|Oct15-07, 10:27 PM||#1|
watts to Kelvins
How might i convert the temperature in kelvins of a 1/2inch 2" x 2" steel plate when i have 28322 watts of sunlight focused on it for 5min? also if i am stating this question wrong i will be focusing 850 2" x 2" spots of sunlight on a single 2" x 2" surface how might i calculate the temperature in K?
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|Oct15-07, 11:02 PM||#2|
The temperature will be the result of the balance between incoming power (your sunlight), and whatever it will lose (by conduction, radiation, reflection, ...).
So the incoming power is only one element. As the lost power is usually a strong function of the temperature (the higher the temperature, the more losses there are), we can write: W_out(T). So the end temperature will be the solution of W_in = W_out(T).
The hardest part is always to figure out W_out(T), that is: the lost power as a function of temperature.
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