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Chapter 13: Problem 49

A long, thin-walled horizontal tube \(100 \mathrm{~mm}\) in diameter is maintained at \(120^{\circ} \mathrm{C}\) by the passage of steam through its interior. A radiation shield is installed around the tube, providing an air gap of \(10 \mathrm{~mm}\) between the tube and the shield, and reaches a surface temperature of \(35^{\circ} \mathrm{C}\). The tube and shield are diffuse, gray surfaces with emissivities of \(0.80\) and \(0.10\), respectively. What is the radiant heat transfer from the tube per unit length?

Short Answer

Expert verified
The radiant heat transfer from the tube per unit length is \(1454.16 \mathrm{W/m}\).

Step by step solution

01

Convert temperatures to Kelvin

First, we need to convert the given temperatures to Kelvin so that we can use them in our equations: \[T_{tube} = 120^{\circ} \mathrm{C} + 273.15 = 393.15 \mathrm{~K}\] \[T_{shield} = 35^{\circ} \mathrm{C} + 273.15 = 308.15 \mathrm{~K}\]
02

Calculate the surface areas of tube and shield per unit length

Next, let's determine the surface areas of the tube and shield per unit length: \[A_{tube} = \pi D_{tube} = \pi(0.1 \mathrm{m}) = 0.314 \mathrm{m^2/m}\] \[A_{shield} = \pi D_{shield} = \pi(0.12 \mathrm{m}) = 0.377 \mathrm{m^2/m}\]
03

Calculate the radiosity of the tube (J1)

To calculate the radiosity of the tube (J1), we'll use the following formula for a gray surface: \[J_1 = \varepsilon_1 \sigma T_{tube}^4 + \frac{(1-\varepsilon_1)}{\varepsilon_1} E_b\] Where \(\sigma\) is the Stefan-Boltzmann constant \((5.67 \times 10^{-8} \mathrm{W/m^2 K^4})\), \(\varepsilon_1\) is the emissivity of the tube, and \(E_b\) is the blackbody emissive power of the tube. First, let's calculate the blackbody emissive power of the tube: \[E_b = \sigma T_{tube}^4 = (5.67 \times 10^{-8}\mathrm{W/m^2 K^4})(393.15\mathrm{~K})^4 = 5885.23 \mathrm{W/m^2}\] Now, let's find the radiosity of the tube (J1): \[J_1 = (0.80)(5885.23\mathrm{W/m^2}) + \frac{(1-0.80)}{0.80}(5885.23\mathrm{W/m^2}) = 4708.18 \mathrm{W/m^2}\]
04

Calculate the net rate of radiant heat transfer between tube and shield per unit length (Q1)_

Finally, we can determine the net rate of radiant heat transfer between tube and shield per unit length (Q1) using the following formula for two gray surfaces: \[Q_1 = \frac{J_1 - J_2}{\frac{1}{\varepsilon_1 A_{tube}} + \frac{1-\varepsilon_2}{\varepsilon_2 A_{shield}}}\] First, we need to find the radiosity of the shield (J2). We can assume that the shield is perfectly insulated on the outside, which means that the radiosity of the shield (J2) is equal to its blackbody emissive power: \[J_2 = E_{b_{shield}} = \sigma T_{shield}^4 = (5.67 \times 10^{-8}\mathrm{W/m^2 K^4})(308.15\mathrm{~K})^4 = 843.35 \mathrm{W/m^2}\] Now, we can find the net rate of radiant heat transfer between tube and shield per unit length (Q1): \[Q_1 = \frac{4708.18 \mathrm{W/m^2} - 843.35 \mathrm{W/m^2}}{\frac{1}{0.80(0.314 \mathrm{m^2/m})} + \frac{1-0.10}{0.10(0.377 \mathrm{m^2/m})}} = 1454.16 \mathrm{W/m}\] Thus, the radiant heat transfer from the tube per unit length is \(1454.16 \mathrm{W/m}\).

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Key Concepts

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

Heat Transfer
Heat transfer is a fundamental concept within thermal physics, involving the movement of heat energy from one place to another. In the context of our exercise, we are particularly interested in radiant heat transfer, which is the transfer of heat by electromagnetic radiation, primarily in the form of infrared waves. Unlike conduction and convection, radiation does not require a medium and can even occur through a vacuum. Radiant heat transfer is driven by the temperature difference between two surfaces and the emissivity of those surfaces.

In our case, the tube and shield, being at different temperatures, exchange heat radiatively. The rate of this heat transfer is determined by not only the temperatures and emissivities of the tube and shield but also their respective surface areas per unit length. A key aspect to remember is that higher temperatures and higher emissivities will typically result in higher rates of heat transfer.

Understanding how radiant heat moves between surfaces is essential for applications ranging from cooling electronics to designing energy-efficient buildings and even in space travel, where objects must cope with the extreme temperatures of the vacuum of space.
Stefan-Boltzmann Constant
At the core of radiant heat transfer calculations is the Stefan-Boltzmann constant denoted by \( \sigma \). With a value of approximately \( 5.67 \times 10^{-8} \text{ W/m}^2 \text{K}^4 \), this constant is essential in determining the total energy radiated per unit surface area of a black body per unit time. A black body is an idealized physical body that absorbs all incident electromagnetic radiation and reflects none.

In our exercise, the Stefan-Boltzmann constant is used to calculate the blackbody emissive power of the tube (\(E_b\)) based on its temperature. This calculation provides the foundation for determining the radiosity, which represents the total energy leaving the surface per unit area, and includes both emitted and reflected radiation.

The implementation of the Stefan-Boltzmann constant in equations makes it possible to predict and calculate the thermal radiation emitted from real-world objects, like our tube and shield, which only approximate the behavior of a blackbody due to their given emissivity values.
Gray Surfaces
A gray surface is an idealized concept used in heat transfer to describe a surface for which the emissivity does not vary with wavelength or temperature. Real materials do not behave exactly as gray surfaces, but the approximation simplifies analysis and is often sufficiently accurate for engineering purposes.

In the exercise, the tube and the shield are described as diffuse, gray surfaces with given emissivities. Diffuse means they radiate energy uniformly in all directions. The tube, with an emissivity of \(0.80\), and the shield, with an emissivity of \(0.10\), demonstrate how different materials can exchange heat at different rates due to this property. Emissivity is a measure of a material's ability to emit thermal radiation compared to a perfect blackbody (which would have an emissivity of \(1\)).

The gray surface assumption implies that when the tube and shield exchange heat radiatively, their emissivity values can be used to adjust the Stefan-Boltzmann relationship to calculate the net radiant heat transfer. In a practical setting, considering an object as a gray surface simplifies the complex problem of real-world heat exchange into a more manageable form while still providing valuable insight into thermal behavior.

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Most popular questions from this chapter

A radiant oven for drying newsprint consists of a long duct \((L=20 \mathrm{~m})\) of semicincular cross section. The newsprint moves through the oven on a conveyor belt at a velocity of \(V=0.2 \mathrm{~m} / \mathrm{s}\). The newsprint has a water content of \(0.02 \mathrm{~kg} / \mathrm{m}^{2}\) as it enters the oven and is completely dry as it exits. To assure quality, the newsprint must be maintained at room temperature \((300 \mathrm{~K})\) during drying. To aid in maintaining this condition, all system components and the air flowing through the oven have a temperature of \(300 \mathrm{~K}\). The inner sarface of the semicaircular duct, which is of emissivity \(0.8\) and temperature \(T_{1}\), provides the radiant heat required to accomplish the drying. The wet surface of the newsprint can be considered to be black. Air entering the oven has a temperature of \(300 \mathrm{~K}\) and a relative humidity of \(20 \%\). Since the velocity of the air is large, its temperature and relative humidity can be assumed to be constant over the entire duct length. Calculate the required evaporation rate, air velocity \(u_{m}\), and temperature \(T_{1}\) that will ensure steady-state conditions for the process.

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 \(D=10 \mathrm{~mm}\).

Consider the right-circular cylinder of diameter \(D\), length \(L\), and the areas \(A_{1}, A_{2}\), and \(A_{3}\) representing the base, inner, and top surfaces, respectively. (a) Show that the view factor between the base of the cylinder and the inner surface has the form \(F_{12}=2 H\left[\left(1+H^{2}\right)^{1 / 2}-H\right]\), where \(H=L D .\) (b) Show that the view factor for the inner surface to itself has the form \(F_{22}=1+H-\left(1+H^{2}\right)^{1 / 2}\).

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?

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.
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