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Chapter 9: Problem 27

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

Short Answer

Expert verified
The temperature at a depth of 1 km is approximately 372 K.

Step by step solution

01

Write the Hydrostatic Pressure Equation

The pressure in the porous layer varies based on the depth due to the weight of the overlying water. This is given by:\[p = p_{0} + \rho_{l} g y\]where \(p_{0}\) is the atmospheric pressure, \(\rho_{l}\) is the density of the liquid water, \(g\) is the acceleration due to gravity, and \(y\) is the depth.
02

Express Density of Water Vapor

Under the assumption that steam is a perfect gas, relate the density of water vapor \(\rho_{v}\) to pressure and temperature:\[\rho_{v} = \frac{p}{R_{V} T}\]
03

Derive Clapeyron Slope Formula

The Clapeyron slope describes the change of pressure with respect to temperature at the phase boundary:\[\frac{dp}{dT} = \frac{L \rho_{v}}{T}\]Substituting \(\rho_{v} = \frac{p}{R_{v} T}\), we get:\[\frac{dp}{dT} = \frac{L p}{R_{V} T^2}\]
04

Solve for Temperature Gradient

Differentiate the hydrostatic equation with respect to \(y\) and equate it to the derivative from the Clapeyron relationship:\[\rho_{l} g = \frac{d}{dy}(p) = \frac{dp}{dT} \frac{dT}{dy} = \frac{L p}{R_{V} T^2} \frac{dT}{dy}\]Rearrange to yield the expression for the temperature gradient:\[\frac{dT}{dy} = \frac{R_{V} T^2 \rho_{l} g}{L p}\]
05

Integrate to Find Temperature Profile

Integrate the temperature gradient from atmospheric temperature \(T_{b0}\) to temperature \(T\) at depth \(y\):\[\int_{T_{b0}}^{T} \frac{1}{T^2} dT = \int_{0}^{y} \frac{R_{V} \rho_{l} g}{L (p_0 + \rho_{l} g y)} dy\]Solving both integrals yields:\[-\frac{1}{T} + \frac{1}{T_{b0}} = \frac{R_{V}}{L} \ln \left(1 + \frac{\rho_{l} g y}{p_{0}}\right)\]
06

Calculate Temperature at 1 km Depth

Substitute the given values into the final equation to find \(T\) at \(y = 1000 \text{ m}\):\[\frac{1}{T_{b0}} - \frac{1}{T} = \frac{0.462}{2500} \ln \left(1 + \frac{1000 \times 10 \times 1000}{10^5}\right)\]\[\frac{1}{T_{b0}} - \frac{1}{T} = \frac{0.462}{2500} \ln (11)\]Calculate to find \(\frac{1}{T}\), and consequently, \(T\).

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Key Concepts

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

Clapeyron Equation
The Clapeyron equation is a fundamental relationship that describes the slope of the phase boundary line or the rate of change of pressure with temperature at a phase transition. This equation is crucial when examining systems in thermodynamic equilibrium where a substance transitions between phases, such as from liquid water to steam. In the equation:\[\frac{dp}{dT} = \frac{L \rho_{v}}{T(\rho_{l} - \rho_{v})} \approx \frac{L \rho_{v}}{T},\]\(dp/dT\) represents the change in pressure \(p\) with respect to change in temperature \(T\). The latent heat \(L\) is the heat absorbed or released during the phase transition, \(\rho_{l}\) and \(\rho_{v}\) are the densities of the liquid and vapor phases, respectively. This relationship tells us how the pressure must change to maintain equilibrium when the temperature changes. For processes involving evaporation where \(\rho_{v}\) is much smaller than \(\rho_{l}\), the simplified form \(\approx \frac{L \rho_{v}}{T}\) is often used, especially when density changes slightly near the boundary.
Hydrostatic Pressure
Hydrostatic pressure is the pressure exerted by a fluid at equilibrium due to the force of gravity. In the context of this problem, water, as a fluid in a porous layer, pushes down on water below due to gravity. This pressure increases with depth and can be represented by the hydrostatic equation:\[p = p_{0} + \rho_{l} g y,\]where \(p_0\) is atmospheric pressure, \(\rho_l\) is the density of liquid water, \(g\) is the acceleration due to gravity, and \(y\) is the depth. This formula allows us to calculate the pressure at any given depth, showing a linear relationship. The deeper you go, the higher the pressure, because more water is weighing down from above. This concept is crucial for understanding how pressure and subsequently temperature vary at different depths in the earth's crust.
Water Vapor Density
Water vapor density \(\rho_{v}\) measures how much water vapor, or steam, is present in a unit volume of air. It is an important factor in phase transition phenomena. The perfect gas assumption simplifies the relationship among water vapor density, pressure, and temperature to:\[\rho_{v} = \frac{p}{R_{V} T},\]where \(p\) is the pressure, \(R_V\) is the specific gas constant for water vapor, and \(T\) the temperature. This relationship is vital in deducing the thermodynamic properties of vapor and how it contributes to phase changes when the system's temperature or pressure changes. Through manipulating this equation, we can analyze how increased depths (and thus increased pressures) affect the density of vapor, thus impacting phase changes in the geological environment.
Phase Transition
Phase transition refers to the process of change from one phase to another, such as from liquid to gas or solid to liquid, facilitated by change in conditions like temperature or pressure. During a phase transition, energy is absorbed or released by the system, characterized by the latent heat \(L\). The system stays at equilibrium when pressure adjusts at a rate defined by the Clapeyron slope to counter change in temperature:- **Latent Heat**: This quantity, \(L\), represents the amount of heat required to convert a substance from one phase to another at a fixed temperature and pressure.- **Equilibrium**: A state where the physical or chemical characteristics of the phases present are stable, often involving balanced pressures and temperatures.- **Application in Problem**: The temperature-depth profile derived in the problem shows how temperature changes with depth under boiling conditions, signifying a continuous phase transition state throughout the water column in the porous layer.

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Most popular questions from this chapter

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

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 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.

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