/*! 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 69 Long, cylindrical bars are heat-... [FREE SOLUTION] | 91Ó°ÊÓ

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Long, cylindrical bars are heat-treated in an infrared oven. The bars, of diameter \(D=50 \mathrm{~mm}\), are placed on an insulated tray and are heated with an overhead infrared panel maintained at temperature \(T_{p}=800 \mathrm{~K}\) with \(\varepsilon_{p}=0.85\). The bars are at \(T_{b}=300 \mathrm{~K}\) and have an emissivity of \(\varepsilon_{b}=0.92\). (a) For a product spacing of \(s=100 \mathrm{~mm}\) and a product length of \(L=1 \mathrm{~m}\), determine the radiation heat flux delivered to the product. Determine the heat flux at the surface of the panel heater. (b) Plot the radiation heat flux experienced by the product and the panel heater radiation heat flux over the range \(50 \mathrm{~mm} \leq s \leq 250 \mathrm{~mm}\).

Short Answer

Expert verified
For a product spacing of \(s = 100\)mm and a product length of \(L = 1\)m, the radiation heat flux delivered to the product (cylindrical bars) is approximately \(3973.01 \text{ W/m}^2\), and the heat flux at the surface of the panel heater is approximately \(12924.34 \text{ W/m}^2\). To plot the radiation heat flux experienced by the product and the panel heater radiation heat flux over the range of product spacings \(s = 50 \text{ mm }\) to \(s = 250 \text{ mm }\), calculate the radiation heat flux to bars (\(q_{\text{to bars}}\)) and the panel heater radiation heat flux (\(q_{\text{from panel}}\)) for each value of \(s\) in the specified range and plot the values on a graph. The radiation heat flux will vary depending on the product spacing.

Step by step solution

01

We are given: - Diameter of the cylindrical bars, \(D = 50 \)mm - Oven temperature (Panel), \(T_{p} = 800 \)K - Panel emissivity, \(\varepsilon_{p} = 0.85\) - Bars temperature, \(T_{b} = 300 \)K - Bars emissivity, \(\varepsilon_{b} = 0.92\) - Product spacing, \(s = 100 \)mm - Product length, \(L = 1 \)m #Step 2: Find the view factor for the bar-panel system#

View factor is the proportion of radiation from one surface that is incident on another surface in a radiation exchange. For the given spacing and dimensions, we can approximate the view factor F as: \(F\approx\frac{s}{\sqrt{D^2+s^2}}\). Now, we can plug in the values: \(F\approx\frac{100}{\sqrt{50^2 + 100^2}}\approx0.894\) #Step 3: Calculate radiation heat flux to bars using the Stefan-Boltzmann Law#
02

The Stefan-Boltzmann Law states: \(q = \sigma \varepsilon_{1} T_{1}^4 - \sigma \varepsilon_{2} T_{2}^4\), where \(\sigma\) is the Stefan-Boltzmann constant, \(\sigma = 5.67\times10^{-8} \)W m^-2 K^-4. Now we can calculate the radiation heat flux, \(q_{\text{to bars}}\), to the bars: \(q_{\text{to bars}} = F(\sigma \varepsilon_{p} T_{p}^4 - \sigma \varepsilon_{b} T_{b}^4)\) Plug in the given parameter values, and we get: \(q_{\text{to bars}} \approx 0.894(5.67\times10^{-8})(0.85\times800^4 - 0.92\times300^4) \approx 3973.01 \text{ W/m}^2\) #Step 4: Calculate heat flux at the surface of the panel heater#

The radiation heat flux at the surface of the panel heater, \(q_{\text{from panel}}\), can be found using the Stefan-Boltzmann Law: \(q_{\text{from panel}} = \sigma \varepsilon_{p} T_{p}^4\) Substitute the values and we get: \(q_{\text{from panel}} = 5.67\times10^{-8}\times0.85\times800^4 \approx 12924.34 \text{ W/m}^2\) #a)#
03

(a) Calculate radiation heat flux delivered to the product and heat flux at the surface of the panel heater#

For a product spacing of \(s = 100\)mm and a product length of \(L = 1\)m, the radiation heat flux delivered to the product (cylindrical bars) is approximately \(3973.01 \text{ W/m}^2\), and the heat flux at the surface of the panel heater is approximately \(12924.34 \text{ W/m}^2\). #b)#
04

(b) Plot the radiation heat flux experienced by the product and the panel heater radiation heat flux over the range of product spacings#

To create the desired plot, we repeat steps 3 and 4 for the given range of product spacings \(s = 50 \text{ mm }\) to \(s = 250 \text{ mm }\). Calculate the radiation heat flux to bars (\(q_{\text{to bars}}\)) and the panel heater radiation heat flux (\(q_{\text{from panel}}\)) for each value of \(s\) in the specified range. Plot the values on a graph, with the product spacing (\(s\)) on the horizontal axis and the radiation heat flux on the vertical axis. The radiation heat flux experienced by the product and the panel heater radiation heat flux will vary depending on the product spacing.

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Key Concepts

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

Understanding the Stefan-Boltzmann Law
The Stefan-Boltzmann Law is pivotal when studying radiation heat transfer. At its core, it expresses the relationship between the thermal radiation emitted from a black body and its temperature. Mathematically, it is represented as:
\[q = \text{\(\sigma\)} \varepsilon T^4\]
where \(q\) is the radiation heat flux, \(\sigma\) is the Stefan-Boltzmann constant approximately equal to \(5.67\times10^{-8} \)W/m^2K^4, \(\varepsilon\) is the emissivity of the material, and \(T\) is the absolute temperature in Kelvin.
  • \(q\): Total energy radiated per unit surface area
  • \(\sigma\): A constant that defines the proportionality
  • \(\varepsilon\): A dimensionless measure of a real object's emissive ability relative to a perfect black body
  • \(T\): Absolute temperature of the body in Kelvin, raised to the fourth power

When objects are not perfect black bodies, the concept of emissivity comes into play, modifying the straightforward calculation done with the Stefan-Boltzmann Law. Understanding and correctly applying this formula is essential in determining the heat transfer between objects, especially in industrial applications such as the heat treatment of cylindrical bars by an infrared oven as shown in our exercise.
Emissivity and Its Role in Heat Transfer
Emissivity, represented by the symbol \(\varepsilon\), is a crucial factor in determining how much radiation an object emits compared to a perfect black body, which is by definition an ideal emitter of thermal radiation. The value of emissivity ranges between 0 and 1, where 1 corresponds to a perfect black body and 0 represents a perfect reflector that does not emit radiation at all.
Materials with high emissivity are excellent radiators, absorbing and emitting the majority of the thermal energy they come in contact with. Conversely, low-emissivity materials reflect more thermal energy than they emit. For instance, in our exercise, the panel's emissivity of \(0.85\) and the bars' emissivity of \(0.92\) signify that they are good but not perfect emitters. The heat flux that is exchanged between the panel and the bars can be calculated by considering these emissivity values. They ensure that the output temperature and heat flux are adjusted to reflect the material's actual capacities to emit radiation.

Significance in Industrial Applications

Emissivity plays a vital role in industrial settings such as manufacturing and thermal processing. Accurate emissivity values allow engineers to predict and control the heating and cooling of materials effectively. For example, the heating of cylindrical bars in an oven, as in our exercise, requires knowledge of their emissivity to accurately determine the heat flux received from the infrared panel.
The View Factor in Radiative Heat Transfer
The view factor, also known as configuration factor or shape factor, is a dimensionless quantity that represents the fraction of radiation leaving one surface that directly reaches another surface. It depends solely on the geometry of the system. In practical terms, it can be seen as a measure of how 'visible' one surface is to another with respect to radiative heat transfer.
For the complex geometries often found in the real world, the calculation of view factors is an essential step in determining the radiation heat exchange between surfaces. The view factor is especially critical when dealing with surfaces that are not directly facing each other, as it will significantly impact the amount of heat transferred.

Implications for the Exercise

In the context of our exercise, understanding the view factor allows us to calculate the radiation heat flux between the infrared panel and cylindrical bars. With the given dimensions and spacing of the bars beneath the panel, approximating the view factor as we did in Step 2 of the solution is necessary to obtain an accurate assessment of the heat flux delivered to the product. This further emphasizes the dual role of geometry and emissivity in dictating the effectiveness of radiative heat treatments.

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Most popular questions from this chapter

Consider the perpendicular rectangles shown schematically. (a) Determine the shape factor \(F_{12}\). (b) For rectangle widths of \(X=0.5,1.5\), and \(5 \mathrm{~m}\), plot \(F_{12}\) as a function of \(Z_{b}\) for \(0.05 \leq Z_{b} \leq 0.4 \mathrm{~m}\).] Compare your results with the view factor obtained from the two-dimensional relation for perpendicular plates with a common edge (Table 13.1).

Most architects know that the ceiling of an ice-skating rink must have a high reflectivity. Otherwise, condensation may occur on the ceiling, and water may drip onto the ice, causing bumps on the skating surface. Condensation will occur on the ceiling when its surface temperature drops below the dew point of the rink air. Your assignment is to perform an analysis to determine the effect of the ceiling emissivity on the ceiling temperature, and hence the propensity for condensation. The rink has a diameter of \(D=50 \mathrm{~m}\) and a height of \(L=10 \mathrm{~m}\), and the temperatures of the ice and walls are \(-5^{\circ} \mathrm{C}\) and \(15^{\circ} \mathrm{C}\), respectively. The rink air temperature is \(15^{\circ} \mathrm{C}\), and a convection coefficient of \(5 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\) characterizes conditions on the ceiling surface. The thickness and thermal conductivity of the ceiling insulation are \(0.3 \mathrm{~m}\) and \(0.035 \mathrm{~W} / \mathrm{m}-\mathrm{K}\), respectively, and the temperature of the outdoor air is \(-5^{\circ} \mathrm{C}\). Assume that the ceiling is a diffuse-gray surface and that the Walls and ice may be approximated as blackbodies. (a) Consider a flat ceiling having an emissivity of \(0.05\) (highly reflective panels) or \(0.94\) (painted panels). Perform an energy balance on the ceiling to calculate the corresponding values of the ceiling temperature. If the relative humidity of the rink air is \(70 \%\), will condensation occur for either or both of the emissivities? (b) For each of the emissivities, calculate and plot the ceiling temperature as a function of the insulation thickness for \(0.1 \leq r \leq 1 \mathrm{~m}\). Identify conditions for which condensation will occur on the ceiling.

Hot coffee is contained in a cylindrical thermos bottle that is of length \(L=0.3 \mathrm{~m}\) and is lying on its side (horizontally). The coffee containet consists of a glass flask of diameter \(D_{1}=0.07 \mathrm{~m}\), separated from an aluminum housing of diameter \(D_{2}=0.08 \mathrm{~m}\) by air at atmospheric pressure. The outer surface of the flask and the inner surface of the housing are silver coated to provide emissivities of \(\varepsilon_{1}=\varepsilon_{2}=0.25\). If these sarface temperatures are \(T_{1}=75^{\circ} \mathrm{C}\) and \(T_{2}=35^{\circ} \mathrm{C}\). what is the heat loss froen the coffee?

The arrangement shown is to be used to calibrate a heat flux gage. The gage has a black surface that is \(10 \mathrm{~mm}\) in diameter and is maintained at \(17^{\circ} \mathrm{C}\) by means of a water-cooled backing plate. The heater, \(200 \mathrm{~mm}\) in diameter, has a black surface that is maintained at \(800 \mathrm{~K}\) and is located \(0.5 \mathrm{~m}\) from the gage. The surroundings and the air are at \(27^{\circ} \mathrm{C}\) and the convection heat transfer coefficient between the gage and the air is \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Determine the net radiation exchange between the heater and the gage. (b) Determine the net transfer of radiation to the gage per unit area of the gage. (c) What is the net heat transfer rate to the gage per unit area of the gage? (d) If the gage is constructed according to the description of Problem 3.107, what heat flux will it indicate?

A radiometer views a small target (1) that is being heated by a ring-shaped disk heater (2). The target has an area of \(A_{1}=0.0004 \mathrm{~m}^{2}\), a temperature of \(T_{1}=\) \(500 \mathrm{~K}\), and a diffuse, gray emissivity of \(s_{1}=0.8\). The heater operates at \(T_{2}=1000 \mathrm{~K}\) and has a black surface. The radiometer views the entire sample area with a solid angle of \(\omega=0.0008 \mathrm{sr}\). (a) Write an expression for the radiant power leaving the target which is collected by the radiometer, in terms of the target radiosity \(J_{1}\) and relevant geometric parameters. Leave in symbolic form. (b) Write an expression for the target radiosity \(J_{1}\) in terms of its irradiation, emissive power, and appropriate radiative properties. Leave in symbolic form. (c) Write an expression for the irradiation on the target, \(G_{1}\), due to emission from the heater in terms of the heater emissive power, the heater area, and an appropriate view factor. Use this expression to numerically evaluate \(G_{1}\). (d) Use the foregoing expressions and results to determine the radiant power collected by the radiometer.

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