PLAIN AND CIVIL: EXAMPLE 6.1.7. HEAD LOSS USING DARCY WEISBACH FORMULA WITH REYNOLD'S NUMBER

Wednesday, 17 April 2019

EXAMPLE 6.1.7. HEAD LOSS USING DARCY WEISBACH FORMULA WITH REYNOLD'S NUMBER

Fluid with Reynold's number = 1825, flows through a 25mm diameter pipe, 100m long. What is the discharge if head lost is 150m?

With the given Reynold's number = 1825, this is laminar flow. So the formula used for f becomes:

   $f=\frac{64}{R_{e}}=\frac{64}{1825}=0.0351$

Thus,

   $h_{f}=0.0826\frac{fLQ^{2}}{D^{5}}$

   $150=0.0826\frac{0.0351\left ( 100 \right )Q^{2}}{0.025^{5}}$

${\color{Red} \mathbf{Q=2.2478x10^{-3}cms\simeq 2.2478\frac{lit}{s}}}$

RELATED TOPIC:

6.0. PIPES

1 comment:

lost_in_woods27 June 2019 at 22:00
thank you for posting very informative for help go to the link Structural Analysis in USA
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Wednesday, 17 April 2019

EXAMPLE 6.1.7. HEAD LOSS USING DARCY WEISBACH FORMULA WITH REYNOLD'S NUMBER

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