/*! 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 55 Two concentric spheres of diamet... [FREE SOLUTION] | 91影视

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Chapter 13: Problem 55

Two concentric spheres of diameter \(D_{1}=0.8 \mathrm{~m}\) and \(D_{2}=1.2 \mathrm{~m}\) are separated by an air space and have surface temperatures of \(T_{1}=400 \mathrm{~K}\) and \(T_{2}=300 \mathrm{~K}\). (a) If the surfaces are black, what is the net rate of radiation exchange between the spheres? (b) What is the net rate of radiation exchange between the surfaces if they are diffuse and gray with \(\varepsilon_{1}=0.5\) and \(\varepsilon_{2}=0.05\) ? (c) What is the net rate of radiation exchange if \(D_{2}\) is increased to \(20 \mathrm{~m}\), with \(\varepsilon_{2}=0.05, \varepsilon_{1}=0.5\), and \(D_{1}=0.8 \mathrm{~m}\) ? What error would be introduced by assuming blackbody behavior for the outer surface \(\left(\varepsilon_{2}=1\right)\), with all other conditions remaining the same? (d) For \(D_{2}=1.2 \mathrm{~m}\) and emissivities of \(s_{1}=0.1,0.5\), and \(1.0\), compute and plot the net rate of radiation exchange as a function of \(\varepsilon_{2}\) for \(0.05 \leq \varepsilon_{2} \leq 1.0\).

Short Answer

Expert verified
In summary: (a) The net rate of radiation exchange between black surfaces is -8595.4 W/m虏. (b) The net rate of radiation exchange between diffuse and gray surfaces is -18492.7 W/m虏. (c) The net rate of radiation exchange for an increased diameter is about -1.434*10^6 W/m虏, and the error introduced by assuming blackbody behavior is approximately 1.480*10^7 W/m虏. (d) To create a plot of net radiation exchange, you need to calculate the values for key combinations of \(\epsilon_1\) and \(\epsilon_2\) and produce separate lines for each value of \(\epsilon_1\), with Q on the y-axis and \(\epsilon_2\) on the x-axis.

Step by step solution

01

Part (a): Finding net rate of radiation exchange between black surfaces

First, we need to calculate the radiative resistance between the two surfaces. For concentric spheres, the radiative resistance R is given by: \(R = \frac{1}{4 \pi r_1 r_2}(\frac{1}{\epsilon_1A_1} + \frac{1}{\epsilon_2 A_2} - 1)\) Since the surfaces are black, their emissivity \(\epsilon_1 =\epsilon_2=1\). Now, let's find the radii r1 and r2, as well as areas A1 and A2. \(r_1 = \frac{D_1}{2} = 0.4m\) \(r_2 = \frac{D_2}{2} = 0.6m\) \(A_1 = 4 \pi r_1^2 = 4 \pi(0.4)^2\) \(A_2 = 4 \pi r_2^2 = 4 \pi(0.6)^2\) Now, we can calculate the radiative resistance R, which is: \(R = \frac{1}{4 \pi (0.4)(0.6)}(0 +0 -1) = \frac{-1}{0.8 \pi}\) The net rate of radiation exchange Q between the surfaces is given by: \(Q = \frac{T_1^4 - T_2^4}{R}\) Plug in the values and solve for Q: \(Q = \frac{(400^4 - 300^4)}{-\frac{1}{0.8 \pi}} = -8595.4 W/m^2\) This is the net rate of radiation exchange for black surfaces.
02

Part (b): Finding net rate of radiation exchange between diffuse and gray surfaces

For gray surfaces with \(\epsilon_1=0.5\) and \(\epsilon_2=0.05\), we can now calculate the radiative resistance R: \(R = \frac{1}{4 \pi r_1 r_2}(\frac{1}{\epsilon_1A_1} + \frac{1}{\epsilon_2 A_2} - 1)\) \(R = \frac{1}{4 \pi (0.4)(0.6)} (\frac{1}{0.5(4\pi(0.4)^2)} + \frac{1}{0.05(4\pi(0.6)^2)} - 1) = -4.648\) Now, let's calculate the net rate of radiation exchange Q between the surfaces: \(Q = \frac{T_1^4 - T_2^4}{R}\) \(Q = \frac{(400^4 - 300^4)}{-4.648} = -18492.7 W/m^2\) So, the net rate of radiation exchange between diffuse, gray surfaces is approximately -18492.7 W/m^2.
03

Part (c): Finding net rate of radiation exchange with increased diameter and error analysis

For this part, let's increase D2 to 20m and use the given emissivities with D1 remaining unchanged. First, calculate r2 and A2: \(r_2 = \frac{D_2}{2} = 10m\) \(A_2 = 4 \pi r_2^2 = 4 \pi(10)^2\) Now, let's calculate the radiative resistance R: \(R = \frac{1}{4 \pi r_1 r_2} (\frac{1}{\epsilon_1A_1} + \frac{1}{\epsilon_2 A_2} - 1) = -0.006\) Now, find the net rate of radiation exchange Q between the surfaces: \(Q = \frac{T_1^4 - T_2^4}{R}\) \(Q = \frac{(400^4 - 300^4)}{-0.006} = -1.434*10^6 W/m^2\) To find the error introduced by assuming blackbody behavior for the outer surface (\(\epsilon_2=1\)), recalculate R and Q with this new emissivity value: \(R = \frac{1}{4 \pi r_1 r_2}(\frac{1}{\epsilon_1A_1} + \frac{1}{\epsilon_2 A_2} - 1) = -0.053\) \(Q = \frac{T_1^4 - T_2^4}{R}\) \(Q = \frac{(400^4 - 300^4)}{-0.053} = -1.623*10^7 W/m^2\) Now, find the error: Error = \(|-1.623*10^7 - (-1.434*10^6)| = 1.480*10^7 W/m^2\) Therefore, the net rate of radiation exchange for an increased diameter is about -1.434*10^6 W/m^2 and the error introduced by assuming blackbody behavior is approximately 1.480*10^7 W/m^2.
04

Part (d): Creating a plot of net radiation exchange for various emissivities

This part requires you to compute and plot the net rate of radiation exchange as a function of \(\epsilon_2\) for a range of values from 0.05 to 1.0 and three different values of \(\epsilon_1 (0.1,0.5,1.0)\). In this text-based response, we cannot create a plot, but we can provide the steps to take and perform calculations for key points in this range. 1. Calculate the radiative resistance R as a function of \(\epsilon_1\) and \(\epsilon_2\). 2. Calculate the net rate of radiation exchange Q for each combination of \(\epsilon_1\) and \(\epsilon_2\). 3. Create a plot of these Q values on the y-axis against \(\epsilon_2\) on the x-axis, producing separate lines for each value of \(\epsilon_1\). A few key combinations to take note of in your calculations are: - \(\epsilon_1=0.1\) and \(\epsilon_2=0.05\) - \(\epsilon_1=0.5\) and \(\epsilon_2=0.05\) (this one we already calculated in part (b)) - \(\epsilon_1=1.0\) and \(\epsilon_2=0.05\) - \(\epsilon_1=1.0\) and \(\epsilon_2=1.0\) (and this one in part (a)) After calculating these key points, you can plot them and interpolate the values accordingly.

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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 in which an idealized object, a blackbody, emits and absorbs thermal radiation. A perfect blackbody is a theoretical concept in which an object absorbs all incoming radiation, without reflecting any. Hence, it has an emissivity of 1. This means it emits radiation at the maximum rate possible for its given temperature.

The Stefan-Boltzmann Law explains the power radiated by a blackbody in terms of its temperature: \[ E = \sigma T^4 \]where
  • \( E \) is the radiative emissive power,
  • \( \sigma \) is the Stefan-Boltzmann constant, and
  • \( T \) is the absolute temperature of the object.
In practical applications, understanding blackbody radiation helps in determining heat transfer between bodies, especially in thermal engineering and climate studies.
Emissivity
Emissivity is a measure of a material's ability to emit thermal radiation relative to that of a perfect blackbody. It ranges from 0 to 1, where 1 represents a perfect blackbody. In reality, no material has an emissivity of exactly 1. Therefore, practical surfaces emit less radiation than a blackbody at the same temperature.

The emissivity of a material affects its radiative heat transfer. In the context of the concentric spheres problem, emissivity values determine the efficiency of radiation emission for each sphere.
  • Diffuse surfaces have consistent emissivity in all directions.
  • Grey bodies have uniform emissivity across all wavelengths.
Tools like the Infrared thermography utilize principles of emissivity to measure temperature or detect defects in materials.
Concentric spheres
Concentric spheres involve two or more spheres that share the same center point but have different radii. In heat transfer applications, this configuration is often used for analyzing radiative heat exchange between the spheres.

The inner sphere in this setup radiates energy to the outer sphere, and vice versa. The challenge lies in the calculation of radiative resistance and the net exchange rate of energy. This setup simplifies many calculations:
  • Radiative exchange depends on the surface area of each sphere.
  • The arrangement helps isolate radiative heat exchange from conductive or convective modes.
Understanding the roles of area, temperature differences, and emissivities in calculations is critical for modeling energy transfer accurately.
Radiative resistance
Radiative resistance is a concept used to simplify the calculation of heat exchange between bodies, analogous to electrical resistance in circuits but for thermal radiation. It helps quantify the obstruction to radiative heat transfer imposed by material properties and geometries involved.

For concentric spheres, the radiative resistance can be calculated using the formula: \[ R = \frac{1}{4 \pi r_1 r_2} \left( \frac{1}{\epsilon_1 A_1} + \frac{1}{\epsilon_2 A_2} - 1 \right) \]where
  • \( r_1 \) and \( r_2 \) are the radii of the spheres,
  • \( \epsilon_1 \) and \( \epsilon_2 \) are the emissivities, and
  • \( A_1 \) and \( A_2 \) are the surface areas of the spheres.
A smaller radiative resistance indicates more efficient heat transfer. Adjusting the geometry, material surface properties, or both can change the value of radiative resistance, impacting the rate of thermal radiation.

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Most popular questions from this chapter

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.

Applying high-emissivity paints to radiating surfaces is a common technique used to enhance heat transfer by radiation. (a) For large parallel plates, determine the radiation heat flux across the gap when the surfaces are at \(T_{1}=350 \mathrm{~K}, T_{2}=300 \mathrm{~K}, \varepsilon_{1}=\varepsilon_{2}=\varepsilon_{x}=0.85 .\) (b) Determine the radiation heat flux when a very thin layer of high- emissivity paint, \(\varepsilon_{p}=0.98\), is applied to both surfaces. (c) Determine the radiation heat flux when the paint layers are each \(L=2 \mathrm{~mm}\) thick and the thermal conductivity of the paint is \(k=0.21 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). (d) Plot the heat flux across the gap for the bare surface as a function of \(\varepsilon_{x}\) with \(0.05 \leq \varepsilon_{x} \leq 0.95\). Show on the same plot the heat flux for the painted surface with very thin paint layers and the painted surface with \(L=2\)-mm-thick paint layers.

Consider the parallel rectangles shown schematically. Show that the view factor \(F_{12}\) can be expressed as $$ F_{12}=\frac{1}{2 A_{1}}\left[A_{(1,4)} F_{(1,4)(2,3)}-A_{1} F_{13}-A_{4} F_{42}\right] $$ where all view factors on the right-hand side of the equation can be evaluated from Figure \(13.4\) (see Table 13.2) for aligned parallel rectangles.

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 ?

2 A composite wall is comprised of two large plates separated by sheets of refractory insulation, as shown in the schematic. In the installation process, the sheets of thickness \(L=50 \mathrm{~mm}\) and thermal conductivity \(k=0.05 \mathrm{~W} / \mathrm{m}+\mathrm{K}\) are separated at 1 -m intervals by gaps of width \(w=10 \mathrm{~mm}\). The hot and cold plates have temperatures and emissivities of \(T_{1}=400^{\circ} \mathrm{C}\). \(\varepsilon_{1}=0.85\) and \(T_{2}=35^{\circ} \mathrm{C}, \varepsilon_{2}=0.5\), respectively. Assume that the plates and insulation are diffuse-gray surfaces. (a) Determine the heat loss by radiation through the gap per unit length of the composite wall (normal to the page). (b) Recognizing that the gaps are located on a 1-m spacing, determine what fraction of the total heat loss through the composite wall is due to transfer by radiation through the insulation gap.
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