Logout Open In App
Open In App
Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Chapter 9: Problem 21

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.

Short Answer

Expert verified
The temperature gradient above is steeper than below by \\(L\rho_s v_s/k\\).

Step by step solution

01

Understanding the Interface

The interface separates two regions: the region above the interface where only upwelling water is found, and the region below the interface where upwelling steam and water coexist. At this boundary, steam condenses into water due to the cooler temperatures.
02

Analyzing the Heat Balance

In a steady-state scenario, the flux of enthalpy, i.e., heat per unit area per unit time, due to the steam must equal the heat flux due to the change in temperature. The condensation of steam releases latent heat, contributing to a change in thermal energy across the interface.
03

Expressing Heat Flux Due to Condensation

The heat flux due to the condensation of steam is given by \[-L \rho_s v_s\]. Here, \(L\) is the latent heat per unit mass of the phase change, \(\rho_s\) is the density of steam, and \(v_s\) is the upwelling velocity of steam. This term represents the energy released to the surrounding as steam condenses.
04

Relating Heat Flux to Temperature Gradient

The temperature gradient on either side of the interface due to thermal conductivity \(k\) is expressed as a heat flux proportional to the temperature gradient: \[-k \frac{dT}{dy}\]. Above the interface, we can write it as \[-k \left(\frac{dT}{dy}\right)_2\] and below the interface as \[-k \left(\frac{dT}{dy}\right)_1\].
05

Applying the Heat Balance Equation

At the interface, the net heat flux due to temperature gradients and latent heat release must balance: \[-k \left(\frac{dT}{dy}\right)_2 - (-k \left(\frac{dT}{dy}\right)_1) = -L \rho_s v_s\]. This states that the difference in conductive heat transfer above and below the interface contributes to the latent heat release.
06

Solving for the Temperature Gradient Difference

Rearranging the balance equation gives \[-k \left(\left(\frac{dT}{dy}\right)_2 - \left(\frac{dT}{dy}\right)_1\right) = -L \rho_s v_s\]. From this, we derive the required gradient relationship: \[(\frac{dT}{dy})_2 - (\frac{dT}{dy})_1 = \frac{L \rho_s v_s}{k}\]. This shows that the gradient above the interface is steeper by this amount due to latent heat release.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Thermal Gradient
In geodynamics, the thermal gradient is a crucial concept often used to describe how temperature changes over a specific distance within a material. This is especially important near a phase transition, like the steam-water interface. The thermal gradient is represented mathematically as \( \frac{dT}{dy} \), where \( T \) denotes temperature and \( y \) the spatial coordinate.

In the given exercise, the steam condenses, releasing latent heat which leads to a steeper thermal gradient immediately above the interface \( \left(\frac{dT}{dy}\right)_2\) than just below it \( \left(\frac{dT}{dy}\right)_1\). This difference is due to the thermal energy adjustment needed at this phase boundary.

Understanding how thermal gradients alter in different regions of a system helps us grasp energy transport processes more comprehensively. Changes in thermal gradients also reflect energy transitions and play an essential role in influencing material properties and reactions.
Latent Heat
Latent heat is the energy released or absorbed during a phase transition without a change in temperature. In the context of the steam-water interface, latent heat refers to the energy released by steam as it condenses into water.

This change in phase from gas to liquid is fundamental due to the thermal dynamics occurring at the interface. The latent heat can be represented as \( L \) and is crucial for calculating how energy is transferred within the system.

When steam condenses, a significant amount of energy is released, contributing to a steeper thermal gradient immediately above the interface. This change is represented as \( -L \rho_s v_s \), showing the density of the steam \( \rho_s \) and its Darcy velocity \( v_s \). Thus, understanding latent heat allows us to quantify and predict how phase changes impact temperature and energy distribution in geodynamic processes.
Darcy Velocity
Darcy velocity is a concept used to describe the flow of a fluid through a porous medium. It is essentially the volumetric flow rate per unit area. In this exercise, \( v_s \) represents the upwelling Darcy velocity of steam moving towards the steam-water interface.

Understanding Darcy velocity helps in quantifying the flow characteristics of steam in the porous medium, reflecting how quickly and how much fluid is likely to move through a material.

The velocity impacts heat transfer dynamics as well, since it affects how much steam condenses on the upper side of the interface, thereby influencing the thermal gradient. This parameter is an important factor when considering how energy and mass transfer through different geological layers, affecting earth processes like geothermal energy extraction and magmatic fluid movements.
Steam-Water Interface
The steam-water interface is the boundary where steam makes a transition to liquid water due to temperature and pressure changes. This interface is vital because it marks a significant phase change in the system.

At this boundary, steam condenses, releasing latent heat, which influences the surrounding thermal gradient. This heat release causes a stronger temperature gradient above the interface compared to below, due to the interplay of conductive heat transfer and phase change energy dynamics.

The interface acts as a thermal and physical barrier, controlling how heat and mass are distributed within the system. As such, it's key in understanding geodynamic phenomena, modeling of thermal fields, and interpreting fluid movements within Earth's crust.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

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

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 pressurecontrolling 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.

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 \(s\).

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.
See all solutions

Recommended explanations on Geography Textbooks

Globalisation

Read Explanation

Living With The Physical Environment

Read Explanation

Living World

Read Explanation

Coasts Geography

Read Explanation

Sustainable Urban Development

Read Explanation

Migration and Identity

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free