300℃.The allowable heat loss through the pipe is 600 W/m length of pipe for the safety; the temperature of the outside surface of insulation must not exceed 100℃. Determine minimum thickness of insulation required.

$r_2$=$\frac{55}{2}$=27.5mm,$r_1$=25mm,k=0.02W⁄mk,$T_1$=300+273=573k,Q=600W⁄m,

$T_∞\lt373k$

Minimum thickness of insulation is $r_cr$=$\frac{k}{h}$

Q=$\frac{(T_1-T_∞)}{(ln\frac{r_2/r_1 }{2πLk+1}\frac{1}{h(2πr_2 L)})}$

600= $\frac{573-373}{(ln⁡\frac{(27.5/25)}{(2π×0.02)}+\frac{1}{h(2π×27.5×10^-3 })}$

600= $\frac{200}{(0.758+\frac{5.7874}{h})}$

454.8h+3472.44-200h=0

h=13.628$w/m^2 $k

$r_cr$=0.02/13.628

$r_cr$=$1.467x10^-3 $ mANS