Question

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

Short answer

Temperature at 1 km depth is approximately 374.7 K.

Step-by-step solution

Understand the Problem

We need to find the temperature-depth profile for a layer saturated with water at boiling temperature using the given equations. The final expression involves calculating the temperature at a depth of 1 km considering hydrostatic pressure, density of water, and vapor equations.

Start with Hydrostatic Equation

We are given that the pressure at a depth is given by: \( p = p_0 + \rho_l g y \). This accounts for the increase in pressure due to the depth \( y \) in the water.

Apply Clapeyron Approximation

The slope of the Clapeyron curve approximates to \( \frac{d p}{d T} = \frac{L \rho_v}{T} \). Since \( \rho_v = \frac{p}{R_v T} \), substitute into the Clapeyron equation to get: \( \frac{d p}{d T} = \frac{L p}{R_v T^2} \).

Relate Pressure and Temperature Change

By equating the expressions from the hydrostatic equation and Clapeyron approximation, we find: \( d p = \frac{L p}{R_v T^2} dT \). Since \( p = p_0 + \rho_l g y \), substitute this into the equation and integrate to find the temperature profile.

Integrate to Find Temperature Profile

The integration involves rearranging the equation: \( \frac{dT}{T^2} = \frac{R_v d(\ln(p))}{L} \). Integrate from the surface (\( p = p_0, T = T_{b0} \)) to any depth (\( p, T \)).

Solve the Integral

Perform the integration: \( \int_{T_{b0}}^T \frac{dT}{T^2} = \frac{R_v}{L} \int_{p_0}^p \frac{1}{p} dp \). The left side integrates to \( \frac{1}{T_{b0}} - \frac{1}{T} \) and the right side to \( \frac{R_v}{L} \ln\left( \frac{p}{p_0} \right) \).

Substitute Back to Original Profile

Substitute \( p = p_0 + \rho_l g y \) into the integrated equation to obtain: \( \frac{1}{T_{b0}} - \frac{1}{T} = \frac{R_v}{L} \ln\left(1 + \frac{\rho_l g y}{p_0}\right) \), which is the required temperature-depth profile.

Calculate Temperature at 1 km Depth

Plug in \( y = 1000 \) m, along with other known values \( R_v = 0.462 \), \( L = 2500 \), \( T_{b0} = 373 \), \( \rho_l = 1000 \), \( g = 10 \), and \( p_0 = 10^5 \) into the equation. Compute the temperature \( T \).

Key concepts

Hydrostatic Equation

The hydrostatic equation is a fundamental principle in fluid dynamics that relates the pressure change in a fluid to its depth and density. This equation can be expressed as: \[ p = p_0 + \rho_l g y \] where:

• \( p_0 \) is the pressure at the surface, or atmospheric pressure,
• \( \rho_l \) is the density of the fluid,
• \( g \) is the acceleration due to gravity,
• \( y \) is the depth below the surface.

This relation indicates that as you go deeper in a fluid, the pressure increases due to the weight of the fluid above. It's essential for understanding how pressure varies with depth in contexts like underwater environments.
This equation is crucial for the study of temperature-depth profiles in saturated layers. It provides the basis for calculating pressure changes that influence the boiling temperature of water at different depths.

Clapeyron Equation

The Clapeyron equation describes the relationship between pressure and temperature during phase transitions between two phases, such as between liquid water and steam. The equation is simplified in this case to: \[ \frac{d p}{d T} = \frac{L \rho_v}{T} \]In this equation:

• \( L \) is the latent heat of vaporization,
• \( \rho_v \) is the density of the vapor phase,
• \( T \) is the absolute temperature.

This formulation helps us understand how changes in temperature can lead to changes in pressure, especially during the boiling process. It implies that as temperature changes, so does the pressure required to maintain equilibrium.
The approximation assumes that the density of the vapor is significantly smaller than that of the liquid (\( \rho_l \)). This makes it useful for calculating the temperature profile in a system where water transitions between liquid and gaseous phases.

Boiling Temperature

Boiling temperature is the temperature at which a liquid changes to vapor under a given pressure. At sea level, water boils at \( T_{b0} = 373 \) K, or 100 °C, under an atmospheric pressure of \( p_0 = 10^5 \) Pa. However, boiling temperature can vary with pressure:

• At greater depths, pressure increases, resulting in higher boiling temperatures.
• This relationship is illustrated in the temperature-depth profile equation.

The equation derived from the process: \[ \frac{1}{T_{b0}} - \frac{1}{T} = \frac{R_v}{L} \ln\left(1 + \frac{\rho_l g y}{p_0}\right) \]shows how boiling temperature varies at different depths. This is due to increased pressure from the water above, affecting the energy required for water to change states.
Understanding boiling temperature changes with depth is essential in fields ranging from engineering geothermal systems to understanding natural aquatic environments.

Water Vapor Gas Constant

The water vapor gas constant (\( R_v \)) is a specific value used in the equation of state for gases like steam. For water vapor, \( R_v = 0.462 \text{ kJ} \text{ kg}^{-1} \text{ K}^{-1} \). This constant helps relate pressure, volume, and temperature for the gaseous phase of water.
The ideal gas law, \( \rho_v = \frac{p}{R_v T} \), is applicable, where:

• \( \rho_v \) is the vapor density,
• \( p \) is the pressure,
• \( T \) is the absolute temperature.

By substituting this into other equations, we can simplify and solve problems regarding phase changes. The water vapor gas constant plays a crucial role in determining how the gaseous phase interacts with its liquid counterpart and predicting behaviors such as boiling and condensation in various environments.