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

Short Answer

Expert verified
The mound's height decreases, and its width increases over time.

Step by step solution

01

Analyze Initial Conditions

We are given that the initial height of the groundwater mound at \(x=0\) and \(t=t_0\) is \(h_0\), and the initial half-width of the mound is \(l_0\). We need to find how these change at \(t = t_0 + t'\).
02

Understand the Problem Formulation

The problem is framed in the context of a spreading mound of groundwater. The changes in height and width over time depend on various parameters including permeability \(k\), density \(\rho\), gravitational acceleration \(g\), dynamic viscosity \(\mu\), and porosity \(\phi\).
03

Apply the Governing Equation

The height of the mound and its width at a future time \(t = t_0 + t'\) can be modeled using Boussinesq-type equations. These equations show that the changes in these quantities depend inversely on a cubic root of a dimensionless parameter involving these physical properties.
04

Derive New Height Expression

Given that initially at \(t_0\), the height is \(h_0\), the new height at time \(t_0 + t'\) is modeled to be \[ h_0 \left( 1 + \frac{6k \rho g h_0 t'}{\mu \phi l_0^2} \right)^{-1/3} \]. This equation indicates that as time advances, the height decreases.
05

Derive New Half-Width Expression

Similarly, the new half-width of the mound can be expressed as \[ l_0 \left( 1 + \frac{6k \rho g h_0 t'}{\mu \phi l_0^2} \right)^{1/3} \]. This demonstrates that the width increases as the mound spreads over time.
06

Introduce Similarity Variables

To solve for the water table height \(h(x, t)\), we use similarity variables \(f\) and \(\xi\) to simplify the problem. These are \[ f = \left( \frac{k \rho g \phi}{Q_1^2 \mu t} \right)^{1/3} h \] and \[ \xi = \left( \frac{\phi^2 \mu}{k \rho g Q_1 t^2} \right)^{1/3} x \]. Such transformations help make the problem independent of time and space directly.
07

Boussinesq Equation with Similarity Variables

Upon substitution of these similarity variables into the Boussinesq equation, we obtain \[ f \frac{d^2 f}{d \xi^2} + \left( \frac{d f}{d \xi} \right)^2 + \frac{2}{3} \xi \frac{d f}{d \xi} - \frac{1}{3} f = 0 \]. This reduced form allows for easier analysis and solution of the behavior of the mound.
08

Analyze Boundary Condition

Applying the boundary condition \((9-73)\) transformed with the similarity variable \(\xi\), the condition at \(x=0+\) becomes \( \left(f \frac{d f}{d \xi} \right)_{\xi=0+} = -\frac{1}{2} \). This condition gives us a point of reference for solving the differential equation for \(f\).

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Key Concepts

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

Boussinesq Equation
The Boussinesq Equation is a fundamental concept in the study of groundwater hydrodynamics. It describes the flow underneath a water table in a porous medium, such as soil or rock. This equation accounts for the movement of groundwater in terms of its height and flow rate.

In simple terms, the Boussinesq Equation helps to reveal how a body of water, like a groundwater mound, spreads over time. This process is affected by factors like the gravitational force, the material’s permeability, and the fluid’s viscosity. The Boussinesq Equation for our specific problem has been modified with similarity variables to make the analysis more manageable, focusing primarily on how the mound changes at certain points in space and time.

The equation can often be quite complex, requiring mathematical tools to solve. However, its general role is to model how pressure differences in groundwater lead to fluid movement, helping us predict changes in groundwater levels.
Groundwater Mounds
A groundwater mound forms when water accumulates over a permeable surface, creating a "mound" shape as it spreads laterally. Imagine pouring water onto a dry sponge and watching it pool into a dome before it slowly seeps through.

These mounds are important phenomena in hydrodynamics as they can influence local water table levels and groundwater flow patterns. They also help in understanding the effects of recharge from rainfall or artificial sources on groundwater systems. In the given problem, the mound starts with a height of \( h_0 \) at \( x=0 \) and a half-width \( l_0 \), changing as water spreads out over time.

The height of the mound reduces while its width increases due to the lateral dispersal of water, a characteristic behavior captured mathematically in the derived equations provided in the original exercise.
Similarity Variables
Similarity variables transform an otherwise complex problem into a more straightforward, dimensionless form. This is particularly useful in problems involving flow dynamics of groundwater mounds.

By introducing these variables, we can simplify the Boussinesq equation to focus on dimensionless forms that can easily be analyzed. In this problem, the functions \( f \) and \( \xi \) have been used to make the problem dimensionless and related to the spatial and time variables, respectively.
  • \( f = \left( \frac{k \rho g \phi}{Q_1^2 \mu t} \right)^{1/3} h \) is the similarity variable that describes the height.
  • \( \xi = \left( \frac{\phi^2 \mu}{k \rho g Q_1 t^2} \right)^{1/3} x \) is the similarity variable that describes the position in space.
These transformations aid in achieving solutions that are more universal, thus allowing insights without directly referencing the scale of the situation.
Flow Dynamics
Flow dynamics in groundwater hydrodynamics refers to how water moves through porous media, influenced by pressure gradients, gravitational forces, and the characteristics of the medium. In our example, flow dynamics are analyzed through the Boussinesq equation, which accounts for the horizontal flow of a mound of groundwater.

The equation enables us to calculate how quickly or slowly the groundwater moves horizontally, either spreading outward or retreating, based on the given initial conditions and parameters. This flow affects both the height and width of the groundwater mound over time. It is imperative to understand this concept as it directly impacts predictions of groundwater behavior, recharge effects, and the management of water resources in different environmental conditions.
Boundary Conditions
Boundary conditions are crucial in solving differential equations in groundwater hydrodynamics. They provide the necessary constraints that determine the solution’s specific behavior within a system.

For the exercise, we encounter a boundary condition at \( x=0+ \), where a specific height and flow rate are given. Here it is described by the equation: \( \left(f \frac{d f}{d \xi} \right)_{\xi=0+} = -\frac{1}{2} \). This condition ensures that the solution is physically meaningful and reflects the reality of groundwater mound behavior.

Using this boundary condition, along with similarity variables, the problem can be dissected into solvable parts, ensuring the resulting model accurately represents the observed phenomenon. By setting these specific conditions, we can precisely predict and analyze how the groundwater mound evolves over time under the given circumstances.

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

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.

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.

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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