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Chapter 13: Problem 131

A flue gas at 1 -atm total pressure and a temperature of \(1400 \mathrm{~K}\) contains \(\mathrm{CO}_{2}\) and water vapor at partial pressures of \(0.05\) and \(0.10 \mathrm{~atm}\), respectively. If the gas flows through a long flue of \(1-\mathrm{m}\) diameter and \(400 \mathrm{~K}\) surface temperature, determine the net radiative heat flux from the gas to the surface. Blackbody behavior may be assumed for the surface.

Short Answer

Expert verified
The net radiative heat flux from the flue gas to the surface is approximately \(359674.09 \mathrm{W/m^2}\).

Step by step solution

01

Find the Emissivity of CO2 and H2O

To find the emissivities of CO2 and H2O, we can refer to standard emissivity tables for these gases at the given temperature and pressure. According to the tables, the emissivities for CO2 at 1400 K and 0.05 atm, and for H2O at 1400 K and 0.10 atm, can be approximated as: \[\epsilon_{CO2} \approx 0.35\] \[\epsilon_{H2O} \approx 0.55\]
02

Calculate the Effective Emissivity of the Gas Mixture

Using the formula for the effective emissivity of the gas mixture, we can find the value of \(\epsilon_{eff}\): \[\epsilon_{eff} = (1 - (1 - \epsilon_{CO_2})(1 - \epsilon_{H_2O}))\] Substituting the values of \(\epsilon_{CO2}\) and \(\epsilon_{H2O}\), we get: \[\epsilon_{eff} = (1 - (1 - 0.35)(1 - 0.55)) = 0.6975\]
03

Calculate the Net Radiative Heat Flux

Now that we have the effective emissivity, we can calculate the net radiative heat flux using the formula: \[q = \sigma\cdot\epsilon_{eff}(T_{gas}^{4} - T_{surface}^{4})\] Substituting the values of \(\sigma = 5.67 \times 10^{-8} \mathrm{W/m^2 K^4}\), \(T_{gas} = 1400 \mathrm{K}\), and \(T_{surface} = 400 \mathrm{K}\), along with the calculated value of \(\epsilon_{eff}\), we get: \[q = (5.67 \times 10^{-8})(0.6975)(1400^4 - 400^4) \mathrm{W/m^2}\] Evaluating the expression: \[q \approx 359674.09 \mathrm{W/m^2}\]
04

Answer

The net radiative heat flux from the flue gas to the surface is approximately \(359674.09 \mathrm{W/m^2}\).

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Key Concepts

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

Emissivity
Emissivity is a measure of a material's ability to emit thermal radiation compared to an ideal blackbody, which is a perfect emitter. Understanding emissivity is essential in radiative heat transfer, as it defines how much energy is radiated by a body's surface.

In this context:
  • The emissivity of a material can range from 0 (no emission) to 1 (perfect emitter, like a blackbody).
  • For real-world materials, emissivity is often less than 1, as they are not perfect emitters.
  • In the example, CO2 and H2O gases have emissivities of 0.35 and 0.55, respectively, under the given conditions.
These values are crucial for determining the radiative heat transfer in gas mixtures, which helps in calculating the net radiative heat flux.
Blackbody Radiation
Blackbody radiation refers to the electromagnetic radiation emitted by an idealized object that absorbs all incident radiation, regardless of the wavelength. This concept is fundamental to thermal radiation studies.

Key points about blackbody radiation include:
  • An ideal blackbody has an emissivity of 1, meaning it is a perfect emitter of radiation.
  • The Stefan-Boltzmann Law governs the intensity of blackbody radiation, stating that the power radiated per unit area is proportional to the fourth power of the absolute temperature, represented by: \( E = \sigma T^4 \).
  • In the example, the surface is assumed to behave as a blackbody, simplifying the calculations for radiative heat transfer.
This concept allows for more manageable calculations when assessing heat transfer between objects, especially in engineered systems like flue gas conduits.
Flue Gas
Flue gas refers to the gas that is emitted as a byproduct of the combustion processes in furnaces, boilers, or steam generators. It typically contains a mixture of gases, including CO2 and water vapor, important factors in radiative heat transfer calculations.

Considerations for flue gas in this context:
  • Flue gases are usually at high temperatures, making them significant contributors to heat transfer through radiation.
  • The mixture composition of flue gases, particularly the types and concentrations of gases, directly affects the radiative properties like emissivity.
  • In the example, the flue gas is at 1400 K, and specific partial pressures are given for CO2 (0.05 atm) and water vapor (0.10 atm), which are used to calculate their individual and collective emissivities.
Understanding flue gas characteristics is crucial for optimizing the efficiency of industrial equipment and reducing emitted heat loss.
Net Radiative Heat Flux
Net radiative heat flux is the rate of heat transfer by radiation from one surface to another per unit area. Calculating this quantity is vital to understanding energy exchanges in thermal systems.

Here's how net radiative heat flux is determined:
  • The formula to compute the net radiative heat flux \( q \) is \( q = \sigma \cdot \epsilon_{eff}(T_{gas}^4 - T_{surface}^4) \), where \( \sigma \) is the Stefan-Boltzmann constant.
  • \( \epsilon_{eff} \), the effective emissivity of the gas mixture, accounts for the contribution of each gas component to the overall radiation.
  • The temperatures \( T_{gas} \) and \( T_{surface} \) are crucial, representing the gas and the surface temperatures in absolute scale (Kelvin) respectively.
In the provided example, these calculations are used to find the net radiative heat flux from the gas to the surface, which is approximately 359674.09 W/m虏, illustrating the significant energy transfer possible through radiation alone.

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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 novel infrared recycler has been proposed for reclaiming the millions of kilograms of waste plastics produced by the dismantling and shredding of automotive vehicles following their refirement. To address the problem of sorting mixed plastics into components such as polypropylene and polycarbonate, a washed stream of the mixed plastics is routed to an infrared heating system, where it is dried and stbsequently heated to a temperature for which one of the components begins to soften, while the others remain rigid. The mixed stream is then routed through steel rollers, to which the softened plastic sticks and is removed from the stream. Heating of the stream is then continued to facilitate removal of a second component, and the heating/removal process is repeated until all of the components are separated. Consider the initial drying stage for a system comprised of a cylindrical heater aligned coaxially with a rotating drum of diameter \(D_{d}=1 \mathrm{~m}\). Shortly after entering the drum, wet plastic pellets may be assumed to fully cover the bottom semicylindrical section and to remain at a temperature of \(T_{p}=\) \(325 \mathrm{~K}\) during the drying process. The surface area of the pellets may be assumed to correspond to that of the semicylinder and to have an emissivity of \(\varepsilon_{p}=0.95\). (a) If the flow of dry air through the drum maintains a convection mass transfer coefficient of \(0.024 \mathrm{~m} / \mathrm{s}\) on the surface of the pellets, what is the evaporation rate per unit length of the drum? (b) Neglecting convection heat transfer, determine the temperature \(T_{b}\) that must be maintained by a heater of diameter \(D_{\mathrm{b}}=0.10 \mathrm{~m}\) and emissivity \(\varepsilon_{h}=0.8\) to sustain the foregoing evaporation rate. What is the corresponding value of the temperature \(T_{d}\) for the top surface of the drum? The outer surface of the dram is well insulated, and its length-to-diameter ratio is large. As applied to the top \((d)\) or bottom \((p)\) surface of the drum, the view factor of an infinitely long semicylinder to itself, in the presence of a concentric, coaxial cylinder, may be expressed as $$ \begin{aligned} F_{U}=& 1-\frac{2}{\pi}\left\\{\left[1-\left(D_{d} / D_{d}\right)^{2}\right]^{1 / 2}\right.\\\ &\left.+\left(D_{k} / D_{d}\right) \sin ^{-1}\left(D_{h} / D_{d}\right)\right\\} \end{aligned} $$

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.

A small disk of diameter \(D_{1}=50 \mathrm{~mm}\) and emissivity \(\varepsilon_{1}=0.6\) is maintained at a temperature of \(T_{1}=900 \mathrm{~K}\). The disk is covered with a hemispherical radiation shield of the same diameter and an emissivity of \(\varepsilon_{2}=0.02\) (both sides). The disk and cap are located at the bottom of a large evacuated refractory container \(\left(\varepsilon_{4}=0.85\right)\), facing another disk of diameter \(D_{5}=D_{1}\), emissivity \(\varepsilon_{3}=0.4\), and temperature \(T_{3}=400 \mathrm{~K}\). The view factor \(F_{2 s}\) of the shield with respect to the upper disk is \(0.3\). (a) Construct an equivalent thermal circuit for the above system. Label all nodes, resistances, and currents. (b) Find the net rate of heat transfer between the hot disk and the rest of the system.

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