stardustsailor
Well-Known Member
"-How hot did the heat sink get for your passive 30W?
was no Problem to go for 50°C
they are now working at 450mA and more or less 40°C ambient @ 20°C .."
That is impressive...
Would you be so kind ,sir,to please me also ,and
give off the full dimensions of the heatsink used ?
Or a link with this model ?
(Probably Fischer ,uh ? ...)
Edit: And one more ...
Since I think I've found the right man....
Well ,does it play a significant role ,for the heatsink to be black anodised ?
(And I don't mean for alum. corrosion protection...I mean for heat dissipation through
"black"('grey" in reality ) body radiation....
According to Stefan-Boltzmann Law ...
q = ε σ T[SUP]4[/SUP] A
σ = 5.6703 10[SUP]-8[/SUP] (W/m[SUP]2[/SUP]K[SUP]4[/SUP])
ε=
Because ...
According to this...
Net Radiation Loss Rate
If an hot object is radiating energy to its cooler surroundings the net radiation heat loss rate can be expressed as
(T[SUB]ambient[/SUB] =Tc=25°C=298.15°K )
It seems so....
Plus ....
was no Problem to go for 50°C
they are now working at 450mA and more or less 40°C ambient @ 20°C .."
That is impressive...
Would you be so kind ,sir,to please me also ,and
give off the full dimensions of the heatsink used ?
Or a link with this model ?
(Probably Fischer ,uh ? ...)
Edit: And one more ...
Since I think I've found the right man....
Well ,does it play a significant role ,for the heatsink to be black anodised ?
(And I don't mean for alum. corrosion protection...I mean for heat dissipation through
"black"('grey" in reality ) body radiation....
According to Stefan-Boltzmann Law ...
q = ε σ T[SUP]4[/SUP] A
σ = 5.6703 10[SUP]-8[/SUP] (W/m[SUP]2[/SUP]K[SUP]4[/SUP])
ε=
Alloy 24ST Polished | 0.09 |
Alumina, Flame sprayed | 0.8 |
Aluminum Commercial sheet | 0.09 |
Aluminum Foil | 0.04 |
Aluminum Commercial Sheet | 0.09 |
Aluminum Heavily Oxidized | 0.2 - 0.31 |
Aluminum Highly Polished | 0.039 - 0.057 |
Aluminum Anodized | 0.77 |
Aluminum Rough | 0.07 |
Because ...
According to this...
Net Radiation Loss Rate
If an hot object is radiating energy to its cooler surroundings the net radiation heat loss rate can be expressed as
q = ε σ (T[SUB]h[/SUB][SUP]4[/SUP] - T[SUB]c[/SUB][SUP]4[/SUP]) A[SUB]c[/SUB]
where
T[SUB]h[/SUB] = hot body absolute temperature (K)
T[SUB]c[/SUB] = cold surroundings absolute temperature (K)
A[SUB]c[/SUB] = area of the object (m[SUP]2[/SUP])
Anodising ,really starts playing a role ,when Th is over 100° C.(373.15°K )...where
T[SUB]h[/SUB] = hot body absolute temperature (K)
T[SUB]c[/SUB] = cold surroundings absolute temperature (K)
A[SUB]c[/SUB] = area of the object (m[SUP]2[/SUP])
(T[SUB]ambient[/SUB] =Tc=25°C=298.15°K )
It seems so....
Plus ....
Heat emission from a surface in an angle β can be expressed with Lambert's cosine law as
q[SUB]β[/SUB] = q cos β
where
q[SUB]β[/SUB] = heat emission in angle β
q = heat emission from the surface
β = angle
Does anodising plays a significant role for heatsinks ,when we are dealling with less than 100° C ?
Or it's just a waste of money ?
(Quite a difference in retail price between anodised & non-anodised aluminium heatsinks....)
q[SUB]β[/SUB] = q cos β
where
q[SUB]β[/SUB] = heat emission in angle β
q = heat emission from the surface
β = angle
Does anodising plays a significant role for heatsinks ,when we are dealling with less than 100° C ?
Or it's just a waste of money ?
(Quite a difference in retail price between anodised & non-anodised aluminium heatsinks....)