Effective Surface Area: Calculating & Understanding

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Discussion Overview

The discussion revolves around the concepts of effective surface area and surface area in the context of thermal radiation. Participants explore how to calculate the effective surface area of various bodies, particularly focusing on a hollow cylinder as a case study for heat loss calculations.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant questions the difference between effective surface area and the surface area of an emitting body, seeking clarification on how to calculate effective surface area.
  • Another participant suggests that the context of the question could involve the effective surface area of the entire universe, contrasting it with the surface area of the emitting body at the time of radiation release.
  • A participant presents a scenario involving a hollow cylinder with one side open, proposing a formula for calculating heat loss over time, but does not finalize the effective surface area term in the equation.
  • Further elaboration on the hollow cylinder scenario includes considerations of heat loss by convection and radiation, noting the need to account for shape factors and the inner surface area for accurate calculations.

Areas of Agreement / Disagreement

Participants express differing views on the definitions and calculations related to effective surface area, with no consensus reached on the correct approach or interpretation of the concepts discussed.

Contextual Notes

Limitations include potential missing assumptions regarding the definitions of effective surface area, the specific conditions of the hollow cylinder scenario, and unresolved mathematical steps in the proposed calculations.

sadhu
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in thermal radiations what is the difference between effective surface area and surface area of emitting body ,and how to calculate effective surface area of ang body?
 
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What is the context of the question? You can assume that the MBR has an effective surface area of the entire universe, but that the surface area of the emitting body would be the size of the universe when the radiation was first released.
 
consider a case of a hollow cylinder with one side open to outside &
external area is covered by insulating material, temperature is T
assuming perfect BB how to calculate heat lost by it in 1 sec,

E=[tex]\sigma[/tex]*T[tex]^{4}[/tex]*?

length is l, radius is r.as i think ?=[tex]\pi[/tex]*r[tex]^{2}[/tex]
 
sadhu said:
consider a case of a hollow cylinder with one side open to outside &
external area is covered by insulating material, temperature is T
assuming perfect BB how to calculate heat lost by it in 1 sec,
E=[tex]\sigma[/tex]*T[tex]^{4}[/tex]*?

length is l, radius is r.

as i think ?=[tex]\pi[/tex]*r[tex]^{2}[/tex]

Your body loses heat by convection and radiation.
For convection you need to estimate H and then you can calculate how much heat you lose by convection.
your body also loses heat by radiation, but because you are dealing with cylinder , some of the radiation is reflected by the inner sides of the cylinder and only some of the radiation escape outside of the body. you need to know what fraction of the energy escapes to the outside (look for shape factor on any heat transfer book).
your heat transfer area by convection and radiation is the inner surface area of the cylinder. A=PI*D*L
 

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