/*! 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 61 At the bottom of a very large va... [FREE SOLUTION] | 91Ó°ÊÓ

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At the bottom of a very large vacuum chamber whose walls are at \(300 \mathrm{~K}\), a black panel \(0.1 \mathrm{~m}\) in diameter is maintained at \(77 \mathrm{~K}\). To reduce the heat gain to this panel, a radiation shield of the same diameter \(D\) and an emissivity of \(0.05\) is placed very close to the panel. Calculate the net heat gain to the panel.

Short Answer

Expert verified
The net heat gain to the panel with the radiation shield is \(45.547~W\).

Step by step solution

01

Calculating area of the panel

First, we need to find the area of the panel. The panel is circular with a diameter of 0.1 m. The area of a circle is calculated as: \[A = \pi(\frac{D}{2})^2\] Plug in the values: \[A = \pi(\frac{0.1}{2})^2\] \[A = 0.007854~m^2\]
02

Calculate heat transfer without the shield

The heat transfer through radiation between two bodies can be calculated using the Stefan-Boltzmann law: \[Q = σAε(T_{c}^4 - T_{p}^4)\] Where: \(σ\) is the Stefan-Boltzmann constant (\(5.67 × 10^{-8}~W/(m^2 K^4)\)) \(A\) is the area of the panel (from Step 1) \(ε\) is the emissivity (for a black panel, ε = 1) \(T_c\) is the chamber temperature, and \(T_p\) is the panel temperature. Calculate the heat transfer without shield: \[Q = 5.67 × 10^{-8} × 0.007854 × 1 × (300^4 - 77^4)\] \[Q = 47.944~W\]
03

Account for the shield effect

The shield reduces the heat transfer between the panel and the chamber walls. We can use the shield emissivity to account for this reduction, by re-calculating the heat transfer: \[Q_{shield} = σAε_{shield}(T_{c}^4 - T_{p}^4)\] Plug in the values: \[Q_{shield} = 5.67 × 10^{-8} × 0.007854 × 0.05 × (300^4 - 77^4)\] \[Q_{shield} = 2.397~W\]
04

Calculate net heat gain to the panel

Now we need to find the net heat gain to the panel by subtracting the heat transfer with the shield from the heat transfer without the shield: \[Q_{net} = Q - Q_{shield}\] \[Q_{net} = 47.944 - 2.397\] \[Q_{net} = 45.547~W\] The net heat gain to the panel with the radiation shield is 45.547 W.

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Key Concepts

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

Stefan-Boltzmann Law
Understanding the Stefan-Boltzmann law is crucial when dealing with thermal radiation issues in physics. This principle states that the total energy radiated per unit surface area of a black body across all wavelengths each second, also known as the black body irradiance, is directly proportional to the fourth power of the black body's absolute temperature.

The law is mathematically represented by the equation \( Q = \sigma A T^4 \), where \( Q \) is the total heat energy emitted by the black body per unit time, \( \sigma \) (the Stefan-Boltzmann constant, approximately \( 5.67 \times 10^{-8} W/(m^2K^4) \)), represents the proportionality constant, \( A \) is the radiating surface area, and \( T \) is the absolute temperature in kelvins.

This formula reflects the essence of blackbody radiation, which is a perfect emitter, but real-world materials with different emissivities require an adjustment to the equation: \( Q = \sigma A \varepsilon T^4 \), incorporating the emissivity (\( \varepsilon \)) of the material.
Emissivity
Emissivity, denoted by \( \varepsilon \), is a measure of a material's ability to emit energy as thermal radiation. The value ranges from 0 to 1, with 0 meaning the material does not emit radiation efficiently (perfect reflector) and 1 indicating maximum emission efficiency (as with a perfect black body).

Materials with high emissivity look dull and black, while those with low emissivity appear shiny. The emissivity plays a crucial role in heat transfer calculations because it modifies the simple blackbody equation to accommodate real-world materials. Its integration into the Stefan-Boltzmann law allows us to accurately determine the heat transfer between surfaces: \( Q = \sigma A \varepsilon (T_c^4 - T_s^4) \), with \( T_c \) and \( T_s \) being the temperatures of the surrounding environment and the surface, respectively.
Thermal Radiation
Thermal radiation is a form of heat transfer that occurs by the emission of infrared radiation from a surface at a finite temperature. Unlike conduction and convection, thermal radiation does not require a medium to travel; it can transfer energy through the vacuum of space. This type of energy transfer is fundamental to understanding how, for example, the Earth receives energy from the Sun.

In the context of the exercise, thermal radiation is the process by which the black panel and the walls of the vacuum chamber exchange heat. Since the panel is colder, it would typically gain heat from the surrounding warm walls. The amount can be calculated using the laws governing thermal radiation, like the Stefan-Boltzmann law, to determine how much energy is radiated towards the panel.
Heat Transfer Calculation
The calculation of heat transfer, particularly by radiation, involves determining the rate at which energy is transmitted from one body to another due to a temperature difference. In our example, we first calculated the amount of heat transferred from the walls to the panel without a shield using the Stefan-Boltzmann law for a perfect black body.

However, the introduction of a radiation shield significantly impacts this calculation. The shield's emissivity changes the effective radiation heat transfer rate. By incorporating the shield's low emissivity, we substantially reduce the amount of heat the panel receives. The actual net heat gain is the difference between the two scenarios, one representing the chamber without a shield and the other with the shield in place, demonstrating how a small change in emissivity can significantly alter heat transfer rates.

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Most popular questions from this chapter

Consider the right-circular cylinder of diameter \(D\), length \(L\), and the areas \(A_{1}, A_{2}\), and \(A_{3}\) representing the base, inner, and top surfaces, respectively. (a) Show that the view factor between the base of the cylinder and the inner surface has the form \(F_{12}=2 H\left[\left(1+H^{2}\right)^{1 / 2}-H\right]\), where \(H=L D .\) (b) Show that the view factor for the inner surface to itself has the form \(F_{22}=1+H-\left(1+H^{2}\right)^{1 / 2}\).

Liquid oxygen is stored in a thin-walled, spherical container \(0.8 \mathrm{~m}\) in diameter, which is enclosed within a second thin-walled, spherical container \(1.2 \mathrm{~m}\) in diameter. The opaque, diffuse, gray container surfaces have an emissivity of \(0.05\) and are separated by an evacuated space. If the outer surface is at \(280 \mathrm{~K}\) and the inner surface is at \(95 \mathrm{~K}\), what is the mass rate of oxygen lost due to evaporation? (The latent heat of vaporization of oxygen is \(2.13 \times 10^{5} \mathrm{~J} / \mathrm{kg}\).)

Consider the cavities formed by a cone, cylinder, and sphere having the same opening size \((d)\) and major dimension \((L)\), as shown in the diagram. (a) Find the view factor between the inner surface of each cavity and the opening of the cavity. (b) Find the effective emissivity of each cavity, \(\varepsilon_{e}\), as defined in Problem 13.43, assuming the inner walls are diffuse and gray with an emissivity of \(\varepsilon_{1 N^{-}}\) (c) For each cavity and wall emissivities of \(\varepsilon_{w^{\prime}}=0.5\), \(0.7\), and \(0.9\), plot \(\varepsilon_{e}\) as a function of the major dimension- to-opening size ratio, \(L /\), over a range from 1 to 10 .

A flat-bottomed hole \(6 \mathrm{~mm}\) in diameter is bored to a depth of \(24 \mathrm{~mm}\) in a diffuse, gray material having an emissivity of \(0.8\) and a uniform temperature of \(1000 \mathrm{~K} .\) (a) Determine the radiant power leaving the opening of the cavity. (b) The effective emissivity \(\varepsilon_{e}\) of a cavity is defined as the ratio of the radiant power leaving the cavity to that from a blackbody having the area of the cavity opening and a temperature of the inner surfaces of the cavity. Calculate the effective emissivity of the cavity described above. (c) If the depth of the hole were increased, would \(\varepsilon_{e}\) increase or decrease? What is the limit of \(s_{\epsilon}\) as the depth increases?

Consider two large, diffuse, gray, parallel surfaces separated by a small distance. If the surface emissivities are \(0.8\), what emissivity should a thin radiation shield have to reduce the radiation heat transfer rate between the two surfaces by a factor of 10 ?

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