Answer:
98,614.82 W/m²
Explanation:
[tex]Q = 2\pi hL(\frac{T_2-T_1}{Ln\frac{r_2}{r_1}})[/tex]
Where;
Q = the amount of heat loss from the pipe
h = the heat transfer coefficient of the pipe = 50 W/m².K
T₁ = the ambient temperature of the pipe = 30⁰C
T₂ = the outside temperature of the pipe = 100⁰C
L= the length of pipe
r₁ = inner radius of the pipe = 20mm
r₂ = outer radius of the pipe = 25mm
To determine the amount of heat loss from the pipe per unit length
From the equation above
[tex]\frac{Q}{L} = 2\pi h(\frac{T_2-T_1}{Ln\frac{r_2}{r_1}})[/tex]
[tex]\frac{Q}{L} = 2\pi* 50(\frac{100-30}{Ln\frac{25}{20}})[/tex]
[tex]\frac{Q}{L} = 314.159(\frac{70}{0.223})[/tex]
[tex]\frac{Q}{L} = 314.159(313.901)[/tex] = 98,614.82 W/m²