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A 20-cm in diameter pipeline with a relative roughness of 0.01 has a total length of 45 m. When water (viscosity is 1x10-7 m2/s)is pumped through it at a rate of 5 m3/min, the major head loss (m) is most nearly
A) 3 mB) 10 mC) 15 mD) 20 m

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Answer:

A) 3 m

Explanation:

The major head loss is calculated by using the expression: [tex]f \dfrac{L}{D}( \dfrac {U^2}{2g})[/tex]

here:

U = Q/A

Using the continuity equation:

[tex]U = \dfrac{Q}{\dfrac{\pi}{4}D^2}[/tex]

[tex]U = \dfrac{5/60 \ m^3/s}{\dfrac{\pi}{4}(0.2)^2}[/tex]

U = 2.65 m/s

Reynolds no = [tex]\dfrac{\rho U D}{\mu}[/tex]

[tex]= \dfrac{998 \ kg/m^3 \times 2.65 \ m/s \times 0.2 m}{8.93 \times 10^{-4} \ Pa^-s}[/tex]

= 594,894

Thus, this implies that the flow is turbulent.

Using Moddy's diagram at 5.94 × 10⁻⁵ &;

the relative roughness of 0.01

f = 0.038

Thus, the major head loss = [tex]f \dfrac{L}{D}( \dfrac {U^2}{2g})[/tex]

[tex]=0.038 \times (\dfrac{45\ m }{0.2\ m } )\times \dfrac{(2.65 \ m/s) ^2}{2(9.81 \ m/s^2)}[/tex]

= 3.06 m

[tex]\simeq[/tex] 3 m

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