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Answer:
6.27 × 10⁻⁴
Explanation:
Relative roughness, k = ε/D where ε = absolute roughness = 0.0025 mm and D = internal pipe diameter = 0.157 in = 0.157 × 25.4 mm = 3.9878 mm
So, k = ε/D
= 0.0025 mm/3.9878 mm
= 6.27 × 10⁻⁴
The relative pipe roughness for a plastic pipe will be:
"6.27 × 10⁻⁴".
According to the question,
Absolute roughness, ε = 0.0025 mm
Internal pipe diameter, D = 0.157 in or,
= 0.157 × 25.4 mm
= 3.9878 mm
We know that,
The relative roughness be:
→ k = [tex]\frac{Absolute \ roughness}{Diameter}[/tex]
or,
k = [tex]\frac{\varepsilon }{D}[/tex]
By substituting the above values,
= [tex]\frac{0.0025}{3.9878}[/tex]
= 6.27 × 10⁻⁴
Thus the above approach is correct.
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