (there exists an \(\mathrm{Ra}_{\mathrm{cr}}\) for each a). Then try a value of \(\delta\). Compute \(\gamma\) from \(\gamma^{2}=2 a^{2}+\) \(\delta^{2} .\) Then compute \(\tan \delta / \delta\) and \(-\tanh \gamma / \gamma\). Iterate on \(\delta\) until these ratios are equal. With \(\delta\) determined \(\mathrm{Ra}_{\mathrm{cr}}\) follows from Equation \((9.139) .\) Repeat the process for different values of \(a\) until \(\min \left(\operatorname{Ra}_{c r}\right)\) is found.

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