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Nov 24, 2019 at 22:04 vote accept some_math_guy
Nov 24, 2019 at 1:13 comment added Chet Miller If it were conduction, it would be $kS\frac{\Delta T}{\Delta x}$, where $\Delta x$ is the distance over which the temperature changes. The convective heat transfer rate can be alternately interpreted as conduction of sorts if we set the heat transfer coefficient equal to $h=k/\delta$, where k is the thermal conductivity of the gas surrounding the wire and $\delta$ is the thickness of the thermal boundary layer. We would then have $\dot{Q}=kS\frac{T_0-T}{\delta}$
Nov 23, 2019 at 20:29 comment added some_math_guy so , the reason why is hS(T0−T) and not hSdT is because it is convective heat transfer instead of conductive heat transfer?
Nov 23, 2019 at 2:24 history answered Chet Miller CC BY-SA 4.0