/*! 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 64 A diffuse, gray radiation shield... [FREE SOLUTION] | 91Ó°ÊÓ

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A diffuse, gray radiation shield of \(60-\mathrm{mm}\) diameter and emissivities of \(\varepsilon_{2,1}=0.01\) and \(\varepsilon_{2,0}=0.1\) on the inner and outer surfaces, respectively, is concentric with a long tube transporting a hot process fluäd. The tube surface is black with a diameter of \(20 \mathrm{~mm}\). The region interior to the shield is evacuated. The exterior surface of the shield is exposed to a large room whose walls are at \(17^{\circ} \mathrm{C}\) and experiences convection with air at \(27^{\circ} \mathrm{C}\) and a convection heat transfer coefficient of \(10 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\). Determine the operating temperature for the inner tube if the shield temperature is maintained at \(42^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The operating temperature of the inner tube, \(T_1\), can be found using the following steps: 1. Calculate the view factors for the radiation exchange, \(F_{2 \to 1} = \frac{1}{3}\). 2. Determine the radiative heat transfer between surfaces with shield emissivities, \(q_{1 \to 2}\). 3. Determine the convection heat transfer between the shield and surrounding air, \(q_{conv}\). 4. Set up the heat balance equation, \(q_{1 \to 2} = q_{conv}\), and solve for \(T_1\). After solving this equation for \(T_1\), we will find the operating temperature for the inner tube.

Step by step solution

01

Calculate the view factors

View factors represent the fraction of the radiation energy leaving one surface and arriving at another. In this case, since surfaces are concentric, we can use the following view factors: - \(F_{1 \to 2}\) (tube surface to shield inner surface) = 1 - \(F_{2 \to 1}\) (shield inner surface to tube surface) = \( \frac{D_1}{D_2} \), where \(D_1\) is the diameter of the tube and \(D_2\) is the diameter of the shield. Using the given diameters, we can calculate the view factor: \(F_{2 \to 1} = \frac{20}{60} = \frac{1}{3}\)
02

Determine radiative heat transfer

To calculate the radiative heat transfer between the tube and the shield, we can use the following equation: \(q_{rad} = \sigma F_{1 \to 2} A_1 (T_1^4 - T_2^4)\), where - \(\sigma\) is the Stefan-Boltzmann constant, - \(F_{1 \to 2}\) is the view factor from the tube surface to the shield inner surface, - \(A_1\) is the surface area of the tube, - \(T_1\) is the temperature of the inner tube (unknown), - \(T_2\) is the temperature of the shield (in Kelvin, \(42 + 273.15 = 315.15K\)). Additionally, we need to consider shield emissivities using the following equation: \(q_{1 \to 2} = \frac{q_{rad}}{1 - \varepsilon_{2,1} + (1 - \varepsilon_{2,0}) (F_{2 \to 1})}\)
03

Determine convection heat transfer

For the convection heat transfer between the shield and surrounding air, we can use the following equation: \(q_{conv} = h A_2 (T_2 - T_{\infty})\), where - \(h\) is the convection heat transfer coefficient, - \(T_{\infty}\) is the temperature of the surrounding air (in Kelvin, \(27 + 273.15 = 300.15K\)), - \(A_2\) is the surface area of the shield.
04

Set up the heat balance equation and solve for the inner tube temperature

In this step, we will assume that heat transfer by radiation equals the heat transfer by convection at the outer surface of the shield: \(q_{1 \to 2} = q_{conv}\) Substituting the expressions from Step 2 and Step 3 and solving for the unknown inner tube temperature, we get: \(\frac{\sigma F_{1 \to 2} A_1 (T_1^4 - T_2^4)}{1 - \varepsilon_{2,1} + (1 - \varepsilon_{2,0}) (F_{2 \to 1})} = h A_2 (T_2 - T_{\infty})\) By solving this equation for \(T_1\), we will find the operating temperature for the inner tube.

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Key Concepts

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

Convection Heat Transfer
Convection heat transfer is a process where heat is transferred between a surface and a fluid that is moving. This can occur in gases or liquids. It's an essential concept in thermal engineering, especially in systems like heating or cooling, where the fluid's movement helps distribute temperature changes.

The rate of heat transfer via convection is captured by the equation: \[ q_{conv} = h A (T_s - T_{ ext{fluid}}) \]where:
  • \( q_{conv} \) is the convective heat transfer rate.
  • \( h \) is the convection heat transfer coefficient, a measure of how effectively heat is transferred.
  • \( A \) is the surface area.
  • \( T_s \) is the surface temperature.
  • \( T_{\text{fluid}} \) is the fluid temperature.
The larger the temperature difference or the greater the surface area, the higher the heat transfer. A high convection coefficient means faster heat transfer and vice versa. In the exercise, convection is analyzed between the shield's exterior and the surrounding air, with a defined heat transfer coefficient \( h \) of \( 10 \; \text{W/m}^2 \cdot \text{K} \). This provides an understanding of how much energy is transferred from the shield to the air.
View Factors
View factors play an important role in radiative heat transfer calculations. They represent how much radiation leaving one surface reaches another surface directly. When surfaces are concentric, like in the exercise with the tube and shield, specific view factors can be calculated easily.

For concentric cylinders or circles, the view factor from the inner surface to the outer one, \( F_{1 \to 2} \), is unity or 1, meaning all radiation from the inner surface can hit the outer surface directly. The reverse, however, depends on the sizes, calculated as:\[ F_{2 \to 1} = \frac{D_1}{D_2} \]where:
  • \( D_1 \) is the diameter of the tube,
  • \( D_2 \) is the diameter of the shield.
This accounts for geometry, determining how effective radiative transfer is between these two. The exercise calculates \( F_{2 \to 1} \) as \( \frac{1}{3} \) using given diameters, emphasizing how shape and size affect energy distribution.
Emissivity
Emissivity refers to a surface's ability to emit thermal radiation. It ranges between 0 and 1, where 1 represents a perfect black body emitting radiation most effectively. Real-life surfaces have lower emissivities, affecting heat transfer efficiency.

In the exercise, the shield has varying emissivities on its surfaces. The inner surface has an emissivity \( \varepsilon_{2,1} = 0.01 \), very low, indicating it barely emits radiation, almost like a reflective surface. The outer surface emissivity is higher at \( \varepsilon_{2,0} = 0.1 \), allowing for more radiation emission, still relatively low but significantly more than the inner surface.

Emissivity alters the radiative heat transfer equation's effectiveness, modifying how surfaces interact thermally:\[ q_{1 \to 2} = \frac{q_{rad}}{1 - \varepsilon_{2,1} + (1 - \varepsilon_{2,0})(F_{2 \to 1})} \]This reflects how the emissivity and view factor together control the net heat transfer rate. Understanding and calculating emissivities accurately is crucial for thermal system design and analysis, as seen in maintaining efficient energy balances.

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Most popular questions from this chapter

A cylindrical cavity of diameter \(D\) and depth \(L\) is machined in a metal block, and conditions are such that the base and side surfaces of the cavity are maintained at \(T_{1}=1000 \mathrm{~K}\) and \(T_{2}=700 \mathrm{~K}\), respectively. Approximating the surfaces as black, determine the emissive power of the cavity if \(L=20 \mathrm{~mm}\) and \(D=10 \mathrm{~mm}\).

Two convex objects are inside a large vacuum enclosure whose walls are maintained at \(T_{3}=300 \mathrm{~K}\). The objects have the same area, \(0.2 \mathrm{~m}^{2}\), and the same emissivity, 0.2. The view factor from object 1 to object 2 is \(F_{12}=0_{3} 3\). Embedded in object 2 is a heater that generates 400 W. The temperature of object 1 is maintained at \(T_{1}=200 \mathrm{~K}\) by circulating a fluid through channels machined into it. At what rate must heat be supplied (or removed) by the fluid to maintain the desired temperature of object 1? What is the temperatare of object 2 ?

An electrically heated sample is maintained at a surface temperature of \(T_{s}=500 \mathrm{~K}\). The sample coating is diffuse but spectrally selective, with the spectral emissivity distribution shown schematically. The sample is irradiated by a furnace located coaxially at a distance of \(L_{\mathrm{s}}=750 \mathrm{~mm}\). The furnace has isothermal walls with an emissivity of \(s_{f}=0.7\) and a uniform temperature of \(T_{f}-3000 \mathrm{~K}\). A radiation detector of area \(A_{d}=8 \times 10^{-5} \mathrm{~m}^{2}\) is positioned at a distance of \(L_{\mathrm{dd}}=1.0 \mathrm{~m}\) from the sample along a direction that is \(45^{\circ}\) from the sample normal. The detector is sensitive to spectral radiant power only in the spectral region from 3 to \(5 \mu \mathrm{m}\). The sample surface experiences convection with a gas for which \(T_{\infty}=300 \mathrm{~K}\) and \(h=20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The surroundings of the sample mount are large and at a uniform temperature of \(T_{\text {sur }}=300 \mathrm{~K}\). (a) Determine the electrical power, \(P_{c}\), required to maintain the sample at \(T_{s}=500 \mathrm{~K}\). (b) Considering both emission and reflected irradiation from the sample, determine the radiant power that is incident on the detector within the spectral region from 3 to \(5 \mu \mathrm{m}\).

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?

Heat transfer by radiation occurs between two large parallel plates, which are maintained at temperatures \(T_{1}\) and \(T_{2}\), with \(T_{1}>T_{2}\). To reduce the rate of heat transfer between the plates, it is proposed that they be separated by a thin shield that has different emissivities on opposite surfaces. In particular, one surface has the emissivity \(\varepsilon_{s}<0.5\), while the opposite surface has an emissivity of \(2 \varepsilon_{s}\). (a) How should the shield be oriented to provide the larger reduction in heat transfer between the plates? That is, should the surface of emissivity \(\varepsilon_{s}\) or that of emissivity \(2 s_{s}\) be oriented toward the plate at \(T_{1}\) ? (b) What orientation will result in the larger value of the shield temperature \(T_{s}\) ?

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