/*! 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 124 Consider a house in Atlanta, Geo... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 124

Consider a house in Atlanta, Georgia, that is maintained at \(22^{\circ} \mathrm{C}\) and has a total of \(20 \mathrm{~m}^{2}\) of window area. The windows are double-door type with wood frames and metal spacers and have a \(U\)-factor of \(2.5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (see Prob. 1-125 for the definition of \(U\)-factor). The winter average temperature of Atlanta is \(11.3^{\circ} \mathrm{C}\). Determine the average rate of heat loss through the windows in winter.

Short Answer

Expert verified
Answer: The average rate of heat loss through the windows during winter in the given house is 535 W.

Step by step solution

01

Understand the U-factor and the formula for heat loss

The U-factor (also called thermal transmittance) represents the rate at which heat is transferred through a building component (in this case, windows). The U-factor is expressed in W/m虏路K. The formula for heat loss through the windows can be given as: Heat Loss = U-factor 脳 Window Area 脳 Temperature Difference Now that we know the U-factor, we can plug in the given values for window area and temperature difference and calculate the average rate of heat loss through the windows in winter.
02

Calculate the temperature difference

The temperature difference between the inside and outside of the house is crucial for calculating heat loss. Given the inside temperature of the house C_{in} = 22掳C and the winter average outdoor temperature, C_{out} = 11.3掳C. We can calculate the temperature difference (鈭员) as: 鈭员 = C_{in} - C_{out} 鈭员 = 22掳C - 11.3掳C 鈭员 = 10.7掳C
03

Calculate the average rate of heat loss through the windows in winter

Now that we have the temperature difference (鈭员), the U-factor (given), and the window area (given), we can plug in the values into the Heat Loss formula: Heat Loss = U-factor 脳 Window Area 脳 Temperature Difference Heat Loss = (2.5 W/m虏路K) 脳 (20 m虏) 脳 (10.7 掳C) Heat Loss = 535 W The average rate of heat loss through the windows in winter is 535 W.

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

U-factor
The U-factor is a key concept in understanding heat transfer through building materials. It measures how well a component, like a window, conducts heat from a warmer area to a colder one. Simply put, the U-factor indicates the efficiency of a material's insulating ability. The lower the U-factor, the better the material is at insulating. It is expressed in units of watts per square meter per Kelvin (
  • U-factor, also known as thermal transmittance, represents how well a window prevents heat from escaping.
  • A window with a high U-factor loses heat more quickly than a window with a low U-factor.
  • In the example, the window's U-factor is 2.5 W/m虏路K, meaning it allows 2.5 watts per square meter of heat to pass through with every degree of temperature difference between inside and outside.
Thermal transmittance
Thermal transmittance is another term for U-factor. It quantifies the rate of heat transfer through a structure. When assessing a building鈥檚 energy efficiency, understanding its thermal transmittance is crucial. It allows us to determine how much heat is being lost through materials, which can directly affect energy costs.
  • Thermal transmittance measures the effectiveness of a material's capacity to conduct heat.
  • A lower thermal transmittance value denotes better insulation and less energy loss.
  • In practical terms, high thermal transmittance means more heat loss during the winter, leading to higher heating costs.
Heat loss calculation
Calculating heat loss involves determining how quickly heat energy passes through materials from the inside to the outside. The formula used is straightforward: Heat Loss = U-factor 脳 Window Area 脳 Temperature Difference. This equation allows us to quantify heat loss and is fundamental in energy assessments.
  • To compute heat loss through windows, identify the U-factor, window area, and temperature difference.
  • Using the given example, the equation becomes: Heat Loss = 2.5 W/m虏路K 脳 20 m虏 脳 10.7 掳C.
  • The calculated heat loss for the example is 535 watts, showing the rate at which heat exits through the windows during winter.
Temperature difference
The temperature difference ( 鈭员) is a critical factor in the heat loss equation. It represents the difference between the warm indoor temperature and the cooler outdoor temperature. A greater difference means more heat loss because the drive for heat to flow from warmer to cooler areas increases.
  • To find the temperature difference, subtract the outdoor temperature from the indoor temperature.
  • In the example, the indoor temperature is 22掳C, and the outdoor temperature is 11.3掳C, resulting in a temperature difference of 10.7掳C.
  • The larger this difference, the more pronounced the heat loss, emphasizing the need for efficient insulation.

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

What is asymmetric thermal radiation? How does it cause thermal discomfort in the occupants of a room?

A concrete wall with a surface area of \(20 \mathrm{~m}^{2}\) and a thickness of \(0.30 \mathrm{~m}\) separates conditioned room air from ambient air. The temperature of the inner surface of the wall \(\left(T_{1}\right)\) is maintained at \(25^{\circ} \mathrm{C}\). (a) Determine the heat loss \(\dot{Q}(\mathrm{~W})\) through the concrete wall for three thermal conductivity values of \((0.75,1\), and \(1.25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) and outer wall surface temperatures of \(T_{2}=-15,-10,-5,0,5,10,15,20,25,30\), and \(38^{\circ} \mathrm{C}\) (a total of 11 data points for each thermal conductivity value). Tabulate the results for all three cases in one table. Also provide a computer generated graph [Heat loss, \(\dot{Q}(\mathrm{~W})\) vs. Outside wall temperature, \(\left.T_{2}\left({ }^{\circ} \mathrm{C}\right)\right]\) for the display of your results. The results for all three cases should be plotted on the same graph. (b) Discuss your results for the three cases.

A room is heated by a \(1.2 \mathrm{~kW}\) electric resistance heater whose wires have a diameter of \(4 \mathrm{~mm}\) and a total length of \(3.4 \mathrm{~m}\). The air in the room is at \(23^{\circ} \mathrm{C}\) and the interior surfaces of the room are at \(17^{\circ} \mathrm{C}\). The convection heat transfer coefficient on the surface of the wires is \(8 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the rates of heat transfer from the wires to the room by convection and by radiation are equal, the surface temperature of the wire is (a) \(3534^{\circ} \mathrm{C}\) (b) \(1778^{\circ} \mathrm{C}\) (c) \(1772^{\circ} \mathrm{C}\) (d) \(98^{\circ} \mathrm{C}\) (e) \(25^{\circ} \mathrm{C}\)

Two surfaces, one highly polished and the other heavily oxidized, are found to be emitting the same amount of energy per unit area. The highly polished surface has an emissivity of \(0.1\) at \(1070^{\circ} \mathrm{C}\), while the emissivity of the heavily oxidized surface is \(0.78\). Determine the temperature of the heavily oxidized surface.

Heat is lost steadily through a \(0.5-\mathrm{cm}\) thick \(2 \mathrm{~m} \times 3 \mathrm{~m}\) window glass whose thermal conductivity is \(0.7 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The inner and outer surface temperatures of the glass are measured to be \(12^{\circ} \mathrm{C}\) to \(9^{\circ} \mathrm{C}\). The rate of heat loss by conduction through the glass is (a) \(420 \mathrm{~W}\) (b) \(5040 \mathrm{~W}\) (c) \(17,600 \mathrm{~W}\) (d) \(1256 \mathrm{~W}\) (e) \(2520 \mathrm{~W}\)
See all solutions

Recommended explanations on Physics Textbooks

Space Physics

Read Explanation

Fundamentals of Physics

Read Explanation

Physics of Motion

Read Explanation

Rotational Dynamics

Read Explanation

Fluids

Read Explanation

Kinematics Physics

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.