Question 1 |

A hydraulic jump takes place in a 6 m wide rectangular channel at a point where
the upstream depth is 0.5 m (just before the jump). If the discharge in the
channel is 30 m^3/s and the energy loss in the jump is 1.6 m, then the Froude
number computed at the end of the jump is ___________. (round off to two
decimal places)
(Consider the acceleration due to gravity as 10 m/s^2.)

0.40 | |

0.85 | |

0.65 | |

0.75 |

Question 1 Explanation:

Q=30m^3/sec

B=6m

y_1=0.5m

E_L=1.6m

q= \frac{Q}{B} =\frac{30}{6}=5m^3/sec/m

We know that,

\begin{aligned} E_L=\frac{(y_2-y_1)^3}{4y_1y_2}&=1.6m\\ \frac{(y_2-0.5)^3}{4 \times 0.5 \times y_2}&=1.6\\ y_2^3-1.5y_2^2+0.75y_2-0.125&=3.2y_2\\ y_2=2.5m,-0.0527m,&-0.0947m \end{aligned}

Hence, y_2=2.5m

Post jump Froude's No.

(F_2)=\frac{V_2}{\sqrt{\sqrt{gy_2}}}=\frac{\left ( \frac{30}{6 \times 2.5} \right )}{\sqrt{10 \times 2.5}}=0.4

Question 2 |

Water is flowing in a horizontal, frictionless, rectangular channel. A smooth
hump is built on the channel floor at a section and its height is gradually increased
to reach choked condition in the channel. The depth of water at this section is y_2 and that at its upstream section is y_1. The correct statement(s) for the choked and
unchoked conditions in the channel is/are

In choked condition, y_1 decreases if the flow is supercritical and increases if the flow is subcritical. | |

In choked condition, y_2 is equal to the critical depth if the flow is supercritical or subcritical. | |

In unchoked condition, y_1 remains unaffected when the flow is supercritical or subcritical. | |

In choked condition, y_1 increases if the flow is supercritical and decreases if the flow is subcritical. |

Question 2 Explanation:

Question 3 |

Depth of water flowing in a 3 m wide rectangular channel is 2 m. The channel carries a discharge of 12 m^3/s . Take g = 9.8 m/s ^2.

The bed width (in m) at contraction, which just causes the critical flow, is _________ without changing the upstream water level. (round off to two decimal places)

The bed width (in m) at contraction, which just causes the critical flow, is _________ without changing the upstream water level. (round off to two decimal places)

2.85 | |

4.25 | |

2.15 | |

1.55 |

Question 3 Explanation:

Given: B = 3m, y = 2m, Q = 12 m^3
/sec

Velocity(v)=\frac{Q}{A} =\frac{12}{2 \times 3}=2m/sec

Specific energy at section (1-1)

E=y+\frac{v^2}{2g}=2+\frac{2^2}{2 \times 9.8}=\frac{108}{49}m

When channel section is contracted to minimum width and for constant discharge Q, the flow over contracted section will be critical flow and under the assumption that no energy loss has taken place.

E=E_c=\frac{108}{49}

We have that, E_c=\frac{3}{2}y_c (for rectangular crosssection)

y_c=\frac{2}{3}E_c=\frac{2}{3} \times \frac{108}{49}=\frac{72}{49}

For critical flow condition,

\begin{aligned} \frac{Q^2T}{gA^3}&=1\\ \frac{Q^2B_{min}}{g(B_{min} \times y_c)^3}&=1\\ (B_{min})^2&=\frac{Q^2}{gy_c^3}\\ B_{min}&=\left ( \frac{(12)^2}{9.8 \times \left ( \frac{72}{49} \right )^3} \right )^{1/2}\\ &=2.152m \end{aligned}

Velocity(v)=\frac{Q}{A} =\frac{12}{2 \times 3}=2m/sec

Specific energy at section (1-1)

E=y+\frac{v^2}{2g}=2+\frac{2^2}{2 \times 9.8}=\frac{108}{49}m

When channel section is contracted to minimum width and for constant discharge Q, the flow over contracted section will be critical flow and under the assumption that no energy loss has taken place.

E=E_c=\frac{108}{49}

We have that, E_c=\frac{3}{2}y_c (for rectangular crosssection)

y_c=\frac{2}{3}E_c=\frac{2}{3} \times \frac{108}{49}=\frac{72}{49}

For critical flow condition,

\begin{aligned} \frac{Q^2T}{gA^3}&=1\\ \frac{Q^2B_{min}}{g(B_{min} \times y_c)^3}&=1\\ (B_{min})^2&=\frac{Q^2}{gy_c^3}\\ B_{min}&=\left ( \frac{(12)^2}{9.8 \times \left ( \frac{72}{49} \right )^3} \right )^{1/2}\\ &=2.152m \end{aligned}

Question 4 |

A rectangular channel with Gradually Varied Flow (GVF) has a changing bed
slope. If the change is from a steeper slope to a steep slope, the resulting GVF
profile is

S_3 | |

S_1 | |

S_2 | |

either S_1 or S_2 , depending on the magnitude of the slopes |

Question 4 Explanation:

Question 5 |

A rectangular open channel of 6 m width is carrying a discharge of 20 \mathrm{~m}^{3} / \mathrm{s}. Consider the acceleration due to gravity as 9.81 \mathrm{~m} / \mathrm{s}^{2} and assume water as incompressible and inviscid. The depth of flow in the channel at which the specific energy of the flowing water is minimum for the given discharge will then be

0.82 m | |

1.04 m | |

2.56 m | |

3.18 m |

Question 5 Explanation:

Minimum specific energy will correspond to a critical flow condition.

The critical depth \begin{aligned} \left(Y_{C}\right) &=\left[\frac{q^{2}}{g}\right]^{1 / 3} \\ Y_{C} &=\left[\frac{(20 / 6)^{2}}{9.81}\right]^{1 / 3}=1.042 \mathrm{~m} \end{aligned}

Question 6 |

If water is flowing at the same depth in most hydraulically efficient triangular and rectangular channel sections then the ratio of hydraulic radius of triangular section to that of rectangular section is

\frac{1}{\sqrt{2}} | |

\sqrt{2} | |

1 | |

2 |

Question 6 Explanation:

Efficient channel section

\begin{aligned} A & =y^{2} & & A=2 y^{2} \\ P & =2 \sqrt{2} y & P & =4 y \\ R_{I} & =\frac{y}{2 \sqrt{2}} & R_{I I}&=\frac{y}{2} \\ \therefore \qquad \qquad \qquad \frac{R_{I}}{R_{I I}} & =\frac{1}{\sqrt{2}} & \end{aligned}

\begin{aligned} A & =y^{2} & & A=2 y^{2} \\ P & =2 \sqrt{2} y & P & =4 y \\ R_{I} & =\frac{y}{2 \sqrt{2}} & R_{I I}&=\frac{y}{2} \\ \therefore \qquad \qquad \qquad \frac{R_{I}}{R_{I I}} & =\frac{1}{\sqrt{2}} & \end{aligned}

Question 7 |

A hydraulic jump occurs, in a triangular (V-shaped) channel with side slopes 1:1 (vertical
to horizontal). The sequent depths are 0.5 m and 1.5 m. The flow rate (in m^3/s, round
off to two decimal places) in the channel is _________.

1.24 | |

1.68 | |

1.73 | |

2.14 |

Question 7 Explanation:

\begin{aligned} A &=\frac{1}{2} \times 2Y \times Y =Y^2 \\ \bar{Y} &=\frac{Y}{3} \end{aligned}

For a horizontal and frictionless channel

Specific Force (F) =A\bar{Y}+\frac{Q^2}{Ag}=constant

\Rightarrow \; Y^2\left ( \frac{Y}{3} \right )+\frac{Q^2}{(Y^2)g}=constant

\Rightarrow \; \frac{Y^3}{3}+\frac{Q^2}{gY^2}=constant

If Y_1 and Y_2 are conjugate depth

\begin{aligned} \frac{Y_1^3}{3}+\frac{Q^2}{gY_1^2} &= \frac{Y_2^3}{3}+\frac{Q^2}{gY_2}\\ \frac{0.5^3}{3}+\frac{Q^2}{g \times 0.5^2}&=\frac{1.5^3}{3}+\frac{Q^2}{g \times 1.5^2} \\ \frac{1.5^3}{3}-\frac{0.5^3}{3}&=\frac{Q^2}{g}\left ( \frac{1}{0.5^2}-\frac{1}{1.5^2} \right ) \\ Q&=1.728 m^3/sec \end{aligned}

Question 8 |

Water flows at the rate of 12 m^3/s in a 6 m wide rectangular channel. A hydraulic jump
is formed in the channel at a point where the upstream depth is 30 cm (just before the
jump). Considering acceleration due to gravity as 9.81 m/s^2 and density of water as 1000
kg/m^3, the energy loss in the jump is

114.2 kW | |

114.2 MW | |

141.2 h.p. | |

141.2 J/s |

Question 8 Explanation:

Assuming channle bed to be horizontal and frictionless.

q=\frac{12}{6}=2 m^3/s/m

Initial Froude No.

\begin{aligned} (F_r) &=\left ( \frac{q^2}{gY_1^3} \right )^{1/2} \\ &= \left ( \frac{2^2}{9.81 \times 0.3^3} \right )^{1/2}\\ &=3.88 \end{aligned}

From Belenger's Momentum equation for a rectangular channel

\begin{aligned} \frac{Y_2}{Y_1}&=\frac{1}{2}(-1+\sqrt{1+8F_1^2}) \\ &=\frac{1}{2}(-1+\sqrt{1+8\times 3.88^2}) \\ &= 5.018\\ Y_2&=5.018 \times 0.3=1.505m \end{aligned}

Heal loss in the jump

\begin{aligned} (h_L) &=\frac{(Y_2-Y_1)^3}{4Y_1Y_2} \\ &= \frac{(1.505-0.3)^3}{4 \times 1.505 \times 0.3}\\ &= 0.968m \end{aligned}

Power lost in the jump

\begin{aligned} &=\gamma _wQh_L \\ &= (9.81 \times 12 \times 0.968)kW\\ &=114.04 kW \end{aligned}

q=\frac{12}{6}=2 m^3/s/m

Initial Froude No.

\begin{aligned} (F_r) &=\left ( \frac{q^2}{gY_1^3} \right )^{1/2} \\ &= \left ( \frac{2^2}{9.81 \times 0.3^3} \right )^{1/2}\\ &=3.88 \end{aligned}

From Belenger's Momentum equation for a rectangular channel

\begin{aligned} \frac{Y_2}{Y_1}&=\frac{1}{2}(-1+\sqrt{1+8F_1^2}) \\ &=\frac{1}{2}(-1+\sqrt{1+8\times 3.88^2}) \\ &= 5.018\\ Y_2&=5.018 \times 0.3=1.505m \end{aligned}

Heal loss in the jump

\begin{aligned} (h_L) &=\frac{(Y_2-Y_1)^3}{4Y_1Y_2} \\ &= \frac{(1.505-0.3)^3}{4 \times 1.505 \times 0.3}\\ &= 0.968m \end{aligned}

Power lost in the jump

\begin{aligned} &=\gamma _wQh_L \\ &= (9.81 \times 12 \times 0.968)kW\\ &=114.04 kW \end{aligned}

Question 9 |

A 4 m wide rectangular channel carries 6 m^3/s of water. The Manning's 'n' of the open
channel is 0.02. Considering g = 9.81 m/s^2, the critical velocity of flow (in m/s, round
off to two decimal places) in the channel, is ________.

1.32 | |

5.65 | |

2.45 | |

4.25 |

Question 9 Explanation:

\begin{aligned} \text{Critical depth} \; (Y_c)&=\left ( \frac{q^2}{g} \right )^{1/3} \\ &=\left ( \frac{1.5^2}{9.81} \right )^{1/3} \\ &= 0.612m\\ \text{Critical velocity}\; (V_c)&=\sqrt{gY_c}=\sqrt{9.81 \times 0.612} \\ &= 2.45 m/s \end{aligned}

Question 10 |

A rectangular open channel has a width of 5 m and a bed slope of 0.001. For a uniform flow of depth 2 m, the velocity is 2 m/s. The Manning's roughness coefficient for the channel is

0.002 | |

0.017 | |

0.033 | |

0.05 |

Question 10 Explanation:

Given,

\begin{aligned} V&= 2 m/sec, \; S=0.001\\ A&=5 \times 2=10m^2\\ P&=5+2+2=9m\\ & \text{Hydraulic mean radius}\\ (R)&=\frac{A}{P}=\frac{10}{9}\\ V&=\frac{1}{n}R^{2/3}S^{1/2}\\ 2&=\frac{1}{n}\left ( \frac{10}{9} \right )^{2/3}(0.001)^{1/2}\\ n&=0.0169\simeq 0.017 \end{aligned}

\begin{aligned} V&= 2 m/sec, \; S=0.001\\ A&=5 \times 2=10m^2\\ P&=5+2+2=9m\\ & \text{Hydraulic mean radius}\\ (R)&=\frac{A}{P}=\frac{10}{9}\\ V&=\frac{1}{n}R^{2/3}S^{1/2}\\ 2&=\frac{1}{n}\left ( \frac{10}{9} \right )^{2/3}(0.001)^{1/2}\\ n&=0.0169\simeq 0.017 \end{aligned}

There are 10 questions to complete.