Question 1 |

Which of the following statements is/are TRUE?

The thickness of a turbulent boundary layer on a flat plate kept parallel to the flow direction is proportional to the square root of the distance from the leading edge | |

If the streamlines and equipotential lines of a source are interchanged with each other, the resulting flow will be a sink. | |

For a curved surface immersed in a stationary liquid, the vertical component of the force on the curved surface is equal to the weight of the liquid above it | |

For flow through circular pipes, the momentum correction factor for laminar flow is larger than that for turbulent flow |

Question 1 Explanation:

P_{H}= Total pressure on the projected are of the curved surface on the vertical plane.

P_{H} is equal to the total pressure exerted by the liquid on an imaginary vertically immersed plane surface which is the vertical projection of the curved surface, it will act at the center of pressure of the lane surface.

P_{v}= The weight o the liquid contained in the portion extending vertically above the curved surface upto the free surface of the liquid.

P_{V} will act through the centre of gravity of the volume of liquid contained in the portion extending above the curved surface upto the free surface of the liquid (represented by the profile ABCDEFA in the present case).

Momentum correction factor (\beta)=\frac{1}{A V^{2}} \int \mathrm{v}^{2} d A

{Sink Flow :}

The sink flow is the flow in which fluid moves radially inwards towards a point where it disappears at a constant rate. This flow is just opposite to the source flow.

Figure shows a sink flow in which the fluid moves radially inwards towards point \mathrm{O}, where it disappears tat a constant rate.

The pattern of stream lines and equipotential lines of a sink flow is the same as that of source flow.

All the equations derived for a source flow shall hold to good for sink flow also except that in sink flow equations, q is to be replaced by (-q)

**Boundary layer thickness (Normal Boundary layer thickness disturbance thickness.**

Distance from the boundary surface in which the velocity reaches 99 \% of the stream velocity is the boundary layer thickness.

Turbulent boundary layer on smooth plate.

\frac{\delta}{x}=\frac{0.376}{\left(R_{e x}\right)^{1 / 5}}, \quad R_{e x}=\frac{v_{0} x}{v}

[Valid for R_{e x} \gt 5 \times 10^{5} and R_{e x} \lt 10^{7} ]

=\frac{0.22}{\left(R_{e x}\right)^{1 / 6}}, 10^{7} \lt R_{e x} \lt 10^{9}

Question 2 |

The critical flow condition in a channel is given by ____

[Note: \alpha - kinetic energy correction factor; Q - discharge; A_{C} - cross-sectional area of flow at critical flow condition; T_{C}- top width of flow at critical flow condition; g- acceleration due to gravity]

[Note: \alpha - kinetic energy correction factor; Q - discharge; A_{C} - cross-sectional area of flow at critical flow condition; T_{C}- top width of flow at critical flow condition; g- acceleration due to gravity]

\frac{\alpha Q^{2}}{g}=\frac{A_{c}^{3}}{T_{C}} | |

\frac{\alpha Q}{g}=\frac{A_{c}^{3}}{T_{C}^{2}} | |

\frac{\alpha Q^{2}}{g}=\frac{A_{c}^{3}}{T_{C}^{2}} | |

\frac{\alpha Q}{g}=\frac{A_{c}^{3}}{T_{C}} |

Question 2 Explanation:

Specific Energy (E)=y+\frac{v^{2}}{2 g}

\Rightarrow \quad E=y+\alpha \cdot \frac{Q^{2}}{2 g A^{2}}\left[\right. as \left.\frac{Q}{A}=v\right]

For critical flow condition,

\begin{aligned} \frac{d E}{d y} & =0 \\ \Rightarrow \quad \frac{d E}{d y} & =1+\frac{\alpha Q^{2}}{2 g} \frac{d}{d y}\left(A^{-2}\right) \\ \Rightarrow \quad \frac{d E}{d y} & =1+\frac{\alpha Q^{2}}{2 g}\left(-\frac{2}{A^{3}}\right) \frac{d A}{d y} \end{aligned}

'dA' can be written as 'Tdy' \Rightarrow \frac{d A}{d y}=T

\Rightarrow \quad \frac{\mathrm{dE}}{\mathrm{dy}}=1-\frac{\alpha \mathrm{Q}^{2} \mathrm{~T}}{\mathrm{gA} \mathrm{A}^{3}}

\Rightarrow 1-\frac{\alpha Q^{2} \mathrm{~T}}{g A^{3}}=0

\Rightarrow \quad \frac{\alpha Q^{2}}{g}=\frac{\mathrm{A}^{3}}{\mathrm{~T}}

At critical flow, A=A_{C};

\mathrm{T}=\mathrm{T}_{\mathrm{C}}

\Rightarrow \quad \frac{\alpha Q^{2}}{g}=\frac{A_{C}^{3}}{T_{C}}

\Rightarrow \quad E=y+\alpha \cdot \frac{Q^{2}}{2 g A^{2}}\left[\right. as \left.\frac{Q}{A}=v\right]

For critical flow condition,

\begin{aligned} \frac{d E}{d y} & =0 \\ \Rightarrow \quad \frac{d E}{d y} & =1+\frac{\alpha Q^{2}}{2 g} \frac{d}{d y}\left(A^{-2}\right) \\ \Rightarrow \quad \frac{d E}{d y} & =1+\frac{\alpha Q^{2}}{2 g}\left(-\frac{2}{A^{3}}\right) \frac{d A}{d y} \end{aligned}

'dA' can be written as 'Tdy' \Rightarrow \frac{d A}{d y}=T

\Rightarrow \quad \frac{\mathrm{dE}}{\mathrm{dy}}=1-\frac{\alpha \mathrm{Q}^{2} \mathrm{~T}}{\mathrm{gA} \mathrm{A}^{3}}

\Rightarrow 1-\frac{\alpha Q^{2} \mathrm{~T}}{g A^{3}}=0

\Rightarrow \quad \frac{\alpha Q^{2}}{g}=\frac{\mathrm{A}^{3}}{\mathrm{~T}}

At critical flow, A=A_{C};

\mathrm{T}=\mathrm{T}_{\mathrm{C}}

\Rightarrow \quad \frac{\alpha Q^{2}}{g}=\frac{A_{C}^{3}}{T_{C}}

Question 3 |

A compound symmetrical open channel section as shown in the figure has a maximum of depth(s).

B_{m}- Bottom width of main channel

B_{\mathrm{f}}- Bottom width of flood channel

\mathrm{y}_{\mathrm{m}} - Depth of main channel

y - Total depth of the channel

\mathrm{n}_{\mathrm{ma}}-Manning's roughness of the main channel

\mathrm{n}_{f}- Manning's roughness of the flood channel

B_{m}- Bottom width of main channel

B_{\mathrm{f}}- Bottom width of flood channel

\mathrm{y}_{\mathrm{m}} - Depth of main channel

y - Total depth of the channel

\mathrm{n}_{\mathrm{ma}}-Manning's roughness of the main channel

\mathrm{n}_{f}- Manning's roughness of the flood channel

3 | |

2 | |

1 | |

4 |

Question 3 Explanation:

Question 4 |

The pressure in a pipe at X is to be measured by an open manometer as shown in figure. Fluid A is oil with a specific gravity of 0.8 and Fluid B is mercury with a specific gravity of 13.6. The absolute pressure at X is \mathrm{kN} / \mathrm{m}^{2} (round off to one decimal place).

[Assume density of water as 1000 \mathrm{~kg} / \mathrm{m}^{3} and acceleration due to gravity as 9.81 \mathrm{~m} / \mathrm{s}^{2} and atmospheric pressure as \left.101.3 \mathrm{kN} / \mathrm{m}^{2}\right]

[Assume density of water as 1000 \mathrm{~kg} / \mathrm{m}^{3} and acceleration due to gravity as 9.81 \mathrm{~m} / \mathrm{s}^{2} and atmospheric pressure as \left.101.3 \mathrm{kN} / \mathrm{m}^{2}\right]

115.2 | |

253.6 | |

254.3 | |

140.5 |

Question 4 Explanation:

Equating pressure at \mathrm{A}-\mathrm{A}^{\prime}

\begin{aligned} & P_{a t m}+\rho_{B} \gamma_{w} \times \frac{25}{100}=P_{x}-\rho_{A} \gamma_{w} \times \frac{75}{100} \\ & \therefore \quad P_{x}=P_{a t m}+\rho_{B} \gamma_{w} \times 0.25+\rho_{A} \gamma_{w} 0.75 \end{aligned}

\begin{gathered} =101.3+13.6 \times 9.81 \times 0.25+0.8 \times 9.81 \times 0.75 \\ =140.54 \mathrm{kN} / \mathrm{m}^{2} \end{gathered}

Question 5 |

A hydraulic jump occurs in a 1.0 \mathrm{~m} wide horizontal, frictionless, rectangular channel, with a pre-jump depth of 0.2 \mathrm{~m} and a post-jump depth of 1.0 \mathrm{~m}. The value of g may be taken as 10 \mathrm{~m} / \mathrm{s}^{2}. The values of the specific force at the pre-jump and post-jump sections are same and are equal to (in \mathrm{m}^{3}, rounded off to two decimal places) ____

0.25 | |

0.38 | |

0.51 | |

0.62 |

Question 5 Explanation:

Given,

\begin{aligned} & B=1.0 \mathrm{~m} \\ & y_{1}=0.2 \mathrm{~m} \\ & y_{2}=1.0 \mathrm{~m} \\ & g=10 \mathrm{~m} / \mathrm{s} \end{aligned}

The value of specific force

\begin{aligned} & =\frac{\mathrm{P}+\mathrm{M}}{\gamma_{\mathrm{w}}}=\text { constant } \\ & =\frac{\mathrm{P}_{1}+\mathrm{M}_{1}}{\gamma_{\mathrm{w}}}=\frac{\mathrm{P}_{2}+\mathrm{M}_{2}}{\gamma_{\mathrm{w}}} \\ & =A \overline{\mathrm{y}}_{1}+\frac{\mathrm{Q}^{2}}{\mathrm{Ag}} \end{aligned}

For horizontal, frictionless rectangular channel

\therefore \quad \frac{2 q^{2}}{g}=y_{1} y_{2}\left(y_{1}+y_{2}\right)

\begin{aligned} Q & =\sqrt{\frac{1 \times 0.2 \times(1.2) \times 10}{2}} \quad(B=1 \mathrm{~m}) \\ & =1.095 \mathrm{~m}^{3} / \mathrm{s} / \mathrm{m} \\ Q & =1.095 \mathrm{~m}^{3} / \mathrm{s} \end{aligned}

Now, specific force =0.2 \times 1 \times \frac{0.2}{2}+\frac{1.095^{2}}{1 \times 0.2 \times 10} =(0.62) \mathrm{m}^{3}

\begin{aligned} & B=1.0 \mathrm{~m} \\ & y_{1}=0.2 \mathrm{~m} \\ & y_{2}=1.0 \mathrm{~m} \\ & g=10 \mathrm{~m} / \mathrm{s} \end{aligned}

The value of specific force

\begin{aligned} & =\frac{\mathrm{P}+\mathrm{M}}{\gamma_{\mathrm{w}}}=\text { constant } \\ & =\frac{\mathrm{P}_{1}+\mathrm{M}_{1}}{\gamma_{\mathrm{w}}}=\frac{\mathrm{P}_{2}+\mathrm{M}_{2}}{\gamma_{\mathrm{w}}} \\ & =A \overline{\mathrm{y}}_{1}+\frac{\mathrm{Q}^{2}}{\mathrm{Ag}} \end{aligned}

For horizontal, frictionless rectangular channel

\therefore \quad \frac{2 q^{2}}{g}=y_{1} y_{2}\left(y_{1}+y_{2}\right)

\begin{aligned} Q & =\sqrt{\frac{1 \times 0.2 \times(1.2) \times 10}{2}} \quad(B=1 \mathrm{~m}) \\ & =1.095 \mathrm{~m}^{3} / \mathrm{s} / \mathrm{m} \\ Q & =1.095 \mathrm{~m}^{3} / \mathrm{s} \end{aligned}

Now, specific force =0.2 \times 1 \times \frac{0.2}{2}+\frac{1.095^{2}}{1 \times 0.2 \times 10} =(0.62) \mathrm{m}^{3}

There are 5 questions to complete.