/*! 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 109 A deep cavity of \(50-\mathrm{mm... [FREE SOLUTION] | 91影视

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Chapter 12: Problem 109

A deep cavity of \(50-\mathrm{mm}\) diameter approximates a blackbody and is maintained at \(250^{\circ} \mathrm{C}\) while exposed to solar irradiation of \(800 \mathrm{~W} / \mathrm{m}^{2}\) and surroundings and ambient air at \(25^{\circ} \mathrm{C}\). A thin window of spectral transmissivity and reflectivity \(0.9\) and 0 , respectively, for the spectral range \(0.2\) to \(4 \mu \mathrm{m}\) is placed over the cavity opening. In the spectral range beyond \(4 \mu \mathrm{m}\), the window behaves as an opaque, diffuse, gray body of emissivity \(0.95\). Assuming that the convection coefficient on the upper surface of the window is \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\),

Short Answer

Expert verified
First, convert the given temperatures from Celsius to Kelvin as \(T_{cavity} = 523.15 K\) and \(T_{ambient} = 298.15 K\). Next, calculate the energy absorbed by the window due to solar radiation as \(Q_{absorbed} = 720 W/m^2\). Then, calculate the energy emitted by the window as a gray body using the Stefan-Boltzmann law and the energy loss due to convection. Apply the energy balance equation as follows: \(720 = 0.95 \cdot 5.67 \cdot 10^{-8} \cdot T_{window}^4 + 10 \cdot (T_{window} - 298.15)\). Solve this nonlinear equation for \(T_{window}\) using a numerical method, such as the Newton-Raphson method or bisection method, to find the temperature of the window's upper surface.

Step by step solution

01

Conversion to Kelvin temperature scale

First, convert the given temperatures from Celsius to Kelvin. For the cavity temperature: \(T_{cavity} = 250 + 273.15 = 523.15 K\) For the ambient air temperature: \(T_{ambient} = 25 + 273.15 = 298.15 K\)
02

Calculate energy absorbed by the window due to solar radiation

\ We are given the solar irradiation as \(800 W/m^2\). To determine the energy absorbed by the window, we need to multiply this by the window's transmissivity. The window's spectral transmissivity is \(0.9\), so the net energy absorbed by the window from the solar radiation is: \(Q_{absorbed} = 0.9 \cdot 800 = 720 W/m^2\)
03

Calculate energy emitted by the window as a gray body

We are given the emissivity of the window in the spectral range beyond \(4\mu m\) to be \(0.95\). Using the Stefan-Boltzmann law to calculate the energy emitted by the window, we have: \(Q_{emitted} = \epsilon \cdot \sigma \cdot T_{window}^4\) Where \(\epsilon\) is the emissivity of the window, \(\sigma\) is the Stefan-Boltzmann constant (\(5.67 \cdot 10^{-8} W/m^2 K^4\)), and \(T_{window}\) is the temperature of the window in Kelvin.
04

Calculate energy loss due to convection

We are given the convection coefficient on the upper surface of the window is \(10 W/m^2 K\). The energy loss due to convection can be calculated using: \(Q_{conv} = h_c \cdot (T_{window} - T_{ambient})\) Where \(h_c\) is the convection coefficient.
05

Apply energy balance equation to find the window's temperature

The energy balance equation states that the energy absorbed by the window equals the sum of the emitted energy plus the convective energy loss: \(Q_{absorbed} = Q_{emitted} + Q_{conv}\) Substituting the equations from steps 2, 3, and 4, we get: \(720 = 0.95 \cdot 5.67 \cdot 10^{-8} \cdot T_{window}^4 + 10 \cdot (T_{window} - 298.15)\) This is a nonlinear equation in \(T_{window}\), and we need to solve for \(T_{window}\). One common approach is to use an iterative method, such as the Newton-Raphson method or bisection method, to numerically solve for \(T_{window}\). By applying a numerical method, we can find the temperature of the window's upper surface.

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Key Concepts

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

Blackbody Radiation
Blackbody radiation refers to the way objects emit thermal radiation. A blackbody is an idealized physical object that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence.
This means a perfect blackbody does not reflect or transmit any radiation, only emitting energy in the form of heat, making it the most efficient emitter of radiation. In practical situations, no object is a perfect blackbody, but many can approximate this behavior in certain conditions. For instance, a deep cavity, such as the one described in the original exercise, can act as a near-perfect blackbody because it absorbs nearly all incoming radiation with minimal reflection and transmission.
Because of this property, the emitted radiation depends solely on the object's temperature.
Solar Irradiation
Solar irradiation is the power per unit area received from the Sun in the form of electromagnetic radiation. This power incidence is a crucial component of Earth's climate system and has a significant influence on both weather patterns and long-term climate change. In the context of the exercise, solar irradiation is described as being 800 W/m虏, which strikes a window over the cavity.
To determine how much energy the window absorbs, we multiply this solar irradiation by the window's transmissivity, which indicates the portion of radiation that passes through.
Thus, calculating the absorbed power helps us understand subsequent energy exchanges between the window and its surroundings.
Emissivity
Emissivity is a measure of an object's ability to emit infrared energy compared to a perfect blackbody and is crucial in thermal radiation calculations. It is defined as the ratio of the radiation emitted by a material to the radiation emitted by a blackbody at the same temperature. Values of emissivity range from 0 to 1:
  • An emissivity of 1 implies the object is a perfect emitter (blackbody).
  • Values less than 1 indicate that the object emits less radiation than a blackbody.
In the exercise, the window has an emissivity value of 0.95 beyond the spectral range of 4 碌m, indicating it is a good, but not perfect, emitter of radiation in that range.
This concept is crucial for calculating the energy emitted by objects as outlined in the Stefan-Boltzmann Law.
Stefan-Boltzmann Law
The Stefan-Boltzmann Law is a fundamental principle of thermal radiation and states that the total energy radiated per unit surface area of a blackbody is directly proportional to the fourth power of the blackbody's temperature.Mathematically, it is expressed as:\[ Q = \epsilon \cdot \sigma \cdot T^4 \]Where:
  • \( Q \) is the energy radiated per unit area.
  • \( \epsilon \) is the emissivity of the material.
  • \( \sigma \) is the Stefan-Boltzmann constant, approximately \( 5.67 \cdot 10^{-8} W/m^2 \, K^4 \).
  • \( T \) is the absolute temperature in Kelvin.
In our exercise, this law is applied to calculate how much energy the window emits based on its temperature and emissivity, enabling us to perform an accurate energy balance and determine the window's temperature.

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Most popular questions from this chapter

One scheme for extending the operation of gas turbine blades to higher temperatures involves applying a ceramic coating to the surfaces of blades fabricated from a superalloy such as inconel. To assess the reliability of such coatings, an apparatus has been developed for testing samples under laboratory conditions. The sample is placed at the bottom of a large vacuum chamber whose walls are cryogenically cooled and which is equipped with a radiation detector at the top surface. The detector has a surface area of \(A_{d}=10^{-5} \mathrm{~m}^{2}\), is located at a distance of \(L_{\text {sl }}=1 \mathrm{~m}\) from the sample, and views radiation originating from a portion of the ceramic surface having an area of \(\Delta A_{c}=10^{-4} \mathrm{~m}^{2}\). An electric heater attached to the bottom of the sample dissipates a uniform heat flux, \(q_{b}^{\prime \prime}\), which is transferred upward through the sample. The bottom of the heater and sides of the sample are well insulated. Consider conditions for which a ceramic coating of thickness \(L_{c}=0.5 \mathrm{~mm}\) and thermal conductivity \(k_{c}=\) \(6 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) has been sprayed on a metal substrate of thickness \(L_{s}=8 \mathrm{~mm}\) and thermal conductivity \(k_{s}=\) \(25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The opaque surface of the ceramic may be approximated as diffuse and gray, with a total, hemispherical emissivity of \(\varepsilon_{c}=0.8\). (a) Consider steady-state conditions for which the bottom surface of the substrate is maintained at \(T_{1}=1500 \mathrm{~K}\), while the chamber walls (including the surface of the radiation detector) are maintained at \(T_{w}=90 \mathrm{~K}\). Assuming negligible thermal contact resistance at the ceramic- substrate interface, determine the ceramic top surface temperature \(T_{2}\) and the heat flux \(q_{b}^{\prime \prime}\). (b) For the prescribed conditions, what is the rate at which radiation emitted by the ceramic is intercepted by the detector?

A furnace with a long, isothermal, graphite tube of diameter \(D=12.5 \mathrm{~mm}\) is maintained at \(T_{f}=2000 \mathrm{~K}\) and is used as a blackbody source to calibrate heat flux gages. Traditional heat flux gages are constructed as blackened thin films with thermopiles to indicate the temperature change caused by absorption of the incident radiant power over the entire spectrum. The traditional gage of interest has a sensitive area of \(5 \mathrm{~mm}^{2}\) and is mounted coaxial with the furnace centerline, but positioned at a distance of \(L=60 \mathrm{~mm}\) from the beginning of the heated section. The cool extension tube serves to shield the gage from extraneous radiation sources and to contain the inert gas required to prevent rapid oxidation of the graphite tube. (a) Calculate the heat flux \(\left(\mathrm{W} / \mathrm{m}^{2}\right)\) on the traditional gage for this condition, assuming that the extension tube is cold relative to the furnace. (b) The traditional gage is replaced by a solid-state (photoconductive) heat flux gage of the same area, but sensitive only to the spectral region between \(0.4\) and \(2.5 \mu \mathrm{m}\). Calculate the radiant heat flux incident on the solid-state gage within the prescribed spectral region. (c) Calculate and plot the total heat flux and the heat flux in the prescribed spectral region for the solidstate gage as a function of furnace temperature for the range \(2000 \leq T_{f} \leq 3000 \mathrm{~K}\). Which gage will have an output signal that is more sensitive to changes in the furnace temperature?

A two-color pyrometer is a device that is used to measure the temperature of a diffuse surface, \(T_{s}\). The device measures the spectral, directional intensity emitted by the surface at two distinct wavelengths separated by \(\Delta \lambda\). Calculate and plot the ratio of the intensities \(I_{\lambda+\Delta, e}\left(\lambda+\Delta \lambda, \theta, \phi, T_{s}\right)\) and \(I_{\lambda, e}\left(\lambda, \theta, \phi, T_{s}\right)\) as a function of the surface temperature over the range \(500 \mathrm{~K} \leq T_{s} \leq 1000 \mathrm{~K}\) for \(\lambda=5 \mu \mathrm{m}\) and \(\Delta \lambda=0.1,0.5\), and \(1 \mu \mathrm{m}\). Comment on the sensitivity to temperature and on whether the ratio depends on the emissivity of the surface. Discuss the tradeoffs associated with specification of the various values of \(\Delta \lambda\). Hint: The change in the emissivity over small wavelength intervals is modest for most solids, as evident in Figure 12.17.

A radiation thermometer is a radiometer calibrated to indicate the temperature of a blackbody. A steel billet having a diffuse, gray surface of emissivity \(0.8\) is heated in a furnace whose walls are at \(1500 \mathrm{~K}\). Estimate the temperature of the billet when the radiation thermometer viewing the billet through a small hole in the furnace indicates \(1160 \mathrm{~K}\).

A spherical aluminum shell of inside diameter \(D=2 \mathrm{~m}\) is evacuated and is used as a radiation test chamber. If the inner surface is coated with carbon black and maintained at \(600 \mathrm{~K}\), what is the irradiation on a small test surface placed in the chamber? If the inner surface were not coated and maintained at \(600 \mathrm{~K}\), what would the irradiation be?
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