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The inner and outer surface temperatures of a glass window \(5 \mathrm{~mm}\) thick are 15 and \(5^{\circ} \mathrm{C}\). What is the heat loss through a \(1 \mathrm{~m} \times 3 \mathrm{~m}\) window? The thermal conductivity of glass is \(1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

Short Answer

Expert verified
The heat loss through the $1\mathrm{~m} \times 3\mathrm{~m}$ glass window with a thickness of 0.005 m, inner and outer surface temperatures of 15 and $5^{\circ} \mathrm{C}$, and a thermal conductivity of $1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$ is 8400 watts or 8.4 kilowatts.

Step by step solution

01

Identify the given values

In this exercise, we are given the following values: - Thickness of the window (d) = \( 5 \mathrm{~mm} \) - Inner surface temperature of the window (T1) = \( 15^{\circ} \mathrm{C} \) - Outer surface temperature of the window (T2) = \( 5^{\circ} \mathrm{C} \) - Window dimensions: \( 1\mathrm{~m} \times 3\mathrm{~m} \) - Thermal conductivity of glass (k) = \( 1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \)
02

Calculate temperature difference across the window

To find the temperature difference across the window, subtract the outer surface temperature from the inner surface temperature: \( \Delta T = T1 - T2 \) \( \Delta T = 15^{\circ} \mathrm{C} - 5^{\circ} \mathrm{C} \) \( \Delta T = 10 \mathrm{K} \) The temperature difference across the window is 10 K.
03

Calculate the window area

To find the area of the window, multiply the length and width of the window: \( A = L \times W \) \( A = 1\mathrm{~m} \times 3\mathrm{~m} \) \( A = 3 \mathrm{~m}^{2} \) The window area is 3 square meters.
04

Convert the thickness to meters

The given thickness is in millimeters. We need to convert it to meters to match the units of the thermal conductivity: \( d = 5 \mathrm{~mm} \times \frac{1\mathrm{~m}}{1000\mathrm{~mm}} \) \( d = 0.005\mathrm{~m} \) The thickness in meters is 0.005 m.
05

Calculate heat loss using the formula

Now, we can use the formula for heat transfer through a simple plane wall based on Fourier's law of heat conduction: \( Q = \frac{k \cdot A \cdot \Delta T}{d} \) Plug the given values into the formula: \( Q = \frac{1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \cdot 3 \mathrm{~m}^{2} \cdot 10 \mathrm{K}}{0.005\mathrm{~m}} \) \( Q = 8400 \mathrm{~W} \) The heat loss through the window is 8400 watts or 8.4 kilowatts.

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Key Concepts

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

Fourier's Law of Heat Conduction
Fourier’s law of heat conduction is a fundamental principle that describes the movement of heat within materials. According to this law, the rate at which heat energy, denoted as Q, transfers through a material is directly proportional to the temperature gradient (represented by the change in temperature, \( \Delta T \) and the area \( A \) through which heat is being transferred, and inversely proportional to the thickness (d) of the material.

In mathematical terms, Fourier’s law is expressed as \( Q = \frac{k \cdot A \cdot \Delta T}{d} \), where k stands for the material's thermal conductivity. This formula is crucial for understanding heat loss or gain in different materials, and it has practical applications in fields such as engineering, construction, and environmental science.

For example, in the context of our glass window, the law allows us to calculate the total heat loss based on the known conductivity of glass, area of the window, temperature difference across the window, and its thickness. Understanding this law helps individuals design better insulation solutions to retain heat, reduce energy costs, and improve overall energy efficiency in buildings.
Thermal Conductivity
Thermal conductivity is a material property that quantifies a material's ability to conduct heat. Represented by the symbol k in Fourier’s law, it is measured in watts per meter-kelvin (\(W/m\cdot K\)). Materials with high thermal conductivity, such as metals, are good heat conductors, meaning they can transfer heat quickly. Conversely, materials with low thermal conductivity, such as wood or fiberglass, are considered insulators because they transfer heat slowly.

For our glass window exercise, the thermal conductivity value of glass is \(1.4 W/m\cdot K\). This value is intrinsic to glass and crucial for determining how much heat the window will lose to the outside environment. When working with thermal conductivity, it's important to consider the environmental impact too. Better insulation, resulting from materials with lower thermal conductivity, leads to energy conservation and is therefore ecologically beneficial.
Temperature Difference in Heat Transfer
The temperature difference \(\Delta T\) is a driving force for heat transfer in many physical scenarios, including through a glass window. It is defined as the difference in temperature between two environments on either side of a barrier. In the case of the window, it is the difference between the indoor temperature and the outdoor temperature.

In our exercise scenario, the inside temperature is \(15^\circ C\) and the outside is \(5^\circ C\), resulting in a temperature difference of \(10 K\) (since the difference in Celsius and Kelvin for a given temperature gradient is the same). This temperature difference is a critical element to consider for heating and cooling purposes in buildings, as it impacts energy consumption and thermal comfort. By understanding the role that temperature difference plays in heat transfer, designers and engineers can make informed decisions about window materials, thicknesses, and coatings to minimize unwanted heat loss or gain.

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Most popular questions from this chapter

A square isothermal chip is of width \(w=5 \mathrm{~mm}\) on a side and is mounted in a substrate such that its side and back surfaces are well insulated; the front surface is exposed to the flow of a coolant at \(T_{\infty}=15^{\circ} \mathrm{C}\). From reliability considerations, the chip temperature must not exceed \(T=85^{\circ} \mathrm{C}\). If the coolant is air and the corresponding convection coefficient is \(h=200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), what is the maximum allowable chip power? If the coolant is a dielectric liquid for which \(h=3000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), what is the maximum allowable power?

A thin electrical heating element provides a uniform heat flux \(q_{o}^{\prime \prime}\) to the outer surface of a duct through which airflows. The duct wall has a thickness of \(10 \mathrm{~mm}\) and a thermal conductivity of \(20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). (a) At a particular location, the air temperature is \(30^{\circ} \mathrm{C}\) and the convection heat transfer coefficient between the air and inner surface of the duct is \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). What heat flux \(q_{o}^{\prime \prime}\) is required to maintain the inner surface of the duct at \(T_{i}=85^{\circ} \mathrm{C}\) ? (b) For the conditions of part (a), what is the temperature \(\left(T_{o}\right)\) of the duct surface next to the heater? (c) With \(T_{i}=85^{\circ} \mathrm{C}\), compute and plot \(q_{o}^{\prime \prime}\) and \(T_{o}\) as a function of the air-side convection coefficient \(h\) for the range \(10 \leq h \leq 200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Briefly discuss your results.

If \(T_{s} \approx T_{\text {sur }}\) in Equation \(1.9\), the radiation heat transfer coefficient may be approximated as $$ h_{r, a}=4 \varepsilon \sigma \bar{T}^{3} $$ where \(\bar{T} \equiv\left(T_{s}+T_{\text {sur }}\right) / 2\). We wish to assess the validity of this approximation by comparing values of \(h_{r}\) and \(h_{r, a}\) for the following conditions. In each case, represent your results graphically and comment on the validity of the approximation. (a) Consider a surface of either polished aluminum ( \(\varepsilon=\) \(0.05)\) or black paint \((\varepsilon=0.9)\), whose temperature may exceed that of the surroundings \(\left(T_{\text {sur }}=25^{\circ} \mathrm{C}\right)\) by 10 to \(100^{\circ} \mathrm{C}\). Also compare your results with values of the coefficient associated with free convection in air \(\left(T_{\infty}=T_{\text {sur }}\right)\), where \(h\left(\mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\right)=0.98 \Delta T^{1 / 3}\). (b) Consider initial conditions associated with placing a workpiece at \(T_{s}=25^{\circ} \mathrm{C}\) in a large furnace whose wall temperature may be varied over the range \(100 \leq\) \(T_{\text {sur }} \leq 1000^{\circ} \mathrm{C}\). According to the surface finish or coating, its emissivity may assume values of \(0.05\), \(0.2\), and \(0.9\). For each emissivity, plot the relative error, \(\left(h_{r}-h_{r, a}\right) / h_{r}\), as a function of the furnace temperature.

Most of the energy we consume as food is converted to thermal energy in the process of performing all our bodily functions and is ultimately lost as heat from our bodies. Consider a person who consumes \(2100 \mathrm{kcal}\) per day (note that what are commonly referred to as food calories are actually kilocalories), of which \(2000 \mathrm{kcal}\) is converted to thermal energy. (The remaining \(100 \mathrm{kcal}\) is used to do work on the environment.) The person has a surface area of \(1.8 \mathrm{~m}^{2}\) and is dressed in a bathing suit. (a) The person is in a room at \(20^{\circ} \mathrm{C}\), with a convection heat transfer coefficient of \(3 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). At this air temperature, the person is not perspiring much. Estimate the person's average skin temperature. (b) If the temperature of the environment were \(33^{\circ} \mathrm{C}\), what rate of perspiration would be needed to maintain a comfortable skin temperature of \(33^{\circ} \mathrm{C}\) ?

Under conditions for which the same room temperature is maintained by a heating or cooling system, it is not uncommon for a person to feel chilled in the winter but comfortable in the summer. Provide a plausible explanation for this situation (with supporting calculations) by considering a room whose air temperature is maintained at \(20^{\circ} \mathrm{C}\) throughout the year, while the walls of the room are nominally at \(27^{\circ} \mathrm{C}\) and \(14^{\circ} \mathrm{C}\) in the summer and winter, respectively. The exposed surface of a person in the room may be assumed to be at a temperature of \(32^{\circ} \mathrm{C}\) throughout the year and to have an emissivity of \(0.90\). The coefficient associated with heat transfer by natural convection between the person and the room air is approximately \(2 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

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