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The cylindrical peephole in a furnace wall of thickness \(L=250 \mathrm{~mm}\) has a diameter of \(D=125 \mathrm{~mm}\). The furnace interior has a temperature of \(1300 \mathrm{~K}\), and the surroundings outside the furnace have a temperature of \(300 \mathrm{~K}\). Determine the heat loss by radiation through the peephole.

Short Answer

Expert verified
The heat loss by radiation through the peephole is approximately \(1227.18 \mathrm{~W}\).

Step by step solution

01

Calculate the area of the peephole

To determine the area of the peephole, we will use the following formula for the area of a circle: \(A = \pi (D/2)^2\) Given the diameter of the peephole (\(D = 125 \mathrm{~mm}\)), we first need to convert it to meters: \(D = 125 \mathrm{~mm} × (1 \mathrm{~m}/1000 \mathrm{~mm}) = 0.125 \mathrm{~m}\) Now, we can calculate the area of the peephole: \(A = \pi (0.125/2)^2 = 0.0122718463 \mathrm{~m^2}\)
02

Calculate the temperature difference

The given temperatures of the furnace interior and surroundings are \(1300 \mathrm{~K}\) and \(300 \mathrm{~K}\), respectively. The temperature difference (\(\Delta T\)) can be calculated as follows: \(\Delta T = T_\mathrm{furnace} - T_\mathrm{surroundings} = 1300 \mathrm{~K} - 300 \mathrm{~K} = 1000 \mathrm{~K}\)
03

Calculate the heat loss

Now, we can calculate the heat loss (\(Q\)) using the Stefan-Boltzmann law: \(Q = \sigma A F_1 \Delta T^4\) Substituting the known values into the equation: \(Q = (5.67×10^{-8} \mathrm{W/m^2K^4})(0.0122718463 \mathrm{~m^2})(1.0)(1000 \mathrm{~K})^4\) \(Q = 1227.18463 \mathrm{~W}\) So, the heat loss through radiation through the peephole is approximately \(1227.18 \mathrm{~W}\).

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Key Concepts

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

Stefan-Boltzmann Law
The Stefan-Boltzmann law is a fundamental principle in physics that describes how an object's temperature affects the amount of thermal radiation it emits. Simply put, this law tells us that the total energy radiated per unit surface area of a black body is directly proportional to the fourth power of the black body's temperature. This is crucial in understanding heat transfer problems related to radiation.

The mathematical expression of the Stefan-Boltzmann law is: \[Q = \text{e} \times \sigma A T^4\], where:
  • \(Q\) is the total heat energy radiated per unit time,
  • \(e\) is the emissivity of the material (ranging from 0 for a perfect reflector to 1 for a perfect black body),
  • \(\sigma\) (the Stefan-Boltzmann constant) is approximately 5.67 x 10^-8 W/m^2K^4,
  • \(A\) is the area through which the heat is being radiated, and
  • \(T\) is the absolute temperature of the body in kelvins (K).
It's important to note that this law applies ideally to black bodies, which are perfect emitters of radiation. In real-world applications, materials are not perfect black bodies, so the emissivity factor \text{e} modifies the equation to reflect this reality.
Thermal Radiation
Thermal radiation represents the energy emitted by a material due to its temperature. This form of radiation is a type of electromagnetic radiation and occurs naturally in any material that is not at absolute zero. Unlike conduction and convection, thermal radiation can occur in a vacuum as it doesn't require a medium to transfer heat.

Objects emit a range of electromagnetic waves, but the amount and spectrum of the radiation depend primarily on their temperature. For instance, the hotter an object is, the more radiation it emits, and the peak wavelength of this radiation shifts towards shorter wavelengths. This phenomenon can be observed in everyday life, such as the glow from hot metal or the warmth felt from sunlight. The understanding of thermal radiation is essential in various applications, including furnace design, thermal insulation, environmental science, and space technology.
Temperature Difference
In the context of thermal radiation and heat loss, the temperature difference, often denoted as \(\Delta T\), is a driving factor for the rate of heat transfer. It is defined as the difference in temperature between two systems or between a system and its surroundings. The greater the temperature difference, the higher the rate of heat transfer tends to be.

In many practical scenarios, including our furnace peephole example, the objective is to calculate the rate of heat transfer due to a temperature difference across a boundary. Here, the inside of the furnace is much hotter than the outside environment, creating a large \(\Delta T\) which, according to the Stefan-Boltzmann law, results in significant heat loss through radiation. Understanding how to quantify this temperature difference and apply it in formulas correctly is pivotal in accurately assessing heat transfer rates in engineering, climate control, and energy efficiency analysis.

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Most popular questions from this chapter

A radiant oven for drying newsprint consists of a long duct \((L=20 \mathrm{~m})\) of semicincular cross section. The newsprint moves through the oven on a conveyor belt at a velocity of \(V=0.2 \mathrm{~m} / \mathrm{s}\). The newsprint has a water content of \(0.02 \mathrm{~kg} / \mathrm{m}^{2}\) as it enters the oven and is completely dry as it exits. To assure quality, the newsprint must be maintained at room temperature \((300 \mathrm{~K})\) during drying. To aid in maintaining this condition, all system components and the air flowing through the oven have a temperature of \(300 \mathrm{~K}\). The inner sarface of the semicaircular duct, which is of emissivity \(0.8\) and temperature \(T_{1}\), provides the radiant heat required to accomplish the drying. The wet surface of the newsprint can be considered to be black. Air entering the oven has a temperature of \(300 \mathrm{~K}\) and a relative humidity of \(20 \%\). Since the velocity of the air is large, its temperature and relative humidity can be assumed to be constant over the entire duct length. Calculate the required evaporation rate, air velocity \(u_{m}\), and temperature \(T_{1}\) that will ensure steady-state conditions for the process.

A cylindrical furnace for heat-treating materials in a spacecraft environment has a \(90-\mathrm{mm}\) diameter and an overall length of \(180 \mathrm{~mm}\). Heating elements in the 135 -mm-long section (1) maintain a refractory lining of \(\varepsilon_{1}=0.8\) at \(800^{\circ} \mathrm{C}\). The linings for the bottom (2) and upper (3) sections are made of the same refractory material, but are insulated.

A fumace having a spherical cavity of \(0.5-\mathrm{m}\) diameter contains a gas mixture at 1 atm and \(1400 \mathrm{~K}\). The mixture consists of \(\mathrm{CO}_{2}\), with a partial pressure of \(0.25 \mathrm{~atm}\) and nitrogen with a partial pressure of \(0.75 \mathrm{arm}\). If the cavity wall is black, what is the cooling rate needed to maintain its temperature at \(500 \mathrm{~K}\) ?

The lower side of a \(400-\mathrm{mm}\)-diameter disk is heated by an electric furnace, while the upper side is exposed to quiescent, ambient air and sumoundings at \(300 \mathrm{~K}\). The radiant furnace (negligible convection) is of circular construction with the bottoen surface \(\left(\alpha_{1}-0.6\right)\) and cylindrical side surface \(\left(\varepsilon_{1}=1.0\right)\) maintained af \(T_{1}=T_{2}=500 \mathrm{~K}\). The surface of the disk facing the radiant furnace is black \(\left(\varepsilon_{d, 1}=1.0\right)\). while the upper surface has an emissivity of \(\varepsilon_{d, 2}=0.8\). Assume the plate and furnace surfaces to be diffuse and gray. (a) Determine the net heat transfer rate to the disk, \(q_{\text {nated, when }} T_{d}=400 \mathrm{~K}\). (b) Plot \(q_{\text {netd as a }}\) a function of the disk temperature for \(300 \leq T_{a} \leq 500 \mathrm{~K}\), with all other conditions remaining the same. What is the steady-state temperature of the disk?

Electrical conductors, in the form of parallel plates of length \(L=40 \mathrm{~mm}\), have one edge mounted to a ceramic insulated base and are spaced a distance \(w=10 \mathrm{~mm}\) apart. The plates are exposed to large isothermal surroundings at \(T_{\text {ar }}=300 \mathrm{~K}\). The conductor (1) and ceramic (2) surfaces are diffuse and gray with emissivities of \(\varepsilon_{1}=0.8\) and \(\varepsilon_{2}=0.6\), respectively. For a prescribed operating current in the conductors, their temperature is \(T_{1}=500 \mathrm{~K}\). (a) Detemine the electrical power dissipated in a conductor plate per unit length, \(4_{1}^{r}\), considering oaly radiation exchange. What is the temperature of the insulated base, \(T_{2}\) ? (b) Detemine \(q_{i}^{\prime}\) and \(T_{2}\) when the surfaces experience convection with an airstream at \(300 \mathrm{~K}\) and a convection coefficient of \(h=25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

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