# Turbulent Flow

 Question 1
Velocity of flow is proportional to the first power of hydraulic gradient in Darcy's law. The law is applicable to
 A laminar flow in porous media B transitional flow in porous media C turbulent flow in porous media D laminar as well as turbulent flow in porous media
GATE CE 2020 SET-1   Fluid Mechanics and Hydraulics
Question 1 Explanation:
Darcy's law is valid for laminar flow condition in porous media.
 Question 2
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 ______
 A 2.475 B 2.8.25 C 4.594 D 8.547
GATE CE 2018 SET-2   Fluid Mechanics and Hydraulics
Question 2 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}

 Question 3
The flow of water (mass density = 100 kg/$m^{3}$ and kinematic viscosity = $10^{-6} m^{2}/s$) in a commercial pipe, having equivalent roughness $k_{s}$ as 0.12 mm, yields an average shear stress at the pipe boundary = 600 N/$m^{2}$. The value of $k_{s}/{\delta }'$ (${\delta} '$ being the thickness of laminar sub-layer) for this pipe is
 A 0.25 B 0.5 C 6 D 8
GATE CE 2008   Fluid Mechanics and Hydraulics
Question 3 Explanation:
We know that,
\begin{aligned} \delta^{\prime}&=\frac{11.6 \mathrm{v}}{V_{*}} \\ \text {But } V_{*}&=\sqrt{\frac{\tau_{0}}{\rho}} \\ \therefore \quad 8^{\prime} &=\frac{11.6 \mathrm{v}}{\sqrt{\tau_{0} / \mathrm{p}}} \\ \Rightarrow \quad & \delta=11.6 \times 10^{-6} \times \sqrt{\frac{1000}{600} \times 1000}\\ \Rightarrow \quad \delta^{\prime}&=0.015 \mathrm{mm} \\ \therefore \quad \frac{k_{s}}{\delta^{\prime}}&=\frac{0.12}{0.015}=8.0 \end{aligned}

There are 3 questions to complete.