/*! 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 299 The sun shines on a \(1500-\math... [FREE SOLUTION] | 91Ó°ÊÓ

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The sun shines on a \(1500-\mathrm{ft}^{2}\) road surface so that it is at \(115 \mathrm{~F}\). Below the 2 -in.-thick asphalt, average conductivity of \(0.035 \mathrm{Btu} / \mathrm{h} \mathrm{ft} \mathrm{F},\) is a layer of compacted rubble at a temperature of \(60 \mathrm{~F}\). Find the rate of heat transfer to the rubble.

Short Answer

Expert verified
The rate of heat transfer is 17325 Btu/h.

Step by step solution

01

Identify the Given Parameters

We need to find the rate of heat transfer through the asphalt layer. We are given:- Surface area, \(A = 1500 \text{ ft}^2\)- Surface temperature, \(T_s = 115\text{ °F}\)- Temperature of the bottom layer, \(T_r = 60\text{ °F}\)- Thickness of the asphalt, \(L = 2\text{ in}\)- Thermal conductivity, \(k = 0.035\text{ Btu/h ft °F}\)
02

Convert Units for Consistency

Convert thickness from inches to feet for consistency with the other units:\(L = 2\text{ in} = \frac{2}{12}\text{ ft} = \frac{1}{6}\text{ ft}\)
03

Apply the Heat Transfer Formula

The formula for heat transfer rate \(Q\) through a material is given by Fourier's Law:\[ Q = \frac{k \cdot A \cdot (T_s - T_r)}{L} \]Where:- \(Q\) is the heat transfer rate in Btu/hTo find \(Q\), substitute the given values into the formula.
04

Calculate the Rate of Heat Transfer

Substitute the given values into the formula:\[Q = \frac{0.035 \cdot 1500 \cdot (115 - 60)}{\frac{1}{6}}\]\[Q = \frac{0.035 \cdot 1500 \cdot 55}{\frac{1}{6}}\]\[Q = \frac{2887.5}{\frac{1}{6}}\]\[Q = 17325\text{ Btu/h}\]
05

State the Final Result

The rate of heat transfer to the rubble is \(17325\text{ Btu/h}\).

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Key Concepts

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

Fourier's Law
In the world of heat transfer, Fourier's Law is a foundational principle that helps us understand how heat moves through materials. At its core, Fourier's Law describes the rate at which heat energy flows through a material. This is dependent on the material's properties, temperature difference across the material, and the distance heat must travel.
To visualize this, imagine a slab of material like asphalt being heated by the sun. Heat moves from the hotter surface to the cooler region underneath. Fourier's Law mathematically expresses this flow as:
  • Heat transfer rate, \( Q \)
  • Material thermal conductivity, \( k \)
  • Surface area through which heat passes, \( A \)
  • Temperature difference between the hot and cold sides, \( T_s - T_r \)
  • Thickness of the material, \( L \)
The steady state form of Fourier's Law can be written as:\[Q = \frac{k \cdot A \cdot (T_s - T_r)}{L}\]This formula helps us calculate the amount of thermal energy transferred over time through a given material, helping engineers design systems to manage heat effectively.
Thermal Conductivity
Thermal conductivity, noted as \( k \), is a measure of how well a material can conduct heat. If a material has high thermal conductivity, it typically transfers heat quickly, making it a good conductor. Conversely, if the thermal conductivity is low, the material is a poor conductor and acts as an insulator.
In the context of the asphalt road in the exercise, the thermal conductivity value was given as \(0.035 \text{ Btu/h ft °F}\). This signifies that for every degree Fahrenheit temperature difference across a one-foot thickness, a square foot of asphalt can transmit 0.035 British thermal units (Btu) per hour.
Materials like metals are known to possess high thermal conductivities, while materials like air and foam have low values. In engineering applications, knowing a material's thermal conductivity helps you select the right material for effective heat management. Whether you're designing a thermal barrier or a heat sink, understanding these properties is essential.
Heat Transfer Rate Calculation
The heat transfer rate calculation helps determine how much thermal energy is moving through a material over time. This is critical in engineering applications where controlling the temperature of a system is essential, such as in building insulation or heating systems.
In the exercise, we calculated the rate for heat moving from a hot surface to a cooler layer beneath. Using the formula derived from Fourier's Law:\[Q = \frac{k \cdot A \cdot (T_s - T_r)}{L}\]We substituted known values:
  • Thermal conductivity, \( k = 0.035 \text{ Btu/h ft °F} \)
  • Surface area, \( A = 1500 \text{ ft}^2 \)
  • Temperature difference, \( T_s - T_r = 115 - 60 = 55 \text{ °F} \)
  • Thickness, \( L = \frac{1}{6} \text{ ft} \)
These calculations reveal that the rate of heat transfer \( Q \) to the rubble underlying the asphalt is \( 17325 \text{ Btu/h} \). Such calculations are invaluable in making informed decisions for efficient thermal management in engineering endeavors.

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Most popular questions from this chapter

A piston/cylinder contains \(1.5 \mathrm{~kg}\) water at 600 \(\mathrm{kPa}, 350^{\circ} \mathrm{C} .\) It is now cooled in a process wherein pressure is linearly related to volume to a state of \(200 \mathrm{kPa}, 150^{\circ} \mathrm{C}\). Plot the \(P-v\) diagram for the process, and find both the work and the heat transfer in the process.

A cylinder fitted with a frictionless piston contains \(5 \mathrm{~kg}\) of superheated \(\mathrm{R}-134 \mathrm{a}\) vapor at \(1000 \mathrm{kPa}\) and \(140^{\circ} \mathrm{C}\). The setup is cooled at constant pressure until the R-134a reaches a quality of \(25 \%\). Calculate the work done in the process.

A piston motion moves a 25 -kg hammerhead vertically down \(1 \mathrm{~m}\) from rest to a velocity of \(50 \mathrm{~m} / \mathrm{s}\) in a stamping machine. What is the change in total energy of the hammerhead?

The brake shoe and steel drum of a car continuously absorb \(75 \mathrm{~W}\) as the car slows down. Assume a total outside surface area of \(0.1 \mathrm{~m}^{2}\) with a convective heat transfer coefficient of \(10 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\) to the air at \(20^{\circ} \mathrm{C}\). How hot does the outside brake and drum surface become when steady conditions are reached?

A steel pot, with conductivity of \(50 \mathrm{~W} / \mathrm{m} \mathrm{K}\) and a 5 -mm-thick bottom, is filled with \(15^{\circ}\) C liquid water. The pot has a diameter of \(20 \mathrm{~cm}\) and is now placed on an electric stove that delivers \(500 \mathrm{~W}\) as heat transfer. Find the temperature on the outer pot bottom surface, assuming the inner surface is at \(15^{\circ} \mathrm{C}\).

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