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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.

Short Answer

Expert verified
Pressure increases with depth in the liquid layer but stays constant in the isothermal steam reservoir.

Step by step solution

01

Understand the Hydrostatic Law

The hydrostatic law states that the pressure at each depth in a fluid at rest is given by the equation \( P = P_0 + \rho gh \), where \( P \) is the pressure at depth, \( P_0 \) is the atmospheric pressure, \( \rho \) is the density of the fluid, \( g \) is the acceleration due to gravity, and \( h \) is the depth below the surface.
02

Pressure in the Liquid Layer

In the liquid layer, the pressure increases linearly with depth due to the weight of the overlying liquid. Assuming the density \( \rho_l \) for the liquid, the pressure at the bottom of this layer (just above the steam reservoir), \( P_{liquid} \), is given by the hydrostatic equation: \[ P_{liquid} = P_0 + \rho_l g h_l \] where \( h_l = 400 \ \text{meters} \).
03

Consider the Steaming Reservoir

For the steam reservoir, since it is isothermal and vapor-dominated, the pressure remains constant with depth. This is due to the fact that vapor equality implies the pressure is primarily dependent on temperature and not depth. Thus, \( P_{steam} = P_{liquid} \).
04

Combine to Express Overall Pressure Profile

For a point below the liquid layer, the pressure remains as determined at the boundary since it's an isothermal vapor system. Therefore, for a depth \( z \) within the liquid layer (up to 400 m), \( P(z) = P_0 + \rho_l g z \), and for a depth beyond 400 m, \( P(z) = P_{liquid} \), where \( P_{liquid} \) is the constant pressure in the steam reservoir.

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Key Concepts

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

Hydrostatic Law
The hydrostatic law is a fundamental principle that describes how pressure increases within a fluid at rest due to the weight of the fluid above. This law can be expressed using the formula: \[ P = P_0 + \rho gh \]where:
  • \(P\) is the pressure at a given depth.
  • \(P_0\) is the atmospheric pressure at the surface.
  • \(\rho\) is the density of the fluid.
  • \(g\) is the acceleration due to gravity (approximately \(9.81 \, \text{m/s}^2\) on Earth).
  • \(h\) is the depth below the fluid's surface.
This equation indicates that pressure within the fluid increases linearly with depth. The deeper you go, the greater the pressure due to the weight of the fluid column above. This principle is vital for calculating how pressure changes within the geothermal systems.
Vapor-Dominated Geothermal System
A vapor-dominated geothermal system is a type of geothermal reservoir where steam is the primary phase that influences the pressure. In these systems, the steam's presence is so dominant that it dictates the pressure behavior, unlike in liquid-dominated systems where both liquid and vapor coexist. These systems are important because:
  • Steam's ability to rise allows for efficient energy extraction.
  • Temperature and pressure conditions within the vapor region are uniform, simplifying calculations.
  • They typically exhibit superheated conditions, enhancing energy output.
In our exercise, the steam reservoir is described as being vapor-dominated, meaning the pressure is assumed constant with depth, primarily controlled by the steam at the boiling point.
Isothermal Steam Reservoir
An isothermal steam reservoir maintains a constant temperature throughout the system. In geothermal contexts, this means that:
  • Temperature does not vary with depth, unlike in liquid-dominated reservoirs.
  • The pressure in the steam remains constant due to its direct dependence on temperature rather than depth.
  • Energy extraction can be more predictable due to stable temperature conditions.
In the given problem, the isothermal condition implies that below the liquid layer, the temperature remains consistent, thus keeping the pressure stable based on the vapor's saturated conditions.
Pressure Function of Depth
Understanding pressure as a function of depth is crucial in hydrostatic and geothermal calculations. In this scenario:
  • In the upper liquid layer (400 m thick), pressure increases steadily with depth according to the hydrostatic law: \[P(z) = P_0 + \rho_l g z\]
  • Beneath this layer, within the steam reservoir, pressure remains constant beyond the depth of 400 m as \[P(z) = P_{liquid}\].
  • This constancy reflects the equalizing effect of steam in an isothermal environment.
This dual behavior—a linear increase in pressure in the liquid followed by a constant level in the vapor—allows for a clear depiction of pressure throughout this geothermal system. Students should understand this profile's implications for geothermal energy extraction.

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

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.

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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