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

Carbon steel balls \(\left(\rho=7833 \mathrm{~kg} / \mathrm{m}^{3}, k=54 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right.\), \(c_{p}=0.465 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\), and \(\left.\alpha=1.474 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right) 8 \mathrm{~mm}\) in diameter are annealed by heating them first to \(900^{\circ} \mathrm{C}\) in a furnace and then allowing them to cool slowly to \(100^{\circ} \mathrm{C}\) in ambient air at \(35^{\circ} \mathrm{C}\). If the average heat transfer coefficient is \(75 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine how long the annealing process will take. If 2500 balls are to be annealed per hour, determine the total rate of heat transfer from the balls to the ambient air.

Consider a 1000-W iron whose base plate is made of \(0.5-\mathrm{cm}\)-thick aluminum alloy \(2024-\mathrm{T} 6\left(\rho=2770 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=\right.\) \(\left.875 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, \alpha=7.3 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\right)\). The base plate has a surface area of \(0.03 \mathrm{~m}^{2}\). Initially, the iron is in thermal equilibrium with the ambient air at \(22^{\circ} \mathrm{C}\). Taking the heat transfer coefficient at the surface of the base plate to be \(12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and assuming 85 percent of the heat generated in the resistance wires is transferred to the plate, determine how long it will take for the plate temperature to reach \(140^{\circ} \mathrm{C}\). Is it realistic to assume the plate temperature to be uniform at all times?

Stainless steel ball bearings \(\left(\rho=8085 \mathrm{~kg} / \mathrm{m}^{3}, k=\right.\) \(15.1 \mathrm{~W} / \mathrm{m} \cdot{ }^{\circ} \mathrm{C}, c_{p}=0.480 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\), and \(\left.\alpha=3.91 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) having a diameter of \(1.2 \mathrm{~cm}\) are to be quenched in water. The balls leave the oven at a uniform temperature of \(900^{\circ} \mathrm{C}\) and are exposed to air at \(30^{\circ} \mathrm{C}\) for a while before they are dropped into the water. If the temperature of the balls is not to fall below \(850^{\circ} \mathrm{C}\) prior to quenching and the heat transfer coefficient in the air is \(125 \mathrm{~W} / \mathrm{m}^{2} \cdot{ }^{\circ} \mathrm{C}\), determine how long they can stand in the air before being dropped into the water.

Consider a 7.6-cm-diameter cylindrical lamb meat chunk \(\left(\rho=1030 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=3.49 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}, k=0.456 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right.\), \(\left.\alpha=1.3 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\right)\). Such a meat chunk intially at \(2^{\circ} \mathrm{C}\) is dropped into boiling water at \(95^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(1200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The time it takes for the center temperature of the meat chunk to rise to \(75{ }^{\circ} \mathrm{C}\) is (a) \(136 \mathrm{~min}\) (b) \(21.2 \mathrm{~min}\) (c) \(13.6 \mathrm{~min}\) (d) \(11.0 \mathrm{~min}\) (e) \(8.5 \mathrm{~min}\)

Long cylindrical AISI stainless steel rods \((k=\) \(7.74 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\) and \(\left.\alpha=0.135 \mathrm{ft}^{2} / \mathrm{h}\right)\) of 4 -in-diameter are heat treated by drawing them at a velocity of \(7 \mathrm{ft} / \mathrm{min}\) through a 21 -ft-long oven maintained at \(1700^{\circ} \mathrm{F}\). The heat transfer coefficient in the oven is \(20 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\). If the rods enter the oven at \(70^{\circ} \mathrm{F}\), determine their centerline temperature when they leave. Solve this problem using analytical one-term approximation method (not the Heisler charts).

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