/*! 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 111 Determine the rate of net heat g... [FREE SOLUTION] | 91Ó°ÊÓ

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Determine the rate of net heat gain (or loss) through a 9 -ft-high, 15 -ft- wide, fixed \(\frac{1}{8}\)-in single-glass window with aluminum frames on the west wall at 3 PM solar time during a typical day in January at a location near \(40^{\circ} \mathrm{N}\) latitude when the indoor and outdoor temperatures are \(70^{\circ} \mathrm{F}\) and \(20^{\circ} \mathrm{F}\), respectively.

Short Answer

Expert verified
Answer: The net heat gain through the window is approximately 34,464 Btu/h.

Step by step solution

01

1. Calculate the area of the window

Given the dimensions, the area of the window can be calculated as: Area = height × width = 9 ft × 15 ft = 135 ft²
02

2. Find the solar heat gain coefficient and solar radiation through the window

The average solar heat gain coefficient (SHGC) for single-pane, clear glass windows is approximately 0.86. The solar radiation through the west wall at 3 PM solar time on a clear day in January at a location near \(40^{\circ} \mathrm{N}\) latitude is about 230 Btu/h·ft². Solar Radiation = 230 Btu/h·ft²
03

3. Calculate the heat transfer due to radiation

The heat transfer through the window due to radiation can be calculated using the following formula: Heat transfer due to radiation = Area × SHGC × Solar Radiation Heat transfer due to radiation = 135 ft² × 0.86 × 230 Btu/h·ft² = 26,499 Btu/h
04

4. Determine the thermal transmittance for the window

The thermal transmittance (U-value) for a single-pane, clear glass window with aluminum frames is approximately 1.1 Btu/h·ft²·°F. U-value = 1.1 Btu/h·ft²·°F
05

5. Calculate the heat transfer due to conduction

The heat transfer through the window due to conduction can be calculated using the following formula: Heat transfer due to conduction = Area × U-value × (indoor temperature - outdoor temperature) Heat transfer due to conduction = 135 ft² × 1.1 Btu/h·ft²·°F × (70°F - 20°F) = 7,965 Btu/h
06

6. Calculate the net heat gain (or loss)

Finally, we can determine the rate of net heat gain (or loss) through the window by adding the heat transfer due to radiation and conduction: Net heat gain (or loss) = Heat transfer due to radiation + Heat transfer due to conduction Net heat gain (or loss) = 26,499 Btu/h + 7,965 Btu/h = 34,464 Btu/h The net heat gain through the 9-ft-high, 15-ft-wide, \(\frac{1}{8}\)-in single-glass window with aluminum frames on the west wall at 3 PM solar time during a typical day in January at a location near \(40^{\circ} \mathrm{N}\) latitude when the indoor and outdoor temperatures are \(70^{\circ} \mathrm{F}\) and \(20^{\circ} \mathrm{F}\), respectively, is approximately 34,464 Btu/h.

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Key Concepts

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

Solar Heat Gain Coefficient
The Solar Heat Gain Coefficient (SHGC) is a measure of how much solar energy passes through a window compared to the total solar energy striking the window. It is expressed as a number between 0 and 1. A higher SHGC means more solar heat is admitted. This is particularly useful in colder climates where solar heat can contribute to warming the interior space.
For single-pane, clear glass windows like the one in the exercise, the SHGC is around 0.86. This value indicates that 86% of the solar radiation striking the window will enter through it.
The formula to calculate heat gain due to solar radiation is:
  • Heat transfer due to radiation = Area \(\times\) SHGC \(\times\) Solar Radiation

Hence, knowing the SHGC allows us to quantify the contribution of sunlight to the heating of a space, aiding in energy efficiency planning as well as in understanding the effect of sunlight on indoor temperatures.
Thermal Transmittance
Thermal transmittance, commonly referred to as the U-value, indicates how well a building element conducts heat. It is calculated in Btu/h·ft²·°F for US customary units. The lower the U-value, the better the material is as an insulator.
This exercise utilized a U-value of 1.1 Btu/h·ft²·°F for a single-glass window with aluminum frames. This relatively high U-value suggests the window is not excellent at preventing heat transfer, characteristic of single-glazed windows with metal frames.
To determine the heat transferred through conduction for the window, the formula is:
  • Heat transfer due to conduction = Area \(\times\) U-value \(\times\) (indoor temperature - outdoor temperature)

By using this formula, we can calculate how much heat will be lost to or gained from the outside environment depending on the temperature difference, which is crucial for maintaining comfortable indoor temperatures efficiently.
Conduction and Radiation Calculations
Understanding how heat transfers through a window involves two key processes: conduction and radiation. Here we are examining both to get a full picture of how energy moves across window structures.

Conduction
Conduction is the transfer of heat through a material without the movement of the material itself. In the context of windows, it describes the heat loss or gain that occurs due to the temperature difference between the inside and outside. The formula used is:
  • Heat transfer due to conduction = Area \(\times\) U-value \(\times\) temperature difference


Radiation
Radiation refers to the transfer of heat in the form of electromagnetic waves, such as sunlight. Key to this is the Solar Heat Gain Coefficient (SHGC), which indicates how effectively the window transmits solar heat energy.
The formula to find heat transfer by radiation is:
  • Heat transfer due to radiation = Area \(\times\) SHGC \(\times\) Solar Radiation

By calculating both the conduction and radiation separately, and then adding them, we achieve a holistic understanding of how much total heat is transferred through the window. This ensures decisions regarding heating, cooling, and window materials are grounded in quantitative analysis.

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Most popular questions from this chapter

Consider an opaque horizontal plate that is well insulated on the edges and the lower surface. The plate is uniformly irradiated from above while air at \(T_{\infty}=300 \mathrm{~K}\) flows over the surface providing a uniform convection heat transfer coefficient of \(40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Under steady state conditions the surface has a radiosity of \(4000 \mathrm{~W} / \mathrm{m}^{2}\), and the plate temperature is maintained uniformly at \(350 \mathrm{~K}\). If the total absorptivity of the plate is \(0.40\), determine \((a)\) the irradiation on the plate, \((b)\) the total reflectivity of the plate, \((c)\) the emissive power of the plate, and \((d)\) the total emissivity of the plate.

What is an electromagnetic wave? How does it differ from a sound wave?

The absorber surface of a solar collector is made of aluminum coated with black chrome ( \(\alpha_{s}=0.87\) and \(\left.\varepsilon=0.09\right)\). Solar radiation is incident on the surface at a rate of \(600 \mathrm{~W} / \mathrm{m}^{2}\). The air and the effective sky temperatures are \(25^{\circ} \mathrm{C}\) and \(15^{\circ} \mathrm{C}\), respectively, and the convection heat transfer coefficient is \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). For an absorber surface temperature of \(70^{\circ} \mathrm{C}\), determine the net rate of solar energy delivered by the absorber plate to the water circulating behind it.

The variation of the spectral absorptivity of a surface is as given in Fig. P12-78. Determine the average absorptivity and reflectivity of the surface for radiation that originates from a source at \(T=2500 \mathrm{~K}\). Also, determine the average emissivity of this surface at \(3000 \mathrm{~K}\).

Consider a surface at \(500 \mathrm{~K}\). The spectral blackbody emissive power at a wavelength of \(50 \mu \mathrm{m}\) is (a) \(1.54 \mathrm{~W} / \mathrm{m}^{2} \cdot \mu \mathrm{m}\) (b) \(26.3 \mathrm{~W} / \mathrm{m}^{2} \cdot \mu \mathrm{m}\) (c) \(108.4 \mathrm{~W} / \mathrm{m}^{2} \cdot \mu \mathrm{m}(d) 2750 \mathrm{~W} / \mathrm{m}^{2} \cdot \mu \mathrm{m}(e) 8392 \mathrm{~W} / \mathrm{m}^{2} \cdot \mu \mathrm{m}\)

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