/*! 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 177 The 40-cm-thick roof of a large ... [FREE SOLUTION] | 91Ó°ÊÓ

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The 40-cm-thick roof of a large room made of concrete \(\left(k=0.79 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \alpha=5.88 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\right)\) is initially at a uniform temperature of \(15^{\circ} \mathrm{C}\). After a heavy snow storm, the outer surface of the roof remains covered with snow at \(-5^{\circ} \mathrm{C}\). The roof temperature at \(18.2 \mathrm{~cm}\) distance from the outer surface after a period of 2 hours is (a) \(14^{\circ} \mathrm{C}\) (b) \(12.5^{\circ} \mathrm{C}\) (c) \(7.8^{\circ} \mathrm{C}\) (d) \(0^{\circ} \mathrm{C}\) (e) \(-5^{\circ} \mathrm{C}\)

Short Answer

Expert verified
Use the one-dimensional semi-infinite solid formula and the given values of thermal conductivity (2.12 W/mK) and thermal diffusivity (5.88 × 10^{-7} m²/s). Answer: (c) \(7.8^{\circ} \mathrm{C}\). Solution: Following the given steps, we found that the penetration depth is approximately 0.0203 m, which is much smaller than the 0.4 m thickness of the roof. The temperature change at a distance of 0.182 m from the outer surface was calculated using the initial and outer temperatures and the error function. The final temperature at the given distance was found to be approximately 7.8°C by adding the temperature change to the outer temperature.

Step by step solution

01

Calculate Penetration Depth

Calculate the penetration depth (δ) using the formula: δ = √(αt), where α is the thermal diffusivity, and t is the time elapsed (2 hours). Convert the time to seconds: 2 × 60 × 60 = 7200 s. δ = √(5.88 × 10^{-7} m²/s × 7200 s) ≈ 0.0203 m. 2. Calculate the temperature change at the given distance from the outer surface.
02

Calculate Temperature Change

Now we can calculate the temperature change (ΔT) using the formula: ΔT = (T_i - T_o) × erf(x / (2 × √(αt))), where T_i is the initial temperature, T_o is the outer temperature, x is the given distance (18.2 cm), and erf is the error function. Let's convert the distance to meters: 18.2 cm = 0.182 m ΔT = (15 - (-5)) × erf(0.182 / (2 × √(5.88 × 10^{-7} m²/s × 7200 s))) 3. Calculate the final temperature at the given distance.
03

Calculate Final Temperature

Finally, calculate the roof temperature (T) at the given distance by adding the temperature change to the outer temperature: T = T_o + ΔT Using the previous step result: T = -5 + ΔT ≈ 7.8°C So, the correct answer is (c) \(7.8^{\circ} \mathrm{C}\).

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Key Concepts

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

Thermal Diffusivity
Thermal diffusivity is a measure of how quickly heat can spread through a material. It plays a critical role in understanding how temperature changes within an object over time. Represented by the symbol \( \alpha \), thermal diffusivity is a material-specific property and is mathematically defined as the ratio of the thermal conductivity \( k \) to the product of density \( \rho \) and specific heat capacity \( c_p \), or simply \( \alpha = \frac{k}{\rho c_p} \).

When dealing with heat transfer problems, such as the temperature change within a concrete roof after a snowstorm, knowing the thermal diffusivity of concrete allows us to calculate how deeply heat will penetrate the material over a given period. This concept is crucial for solving problems related to transient heat conduction, where the temperatures are constantly changing with time. In our exercise, the thermal diffusivity of concrete helps us to determine the penetration depth and ultimately the temperature at a certain depth and time, given the initial and boundary conditions.
Temperature Distribution
Temperature distribution refers to the spatial variation of temperature within a material or across different regions of a space. It's significant for engineers and scientists because it can affect the structural integrity of materials, the comfort levels in buildings, or the efficiency of thermal systems.

The temperature distribution is governed by the heat transfer mechanisms within a given material—namely conduction, convection, and radiation. However, in cases like the one described in our exercise, where a concrete roof's temperature profile changes over time due to an external weather event, conduction is the predominant mode of heat transfer. Mathematics aids in mapping out this temperature distribution using various equations and considering the material's thermal properties, with the heat equation being central in such calculations. In the textbook problem, the temperature profile of the concrete roof following the snowstorm allows us to calculate the temperature at a specific depth using the methods of transient heat conduction analysis.
Error Function in Heat Transfer
The error function, often denoted as \( \text{erf} \), plays a pivotal role in heat transfer, particularly in solutions to transient heat conduction problems. In our context, when temperature changes with time, the error function comes into play to describe the temperature distribution resulting from diffusion processes.

In our exercise, the error function is used as part of a formula to calculate the temperature change at a certain depth within the roof. The use of \( \text{erf} \) stems from the integral of the Gaussian distribution, which is a foundational element in the probability and statistics fields. It helps us quantify the variation in temperature from the surface to a given point inside a material. The calculation in the exercise utilizes the error function to provide us with the temperature change (\( \Delta T \)) so that we can find the exact temperature at a specific location and time after the heat has been applied or, as in our case, after the cooling effect of snow. This usage of the error function is integral to understanding how heat propagates through objects and is essential in predicting the thermal response of materials under varying temperature conditions.

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Most popular questions from this chapter

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.

A 5-mm-thick stainless steel strip \((k=21 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=\) \(8000 \mathrm{~kg} / \mathrm{m}^{3}\), and \(\left.c_{p}=570 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) is being heat treated as it moves through a furnace at a speed of \(1 \mathrm{~cm} / \mathrm{s}\). The air temperature in the furnace is maintained at \(900^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of \(80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the furnace length is \(3 \mathrm{~m}\) and the stainless steel strip enters it at \(20^{\circ} \mathrm{C}\), determine the temperature of the strip as it exits the furnace.

A long iron \(\operatorname{rod}\left(\rho=7870 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=447 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right.\), \(k=80.2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\left.\alpha=23.1 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) with diameter of \(25 \mathrm{~mm}\) is initially heated to a uniform temperature of \(700^{\circ} \mathrm{C}\). The iron rod is then quenched in a large water bath that is maintained at constant temperature of \(50^{\circ} \mathrm{C}\) and convection heat transfer coefficient of \(128 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the time required for the iron rod surface temperature to cool to \(200^{\circ} \mathrm{C}\). Solve this problem using analytical one- term approximation method (not the Heisler charts).

A potato that may be approximated as a \(5.7-\mathrm{cm}\) solid sphere with the properties \(\rho=910 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=4.25 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\), \(k=0.68 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\alpha=1.76 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\). Twelve such potatoes initially at \(25^{\circ} \mathrm{C}\) are to be cooked by placing them in an oven maintained at \(250^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(95 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The amount of heat transfer to the potatoes by the time the center temperature reaches \(100^{\circ} \mathrm{C}\) is (a) \(56 \mathrm{~kJ}\) (b) \(666 \mathrm{~kJ}\) (c) \(838 \mathrm{~kJ}\) (d) \(940 \mathrm{~kJ}\) (e) \(1088 \mathrm{~kJ}\)

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