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How does radiation transfer through a participating medium differ from that through a nonparticipating medium?

Short Answer

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Question: Explain the difference between radiation transfer through a participating medium and a nonparticipating medium. Answer: In a participating medium, radiation interacts with the medium's particles, resulting in absorption, emission, and scattering phenomena, leading to changes in the radiation's characteristics. In contrast, a nonparticipating medium allows radiation to pass through without significant interactions, causing little to no changes in its properties.

Step by step solution

01

Definition of a Participating Medium

A participating medium is a type of medium that affects the properties and interactions of electromagnetic radiation, such as absorption, emission, and scattering phenomena. In other words, the radiation interacts with the medium's particles, causing changes in its characteristics.
02

Definition of a Nonparticipating Medium

A nonparticipating medium, on the other hand, is a medium in which radiation passes through without any significant interaction with the medium's particles. There is negligible absorption, emission, or scattering of radiation in a nonparticipating medium.
03

Absorption in Participating vs. Nonparticipating Mediums

In a participating medium, the radiation is absorbed, causing the medium's particles to gain energy. This energy is then converted into other forms, such as thermal or vibrational energy. In a nonparticipating medium, there is little to no absorption of radiation, so the radiation passes through mostly unchanged.
04

Emission in Participating vs. Nonparticipating Mediums

A participating medium can emit radiation due to the energy gained from previous absorptions. This emission results in new radiation originating from the medium itself. In a nonparticipating medium, there is no significant emission of radiation, meaning the radiation passing through the medium is not altered by the process.
05

Scattering in Participating vs. Nonparticipating Mediums

Scattering occurs in a participating medium when radiation interacts with the particles in the medium and changes direction. This can cause the radiation to be redirected several times, leading to a distribution of radiation in multiple directions. In a nonparticipating medium, the absence of scattering means radiation maintains its original direction as it passes through the medium.
06

Conclusion

In summary, radiation transfer through a participating and a nonparticipating medium differ in terms of absorption, emission, and scattering. In a participating medium, radiation interacts with the medium's particles, resulting in absorption, emission, and scattering phenomena. In contrast, a nonparticipating medium causes little to no interaction with the radiation, allowing it to pass through without significant changes to its characteristics.

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Most popular questions from this chapter

A 90 -cm-diameter flat black disk is placed in the center of the top surface of a \(1-m \times 1-m \times 1-m\) black box. The view factor from the entire interior surface of the box to the interior surface of the disk is (a) \(0.07\) (b) \(0.13\) (c) \(0.26\) (d) \(0.32\) (e) \(0.50\)

Consider two infinitely long concentric cylinders with diameters 20 and $25 \mathrm{~cm}\(. The inner surface is maintained at \)700 \mathrm{~K}$ and has an emissivity of \(0.40\), while the outer surface is black. If the rate of radiation heat transfer from the inner surface to the outer surface is $2400 \mathrm{~W}$ per unit area of the inner surface, the temperature of the outer surface is (a) \(605 \mathrm{~K}\) (b) \(538 \mathrm{~K}\) (c) \(517 \mathrm{~K}\) (d) \(451 \mathrm{~K}\) (e) \(415 \mathrm{~K}\)

Consider two rectangular surfaces perpendicular to each other with a common edge which is \(1.6 \mathrm{~m}\) long. The horizontal surface is $0.8 \mathrm{~m}\( wide, and the vertical surface is \)1.2 \mathrm{~m}$ high. The horizontal surface has an emissivity of \(0.75\) and is maintained at $450 \mathrm{~K}\(. The vertical surface is black and is maintained at \)700 \mathrm{~K}$. The back sides of the surfaces are insulated. The surrounding surfaces are at \(290 \mathrm{~K}\) and can be considered to have an emissivity of \(0.85\). Determine the net rate of radiation heat transfer between the two surfaces and between the horizontal surface and the surroundings.

A large ASTM A992 carbon steel plate is $(k=10 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\(. The ceramic plate has a thickness of \)10 \mathrm{~cm}$, with its lower surface at \(T_{0}=800^{\circ} \mathrm{C}\) and upper surface at \(T_{1}=700^{\circ} \mathrm{C}\). The upper surface of the ceramic plate faces the carbon steel plate. Convection occurs on the upper surface of the ceramic plate with air at \(20^{\circ} \mathrm{C}\) and a convection heat transfer coefficient of \(12 \mathrm{~W} / \mathrm{m}^{2}, \mathrm{~K}\). The ceramic and steel plates have emissivity values of \(0.93\) and \(0.75\), respectively. The ASME Code for Process Piping specifies the maximum use temperature suitable for ASTM A992 carbon steel to be \(427^{\circ} \mathrm{C}\) (ASME B31.3-2014, Table A-1M). A radiation shield is to be placed in parallel between the two plates to keep the temperature of the steel plate from exceeding its maximum use temperature. Determine the emissivity that the radiation shield needs to keep the steel plate surface from exceeding \(427^{\circ} \mathrm{C}\).

Two concentric spheres of diameters \(D_{1}=0.3 \mathrm{~m}\) and $D_{2}=0.6 \mathrm{~m}\( are maintained at uniform temperatures \)T_{1}=800 \mathrm{~K}$ and \(T_{2}=500 \mathrm{~K}\) and have emissivities \(\varepsilon_{1}=0.5\) and \(\varepsilon_{2}=0.7\), respectively. Determine the net rate of radiation heat transfer between the two spheres. Also, determine the convection heat transfer coefficient at the outer surface if both the surrounding medium and the surrounding surfaces are at \(30^{\circ} \mathrm{C}\). Assume the emissivity of the outer surface is \(0.35\).

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