1lm is about 0.0015W (683lm=1W). If you use 9A current total lumens are about 3x2000lm=6000lm. This is same as 9W, not more.
1.8m3/h=0.5dm3/s.Thanks for responding people!
So it seems that I have to get rid of about 105 watt.
assuming a thermal resistance of total 2,0 C/watt (0,64 for the junction to LED-body plus 1,36 for the cooling body) and a maximum T-junction of 100 C, I can use the thermal difference between 30 C (t-ambient) and the cooling finns (at about 77 C) to transfer the heat to the air around the flashlight.
deltaT = 47K
Power = 0,034kW
mass of air = 1,2
calculated airvolume will be about 1,8 m3/h times 3 (3 leds) = 5.4m3/h
This should be possible for a small fan :thumbsup:
Can anybody correct me if I'm wrong?
You are right. Led emit invisible wavelenght too.But you need to know what this means - Luminous Efficacy of RADIATION = how good the eye is at converting OPTICAL watts to units of visibity. 555nm radiant energy produces lots of lumens, 755nm radiation produces zero lumens.
The OP is interested in Heat Output Watts, - the eye is NOT sensitive to HEAT i.e energy at less than 700nm - this Luminous Efficacy info is totally irrelevant.