Question 1 |
Velocity of flow is proportional to the first power of hydraulic gradient in Darcy's law. The
law is applicable to
laminar flow in porous media | |
transitional flow in porous media | |
turbulent flow in porous media | |
laminar as well as turbulent flow in porous media |
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 ______
2.475 | |
2.8.25 | |
4.594 | |
8.547 |
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
0.25 | |
0.5 | |
6 | |
8 |
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}
\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.