Question 1 |
The dimension of dynamic viscosity is:
ML^{-1}T^{-1} | |
ML^{-1}T^{-2} | |
ML^{-2}T^{-2} | |
ML^{0}T^{-1} |
Question 1 Explanation:
Unit of dynamic viscosity =\frac{kg}{m.s} \; or \; \frac{Ns}{m^2}=ML^{-1}T^{-1}
Question 2 |
Kinematic viscosity' is dimensionally represented as
\frac{M}{LT} | |
\frac{M}{L^{2} T} | |
\frac{T^{2}}{L} | |
\frac{L^{2}}{T} |
Question 2 Explanation:
Kinematic viscosity
v=\frac{\mu }{\rho }=\frac{kg/m\cdot s}{kg/m^3}=m^2/s
[v]=\frac{m^2}{s }=\frac{L^2}{T}
v=\frac{\mu }{\rho }=\frac{kg/m\cdot s}{kg/m^3}=m^2/s
[v]=\frac{m^2}{s }=\frac{L^2}{T}
Question 3 |
In a laboratory, a flow experiment is performed over a hydraulic structure. The measured values of discharge and velocity are 0.05 m^{3}/s and 0.25 m/s, respectively. If the full scale structure (30 times bigger) is subjected to a discharge of 270 m^{3}/s, then the time scale (model to full scale) value (up to two decimal places) is ______
0 | |
0.18 | |
0.55 | |
0.75 |
Question 3 Explanation:
\begin{aligned} \text { Froude Law }(F r)_{m}&=(F r)_{p} \\ \left(\frac{V}{\sqrt{L \mathrm{g}}}\right)_{m} &=\left(\frac{v}{\sqrt{L g}}\right)_{p} \quad\left(g_{m}=g_{p}\right) \\ v_{r} &=\sqrt{L_{r}} \\ \text{or}\quad \frac{L_{r}}{T_{r}} &=\sqrt{L} \\ T_{r} &=\sqrt{L_{r}} \\ T_{r} &=\sqrt{\frac{1}{30}}=0.1826 \end{aligned}
Question 4 |
A 1:50 model of a spillway is to be tested in the laboratory. The discharge in the prototype spillway is 1000 m^{3}/s. The corresponding discharge (in m^{3}/s, up to two decimal places) to be maintained in the model, neglecting variation in acceleration due to gravity, is ______
0.01 | |
0.06 | |
0.50 | |
0.10 |
Question 4 Explanation:
Froude law is valid
\begin{aligned} Q_{r} &=L_{r}^{25} \\ \frac{Q_{m}}{Q_{p}} &=\left(\frac{1}{50}\right)^{25} \\ \frac{Q_{m}}{1000} &=\left(\frac{1}{50}\right)^{25} \\ Q_{m} &=0.0566 \mathrm{m}^{3 / \mathrm{s}} \\ \text{So},\quad Q_{m} & \simeq 0.06 \mathrm{m}^{3 / \mathrm{s}} \end{aligned}
\begin{aligned} Q_{r} &=L_{r}^{25} \\ \frac{Q_{m}}{Q_{p}} &=\left(\frac{1}{50}\right)^{25} \\ \frac{Q_{m}}{1000} &=\left(\frac{1}{50}\right)^{25} \\ Q_{m} &=0.0566 \mathrm{m}^{3 / \mathrm{s}} \\ \text{So},\quad Q_{m} & \simeq 0.06 \mathrm{m}^{3 / \mathrm{s}} \end{aligned}
Question 5 |
The relationship between the length scale ratio (L_{r}) and the velocity scale ratio (V_{r}) in hydraulic models, in which Froude dynamic similarity is maintained, is:
V_{r}=L_{r} | |
L_{r}=\sqrt{V_{r}} | |
V_{r}=L_{r}^{1.5} | |
V_{r}=\sqrt{L_{r}} |
Question 5 Explanation:
As per Froude Law
\begin{aligned} \left(\frac{V}{\sqrt{L g}}\right)_{m}&=\left(\frac{V}{\sqrt{L g}}\right)_{p} \\ \Rightarrow \quad V_{r}&=\sqrt{L_{r}} \end{aligned}
\begin{aligned} \left(\frac{V}{\sqrt{L g}}\right)_{m}&=\left(\frac{V}{\sqrt{L g}}\right)_{p} \\ \Rightarrow \quad V_{r}&=\sqrt{L_{r}} \end{aligned}
Question 6 |
The drag force F_{D}, on a sphere kept in a uniform flow field depends on the diameter of the sphere, D; flow velocity, V; fluid density, \rho; and dynamic viscosity, \mu. Which of the following options represents the non-dimensional parameters which could be used to analyze this problem?
\frac{F_{D}}{VD}\:\: and\: \:\: \: \frac{\mu }{\rho VD} | |
\frac{F_{D}}{\rho VD^{2}}\:\: and\: \:\: \: \frac{\rho VD }{\mu} | |
\frac{F_{D}}{\rho V^{2}D^{2}}\:\: and\: \:\: \: \frac{\rho VD }{\mu} | |
\frac{F_{D}}{\rho V^{3}D^{3}}\:\: and\: \:\: \: \frac{\mu }{\rho VD} |
Question 6 Explanation:
\begin{aligned} \text{For sphere,}\quad A&=\frac{\pi}{4} D^{2} \\ F_{D} &=C_{D} \times \frac{1}{2} \times p \times A \times V^{2} \\ F_{D} &=C_{D} \times \frac{1}{2} \times \rho \times\left(\frac{\pi}{4} D^{2}\right) \times V^{2} \\ \frac{F_{D}}{\rho D^{2} V^{2}} &=\frac{1}{2} \times\left(\frac{\pi}{4}\right) \times C_{D} \end{aligned}
Hence, right hand side is dimensionless, so \frac{F_{0}}{\rho D^{2} V^{2}} should also be a dimensionless parameter.
Option (C) \frac{F_{0}}{\rho V^{2} D^{2}}, \frac{\rho V D}{\mu}
Hence, right hand side is dimensionless, so \frac{F_{0}}{\rho D^{2} V^{2}} should also be a dimensionless parameter.
Option (C) \frac{F_{0}}{\rho V^{2} D^{2}}, \frac{\rho V D}{\mu}
Question 7 |
Group-I contains dimensionless parameters and Group- II contains the ratios.
The correct match of dimensionless parameters in Group- I with ratios in Group-II is:
The correct match of dimensionless parameters in Group- I with ratios in Group-II is:
P-3, Q-2, R-4, S-1 | |
P-3, Q-4, R-2, S-1 | |
P-2, Q-3, R-4, S-1 | |
P-1, Q-3, R-2, S-4 |
Question 8 |
A river reach of 2.0 km long with maximum flood discharge of 10000m^{3}/s is to be physically modeled in the laboratory where maximum available discharge is 0.20 m^{3}/s. For a geometrically similar model based on equally of Froude number, the length of the river reach (m) in the model is
26.4 | |
25 | |
20.5 | |
18 |
Question 8 Explanation:
According to Froude model law
\begin{aligned} \therefore \quad \frac{V_{t}}{\sqrt{L_{r} g_{1}}}&=1 \quad\left[\because F_{r}=1\right]\\ \text{Now, we have}\\ \frac{Q_{m}}{Q_{p}}&=L_{1}^{2} \times V_{r}\\ \Rightarrow \quad \frac{Q_{m}}{Q_{p}}&=L_{r}^{2} \times \sqrt{L_{r}} \\ \Rightarrow \quad \frac{Q_{m}}{Q_{0}}&=\left(L_{r}\right)^{5 / 2} \\ \Rightarrow \quad \frac{0.20}{10000}&=\left(\frac{L_{m}}{2 \times 1000}\right)^{5 / 2} \\ \Rightarrow \quad L_{m}&=26.4 \mathrm{m} \end{aligned}
\begin{aligned} \therefore \quad \frac{V_{t}}{\sqrt{L_{r} g_{1}}}&=1 \quad\left[\because F_{r}=1\right]\\ \text{Now, we have}\\ \frac{Q_{m}}{Q_{p}}&=L_{1}^{2} \times V_{r}\\ \Rightarrow \quad \frac{Q_{m}}{Q_{p}}&=L_{r}^{2} \times \sqrt{L_{r}} \\ \Rightarrow \quad \frac{Q_{m}}{Q_{0}}&=\left(L_{r}\right)^{5 / 2} \\ \Rightarrow \quad \frac{0.20}{10000}&=\left(\frac{L_{m}}{2 \times 1000}\right)^{5 / 2} \\ \Rightarrow \quad L_{m}&=26.4 \mathrm{m} \end{aligned}
Question 9 |
A 1:50 scale model of a spillway is to be tested in the laboratory. The discharge in the prototype is 1000 m^{3}/s. The discharge to be maintained in the model test is
0.05m^{3}/s | |
0.08m^{3}/s | |
0.57m^{3}/s | |
5.7m^{3}/s |
Question 9 Explanation:
Froude model law will be applicable in this case
\begin{aligned} \frac{Q_{m}}{Q_{p}} &=\left(\frac{L_{m}}{L_{p}}\right)^{5 / 2} \\ \Rightarrow \quad Q_{m} &=1000 \times\left(\frac{1}{50}\right)^{25} \\ &=0.057 \mathrm{m}^{3} / \mathrm{sec} \end{aligned}
\begin{aligned} \frac{Q_{m}}{Q_{p}} &=\left(\frac{L_{m}}{L_{p}}\right)^{5 / 2} \\ \Rightarrow \quad Q_{m} &=1000 \times\left(\frac{1}{50}\right)^{25} \\ &=0.057 \mathrm{m}^{3} / \mathrm{sec} \end{aligned}
Question 10 |
The flow of glycerin (kinematic viscosity v = 5 \times 10^{-4}m^{2}/s) in an open channel
is to be modeled in a laboratory flume using water (v = 10^{-6}m^{2}/s) as the flowing fluid. If both gravity and viscosity are important, what should be the length scale
(i.e. ratio of prototype to model dimensions) for maintaining dynamic similarity ?
1 | |
22 | |
63 | |
500 |
Question 10 Explanation:
Equating Reynolds number and Froude number, we get
\begin{aligned} \frac{V_{t} L}{v_{r}} &=\frac{V_{r}}{\sqrt{L}} \\ \therefore\quad L_{r} &=\left(v_{r}\right)^{2 / 3} \\ v_{r} &=\frac{v_{m}}{v_{p}}=\frac{10^{-6}}{5 \times 10^{-4}}=2 \times 10^{-3} \\ \Rightarrow\quad L_{r} &=\left(2 \times 10^{-3}\right)^{2 / 3} \\ \Rightarrow\quad L_{r} &=0.0159 \quad \text { But } \cdot L_{r}=\frac{L_{m}}{L_{p}} \\ \Rightarrow\quad \frac{L_{p}}{L_{m}} &=\frac{1}{L_{r}}=\frac{1}{0.0159}=62.99 \approx 63 \end{aligned}
\begin{aligned} \frac{V_{t} L}{v_{r}} &=\frac{V_{r}}{\sqrt{L}} \\ \therefore\quad L_{r} &=\left(v_{r}\right)^{2 / 3} \\ v_{r} &=\frac{v_{m}}{v_{p}}=\frac{10^{-6}}{5 \times 10^{-4}}=2 \times 10^{-3} \\ \Rightarrow\quad L_{r} &=\left(2 \times 10^{-3}\right)^{2 / 3} \\ \Rightarrow\quad L_{r} &=0.0159 \quad \text { But } \cdot L_{r}=\frac{L_{m}}{L_{p}} \\ \Rightarrow\quad \frac{L_{p}}{L_{m}} &=\frac{1}{L_{r}}=\frac{1}{0.0159}=62.99 \approx 63 \end{aligned}
There are 10 questions to complete.