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The \(R\) factor for housing insulation gives the thermal resistance in units of \(\mathrm{ft}^{2}{ }^{\circ} \mathrm{F} \mathrm{h} / \mathrm{BTU}\). A good wall for harsh climates, corresponding to about 10.0 in of fiberglass, has \(R=40.0 \mathrm{ft}^{2}{ }^{\circ} \mathrm{F} \mathrm{h} / \mathrm{BTU}\) a) Determine the thermal resistance in SI units. b) Find the heat flow per square meter through a wall that has insulation with an \(R\) factor of 40.0 , with an outside temperature of \(-22.0^{\circ} \mathrm{C}\) and an inside temperature of \(23.0^{\circ} \mathrm{C}\)

Short Answer

Expert verified
Answer: The thermal resistance in SI units is approximately 7.04 (m²·K·h/J), and the heat flow per square meter through the wall is approximately 6.39 (J/m²·h).

Step by step solution

01

Identify the necessary conversion factors

We'll need to convert the given R-factor from the Imperial unit of \((\mathrm{ft}^{2}{ }^{\circ} \mathrm{F} \mathrm{h} / \mathrm{BTU})\) to the SI unit of \((\mathrm{m}^2 . \mathrm{K} . \mathrm{h} / \mathrm{J})\). The conversion factors we need are: 1 ft = 0.3048 m 1 °F = 5/9 K (temperature difference only) 1 h = 3600 s 1 BTU = 1055 J
02

Conversion to SI units

Apply the conversion factors to the given R-factor to find the thermal resistance in SI units: \(R = 40.0 \frac{\mathrm{ft}^2 . ^{\circ} \mathrm{F} . \mathrm{h}}{\mathrm{BTU}} \times \frac{0.3048^2\,\mathrm{m}^2}{\mathrm{ft}^2} \times \frac{5}{9}\,\mathrm{K} /^{\circ} \mathrm{F} \times \frac{\mathrm{h}}{\mathrm{h}} \times \frac{1}{1055}\,\dfrac{\mathrm{J}}{\mathrm{BTU}} \) \(R \approx 7.04 \frac{\mathrm{m}^2 . \mathrm{K} . \mathrm{h}}{\mathrm{J}}\) The thermal resistance in SI units is approximately 7.04 \((\mathrm{m}^2 . \mathrm{K} . \mathrm{h} / \mathrm{J})\). #b) Find the heat flow per square meter through a wall that has insulation with an \(R\) factor of 40.0 , with an outside temperature of \(-22.0^{\circ} \mathrm{C}\) and an inside temperature of \(23.0^{\circ} \mathrm{C}\)#
03

Use the heat flow formula

The formula for heat flow per unit area is given by \(q = \frac{\Delta T}{R}\) where \(q\) is the heat flow per unit area, \(\Delta T\) is the temperature difference across the wall, and \(R\) is the thermal resistance.
04

Calculate the temperature difference

The temperature difference across the wall is given by the difference between the inside and outside temperatures. \(\Delta T = T_{inside} - T_{outside} = 23.0 - (-22.0) = 45.0^{\circ} \mathrm{C}\)
05

Convert temperature difference to Kelvin

Since the units of R are in Kelvin, we need to convert the temperature difference from Celsius to Kelvin. A difference of 1 °C is equal to a difference of 1 K, so: \(\Delta T = 45.0\, \mathrm{K}\)
06

Calculate the heat flow per square meter

Now, we can use the formula for heat flow per unit area to find the heat flow per square meter. \(q = \frac{\Delta T}{R} = \frac{45.0\, \mathrm{K}}{7.04\, \frac{\mathrm{m}^2 . \mathrm{K} . \mathrm{h}}{\mathrm{J}}}\) \(q \approx 6.39 \frac{\mathrm{J}}{\mathrm{m}^2 . \mathrm{h}}\) The heat flow per square meter through the wall is approximately 6.39 \(\dfrac{\mathrm{J}}{\mathrm{m}^2 .\mathrm{h}}\).

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Key Concepts

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

Understanding R-factor in Insulation
The R-factor is a measure of thermal resistance used in the building and construction industry. It indicates how well a material prevents heat from passing through it. When we talk about insulation, a higher R-factor means better insulation because it resists heat flow more effectively. The R-factor is particularly critical in maintaining energy efficiency and comfort in buildings, especially in areas where temperatures can vary widely. For instance, a wall with an R-factor of 40, as mentioned in the exercise, provides substantial resistance to heat flow, making it suitable for harsh climates.

In simple terms, think of the R-factor as a blanket's thickness: the thicker the blanket (higher R-factor), the warmer you stay. Insulation works similarly by keeping warm air inside during winter and hot air outside during summer. Calculating the thermal resistance in SI units involves converting from imperial units to units commonly used internationally, which allows for standardization and easier comparison across different materials and products.
Heat Flow Calculation Simplified
The concept of heat flow is central to understanding thermal dynamics and energy efficiency in buildings. Heat flow, expressed in terms of energy per unit time per unit area, illustrates how much heat is transmitted through a certain area, such as a wall, in one hour. To control and manage this heat transfer, insulation with a specific R-factor is used.

To calculate the actual heat flow through an insulated wall, we first establish the temperature difference between the inside and outside. Then, using the insulation's R-factor, we apply a simple formula: \(q = \frac{\Delta T}{R}\), where the heat flow per square meter \(q\) is inversely proportional to the R-factor \(R\). This calculation helps us determine the effectiveness of insulation and its impact on energy consumption. For energy efficiency in buildings, the goal is to minimize heat flow by choosing the right insulation materials with a suitable R-factor.
Temperature Conversion in Practical Terms
Temperature plays a significant role in calculating heat flow; therefore, it's essential to understand how to convert temperatures correctly. Although temperature is commonly measured in degrees Celsius (°C) or Fahrenheit (°F), scientific calculations often require temperatures in Kelvin (K), the SI base unit of thermodynamic temperature.

When converting temperature for thermal calculations, we only consider the difference in temperature, not the absolute values. This means that a change in temperature of 1 °C is equivalent to a change of 1 K. For example, in the exercise, the difference in Celsius was converted directly to Kelvin without adjusting for the zero points of each scale. This seemingly simple concept is crucial for accurate calculations in thermal dynamics and contributes to our understanding of heat transfer across different media.
The Significance of SI Units in Scientific Measurements
SI units, or the International System of Units, serve as the universal language for scientists, engineers, and educators around the world. They facilitate the comparison and communication of measurements in a clear, standardized manner. Commonly used SI units such as meters (m) for length, joules (J) for energy, and Kelvins (K) for thermodynamic temperature allow for consistent and precise scientific calculations.

In the context of insulation and heat flow, as seen in the exercise, using SI units eliminates confusion that may arise from regional measurement systems. When converting R-factor values and calculating heat flow in SI units, it ensures that professionals across different regions and disciplines can effectively collaborate and assess the energy performance of building materials and designs.

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Most popular questions from this chapter

The radiation emitted by a blackbody at temperature \(T\) has a frequency distribution given by the Planck spectrum: $$ \epsilon_{T}(f)=\frac{2 \pi h}{c^{2}}\left(\frac{f^{3}}{e^{h f / k_{\mathrm{B}} T}-1}\right) $$ where \(\epsilon_{T}(f)\) is the energy density of the radiation per unit increment of frequency, \(v\) (for example, in watts per square meter per hertz), \(h=6.626 \cdot 10^{-34} \mathrm{~J} \mathrm{~s}\) is Planck's constant, \(k_{\mathrm{B}}=1.38 \cdot 10^{-23} \mathrm{~m}^{2} \mathrm{~kg} \mathrm{~s}^{-2} \mathrm{~K}^{-1}\) is the Boltzmann constant, and \(c\) is the speed of light in vacuum. (We'll derive this distribution in Chapter 36 as a consequence of the quantum hypothesis of light, but here it can reveal something about radiation. Remarkably, the most accurately and precisely measured example of this energy distribution in nature is the cosmic microwave background radiation.) This distribution goes to zero in the limits \(f \rightarrow 0\) and \(f \rightarrow \infty\) with a single peak in between those limits. As the temperature is increased, the energy density at each frequency value increases, and the peak shifts to a higher frequency value. a) Find the frequency corresponding to the peak of the Planck spectrum, as a function of temperature. b) Evaluate the peak frequency at temperature \(T=6.00 \cdot 10^{3} \mathrm{~K}\), approximately the temperature of the photosphere (surface) of the Sun. c) Evaluate the peak frequency at temperature \(T=2.735 \mathrm{~K}\), the temperature of the cosmic background microwave radiation. d) Evaluate the peak frequency at temperature \(T=300 . \mathrm{K}\), which is approximately the surface temperature of Earth.

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