Respuesta :
Answer:
heat loss per 1-m length of this insulation is 4368.145 W
Explanation:
given data
inside radius r1 = 6 cm
outside radius r2 = 8 cm
thermal conductivity k = 0.5 W/m°C
inside temperature t1 = 430°C
outside temperature t2 = 30°C
to find out
Determine the heat loss per 1-m length of this insulation
solution
we know thermal resistance formula for cylinder that is express as
Rth = [tex]\frac{ln\frac{r2}{r1}}{2 \pi *k * L}[/tex] .................1
here r1 is inside radius and r2 is outside radius L is length and k is thermal conductivity
so
heat loss is change in temperature divide thermal resistance
Q = [tex]\frac{t1- t2}{\frac{ln\frac{r2}{r1}}{2 \pi *k * L}}[/tex]
Q = [tex]\frac{(430-30)*(2 \pi * 0.5 * 1}{ln\frac{8}{6} }[/tex]
Q = 4368.145 W
so heat loss per 1-m length of this insulation is 4368.145 W
