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Chapter 9: Problem 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.

Short Answer

Expert verified
The upward Darcy velocity \(|V|\) is derived as \(\frac{(1-\phi) k g (\rho_{s} - \rho_{w})}{\mu}\).

Step by step solution

01

Understanding Fluidization

Fluidization occurs when the upward fluid flow counteracts the gravitational pull on the soil particles. This means that the pressure exerted by the fluid is equal to the weight of the soil structure, effectively lifting it.
02

Establishing the Balance Equation

Begin by considering the vertical forces acting on a cubic meter of soil. The gravitational force trying to hold the soil is proportional to the effective weight of the soil, given by: \[W_{soil} = (1-\phi)\rho_{s}g\]where \(\phi\) is the porosity.
03

Formulating Fluid Pressure Equation

The upward force exerted by the fluid flow is a result of the pressure difference between the fluid at the base of the soil and the fluid above. To fluidize, this pressure difference must equal the effective weight per unit volume of the soil:\[\Delta P = \rho_{f}gh = \mu\frac{V}{k}\]where \(\rho_{f}\) equals \(\rho_{w}\), assuming the fluid is water.
04

Solving for Darcy Velocity

To find the Darcy velocity \(|V|\), equate the upward fluid pressure to the weight of the soil structure, taking into account the density difference between the soil particles and the water:\[(1-\phi)\rho_{s}g = \frac{\mu V}{k} \times \rho_{w}\]Rearranging the equation gives:\[|V|=\frac{(1-\phi) k g\left(\rho_{s}-\rho_{w}\right)}{\mu}\]
05

Confirmation of Derived Formula

The derived equation shows the conditions for the onset of fluidization, where the velocity \(|V|\) is determined by soil properties \(\phi\), \(\rho_{s}\), and fluid properties \(\rho_{w}\) and \(\mu\), and the permeability \(k\). This confirms the given formula.

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Key Concepts

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

Fluidization
Fluidization is a fascinating phenomenon in soil mechanics where soil particles become suspended in a fluid, usually due to upward fluid flow that counteracts gravitational forces. This results in the soil behaving more like a liquid than a solid. Imagine particles lifted by a flowing stream. That’s similar to what happens in fluidization.
Fluidization occurs when the pushing force of the rising fluid matches the gravitational pull on the particles. The fluid essentially "lifts" the soil structure, allowing for easy movement, much like boiling water creates movement in grains of rice. Understanding fluidization is crucial for fields like chemical engineering and geology, where controlling particle movement in fluids is essential.
In practical applications, proper fluidization can lead to improved processes such as mixing, reaction efficiency, and separation techniques in industrial settings.
Porosity
Porosity is a key property of any porous material, representing the fraction of the volume of void spaces over the total volume. In simple terms, it tells us how much open space exists within a material.
In soil, porosity determines how much water or air it can hold. It is calculated as \[\phi = \frac{V_{void}}{V_{total}}\]where \( V_{void} \) is the volume of the void spaces and \( V_{total} \) is the total volume of the material.
Porosity is crucial because it influences fluid flow, soil strength, and storage capacity. High porosity means more fluid can be stored or pass through, but it can also mean lower soil stability. Engineers and geologists often measure porosity to predict how a soil or rock layer might behave in natural or artificial processes like groundwater movement or oil extraction.
Soil Density
Soil density is a measure of how compact the soil particles are. It is usually expressed as the mass per unit volume, taking into account both the soil particles and the air-filled voids between them.
Density affects the soil's ability to support structures and how water moves through it. Higher density means less space for air and water, potentially leading to poor drainage. Here's the formula for soil density:\[\rho = \frac{M}{V}\]where \( M \) is the mass of the soil, and \( V \) is the volume.
Understanding soil density is important for construction, agriculture, and environmental applications. In these fields, knowing the density helps ensure that buildings remain stable, plants grow well, and pollutants do not move too quickly through the soil.
Permeability
Permeability refers to the ease with which a fluid can flow through a porous material. It is a measure of how connected the void spaces in a material are, allowing a fluid to pass through.
In soils, high permeability means that water can move quickly through the soil, helping plants access water and avoiding flooding. Permeability is influenced by the size, shape, and arrangement of the soil particles.
The permeability equation in Darcy's Law is:\[k = \frac{Q \cdot L}{A \cdot \Delta P}\]where \( k \) is the permeability, \( Q \) is the flow rate, \( L \) is the length of the flow path, \( A \) is the cross-sectional area, and \( \Delta P \) is the pressure difference driving the flow. Understanding permeability is crucial for projects involving water supply, waste management, and environmental protection. It helps predict how substances move through soils, crucial for securing water resources and managing pollutants.

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

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

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

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 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.
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