/*! 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 30 A furnace with a long, isotherma... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 30

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?

Short Answer

Expert verified
The heat flux on the traditional gage is approximately \(3.97 脳 10^5\) W/m虏. To calculate the heat flux on the solid-state gage, use Planck's law to estimate the total radiant heat flux within the given spectral range (mean of \(E(0.4 渭m)\) and \(E(2.5 渭m)\)). Create a function that computes the heat flux for a given temperature and an optional spectral range. Calculate the heat flux for a range of furnace temperatures from \(2000 K\) to \(3000 K\) for both gage types and plot the results. The gage with a steeper curve as a function of temperature will be more sensitive to changes in the furnace temperature.

Step by step solution

01

(a) Calculate the heat flux on the traditional gage

To calculate the heat flux on the traditional gage, we need the blackbody emissive power which is given by the Stefan-Boltzmann law: \(E = 蟽T_{f}^{4} \textrm{, where }\) \(蟽 = 5.67 脳 10^{-8} \frac{W}{m^2K^4}\) is the Stefan-Boltzmann constant. The heat flux will be the product of the blackbody emissive power and the view factor for the geometry between the tube and gage. The view factor can be calculated using a coaxial cylinder geometric configuration, \(F_{gage-tube} = \frac{D}{2\pi L}\) The heat flux will be: \(q = E \times F_{gage-tube} \) Calculating the heat flux: \(E = 5.67 脳 10^{-8} \times (2000)^4 = 3.61 脳 10^7 \frac{W}{m^2}\) \(F_{gage-tube} = \frac{0.0125}{2\pi 脳 0.06} = 0.01099\) \(q = 3.61 脳 10^7 \times 0.01099 = 3.97 脳 10^5\) W/m虏. The heat flux on the traditional gage is approximately \(3.97 脳 10^5\) W/m虏.
02

(b) Calculate the heat flux on the solid-state gage

For the solid-state gage, we need to find the radiant heat flux within the prescribed spectral region between \(0.4\) and \(2.5 渭m\). To do this, we need to use Planck's law and integrate from the given wavelength range. \(E(位_1, 位_2) = \int_{位_1}^{位_2} E(位)d位\) Planck's law of blackbody radiation is: \(E(位)d位 = \frac{2\pi hc^2}{位^5} \frac{1}{e^{\frac{hc}{位kT}}-1} d位\) where, \(h = 6.626 脳 10^{-34}\) Js is the Planck's constant, \(c = 3 脳 10^8 \frac{m}{s}\) is the speed of light, and \(k = 1.38 脳 10^{-23}\) JK鈦宦� is the Boltzmann constant. We can estimate the integral using the mean emissive power within the given spectral range (mean of \(E(0.4 渭m)\) and \(E(2.5 渭m)\). To calculate the mean emissive power, first convert wavelengths to meters. \(位_1 = 0.4 脳 10^{-6} m\) \(位_2 = 2.5 脳 10^{-6} m\) Now, calculate \(E(位_1)\) and \(E(位_2)\): \(E(位_1) = \frac{2\pi hc^2}{位_1^5} \frac{1}{e^{\frac{hc}{位_1kT}}-1}\) \(E(位_2) = \frac{2\pi hc^2}{位_2^5} \frac{1}{e^{\frac{hc}{位_2kT}}-1}\) To estimate the total radiant heat flux within the given spectral range, \(E(位_1, 位_2) \approx \frac{E(位_1) + E(位_2)}{2} \times (位_2 - 位_1)\) Now, we can calculate the heat flux on the solid-state gage: \(q_{solid\_state} = E(位_1, 位_2) \times F_{gage-tube}\) So, using the temp of \(2000 K\), find the \(E(位_1, 位_2)\) and the heat flux \(q_{solid\_state}\).
03

(c) Calculate and plot the total heat flux and the heat flux in the prescribed spectral region for the solid-state gage

To calculate the total heat flux and the heat flux in the prescribed spectral region for the solid-state gage, we can create a function that computes the heat flux for a given temperature and an optional spectral range. We can then use this function to calculate the heat flux for a range of furnace temperatures from \(2000 K\) to \(3000 K\). Here's a step-by-step plan to create the plot: 1. Create a function that calculates the heat flux given the temperature \(T_f\), the gage type (solid-state or traditional), and the wavelength range. In the case of the traditional gage, don't pass the wavelength range. - Use the Stefan-Boltzmann law, geometric view factor, and Planck's law as described in parts (a) and (b) to calculate the heat flux for the given temperature and wavelength range. 2. Create a list of furnace temperatures from \(2000 K\) to \(3000 K\). 3. Calculate the heat flux for each temperature in the list for both gage types (traditional and solid-state) and store these values in separate lists. 4. Plot the total heat flux and the heat flux in the prescribed spectral region for the solid-state gage as a function of furnace temperature using the lists obtained in step 3. 5. Compare the sensitivity of the output signal for the two gages to changes in furnace temperature, based on the shape of the curves in the plot. The gage with a steeper curve as a function of temperature will be more sensitive to changes in the furnace temperature. By following the above steps, you should be able to create a plot of heat flux for both the traditional and solid-state gages over a range of furnace temperatures and determine which gage is more sensitive to changes in the furnace temperature.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Blackbody Radiation
Blackbody radiation refers to the theoretical concept of an idealized physical body, known as a blackbody, which absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence. Such a body does not reflect or transmit any radiation. Instead, a blackbody in thermal equilibrium emits thermal radiation with a characteristic spectrum that depends solely on its temperature.
A perfect blackbody is a useful model in physics, as it provides a baseline for predicting the spectral and total energy distribution from objects based on their temperature. This model assumes no energy loss through reflection or transmission, thus simplifying analytical calculations.
In our exercise context, the furnace acts as a blackbody radiator. This assumption not only facilitates the calculations but also represents a real-world application in which heat flux gages are calibrated using such sources. By treating the graphite tube at 2000 K within the furnace as a blackbody, we can rely on well-established laws of physics to compute the energy flux received by the gage.
Stefan-Boltzmann Law
The Stefan-Boltzmann law is a fundamental principle in thermal physics that describes how the power radiated by a blackbody increases with temperature. It states that the total emissive power (E) of a blackbody is directly proportional to the fourth power of its absolute temperature (T_f). The law is mathematically described as: \[ E = \sigma T_{f}^{4} \] where \( \sigma \) is the Stefan-Boltzmann constant, valued at approximately \( 5.67 \times 10^{-8} \frac{W}{m^2K^4} \). The law is crucial when calculating the total energy emitted by blackbody sources like our graphite tube furnace. It illustrates an important fact: as temperature rises, the radiative power output increases very rapidly due to the fourth power dependence. This behavior underlies why high-temperature sources emit such significant amounts of radiant power.
In our calculation, we use the Stefan-Boltzmann law to determine the power emitted per unit area from the furnace. This value, known as the blackbody emissive power, becomes a key factor in finding the heat flux impinging upon the heat flux gage.
Planck's Law
Planck's Law provides a detailed account of the spectral distribution of electromagnetic radiation emitted by a blackbody at a given temperature. It encapsulates how the energy emitted at different wavelengths changes with temperature, making it fundamental for analyzing heat flux over specific spectral regions. The law is expressed as: \[ E(\lambda)d\lambda = \frac{2\pi hc^2}{\lambda^5} \frac{1}{e^{\frac{hc}{\lambdakT}}-1} d\lambda \] where \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \text{J\cdots} \)), \( c \) is the speed of light (\( 3 \times 10^8 \text{m/s} \)), and \( k \) is the Boltzmann constant (\( 1.38 \times 10^{-23} \text{J/K} \)). This formula enables us to calculate the intensity of radiation at specific wavelengths, and is particularly useful for analyzing devices sensitive to particular spectral ranges.
For our solid-state gage, which only responds to wavelengths between 0.4 渭m and 2.5 渭m, Planck's law is essential to accurately compute the radiant power received in this narrow band. It allows us to focus on the relevant spectral region, integrating Planck's function within the given bounds to find the incident radiant heat flux on the gage.
Spectral Region Analysis
Analyzing different spectral regions allows scientists to understand how heat flux varies across different segments of the electromagnetic spectrum. This is particularly important in applications involving selective absorption or emission of radiation, such as photoconductive gages that respond only to specific wavelengths.
Spectral region analysis involves breaking down the total emitted radiation into distinct wavelength bands and examining radiation characteristics within those bands. The exercise highlights this by requiring the calculation of heat flux for a solid-state gage within a specific spectral range.
For the analysis, using Planck's law over a specified wavelength range (0.4 to 2.5 渭m) helps to deduce the mean emissive power in this region. By performing an integration over these wavelengths, we identify only the energy that the solid-state gage can capture. This is essential to address the changes that occur as temperature varies, thus leading to effective calibration and enhancing the gage sensitivity evaluation.
This focused approach reveals the sensitivity of certain materials or devices to specific parts of the spectrum, potentially allowing more precise measurements and a better understanding of material properties or environmental conditions affecting sensor outputs.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Four diffuse surfaces having the spectral characteristics shown are at \(300 \mathrm{~K}\) and are exposed to solar radiation. Which of the surfaces may be approximated as being gray?

The 50 -mm peephole of a large furnace operating at \(450^{\circ} \mathrm{C}\) is covered with a material having \(\tau=0.8\) and \(\rho=0\) for irradiation originating from the furnace. The material has an emissivity of \(0.8\) and is opaque to irradiation from a source at room temperature. The outer surface of the cover is exposed to surroundings and ambient air at \(27^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of \(50 \mathrm{~W} / \mathrm{m}^{2}=\mathrm{K}\). Assuming that convection effects on the inner surface of the cover are negligible, calculate the heat loss by the furnace and the temperature of the cover.

A circular metal disk having a diameter of \(0.4 \mathrm{~m}\) is placed firmly against the ground in a barren horizontal region where the earth is at a temperature of \(280 \mathrm{~K}\). The effective sky temperature is also \(280 \mathrm{~K}\). The disk is exposed to quiescent ambient air at \(300 \mathrm{~K}\) and direct solar irradiation of \(745 \mathrm{~W} / \mathrm{m}^{2}\). The surface of the disk is diffuse with \(\varepsilon_{\lambda}=0.9\) for \(0<\lambda<1 \mu \mathrm{m}\) and \(\varepsilon_{\lambda}=0.2\) for \(\lambda>1 \mu \mathrm{m}\). After some time has elapsed, the disk achieves a uniform, steady-state temperature. The thermal conductivity of the soil is \(0.52 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). (a) Determine the fraction of the incident solar irradiation that is absorbed. (b) What is the emissivity of the disk surface? (c) For a steady-state disk temperature of \(340 \mathrm{~K}\), employ a suitable correlation to determine the average free convection heat transfer coefficient at the upper surface of the disk. (d) Show that a disk temperature of \(340 \mathrm{~K}\) does indeed yield a steady-state condition for the disk.

Square plates freshly sprayed with an epoxy paint must be cured at \(140^{\circ} \mathrm{C}\) for an extended period of time. The plates are located in a large enclosure and heated by a bank of infrared lamps. The top surface of each plate has an emissivity of \(\varepsilon=0.8\) and experiences convection with a ventilation airstream that is at \(T_{\infty}=27^{\circ} \mathrm{C}\) and provides a convection coefficient of \(h=20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The irradiation from the enclosure walls is estimated to be \(G_{\text {wall }}=450 \mathrm{~W} / \mathrm{m}^{2}\), for which the plate absorptivity is \(\alpha_{\text {wall }}=0.7\). (a) Determine the irradiation that must be provided by the lamps, \(G_{\text {lamp. }}\). The absorptivity of the plate surface for this irradiation is \(\alpha_{\text {Lamp }}=0.6\). (b) For convection coefficients of \(h=15,20\), and \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), plot the lamp irradiation, \(G_{\text {lamp, as a }}\) function of the plate temperature, \(T_{s}\), for \(100 \leq\) \(T_{x} \leq 300^{\circ} \mathrm{C}\). (c) For convection coefficients in the range from 10 to \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and a lamp irradiation of \(G_{\text {lmp }}=\) \(3000 \mathrm{~W} / \mathrm{m}^{2}\), plot the airstream temperature \(T_{x}\) required to maintain the plate at \(T_{x}=140^{\circ} \mathrm{C}\).

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?
See all solutions

Recommended explanations on Physics Textbooks

Scientific Method Physics

Read Explanation

Torque and Rotational Motion

Read Explanation

Rotational Dynamics

Read Explanation

Linear Momentum

Read Explanation

Electric Charge Field and Potential

Read Explanation

Mechanics and Materials

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.