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Infrared thermometry remote sensing
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[QUOTE="ag123, post: 6029510, member: 648622"] thanks tom, I've just ordered that, but I'm looking at this out of curiosity :smile: fine tuning the equations : $$\text {power received at the probe} \space P = a \varepsilon \sigma A T^4 / 4 \pi r^2 $$ heat transfer (out) from the probe: $$ \text {conduction + convection + radiation = power received at the probe } $$ conductive heat transfer (out) from probe: [URL]https://en.wikipedia.org/wiki/Thermal_conduction#Integral_form[/URL] $$ Q_{cond} = \big. \frac{Q}{\Delta t} = k A \frac{\Delta T}{\Delta x} $$ convective heat transfer (out) from probe: [URL='https://en.wikipedia.org/wiki/Newton%27s_law_of_cooling#Heat_transfer_version_of_the_law']https://en.wikipedia.org/wiki/Newton's_law_of_cooling#Heat_transfer_version_of_the_law[/URL] $$ Q_{conv} = h \cdot A \Delta T $$ radiative head transfer (out) from probe: $$ Q_{rad} = \varepsilon \sigma 4 \pi r_p ^ 2 T^4 $$ so equating both sides $$ Q_{cond} + Q_{conv} + Q_{rad} = a \varepsilon \sigma A T_s^4 / 4 \pi r^2 \\ k A \frac{\Delta T_p}{\Delta x} + h \cdot A \Delta T_p + \varepsilon \sigma 4 \pi r_p ^ 2 T_p^4 = a \varepsilon \sigma A T_s^4 / 4 \pi r^2 $$ aggregating the constant terms $$ C_1 \Delta T_p + C_2 \Delta T_p + C_3 T_p^4 = C_4 T_s^4 / r^2 $$ now this gets complicated as presumbly at lower temperatures conduction and convection dominates and the radiative heat loss at the probe $$ C_3 T_p^4 $$ is small, while at higher temperatures it is large. [/QUOTE]
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