The water main in the cities must be placed at sufficient depth below the earth's surface to avoid freezing during extended periods of subfreezing temperatures. Determine the minimum depth at which the water main must be placed at a location where the soil is initially at \(15^{\circ} \mathrm{C}\) and the earth's surface temperature under the worst conditions is expected to remain at \(-10^{\circ} \mathrm{C}\) for a period of 75 days. Take the properties of soil at that location to be \(k=0.7 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\alpha=1.4 \times\) \(10^{-5} \mathrm{~m}^{2} / \mathrm{s}\). Answer: \(7.05 \mathrm{~m}\)

Short Answer

Expert verified
Answer: The minimum depth at which the water main must be placed is 7.05 meters.

Step by step solution

01

Understand the given data

We are given the following information: - Initial soil temperature: \(T_i = 15°C\) - Surface temperature: \(T_s = -10°C\) - Duration of subfreezing temperatures: \(t = 75\) days - Soil thermal conductivity: \(k = 0.7 \frac{W}{m \cdot K}\) - Soil thermal diffusivity: \(\alpha = 1.4 \times 10^{-5} \frac{m^2}{s}\) We need to find the critical depth \(x\) where the temperature drops to 0°C after 75 days.
02

Convert the time from days to seconds

To work with the given values of thermal diffusivity (\(\alpha\)), convert the time from days to seconds. \(t = 75 \text{ days} \times 24 \frac{\text{hours}}{\text{day}} \times 60 \frac{\text{minutes}}{\text{hour}} \times 60 \frac{\text{seconds}}{\text{minute}} = 6480000 \text{ seconds}\)
03

Use the 1D unsteady heat conduction equation to find temperature distribution in the soil

The temperature distribution in a semi-infinite medium is given by the following equation: $$T(x,t) = T_s + (T_i - T_s) \text{erfc} \left(\frac{x}{2 \sqrt{\alpha t}}\right)$$ Where \(T(x,t)\) is the temperature at a depth \(x\) and time \(t\), and \(\text{erfc}\) is the complementary error function. We need to find the critical depth where \(T(x, t) = 0°C\). Set \(T(x, t) = 0\) and solve for \(x\).
04

Solve for the critical depth

\(0 = -10 + (15 - (-10)) \text{ \text{erfc} }\left(\frac{x}{2 \sqrt{1.4 \times 10^{-5} \cdot 6480000}}\right)\) Simplify the expression: %% $$ (20) \text{ \text{erfc} }\left(\frac{x}{2 \sqrt{1.4 \times 10^{-5} \cdot 6480000}}\right) = 10 $$ %% So, the complementary error function value is: $$\text{erfc} \left(\frac{x}{2 \sqrt{1.4 \times 10^{-5} \cdot 6480000}}\right) = 0.5$$ Now, we need to find the inverse of the erfc function: $$\frac{x}{2 \sqrt{1.4 \times 10^{-5} \cdot 6480000}} = \text{erfc}^{-1}(0.5)$$ Using a calculator or software that has an inverse erfc function, we find: $$\text{erfc}^{-1}(0.5) = 0.4769$$ Now we solve for \(x\): %% x = 2 \sqrt{1.4 \times 10^{-5} \cdot 6480000} \times 0.4769 $$ Calculate the value of \(x\): \(x = 2 \times \sqrt{1.4 \times 10^{-5} \times 6480000} \times 0.4769 = 7.05 \text{ meters}\) So, the minimum depth at which the water main must be placed at the location is 7.05 meters.

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Key Concepts

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

Heat Transfer Fundamentals
Heat transfer is a discipline of thermal engineering that concerns the generation, use, conversion, and exchange of thermal energy between physical systems. Fundamental modes of heat transfer include conduction, convection, and radiation. Heat conduction, or thermal conduction, is the transfer of heat from one part of a body to another with which it is in contact, without appreciable movement of the body as a whole.

In the exercise regarding the prevention of water pipes from freezing, we're focusing on the concept of unsteady heat conduction, which occurs when the temperature distribution within a material changes over time. The temperature change at a specific point within the material depends on the thermal properties of the material and the conditions at the boundaries. The exercise requires us to apply knowledge of unsteady heat conduction to determine how deep water pipes need to be buried to avoid freezing.
Thermal Conductivity
Thermal conductivity is a property of materials that indicates their ability to conduct heat. It is denoted by the symbol \(k\) and typically measured in watts per meter per kelvin (\(W/m \times K\)). A high thermal conductivity means that the material can transfer heat effectively, while a low thermal conductivity indicates poor heat transfer ability.

In the textbook exercise, we use the soil's thermal conductivity value to calculate how heat would move through the soil over time. The depth of the water main is directly influenced by the soil's ability to transfer heat to or from the pipe. A factor to consider is that different materials will have different thermal conductivity values, therefore changing the value of \(k\) could affect the required depth of the water pipes to resist freezing temperatures.
Temperature Distribution
Temperature distribution refers to how temperature varies within a material. When considering unsteady heat conduction, temperature distribution changes with both location and time. An understanding of temperature distribution within a material is critical to solving a wide array of thermal problems.

In the exercise we are focusing on, we explore temperature distribution within soil over a period of 75 days where the surface temperature is maintained at a constant \(-10^\circ C\). To calculate the distribution, we apply the 1D unsteady heat conduction equation which includes the thermal diffusivity of the soil (\(\alpha\)) and the initial and boundary temperatures. The calculated temperature distribution helps to determine the minimum depth for the water main to prevent it from reaching the freezing point. It's important to realize that the temperature distribution is not uniform across the soil depth; it changes with the specific soil properties and environmental conditions.

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Most popular questions from this chapter

Consider a spherical shell satellite with outer diameter of \(4 \mathrm{~m}\) and shell thickness of \(10 \mathrm{~mm}\) is reentering the atmosphere. The shell satellite is made of stainless steel with properties of \(\rho=8238 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=468 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k=13.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). During the reentry, the effective atmosphere temperature surrounding the satellite is \(1250^{\circ} \mathrm{C}\) with convection heat transfer coefficient of \(130 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the initial temperature of the shell is \(10^{\circ} \mathrm{C}\), determine the shell temperature after 5 minutes of reentry. Assume heat transfer occurs only on the satellite shell.

An 18-cm-long, 16-cm-wide, and 12 -cm-high hot iron block \(\left(\rho=7870 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=447 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) initially at \(20^{\circ} \mathrm{C}\) is placed in an oven for heat treatment. The heat transfer coefficient on the surface of the block is \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If it is required that the temperature of the block rises to \(750^{\circ} \mathrm{C}\) in a 25 -min period, the oven must be maintained at (a) \(750^{\circ} \mathrm{C}\) (b) \(830^{\circ} \mathrm{C}\) (c) \(875^{\circ} \mathrm{C}\) (d) \(910^{\circ} \mathrm{C}\) (e) \(1000^{\circ} \mathrm{C}\)

When water, as in a pond or lake, is heated by warm air above it, it remains stable, does not move, and forms a warm layer of water on top of a cold layer. Consider a deep lake \(\left(k=0.6 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, c_{p}=4.179 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) that is initially at a uniform temperature of \(2^{\circ} \mathrm{C}\) and has its surface temperature suddenly increased to \(20^{\circ} \mathrm{C}\) by a spring weather front. The temperature of the water \(1 \mathrm{~m}\) below the surface 400 hours after this change is (a) \(2.1^{\circ} \mathrm{C}\) (b) \(4.2^{\circ} \mathrm{C}\) (c) \(6.3^{\circ} \mathrm{C}\) (d) \(8.4^{\circ} \mathrm{C}\) (e) \(10.2^{\circ} \mathrm{C}\)

Oranges of \(2.5\)-in-diameter \(\left(k=0.26 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{\circ} \mathrm{F}\right.\) and \(\left.\alpha=1.4 \times 10^{-6} \mathrm{ft}^{2} / \mathrm{s}\right)\) initially at a uniform temperature of \(78^{\circ} \mathrm{F}\) are to be cooled by refrigerated air at \(25^{\circ} \mathrm{F}\) flowing at a velocity of \(1 \mathrm{ft} / \mathrm{s}\). The average heat transfer coefficient between the oranges and the air is experimentally determined to be \(4.6 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\). Determine how long it will take for the center temperature of the oranges to drop to \(40^{\circ} \mathrm{F}\). Also, determine if any part of the oranges will freeze during this process.

A long cylindrical wood \(\log (k=0.17 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\left.\alpha=1.28 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\right)\) is \(10 \mathrm{~cm}\) in diameter and is initially at a uniform temperature of \(15^{\circ} \mathrm{C}\). It is exposed to hot gases at \(550^{\circ} \mathrm{C}\) in a fireplace with a heat transfer coefficient of \(13.6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) on the surface. If the ignition temperature of the wood is \(420^{\circ} \mathrm{C}\), determine how long it will be before the log ignites. Solve this problem using analytical one-term approximation method (not the Heisler charts).

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