/*! 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 2 A concrete wall, which has a sur... [FREE SOLUTION] | 91Ó°ÊÓ

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A concrete wall, which has a surface area of \(20 \mathrm{~m}^{2}\) and is \(0.30 \mathrm{~m}\) thick, separales conditioned room air from ambient air. The temperature of the inner surface of the wall is maintained at \(25^{\circ} \mathrm{C}\), and the thermal conductivity of the concrete is \(1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). (a) Determine the heat loss through the wall for outer surface temperatures ranging from \(-15^{\circ} \mathrm{C}\) to \(38^{\circ} \mathrm{C}\), which correspond to winter and summer extremes, respectively. Display your results graphically.

Short Answer

Expert verified
Heat loss ranges from 2666.67 W in winter to -866.67 W in summer.

Step by step solution

01

Understand the Heat Transfer Equation

To calculate the heat loss through the wall, we will use the formula derived from Fourier's law of thermal conduction: \ \[ Q = \frac{k \cdot A \cdot (T_{inside} - T_{outside})}{d} \] \ where:- \( Q \) is the heat loss (W),- \( k \) is the thermal conductivity of the material (1 W/m·K),- \( A \) is the surface area of the wall (20 m²),- \( T_{inside} \) and \( T_{outside} \) are the temperatures of the inside and outside surfaces (°C),- \( d \) is the wall thickness (0.30 m).
02

Calculate Heat Loss for Winter Conditions

Plug in the values into the formula for the winter condition, where \( T_{outside} = -15^{\circ}C \): \ \[ Q_{winter} = \frac{1 \cdot 20 \cdot (25 - (-15))}{0.30} \] \ Simplifying, we find: \ \[ Q_{winter} = \frac{1 \cdot 20 \cdot (25 + 15)}{0.30} = \frac{1 \cdot 20 \cdot 40}{0.30} = \frac{800}{0.30} = 2666.67 \text{ W} \]
03

Calculate Heat Loss for Summer Conditions

Now, consider the summer condition, where \( T_{outside} = 38^{\circ}C \). Substitute the values into the formula: \ \[ Q_{summer} = \frac{1 \cdot 20 \cdot (25 - 38)}{0.30} \] \ Simplifying, we get: \ \[ Q_{summer} = \frac{1 \cdot 20 \cdot (-13)}{0.30} = \frac{-260}{0.30} = -866.67 \text{ W} \] \ The negative sign indicates heat is entering the room.
04

Prepare a Graph of Results

For graphical representation, plot heat loss \( Q \) along the y-axis against the outside temperature \( T_{outside} \) along the x-axis. The x-axis should range from \(-15^{\circ}C\) to \(38^{\circ}C\), with corresponding \( Q \) values from the calculations showing a linear decrease.

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Key Concepts

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

Fourier's Law of Thermal Conduction
Fourier's Law of Thermal Conduction is essential for understanding how heat moves through materials, such as walls or other solid objects. This fundamental principle explains heat transfer, which occurs from warmer areas to cooler ones.
This process takes place due to the temperature difference between two surfaces, causing heat to flow until equilibrium is reached. The law is mathematically described by the equation: \[ Q = -k abla T\]
Here, \( Q \) is the rate of heat transfer in watts, \( k \) is the thermal conductivity of the material, and \( abla T \) represents the temperature gradient which is the change in temperature divided by the distance over which the change occurs.
For practical purposes, walls or structures often have a simplified version of this equation when they have constant thermal properties and uniform thickness. This becomes: \[ Q = \frac{k \cdot A \cdot (T_{inside} - T_{outside})}{d} \]
Where:
  • \( A \) is the area through which heat is being transferred.
  • \( T_{inside} \) and \( T_{outside} \) are the temperatures on either side of the barrier.
  • \( d \) is the thickness of the material.
Understanding Fourier's Law allows us to calculate how much heat is either lost or gained through walls based on the temperature differences and the material properties.
Thermal Conductivity
Thermal conductivity is a property of materials that says how well they can conduct heat. It's denoted by the symbol \( k \) and measured in watts per meter per Kelvin (W/m·K). This value tells us how much heat will pass through a material over a certain time when there's a temperature difference across the material.
Materials with high thermal conductivity are great conductors of heat. For example, metals like copper and aluminum can transfer heat quickly. Meanwhile, materials like wood or fiberglass have low thermal conductivity, meaning they are good insulators.
In exercises involving heat transfer, knowing a material's thermal conductivity helps determine how efficient it will be at allowing heat to pass through. In our case, the concrete wall has a thermal conductivity of 1 W/m·K. This indicates that concrete is a moderate conductor, not as efficient as a metal but better than most insulators.
Heat Loss Calculation
The calculation of heat loss through a wall involves determining the amount of heat energy that leaves a space due to differences in temperature inside and outside a building. Using Fourier's law and knowledge of the material's thermal conductivity allows for precise calculation.
Let's consider a wall with given dimensions:
  • Surface area \( A = 20 \, \text{m}^2 \)
  • Material thickness \( d = 0.30 \, \text{m} \)
  • Inside temperature \( T_{inside} = 25^{\circ} \text{C} \)
  • Outside temperatures range from \(-15^{\circ} \text{C}\) to \(38^{\circ} \text{C}\).
For each outside temperature, substitute the values into Fourier's equation:
\[ Q = \frac{k \cdot A \cdot (T_{inside} - T_{outside})}{d} \]
This will give the rate of heat transfer in watts (W). For example, in winter when \( T_{outside} = -15^{\circ}C \), the heat loss \( Q_{winter} \) is calculated at 2666.67 W. In summer, when \( T_{outside} = 38^{\circ}C \), the equation shows \( Q_{summer} \) as -866.67 W, indicating heat entering the building.
This calculated result helps understand not just the magnitude of heat lost or gained but also guides in designing wall insulations and improving energy efficiency in buildings.

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Most popular questions from this chapter

An instrumentation package has a spherical outer surface of diameter \(D=100 \mathrm{~nm}\) and emissivity \(c=0.25\). The package is placed in a large space simulation chamber whose walls are maintained at \(77 \mathrm{~K}\). If operation of the electronic components is restricted to the lemperature range \(40 \leq 7 \leq 85^{\circ} \mathrm{C}\), what is the range of acceptable power dissipation for the package? Display your results graphically, showing also the effect of variations in the emissivity by censidering values of \(0.20\) and \(0.30\).

A spherical interplanetary probe of \(0.5\) - m diameter contains electronics that dissipate \(150 \mathrm{~W}\), If the probe surface has an emissivity of \(0.8\) and the probe does not receive radiation from other surfaces, as, for example, from the sun, what is its surface temperature?

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 \(2 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

In considcring the following jroblems involving heat transfer in the natural environment (outdoors), recog. nize that solar radiation is comprised of long and shon wavelength components. If this radiation is incident on a semitrarsporent medium, such as water or glass, two things will happen to the noneflecied porticn of the radiation. The long wavelength component will be aboubed at the surface of the medium, whereas the short wavelength component will be transmitted by the surface. (a) The number of panes in a window can strongly inftuence the heat loss from a heated room to the outside ambient air, Compare the single- and doublepaned units shown by identifying relevant healt transfer processes for each case. (b) In a typical fat-plate solar collector, energy is collected by a working fluid thut is circulated through fubes that are in good contact with the hack face of an absorber plate. The back. face is insulated frum the surroundings, and the absuber plate receives solar radiation on its front face, which is typically covered by one or more transparent plates. Idensify the relevant heat transfer processes, first for the absorber plate with no cover plate and then for the absorber plate with a single cover plate. (c) The solar energy collector design stoun below has been used for agricultural applications. Air is blown through a long duct whone cross section is in the form of an equilateral triangle. One side of the triangle is comprised of a double-paned, semitransparent cover, while the ceher two sides are constructed from aluminum sheets painted flat black on the inside and covered on the outside with a Layer of styroform insulation. During sunny periods, air entering the system is heated for delivery to cither a greenhouse, grain drying unit, or a storage system. Identify all heat transfer processes associated with the cover plates, the absorter plate(s), and the air. (d) Evacuated-tube solar collectors are capable of improved performance relative to flat-plate collectors. The design consists of an inner tube enclosed in an outer tube that is transparent to solar radiation. The annular space between the tubes is evacuated. The ouler, opaque surface of the inner tube absorbs solar raciation, and a working fluid is passed through the tube to collect the solar energy. The collector design generally consists of a row of such tubes arranged in frunt of a reflecting panel. Identify all heat transfer processes relevant to the performance of this device.

The heat flux through a wood slab \(50 \mathrm{~mm}\) thick, whose inner and outer surface temperatures are 40 and \(20^{\circ} \mathrm{C}\), respectively, has been determined to be \(40 \mathrm{~W} / \mathrm{m}^{2}\). What is the thermal conductivity of the wood?

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