/*! 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 108 The roof of a house consists of ... [FREE SOLUTION] | 91影视

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Chapter 1: Problem 108

The roof of a house consists of a 22-cm-thick (st) concrete slab \((k=2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) that is \(15 \mathrm{~m}\) wide and \(20 \mathrm{~m}\) long. The emissivity of the outer surface of the roof is \(0.9\), and the convection heat transfer coefficient on that surface is estimated to be \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The inner surface of the roof is maintained at \(15^{\circ} \mathrm{C}\). On a clear winter night, the ambient air is reported to be at \(10^{\circ} \mathrm{C}\) while the night sky temperature for radiation heat transfer is \(255 \mathrm{~K}\). Considering both radiation and convection heat transfer, determine the outer surface temperature and the rate of heat transfer through the roof. If the house is heated by a furnace burning natural gas with an efficiency of 85 percent, and the unit cost of natural gas is \(\$ 1.20\) / therm ( 1 therm \(=105,500 \mathrm{~kJ}\) of energy content), determine the money lost through the roof that night during a 14-hour period.

Short Answer

Expert verified
Answer: The key factors taken into consideration when calculating the money lost through the roof during a 14-hour period include the conduction, radiation, and convection heat transfers, the efficiency of the furnace, the unit cost of natural gas, and the duration of time.

Step by step solution

01

Find conduction heat transfer through the slab

For this, we will use Fourier's law of heat conduction: \(q_{cond} = k * A * \frac{T_{in} - T_{out}}{d}\) Where \(k = 2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is the thermal conductivity, \(A = 15 \mathrm{~m} * 20 \mathrm{~m}\) is the area of the slab, \(T_{in} = 15 ^{\circ} \mathrm{C}\) is the inner surface temperature, \(T_{out}\) is the outer surface temperature, and \(d = 0.22 \mathrm{~m}\) is the thickness of the slab.
02

Find the radiation heat transfer from the slab to the sky

We will use the radiation heat transfer equation: \(q_{rad} = \epsilon * A * \sigma * (T_{out}^{4} - T_{sky}^{4})\) Where \(\epsilon = 0.9\) is the emissivity, \(\sigma = 5.67 脳 10^{-8} \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}^{4}\) is the Stefan-Boltzmann constant, and \(T_{sky} = 255 \mathrm{~K}\) is the sky temperature during the night.
03

Find the convection heat transfer from the slab to the ambient air

Using Newton's law of cooling: \(q_{conv} = h * A * (T_{out} - T_{amb})\) Where \(h = 15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) is the convection heat transfer coefficient and \(T_{amb} = 10 ^{\circ} \mathrm{C}\) is the ambient temperature.
04

Combine the heat transfer equations and find the outer surface temperature

Add the conduction, radiation, and convection heat transfer equations together: \(q_{cond} = q_{rad} + q_{conv}\) Substitute the equations from steps 1, 2, and 3 into this equation and solve for \(T_{out}\).
05

Find the total heat transfer through the roof

Plug the value of \(T_{out}\) obtained in step 4 into the conduction heat transfer equation obtained in step 1 to calculate the heat transfer rate, \(q_{cond}\).
06

Determine the energy lost during a 14-hour period

Multiply the heat transfer rate by the duration of time: \(E_{lost} = q_{cond} * (14 \mathrm{~hours} * 3600 \mathrm{~s} / \mathrm{hour})\)
07

Determine the amount of natural gas energy required to compensate for the energy lost

We're given the efficiency of the furnace, which is 85 percent, so divide the energy lost by the efficiency to find the amount of natural gas energy required: \(E_{ng} = \frac{E_{lost}}{0.85}\)
08

Determine the cost of natural gas energy

Finally, using the unit cost of natural gas, we can calculate the money lost through the roof for a 14-hour period: \(Cost_{ng} = \frac{E_{ng}}{105,500 \mathrm{~kJ / therm}} * \$1.20 / \mathrm{therm}\) Now you have the outer surface temperature, rate of heat transfer through the roof, and the money lost through the roof during a 14-hour period.

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Key Concepts

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

Fourier's Law of Heat Conduction
At the heart of understanding how heat moves through solids like the concrete slab roof in the problem is Fourier's Law of Heat Conduction. This principle provides us with the mathematical formulation to calculate the rate of heat transfer, or heat flux, through a material. The law states that the heat flux (\(q_{cond}\)) is proportional to the negative gradient in temperature and the area through which heat is being transferred.

More simply, it can be expressed by the equation: \[q_{cond} = k \cdot A \cdot \frac{T_{in} - T_{out}}{d}\] where:
  • \(k\) is the thermal conductivity of the material, indicating how well it conducts heat,
  • \(A\) is the area perpendicular to the direction of heat flow,
  • \(T_{in} - T_{out}\) is the temperature difference between the hot and cold sides of the material,
  • and \(d\) is the thickness of the material.
In our problem, by rearranging and solving the equation for \(T_{out}\), we can determine the outside temperature of the concrete roof slab through which the heat is being transferred.
Radiation Heat Transfer
When it comes to the roof losing heat to the cold night sky, radiation heat transfer is the key mechanism at play. This mode of heat transfer doesn't require any medium and can occur across the vacuum of space, which is why the roof radiates heat to the sky even though the air temperature is different. \

Radiation Heat Transfer Equation\

The formula to calculate radiative heat transfer is: \[q_{rad} = \epsilon \cdot A \cdot \sigma \cdot (T_{out}^{4} - T_{sky}^{4})\] where:
  • \(\epsilon\) is the emissivity of the material's surface,
  • \(A\) is the area that is emitting or absorbing radiation,
  • \(\sigma\) is the Stefan-Boltzmann constant, a fundamental constant of nature,
  • and \(T_{out}^{4} - T_{sky}^{4}\) represents the difference in the fourth power of absolute temperature between the roof surface and the sky temperature.
By considering both \(T_{out}\) and \(T_{sky}\), and knowing all other variables, this part of the equation lets us calculate the energy radiating away into space.
Convection Heat Transfer
Finally, convection heat transfer is the process that describes how heat is exchanged between a surface and a fluid moving over it鈥攊n our case, the cool air brushing against the roof. This is represented in Newton's Law of Cooling which is often used to approximate the heat transfer rate due to convection.

The formula for convection heat transfer is: \[q_{conv} = h \cdot A \cdot (T_{out} - T_{amb})\] where:
  • \(h\) is the convection heat transfer coefficient, which characterizes the convective heat transfer behavior between the surface and the fluid,
  • \(A\) is the surface area,
  • \(T_{out}\) is the surface temperature,
  • and \(T_{amb}\) is the ambient temperature of the fluid.
By computing \(T_{out}\) with considerations from both radiation and convection heat loss, we can estimate how much heat is being lost from the house into the cooler air of the environment.

Merging these calculations gives us a wholesome picture of how the roof temperature affects and is affected by the surrounding environment, allowing us to determine the outer surface temperature and the rate of heat loss from the roof.

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Most popular questions from this chapter

An aluminum pan whose thermal conductivity is \(237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) has a flat bottom with diameter \(15 \mathrm{~cm}\) and thickness \(0.4 \mathrm{~cm}\). Heat is transferred steadily to boiling water in the pan through its bottom at a rate of \(1400 \mathrm{~W}\). If the inner surface of the bottom of the pan is at \(105^{\circ} \mathrm{C}\), determine the temperature of the outer surface of the bottom of the pan.

A soldering iron has a cylindrical tip of \(2.5 \mathrm{~mm}\) in diameter and \(20 \mathrm{~mm}\) in length. With age and usage, the tip has oxidized and has an emissivity of \(0.80\). Assuming that the average convection heat transfer coefficient over the soldering iron tip is \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and the surrounding air temperature is \(20^{\circ} \mathrm{C}\), determine the power required to maintain the tip at \(400^{\circ} \mathrm{C}\).

Determine a positive real root of this equation using \(E E S\) : $$ 3.5 x^{3}-10 x^{0.5}-3 x=-4 $$

A flat-plate solar collector is used to heat water by having water flow through tubes attached at the back of the thin solar absorber plate. The absorber plate has a surface area of \(2 \mathrm{~m}^{2}\) with emissivity and absorptivity of \(0.9\). The surface temperature of the absorber is \(35^{\circ} \mathrm{C}\), and solar radiation is incident on the absorber at \(500 \mathrm{~W} / \mathrm{m}^{2}\) with a surrounding temperature of \(0^{\circ} \mathrm{C}\). Convection heat transfer coefficient at the absorber surface is \(5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), while the ambient temperature is \(25^{\circ} \mathrm{C}\). Net heat rate absorbed by the solar collector heats the water from an inlet temperature \(\left(T_{\text {in }}\right)\) to an outlet temperature \(\left(T_{\text {out }}\right)\). If the water flow rate is \(5 \mathrm{~g} / \mathrm{s}\) with a specific heat of \(4.2 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\), determine the temperature rise of the water.

A solid plate, with a thickness of \(15 \mathrm{~cm}\) and a thermal conductivity of \(80 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), is being cooled at the upper surface by air. The air temperature is \(10^{\circ} \mathrm{C}\), while the temperatures at the upper and lower surfaces of the plate are 50 and \(60^{\circ} \mathrm{C}\), respectively. Determine the convection heat transfer coefficient of air at the upper surface and discuss whether the value is reasonable or not for force convection of air.
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