/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 96 The soil temperature in the uppe... [FREE SOLUTION] | 91Ó°ÊÓ

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The soil temperature in the upper layers of the earth varies with the variations in the atmospheric conditions. Before a cold front moves in, the earth at a location is initially at a uniform temperature of \(10^{\circ} \mathrm{C}\). Then the area is subjected to a temperature of \(-10^{\circ} \mathrm{C}\) and high winds that resulted in a convection heat transfer coefficient of \(40 \mathrm{~W} / \mathrm{m}^{2}\). \(\mathrm{K}\) on the earth's surface for a period of \(10 \mathrm{~h}\). Taking the properties of the soil at that location to be \(k=0.9 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\alpha=1.6 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\), determine the soil temperature at distances \(0,10,20\), and \(50 \mathrm{~cm}\) from the earth's surface at the end of this \(10-\mathrm{h}\) period.

Short Answer

Expert verified
Answer: The soil temperatures after 10 hours are approximately: - At 0 cm: \(-10^{\circ}\mathrm{C}\) - At 10 cm: \(9.17^{\circ}\mathrm{C}\) - At 20 cm: \(9.34^{\circ}\mathrm{C}\) - At 50 cm: \(9.75^{\circ}\mathrm{C}\)

Step by step solution

01

Convert Time to Seconds

First, we need to convert the time from 10 hours to seconds: \(t = 10 \cdot 3600 \text{ seconds}\)
02

Calculate Temperature at Depth 0 cm (Surface)

At the surface, \(x = 0\): \(T(0,t) = 10 + (-10 - 10)[1 - \operatorname{erf}(0)] = -10^{\circ}\mathrm{C}\)
03

Calculate Temperature at Depth 10 cm

At depth 10 cm, \(x=0.1 \mathrm{~m}\): \(T(0.1,t) = 10 + (-10 - 10)[1 - \operatorname{erf}(\frac{0.1}{2\sqrt{1.6 \times 10^{-5} \cdot 36000}})] \approx 9.17^{\circ}\mathrm{C}\)
04

Calculate Temperature at Depth 20 cm

At depth 20 cm, \(x=0.2 \mathrm{~m}\): \(T(0.2,t) = 10 + (-10 - 10)[1 - \operatorname{erf}(\frac{0.2}{2\sqrt{1.6 \times 10^{-5} \cdot 36000}})] \approx 9.34^{\circ}\mathrm{C}\)
05

Calculate Temperature at Depth 50 cm

At depth 50 cm, \(x=0.5 \mathrm{~m}\): \(T(0.5,t) = 10 + (-10 - 10)[1 - \operatorname{erf}(\frac{0.5}{2\sqrt{1.6 \times 10^{-5} \cdot 36000}})] \approx 9.75^{\circ}\mathrm{C}\) The soil temperature at distances 0 cm, 10 cm, 20 cm, and 50 cm from the earth's surface after 10 hours are approximately: - At 0 cm: \(-10^{\circ}\mathrm{C}\) - At 10 cm: \(9.17^{\circ}\mathrm{C}\) - At 20 cm: \(9.34^{\circ}\mathrm{C}\) - At 50 cm: \(9.75^{\circ}\mathrm{C}\)

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Key Concepts

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

Temperature Profile in Soil
Understanding how the temperature changes within soil is essential for various environmental and engineering applications, from agriculture to geotechnical engineering. Temperature within the soil doesn't remain constant but instead varies with depth and time, influenced by surface conditions and the soil's thermal properties. This phenomenon is depicted by what we call the temperature profile in soil.

In the given exercise, we examine how the upper layers of the earth respond to a sudden drop in air temperature. Initially, the soil temperature is uniform, but exposure to lower temperatures causes heat to transfer from the soil to the air, thereby creating a temperature gradient within the soil. The soil surface's temperature changes more rapidly, while deeper layers take longer to respond to this change, forming a temperature profile that shows temperatures at various depths after a certain period. When solving the problem, we calculate the temperatures at different depths using thermal properties and time, yielding a snapshot of this temperature profile after 10 hours of exposure to colder air.
Convection Heat Transfer Coefficient
The convection heat transfer coefficient is a measure of the heat transfer rate between a solid surface and a fluid (like air or water) in motion. It quantifies how effectively heat is being carried away from or towards the surface by the fluid's motion.

In our scenario, high winds significantly impact heat transfer at the earth's surface, characterized by a given convection heat transfer coefficient of 40 W/m²·K. This value tells us how much heat per unit area is transferred for every degree of temperature difference between the soil surface and the air. A higher coefficient implies faster heat transfer. This is a crucial factor when calculating the surface temperature at time t, as it affects how quickly the soil loses heat to the atmosphere.
Thermal Properties of Soil
The behavior of soil in response to temperature changes is determined by its thermal properties, which include thermal conductivity and thermal diffusivity.

Thermal Conductivity (k)

Thermal conductivity (k) indicates how well the soil can conduct heat. It's given in the units W/m·K. In our case, the soil's thermal conductivity is 0.9 W/m·K, implying that the soil does not transfer heat very efficiently.

Thermal Diffusivity (α)

Thermal diffusivity (α) measures the rate at which temperatures can equalize within a material due to thermal motion. It's expressed in m²/s. For the soil in this problem, α is 1.6 x 10^-5 m²/s, a property essential for determining how fast thermal waves propagate into the soil. These properties are required for analyzing the soil's response to thermal events and for predicting temperature variations at different depths.
Transient Heat Conduction
When a temperature change occurs, the process by which heat moves through a material like soil is known as transient heat conduction. Unlike steady-state conduction, transient heat conduction involves time-dependent temperature variations, meaning heat transfer changes with time.

The temperature distribution during such a process doesn't remain constant but changes until a new steady state (if ever) is achieved. The exercise problem focuses exactly on this type of heat conduction, considering how the soil temperature evolves over a period of 10 hours. By employing the thermal diffusivity and thermal conductivity of the soil, along with the surface boundary conditions provided by the convection heat transfer coefficient, we determine the soil temperature at various depths using mathematical solutions to the transient heat conduction equation, like the error function (erf) in the context of this exercise.

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Most popular questions from this chapter

What is an infinitely long cylinder? When is it proper to treat an actual cylinder as being infinitely long, and when is it not? For example, is it proper to use this model when finding the temperatures near the bottom or top surfaces of a cylinder? Explain.

An electronic device dissipating \(20 \mathrm{~W}\) has a mass of \(20 \mathrm{~g}\), a specific heat of \(850 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and a surface area of \(4 \mathrm{~cm}^{2}\). The device is lightly used, and it is on for \(5 \mathrm{~min}\) and then off for several hours, during which it cools to the ambient temperature of \(25^{\circ} \mathrm{C}\). Taking the heat transfer coefficient to be \(12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the temperature of the device at the end of the 5 -min operating period. What would your answer be if the device were attached to an aluminum heat sink having a mass of \(200 \mathrm{~g}\) and a surface area of \(80 \mathrm{~cm}^{2}\) ? Assume the device and the heat sink to be nearly isothermal.

Refractory bricks are used as linings for furnaces, and they generally have low thermal conductivity to minimize heat loss through the furnace walls. Consider a thick furnace wall lining with refractory bricks \(\left(k=1.0 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right.\) and \(\left.\alpha=5.08 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\right)\), where initially the wall has a uniform temperature of \(15^{\circ} \mathrm{C}\). If the wall surface is subjected to uniform heat flux of \(20 \mathrm{~kW} / \mathrm{m}^{2}\), determine the temperature at the depth of \(10 \mathrm{~cm}\) from the surface after an hour of heating time.

Under what conditions can a plane wall be treated as a semi-infinite medium?

Obtain relations for the characteristic lengths of a large plane wall of thickness \(2 L\), a very long cylinder of radius \(r_{o}\), and a sphere of radius \(r_{o}\).

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