Answer:
The heat loss per unit length is [tex]\frac{Q}{L} = 2981 W/m[/tex]
Explanation:
From the question we are told that
The outer diameter of the pipe is [tex]d = 104mm = \frac{104}{1000} = 0.104 m[/tex]
The thickness is [tex]D = 2mm = \frac{2}{1000} = 0.002m[/tex]
The temperature of water is [tex]T = 90^oC = 90 + 273 = 363K[/tex]
The outside air temperature is [tex]T_a = -10^oC = -10 +273 = 263K[/tex]
The water side heat transfer coefficient is [tex]z_1 = 300 W/ m^2 \cdot K[/tex]
The heat transfer coefficient is [tex]z_2 = 20 W/m^2 \cdot K[/tex]
The heat lost per unit length is mathematically represented as
[tex]\frac{Q}{L} = \frac{2 \pi (T - Ta)}{ \frac{ln [\frac{d}{D} ]}{z_1} + \frac{ln [\frac{d}{D} ]}{z_2}}[/tex]
Substituting values
[tex]\frac{Q}{L} = \frac{2 * 3.142 (363 - 263)}{ \frac{ln [\frac{0.104}{0.002} ]}{300} + \frac{ln [\frac{0.104}{0.002} ]}{20}}[/tex]
[tex]\frac{Q}{L} = \frac{628}{0.2107}[/tex]
[tex]\frac{Q}{L} = 2981 W/m[/tex]
