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Chapter 9: Problem 28

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.

Short Answer

Expert verified
Pressure increases linearly with depth in the liquid layer and remains constant in the steam reservoir.

Step by step solution

01

Introduction to the Problem

We need to calculate pressure as a function of depth in a geothermal system with two layers: a liquid layer and a steam reservoir. We will use the hydrostatic law to derive the pressure profile.
02

Pressure in the Liquid Layer

For the liquid layer, pressure increases with depth due to the weight of the liquid above. The pressure at any depth within this layer can be calculated using the hydrostatic pressure formula: \( P(z) = P_0 + \rho g z \), where \( P_0 \) is the atmospheric pressure, \( \rho \) is the density of the liquid, \( g \) is the acceleration due to gravity, and \( z \) is the depth.
03

Pressure at the Base of the Liquid Layer

At the base of the liquid layer (\( z = 400 \, \text{m} \)), the pressure can be expressed as \( P(400) = P_0 + \rho g (400) \). This gives the total pressure just above the steam reservoir.
04

Isothermal Steam Reservoir

In the steam reservoir, the pressure is mainly controlled by vapor. Since the reservoir is isothermal, the pressure at any depth remains constant because temperature does not change with depth and the phase is mainly vapor.
05

Final Pressure Profile

The pressure profile in the system can be summarized as follows: In the liquid layer, \( P(z) = P_0 + \rho g z \) for \( 0 \leq z \leq 400 \, \text{m} \). For the steam reservoir, the pressure remains constant as equal to \( P(400) \) determined at the base of the liquid layer.

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Key Concepts

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

Pressure Calculation in Geothermal Systems
Understanding pressure calculation in geothermal systems involves recognizing how different factors affect the pressure at various depths. Geothermal systems typically have complex structures consisting of different phases, such as liquid, and vapor. To calculate the pressure accurately, we apply different scientific principles depending on the specific characteristics of each layer.

Using the hydrostatic law, pressure in the fluid layer above a steam reservoir can be deduced using the equation:
  • \( P(z) = P_0 + \rho g z \)
Here, \( P(z) \) is the pressure at depth \( z \), \( P_0 \) is the initial pressure, typically atmospheric pressure, \( \rho \) is fluid density, and \( g \) is the gravitational force. Understanding this relationship is key when calculating how pressure changes throughout a geothermal system. Analyzing these variations allows scientists and engineers to better manage geothermal energy resources effectively, leading to improved operational efficiency and safety.
Hydrostatic Pressure
Hydrostatic pressure is a fundamental concept in understanding how pressure changes within a fluid that is at rest. It is the pressure exerted by a fluid as a result of its weight. In the context of geothermal systems, particularly those involving a liquid layer, hydrostatic pressure explains why pressure increases with depth.

The formula \( P(z) = P_0 + \rho g z \) illustrates that hydrostatic pressure depends on three main factors:
  • Density of the liquid \( (\rho) \)
  • Acceleration due to gravity \( (g) \)
  • Depth \( (z) \) below the surface
These variables directly dictate the pressure exerted at any given depth. The deeper into a liquid, the greater the pressure due to the weight of the fluid above. This concept is crucial for predicting pressure variations in any fluid layer, particularly in geothermal systems where it influences reservoir operations and safety measures.
Vapor-Dominated Geothermal Systems
Vapor-dominated geothermal systems are unique types of systems where steam is the primary phase controlling pressure and energy extraction. These systems are generally marked by their capacity to produce dry steam, making them highly desirable for power generation.

Such systems have the following characteristics:
  • High-temperature steam within the reservoir
  • Pressure within these systems is largely constant at given depths due to the absence of significant liquid phase changes
  • Heat is primarily carried by vapor instead of liquid
Understanding vapor-dominated systems helps in efficiently harnessing geothermal heat. Because the steam is the primary contributor to pressure in these systems, it simplifies the pressure profile. In the context of the problem solved, the steam reservoir was identified as vapor-dominated, where the pressure remained constant due to isothermal conditions.
Isothermal Steam Reservoir
An isothermal steam reservoir in geothermal systems is a reservoir where temperature remains the same throughout. This characteristic leads to consistent steam pressure, making calculations and predictions of energy output more straightforward.

Key features of isothermal steam reservoirs include:
  • Uniform temperature throughout the reservoir
  • Constant pressure, since temperature doesn't vary with depth
  • Through having stable thermal conditions, these reservoirs tend to provide a steady, reliable stream of geothermal energy
In an isothermal steam reservoir, because the temperature and hence pressure remain constant throughout, calculations become easier since there aren't variables such as temperature gradients to account for. This stability facilitates consistent energy extraction, highlighting their importance in geothermal energy projects.

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Most popular questions from this chapter

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 one-dimensional flow through a confined porous aquifer of total thickness \(b\) and crosssectional area \(A\). Suppose the aquifer consists of \(N\) layers, each of thickness \(b_{i}(i=1, \ldots, N)\) and permeability \(k_{i}(i=1, \ldots, N) .\) Determine the total flow rate through the aquifer if all the layers are subjected to the same driving pressure gradient. What is the uniform permeability of an aquifer of thickness \(b\) that delivers the same flow rate as the layered aquifer when the two are subjected to the same pressure gradient?

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

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.

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.
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