/*! 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 16 A TV advertisement by a well-kno... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 16

A TV advertisement by a well-known insulation manufacturer states: it isn't the thickness of the insulating material that counts, it's the R-value. The ad shows that to ottain an R-value of 19, you need \(18 \mathrm{ft}\) of rock, 15 in. of wood. or just 6 in. of the manufacturer's insulation. Is this advertisement techinically reasonable? If you are like most TV viewers, you don't know the R-value is defined as \(L / k\). where \(L\). \(\mathrm{in}\).) is the thickness of the insulation and \(k\) (Btu * in \(/ \mathrm{hr}-\mathrm{ft}^{2}+{ }^{\mathrm{F}}\) ) is the thermal conductivity of the muterial.

Short Answer

Expert verified
Yes, the ad is technically reasonable based on R-values.

Step by step solution

01

Understand R-value definition

The R-value is expressed by the formula \( R = \frac{L}{k} \) where \( L \) is the thickness of the insulating material in inches and \( k \) is the thermal conductivity of the material. A higher R-value indicates better insulation.
02

Compare R-values from advertisement

According to the advertisement, the R-value of 19 can be achieved with 18 ft of rock, 15 inches of wood, or 6 inches of the manufacturer's insulation. Convert all thicknesses to inches: \(18 \text{ ft} = 216 \text{ in}\).
03

Calculate k for each material

We have \( R = \frac{L}{k} \). Therefore, \( k = \frac{L}{R} \). For each material:- Rock: \( k = \frac{216}{19} \approx 11.37 \)- Wood: \( k = \frac{15}{19} \approx 0.79 \)- Manufacturer's insulation: \( k = \frac{6}{19} \approx 0.32 \)
04

Analyze technical reasonability

The manufacturer's insulation has a lower thermal conductivity \( k = 0.32 \) compared to rock and wood, which means it is a more efficient insulator and requires less thickness for the same R-value. This supports the claim in the advertisement.

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.

Thermal Conductivity
Thermal conductivity, denoted by the symbol \( k \), measures how well a material conducts heat. It is a crucial property in selecting materials for insulation.The lower the thermal conductivity, the better the material is at insulating.
Materials with high thermal conductivity allow heat to pass through them easily, while those with low thermal conductivity do the opposite, effectively trapping heat.
Think about it like this: if you have metal and wood, metal usually feels colder because it conducts heat away from your hand faster, due to its higher thermal conductivity.Wood doesn鈥檛 conduct heat as well, so it feels warmer.If you were asked to choose between these materials to make a mug you can hold while sipping hot coffee, wood would keep your hands safer because it has lower thermal conductivity.In the original exercise, we calculated the \( k \) value for different materials to understand how efficiently each insulates:
  • Rock: \( k \approx 11.37 \)
  • Wood: \( k \approx 0.79 \)
  • Manufacturer's insulation: \( k \approx 0.32 \)
Notice how the manufacturer's insulation has the lowest \( k \).
This tells us it is less prone to letting heat through, making it an excellent choice for insulation.
Insulating Materials
Insulating materials are substances used to slow down the transfer of heat. They are essential in our homes to keep warm air inside during winter and outside during summer.
Not all insulators work the same way, though. Their performance is often evaluated by their R-value鈥攖he higher the R-value, the better the insulation. In the exercise, we see different materials like rock, wood, and a special insulation material from a manufacturer:
  • Rock needs to be really thick (18 ft) to achieve the same R-value as thinner layers of other materials.
  • Wood, although not as thick, requires 15 inches to match the desired R-value.
  • The manufacturer's insulation needs only 6 inches to reach an R-value of 19.
The difference in the performance of these materials comes down to their thermal conductivity and how well they trap heat.
When selecting insulating materials, consider not just the thickness, but especially the thermal conductivity and resulting R-value.
The most efficient insulators require less thickness to achieve high R-values.
Heat Transfer
Heat transfer involves the movement of thermal energy from one place to another. It can occur in three main ways: conduction, convection, and radiation. In the context of insulation, we are primarily concerned with conduction, which happens when heat is passed through materials.
In insulating materials, we aim to prevent conduction. The goal is to keep heat from moving freely between the inside and outside of a building.
Insulation materials work by creating barriers to slow down this process. For example, in the exercise, the manufacturer's insulation is effective because its low thermal conductivity reduces the rate of conductive heat transfer. Heat naturally moves from warmer to cooler areas. In a heated house during winter, it tries to escape to the cooler outdoor environment. Using materials with low thermal conductivity minimizes this escape. When evaluating insulation, consider factors such as:
  • The environment and temperature differences you are dealing with.
  • Availability and properties of materials you can use.
  • Long-term cost savings related to energy efficiency.
The efficient limiting of heat transfer is key to maintaining comfort within a building while conserving energy.

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 spherical particle of radius \(r_{1}\) experiences uniform thermal generation at a rate of \(\dot{q}\). The particle is encapoulated by a spherical shell of cutside radius \(r_{2}\) that is cooled by ambient air. The thermal conductivities of the particle and shell are \(k_{1}\) and \(k_{2}\), respectively, where \(k_{1}=2 k_{2}\) (a) By applying the conservation of encrgy principle to spherical control volume \(A\), which is placed at an arbitrary location within the sphere, determine a relationship between the temperature gradient, \(d T / d r\), and the local radius, \(r\), for \(0 \leq r \leq r_{1}\). (b) By applying the conservation of energy principle to ppherical control volume \(B\), which is placed at an artitrary location within the spherical shell, determine a relationship between the temperature gradient, dT/Lr, and the local radius, \(r_{\text {, for }} r_{1} \leq r \leq r_{2}\). (c) On \(T-r\) coordinates, sketch the temperature distribution over the range \(0 \leq r \leq r_{2}\).

A salt-gradient solar pond is a shallow body of water that consists of three distinct fluid layers and is used to collect solar encrgy. The upper- and lower-most layers are well mixed and serve to maintain the upper and lower surfaces of the central layer at uniform temperatures \(T_{1}\) and \(T_{2}\). where \(T_{2}>T_{1}\). Although there is bulk fluid motion in the mixed layers, there is no such motion in the central layer. Consider conditions for which solar radiation absorption in the central layer provides nonuniform heat generation of the form \(\dot{q}=A e^{-u ;}\), and the lemperature distribution in the central layer is $$ T(x)=-\frac{A}{k a^{2}} e^{-a x}+B x+C $$ The quantities \(A\left(\mathrm{~W} / \mathrm{m}^{3}\right), a(1 / \mathrm{m}), B(\mathrm{~K} / \mathrm{m})\), and \(C(\mathrm{~K})\) are known constants having the prescribed units, and \(k\) is the thermal conductivity, which is also constant. (a) Obtain expressions for the rate at which heat is transferred per unit area from the lower mixed layer to the central layer and from the central layer to the upper mixed layer. (b) Determine whether cunditions are steady or transicnt. (c) Obtain an expression for the rate at which thermal cnergy is generated in the entire central layer, per unit surface area.

A young engineer is asked to design a themal protection barricr for a sensitive clectronic device that might be exposed to irradiation from a high- powered infrarcd laser. Having learmed as a student thut a low thermal conductivity material provides good insulating characteristics, the enginecr specifies use of a nanostructured aerogel, characterized by a thermal conductivity of \(k_{e}=\) 0.0ns W/m - \(\mathrm{K}\), for the protective harrier. The engineer's boss questions the wisdom of selecting the acrogel because it has a low themal conductivity. Consider the sudden laser irradiation of (a) peire aluminum, (b) glass, and (c) aerogel. The laser provides irradiation of \(G=10 \times 10^{6}\) Whm \({ }^{2}\). The absorptivities of the materials are \(a=0.2,0.9\), and \(0.8\) for the aluminum, glass, and acrogel, rerpectively, and the initial ternperature of the barrier is \(T_{i}=300 \mathrm{~K}\). Explain why the boss is concerncd. Hint: All materials experience thetmal expunsion (or contraction), and local stresses that develop within a material are, to a first approximation, proportional io the local temperahure gradicnt.

At a given instant of time the temperature distribution within an infinite homogeneous body is given by the function $$ T(x, y, z)=x^{2}-2 y^{2}+z^{2}-x y+2 y t $$ Assuming constant propertics and no internal heat generation, determine the regions where the temperature changes with time.

Sections of the trans-Alaska pipeline run above the ground and ate supported by vertical steel shafts \((k=25\) W/m \(-\mathrm{K})\) that are \(1 \mathrm{~m}\) long and have a cross-sectional area of \(0.005 \mathrm{~m}^{2}\). Under normal cperating conditions, the temperature variation along the length of a shaft is known to be governed by an expression of the form $$ T=100-150 x+10 x^{2} $$ where \(T\) and \(x\) have units of \({ }^{\circ} \mathrm{C}\) and meters, respectively. Temperature variations are small over the shaft cross section. Evaluate the temperature and conduction heat rate at the shaft-pipeline joint \((x=0)\) and at the shaft-ground interface \((x=1 \mathrm{~m})\). Explain the difference in the heat rates.
See all solutions

Recommended explanations on Physics Textbooks

Solid State Physics

Read Explanation

Physics of Motion

Read Explanation

Energy Physics

Read Explanation

Astrophysics

Read Explanation

Electrostatics

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.