Q1GATE 2020MCQ

Velocity of flow is proportional to the first power of hydraulic gradient in Darcy's law. The law is applicable to

Q2GATE 2018NAT

A rough pipe of 0.5 m diameter, 300 m length and roughness height of 0.25 mm, carries water (kinematic viscosity $0.9\times 10^{-6}\: \: m^{2}/s$ ) with velocity of 3 m/s. Friction factor (f) for laminar flow is given by $f=64/R_{e}$ , and for turbulent flow it is given by $\frac{1}{\sqrt{f}}=2\log_{10}\left ( \frac{r}{k} \right )+1.74$ where, $R_e$ = Reynolds number, r = radius of pipe, k = roughness height and $g=9.81\: \: m/s^{2}$ . The head loss (in m, up to three decimal places) in the pipe due to friction is ______

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📖 Explanation

\begin{aligned} R e &=\frac{\rho \cdot V \cdot D}{\mu}=\frac{V \cdot D}{v}=\frac{3 \times (0.5)}{0.9 \times 10^{-6}} \\ &=1.67 \times 10^{6}\\ \text{Means }R e&>2000 \text{ turbulent flow.}\\ \text{So, }\quad \frac{1}{\sqrt{f}}&=2 \log _{10} \frac{D}{2 k_{s}}+1.74 \\ \frac{1}{\sqrt{f}} &=2 \log _{10} \frac{0.5}{2 \times 0.25 \times 10^{-3}}+1.74 \\ f &=0.01669 \\ h_{f} &=\frac{f \cdot L \cdot V^{2}}{2 g D}=\frac{(0.01669)(300)(3)^{2}}{2 \times 9.81 \times 0.5} \\ &=4.594 \mathrm{m} \end{aligned}

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