$h=\frac{Nu_{D}k}{D}=\frac{10 \times 0.025}{0.004}=62.5W/m^{2}K$
The convective heat transfer coefficient for a cylinder can be obtained from:
The outer radius of the insulation is:
$\dot{Q}=h \pi D L(T_{s}-T
The heat transfer due to radiation is given by:
Assuming $k=50W/mK$ for the wire material,
The heat transfer from the insulated pipe is given by:
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