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

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 \(r=b\).

Short Answer

Expert verified
The central temperature is: \( T(0) = T_0 + \frac{H(a^2 - b^2)}{6k} - \Gamma(b-a) \).

Step by step solution

01

Understand the Problem

The problem involves two regions of heat transfer within the planet: convection in the core and conduction in the lithosphere. We need to derive the temperature at the center considering these heat transfer modes. Given conditions such as surface temperature and continuity at the boundary, we will establish the necessary equations.
02

Apply Heat Generation Equation to the Lithosphere

For the lithosphere, the equation of heat conduction with internal generation in spherical coordinates is \\[ \frac{1}{r^2} \frac{d}{dr} \left( r^2 \frac{dT}{dr} \right) + \frac{H}{k} = 0.\] Integrating this equation helps find the temperature distribution over the lithosphere. Here, \(k\) is the thermal conductivity.
03

Solve the Integrated Equation

Integrate the conduction equation once using the boundary condition at the surface (\(r = a\)), where temperature is \(T_0\), to find the general solution for \(T(r)\): \\[ r^2 \frac{dT}{dr} = -\frac{Hr^3}{3k} + C_1, \]where \(C_1\) is an integration constant.
04

Integrate Again and Apply Boundary Conditions

Integrate the result again to obtain: \\[ T(r) = -\frac{H r^2}{6k} + \frac{C_1}{r} + C_2. \]Apply the boundary condition at the outer surface (\(r = a\)), where \(T = T_0\):\\[ T_0 = -\frac{H a^2}{6k} + \frac{C_1}{a} + C_2. \]
05

Find Constants Using Continuity at Convection Boundary

At \(r = b\), convective heat transfer is maintained with \(\frac{dT}{dr} = -\Gamma\): \\[ -\frac{Hb}{3k} + \frac{C_1}{b^2} = -\Gamma. \]This allows solving for \(C_1\) and \(C_2\) using connection with temperatures from convection.
06

Formulate Expression for Temperature at the Center

Finally, from \(T(r)\), solve for \(T(0)\). Use the boundary condition at \(r=b\) and substitute calculated constants into the integrated equation for the center temperature. \Simplify this to derive explicit expression: \\[ T(0) = T_0 + \frac{H(a^2 - b^2)}{6k} - \Gamma(b-a). \]
07

Final Step: Verify and Confirm Derivation

Check the expression for correctness by ensuring 1. The use of boundary conditions is consistent. 2. The final result matches physical expectations for cases with zero heat generation or zero adiabatic gradient.

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.

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.

Heat Conduction
Heat conduction is a form of heat transfer where heat flows through a material without the material itself moving. It occurs due to the thermal motion of molecules within the material. In the context of a planet, the lithosphere (outer shell) conducts heat produced internally towards the surface.

The process can be modeled by a differential equation that describes how temperature varies with distance inside a spherical object, like a planet. This equation accounts for the heat flow between different layers of the lithosphere. It is given by:
  • \( \frac{1}{r^2} \frac{d}{dr} \left( r^2 \frac{dT}{dr} \right) = 0 \)
where \(T\) is the temperature, and \(r\) is the distance from the center of the planet. This equation suggests that the heat spread depends on the changing radius and is subject to specific boundary conditions such as continuity of temperature and heat flux at certain radii in the planet.

In simpler terms, think of conduction as the planet trying to equalize temperatures within its layers, but the rate and efficiency of heat flow vary with depth due to factors like thickness and material properties.
Internal Heat Generation
Internal heat generation refers to the heat produced within the planet. It can arise from various sources, such as radioactive decay or gravitational energy release. This generation of heat is crucial for understanding the planet's temperature distribution.

In our problem, the internal heat generation rate is denoted by \(H\). This value significantly affects how the temperature changes through the lithosphere. The higher the internal heat generation, the warmer the planet will be, and it must be conducted or convected towards the surface.

The heat generation affects the planet's temperature profile by contributing an additional source term in the heat conduction equation:
  • \( \frac{1}{r^2} \frac{d}{dr} \left( r^2 \frac{dT}{dr} \right) + \frac{H}{k} = 0 \)
Here, \(k\) indicates thermal conductivity, which means internal heat generation is balanced against how quickly or slowly a material conducts heat. This balance is key to predicting planetary temperature changes.
Thermal Conductivity
Thermal conductivity is a material property that describes how well a material can conduct heat. In simple terms, it measures the ease with which heat can travel through a substance.

In our spherical planet model, thermal conductivity (denoted as \(k\)) plays a crucial role in determining how efficiently the heat generated internally can be conducted to the planet's surface. A high thermal conductivity means the planet can easily conduct heat, resulting in a more uniform temperature distribution. Conversely, low thermal conductivity means that the planet might have significant temperature differences between its core and surface.

Knowing the thermal conductivity helps solve the heat conduction equation, as it appears in the expression for internal heat generation:
  • \( \frac{1}{r^2} \frac{d}{dr} \left( r^2 \frac{dT}{dr} \right) + \frac{H}{k} = 0 \)
Here, \(k\) acts as a divisor in the equation, meaning it limits or facilitates the temperature spread triggered by internal heat generation, ultimately impacting the planet's thermal evolution.
Adiabatic Gradient
The adiabatic gradient represents the temperature change rate with respect to altitude or depth, assuming no heat is exchanged with the surroundings. In our planetary model, this concept helps describe the temperature change in the convecting core region in terms of pressure rather than direct heat exchange.

The adiabatic gradient is significant because it tells us how temperature naturally changes without external influences, driven by pressure differences as you move through layers of different depths in a planet. At the boundary layer where convection gives way to conduction, the adiabatic gradient provides a condition for continuity.

In terms of mathematics, the adiabatic gradient \(-\Gamma\) is used when setting boundary conditions at the depth \(r=b\), where conduction in the lithosphere abuts convection in the core:
  • \( \frac{dT}{dr} = -\Gamma \)
This condition ensures temperatures remain continuous across the boundary, aligning the conductive and convective regimes smoothly. It notably influences how temperature is distributed at different depths within the planet, creating a realistic model of the planet’s thermal landscape.

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

Assume that the continental lithosphere satisfies the half-space cooling model. If a continental region has an age of \(1.5 \times 10^{9}\) years, how much subsidence would have been expected to occur in the last 300 Ma? Take \(\rho_{m}=3300 \mathrm{~kg} \mathrm{~m}^{-3}, \kappa=1 \mathrm{~mm}^{2} \mathrm{~s}^{-1}\), \(T_{m}-T_{0}=1300 \mathrm{~K},\) and \(\alpha_{v}=3 \times 10^{-5} \mathrm{~K}^{-1}\). Assume that the subsiding lithosphere is being covered to sea level with sediments of density \(\rho_{s}=2500 \mathrm{~kg} \mathrm{~m}^{-3}\).

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 \(L=320 \mathrm{~kJ} \mathrm{~kg}^{-1}\), \(k=2 \mathrm{~J} \mathrm{~m}^{-1} \mathrm{~s}^{-1} \mathrm{~K}^{-1}, c=4 \mathrm{~kJ} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}, \rho=\) \(1000 \mathrm{~kg} \mathrm{~m}^{-3}\).

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 \(q_{m}=28 \mathrm{~mW} \mathrm{~m}^{-2}\).

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 \(0.2 \mathrm{~mm}^{2} \mathrm{~s}^{-1}\).

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.

See all solutions

Recommended explanations on Geography Textbooks

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