Consider two flat infinite plates, surface A and surface B, both emitting radiation towards one another. Surface B is assumed to be an ideal emitter, εB= 1.
Surface A will emit radiation according to Stefan’s Boltzmann law as
(1)
And will receive radiation as
(2)
Net heat flow from surface A will be
(3)
Now suppose that the two surfaces are at exactly same temperature then, εA = αA
Emissivity of surface will depend on the material of which it is composed.
The radiation emitted per unit area per unit time from the surface of a body is called its emissive power. The ratio of emissive power of a body to the emissive power of a black body is called emissivity.
Heat emitted by the black body per unit area, (4)
Heat emitted by the test plate per unit area, (5)
εb Emissivity of the black plate.
εp Emissivity of the test plate
σ Stefan-Boltzmann constant = 5.67×10-8 W m-2K-4
Tb Black body temperature in Kelvin
Tc Chamber temperature in Kelvin
Tp Test plate temperature in Kelvin
Qb = Qp since input power to the two plates is same and conduction heat loss are also same.
Emissivity, (6)
Consider two flat infinite plates, surface A and surface B, both emitting radiation towards one another. Surface B is assumed to be an ideal emitter, εB= 1.Surface A will emit radiation according to Stefan’s Boltzmann law as (1)And will receive radiation as (2)Net heat flow from surface A will be (3)Now suppose that the two surfaces are at exactly same temperature then, εA = αAEmissivity of surface will depend on the material of which it is composed. The radiation emitted per unit area per unit time from the surface of a body is called its emissive power. The ratio of emissive power of a body to the emissive power of a black body is called emissivity.Heat emitted by the black body per unit area, (4)Heat emitted by the test plate per unit area, (5)εb Emissivity of the black plate.εp Emissivity of the test plateσ Stefan-Boltzmann constant = 5.67×10-8 W m-2K-4 Tb Black body temperature in KelvinTc Chamber temperature in KelvinTp Test plate temperature in KelvinQb = Qp since input power to the two plates is same and conduction heat loss are also same.Emissivity, (6)
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