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Suppose that upon entering the Earth's atmosphere, the surface of a meteorite has been heated to the melting point and the molten material is carried away by the flow. It is of interest to calculate the rate at which melting removes material from the meteorite. For this purpose, consider the following problem. The surface of a semi-infinite half-space moves downward into the half-space with constant velocity \(V\), as indicated in Figure \(4.39 .\) The surface is always at the melting temperature \(T_{m},\) and melted material above the instantaneous surface is removed from the problem. Assume that the surface of the half-space is melted by a constant heat flux \(q_{m}\) into the half-space from above the surface. Assume also that far from the melting surface the temperature is \(T_{0}\); that is, \(T \rightarrow T_{0}\) as \(\zeta \rightarrow \infty\). Find the temperature distribution in the half-space as a function of time \(T(\zeta, t),\) and determine \(V\) in terms of \(q_{m}\) and the thermodynamic properties of the rock. Account for the latent heat \(L\) required to melt the material.

Short Answer

Expert verified
The temperature distribution is found using \( T(\zeta, t) = T_0 + (T_m - T_0) \text{ erf }(\frac{\zeta}{\sqrt{4\alpha t}}) \), and velocity \( V \) is \( \frac{q_m}{\rho L} \) in a steady state.

Step by step solution

01

Set up the Problem

We need to find the temperature distribution, \( T(\zeta, t) \), in the semi-infinite body, and the velocity \( V \) at which the surface moves. The key is to consider the heat conducted into the semi-infinite body and the heat required to melt the material. Assume the heat conduction follows one-dimensional heat conduction principles with constant properties.
02

Use the Heat Equation

For a semi-infinite body, the governing one-dimensional heat conduction equation is \(\frac{\partial T}{\partial t} = \alpha \frac{\partial^2 T}{\partial \zeta^2} \), where \( \alpha \) is the thermal diffusivity \( \alpha = \frac{k}{\rho c_p} \). This equation must be solved with the boundary and initial conditions.
03

Define the Boundary and Initial Conditions

1. At \( \zeta = 0 \), the temperature is constantly at the melting temperature: \( T(0, t) = T_m \).2. Far from the surface, the temperature remains at the initial temperature: \( T(\zeta, t) \to T_0 \) as \( \zeta \to \infty \).3. Initially, the entire material is at uniform temperature \( T_0 \).
04

Apply the Stefan Condition at the Melting Interface

The Stefan condition gives the relationship between the velocity \( V \) and the heat flux \( q_m \), accounting for latent heat \( L \):\[ q_m = \rho V L + k \left. \frac{\partial T}{\partial \zeta} \right|_{\zeta=0} \]Where \( k \) is the thermal conductivity. This balances the heat conducted into the semi-infinite domain and the heat used for melting.
05

Solve for the Temperature Distribution

Knowing that the problem satisfies a semi-infinite domain under initial conditions, the solution can be expressed using the similarity variable \( \eta = \frac{\zeta}{\sqrt{4\alpha t}} \). The temperature distribution thus follows:\[ T(\zeta, t) = T_0 + (T_m - T_0) \text{ erf }\left(\eta\right) \]
06

Calculate Velocity from Heat Flux

Substitute the boundary condition into the Stefan condition to solve for \( V \). Integrate the error function to get the gradient at \( \zeta=0 \):\[ V = \frac{q_m - k\frac{\partial T}{\partial \zeta}|_{\zeta=0}}{\rho L} = \frac{q_m - k\frac{T_m - T_0}{\sqrt{\pi \alpha t}}}{\rho L} \] As \( t \to \infty \), the velocity \( V \) reaches a steady state under constant \( q_m \).

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Key Concepts

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

Heat Conduction
Heat conduction is a process where heat energy transfers from the warmer part of a material to the cooler part without any movement of the material itself. In the case of meteorite melting, this concept is crucial. Here, heat is conducted from the surface of the meteorite, which is at the melting temperature, into the cooler interior of the meteorite.
This scenario can be modeled using the heat conduction equation:
  • \( \frac{\partial T}{\partial t} = \alpha \frac{\partial^2 T}{\partial \zeta^2} \)
where \( T \) is the temperature, \( t \) is time, \( \zeta \) is the depth into the meteorite, and \( \alpha \) is the thermal diffusivity.
Heat conduction in this context needs a steady flow to melt the meteorite effectively, as the heat moves from the surface deeper into the meteorite, changing the temperature distribution over time.
Latent Heat
Latent heat refers to the energy required for a substance to change its phase without changing its temperature. In the context of a melting meteorite, it’s the amount of energy needed to turn the solid rock into a liquid form.
In our problem, this comes into play when the solid rock melts due to the heat flux \( q_m \). The energy needed to turn the solid meteorite material into liquid without increasing its temperature is called latent heat \( L \).
Latent heat is vital because it determines how much energy is consumed in the phase change. This is crucial for understanding the meteorite's melting rate, as this energy subtraction from the total heat available affects how much further heat is conducted into the semi-infinite body.
Thermal Diffusivity
Thermal diffusivity \( \alpha \) is a material-specific property that measures how quickly heat spreads through the material. It's calculated as:
  • \( \alpha = \frac{k}{\rho c_p} \)
where \( k \) is the thermal conductivity, \( \rho \) is the density, and \( c_p \) is the specific heat capacity of the meteorite.
This property combines the effects of thermal conductivity and heat capacity, indicating how effectively a material can conduct thermal energy relative to storing it. Higher thermal diffusivity means the material can conduct the heat quickly through itself, which implies that the meteorite surface will cool rapidly as the heat is distributed through the rock body.
The understanding of thermal diffusivity helps predict how the interior of the meteorite will warm up when exposed to the same heat flux continuously.
Stefan Condition
The Stefan condition is a principle used to relate phase changes involving latent heat with heat transfer. It gives us a way to calculate the velocity \( V \) of the moving interface where melting occurs. In our scenario, this interface is the surface of the meteorite.
The Stefan condition is expressed by the equation:
  • \( q_m = \rho V L + k \left. \frac{\partial T}{\partial \zeta} \right|_{\zeta=0} \)
This equation states that the heat flux \( q_m \) used for melting is split between the latent heat used to melt the material and the conductive heat past the melting interface.
Using this condition involves balancing the energy to find \( V \), which indicates how fast the surface melts into the meteorite. Without considering this, one might miscalculate how fast the meteorite melts or fails to explain how certain properties affect the rate of melting.

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Most popular questions from this chapter

Derive an expression for the temperature at the center of a planet of radius \(a\) with uniform density \(\rho\) and internal heat generation \(H .\) Heat transfer in the planet is by conduction only in the lithosphere, which extends from \(r=b\) to \(r=a\). For \(0 \leq r \leq b\) heat transfer is by convection, which maintains the temperature gradient \(d T / d r\) constant at the adiabatic value \(-\Gamma\). The surface temperature is \(T_{0}\). To solve for \(T(r),\) you need to assume that \(T\) and the heat flux are continuous at \(r=b\).

Assume that the continental crust and lithosphere have been stretched by a factor \(\alpha=2\). Taking \(h_{c c}=35 \mathrm{~km}, y_{L 0}=125 \mathrm{~km}, \rho_{m}=3300\) \(\begin{array}{lllll}\mathrm{kg} \mathrm{m}^{-3}, & \rho_{c c}=2750 & \mathrm{~kg} \mathrm{~m}^{-3}, & \rho_{s}=2550 & \mathrm{~kg} \mathrm{~m}^{-3}\end{array}\) \(\alpha_{v}=3 \times 10^{-5} \mathrm{~K}^{-1}\), and \(T_{1}-T_{0}=1300 \mathrm{~K},\) deter- mine the depth of the sedimentary basin. What is the depth of the sedimentary basin when the thermal lithosphere has thickened to its original thickness?

The exponential depth dependence of heat production is preferred because it is self-preserving upon erosion. However, many alternative models can be prescribed. Consider a two-layer model with \(H=\) \(H_{1}\) and \(k=k_{1}\) for \(0 \leq y \leq h_{1},\) and \(H=H_{2}\) and \(k=k_{2}\) for \(h_{1} \leq y \leq h_{2} .\) For \(y>h_{2}, H=0\) and the upward heat flux is \(q_{m} .\) Determine the surface heat flow and temperature at \(y=h_{2}\) for \(\rho_{1}=2600 \mathrm{~kg} \mathrm{~m}^{-3}\), \(\rho_{2}=3000 \mathrm{~kg} \mathrm{~m}^{-3}, k_{1}=k_{2}=2.4 \mathrm{~W} \mathrm{~m}^{-1} \mathrm{~K}^{-1}\) \(h_{1}=8 \mathrm{~km}, h_{2}=40 \mathrm{~km}, \rho_{1} H_{1}=2 \mu W \mathrm{~m}^{-3}, \rho_{2} H_{2}=\) \(0.36 \mu W \mathrm{~m}^{-3}, T_{0}=0^{\circ} \mathrm{C},\) and \(q_{m}=28 \mathrm{~mW} \mathrm{~m}^{-2}\).

A body of water at \(0^{\circ} \mathrm{C}\) is subjected to a constant surface temperature of \(-10^{\circ} \mathrm{C}\) for 10 days. How thick is the surface layer of ice? Use \(L=320 \mathrm{~kJ} \mathrm{~kg}^{-1}\), \(k=2 \mathrm{~J} \mathrm{~m}^{-1} \mathrm{~s}^{-1} \mathrm{~K}^{-1}, c=4 \mathrm{~kJ} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}, \rho=\) \(1000 \mathrm{~kg} \mathrm{~m}^{-3}\).

Estimate the effects of variations in bottom water temperature on measurements of oceanic heat flow by using the model of a semi-infinite half-space subjected to periodic surface temperature fluctuations. Such water temperature variations at a specific location on the ocean floor can be due to, for example, the transport of water with variable temperature past the site by deep ocean currents. Find the amplitude of water temperature variations that cause surface heat flux variations of \(40 \mathrm{~mW} \mathrm{~m}^{-2}\) above and below the mean on a time scale of 1 day. Assume that the thermal conductivity of sediments is \(0.8 \mathrm{~W} \mathrm{~m}^{-1} \mathrm{~K}^{-1}\) and the sediment thermal diffusivity is \(0.2 \mathrm{~mm}^{2} \mathrm{~s}^{-1}\).

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