Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 12: Problem 111

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A flame from a match may be approximated as a blackbody at the effective surface temperature of \(1700 \mathrm{~K}\), while moonlight may be approximated as a blackbody at the effective surface temperature of \(4000 \mathrm{~K}\), respectively. Determine the peak spectral blackbody emissive power for both lighting sources (match flame and moonlight).

Determine the equilibrium temperature of the absorber surface in Prob. 12-98 if the back side of the absorber is insulated.

A furnace that has a \(40-\mathrm{cm} \times 40-\mathrm{cm}\) glass window can be considered to be a blackbody at \(1200 \mathrm{~K}\). If the transmissivity of the glass is \(0.7\) for radiation at wavelengths less than \(3 \mu \mathrm{m}\) and zero for radiation at wavelengths greater than \(3 \mu \mathrm{m}\), determine the fraction and the rate of radiation coming from the furnace and transmitted through the window.

What is a blackbody? Does a blackbody actually exist?

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 \(3000 \mathrm{~K}\).
See all solutions

Recommended explanations on Physics Textbooks

Electrostatics

Read Explanation

Linear Momentum

Read Explanation

Work Energy and Power

Read Explanation

Mechanics and Materials

Read Explanation

Fundamentals of Physics

Read Explanation

Wave Optics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free