Post by foof on Oct 9, 2016 12:27:23 GMT
I tried do some calculations for radiators. I want to use this formula:
Source
σ is given and I get P and T from the module I want to cool. However I need ε which is how effective a material can get rid of heat in form of thermal radiation.
It is a dimensionless number between 0 (for a perfect reflector) and 1 (for a perfect emitter).
But ε is not listed as one of the material properties given by the CoaDE, can I derive it somehow or is there another way to calculate the area of a radiator with only the material as a variable?
Any advice would be helpful.
P = A * ε * σ * T4
A = P / (ε * σ * T4)
where
P = the power of waste heat the radiator can get rid of (watts)
σ = 5.670373×10-8 = Stefan-Boltzmann constant (W m-2K-4)
ε = emissivity of radiator (theoretical maximum is 1.0 for a perfect black body, real world radiator will be less. Should be at least 0.8 or above to be worth-while)
A = area of radiator (m2)
T = temperature of radiator, this assumes temperature of space is zero degrees (degrees K)
x4 = raise x to the fourth power, i.e, x * x * x * x
A = P / (ε * σ * T4)
where
P = the power of waste heat the radiator can get rid of (watts)
σ = 5.670373×10-8 = Stefan-Boltzmann constant (W m-2K-4)
ε = emissivity of radiator (theoretical maximum is 1.0 for a perfect black body, real world radiator will be less. Should be at least 0.8 or above to be worth-while)
A = area of radiator (m2)
T = temperature of radiator, this assumes temperature of space is zero degrees (degrees K)
x4 = raise x to the fourth power, i.e, x * x * x * x
σ is given and I get P and T from the module I want to cool. However I need ε which is how effective a material can get rid of heat in form of thermal radiation.
It is a dimensionless number between 0 (for a perfect reflector) and 1 (for a perfect emitter).
But ε is not listed as one of the material properties given by the CoaDE, can I derive it somehow or is there another way to calculate the area of a radiator with only the material as a variable?
Any advice would be helpful.