Question 1 |
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 1 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 2 |
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 2 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 3 |
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 3 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 4 |
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 4 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 5 |
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 5 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 6 |
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 6 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}
Question 7 |
In a rectangular channel, the ratio of the velocity head to the flow depth for critical flow condition, is
\frac{1}{2} | |
\frac{2}{3} | |
\frac{3}{2} | |
2 |
Question 7 Explanation:
For Critical flow \Rightarrow Velocity Head is equal to half of hydraulic depth.
\frac{V^2}{2g}=\frac{D}{2}
for rectangular channel D = y
\frac{\frac{V^2}{2g}}{y}=\frac{1}{2}
\frac{V^2}{2g}=\frac{D}{2}
for rectangular channel D = y
\frac{\frac{V^2}{2g}}{y}=\frac{1}{2}
Question 8 |
A 5 m wide rectangular channel, the velocity u distribution in the vertical direction y is given by u=1.25y^{\frac{1}{6}}. The distance y is measured from the channel bed. If the flow depth is 2 m, the discharge per unit width of the channel is
2.40m^{3}/s/m | |
2.80m^{3}/s/m | |
3.27m^{3}/s/m | |
12.02m^{3}/s/m |
Question 8 Explanation:
Discharge through elementary strip,
d Q=5 d y \times u
\therefore Total discharge,
\begin{aligned} Q&=\int_{0}^{2}(5 d y)\left(1.25 y^{\sqrt{6}}\right) \\ \Rightarrow Q&=6.25 \int_{0}^{2} y^{1 / 6} d y=6.25 \times\left[\frac{y^{1 / 6+1}}{\frac{1}{6}+1}\right]_{0}^{2}\\ \because \quad Q&=12.026 \mathrm{m}^{3} \\ \end{aligned}
\therefore Discharge per unit width,
q=\frac{12.026}{5}=2.405 \mathrm{m}^{3} / \mathrm{s} / \mathrm{m}
Question 9 |
For a given discharge in an open channel, there are two depths which have the same specific energy. These two depths are known as
alternate depths | |
critical depths | |
normal depths | |
sequent depths |
Question 9 Explanation:
Depth with same specific energy are called Alternate depths of flow. It represents a subcritical depth of flow \left(Y_{2}\right) and a supercritical depth of flow \left(Y_{1}\right)
Question 10 |
Variation of water depth(y) in a gradually varied open channel flow is given by the first order differential equation
\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{1-e^{-\frac{10}{3}\ln(y)} }{250-45e^{-3\ln(y)}}
Given initial condition: y(x=0)=0.8 m. The depth (in m, up to three decimal places) of flow at a downstream section at x = 1 m from one calculation step of Single Step Euler Method is ______
\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{1-e^{-\frac{10}{3}\ln(y)} }{250-45e^{-3\ln(y)}}
Given initial condition: y(x=0)=0.8 m. The depth (in m, up to three decimal places) of flow at a downstream section at x = 1 m from one calculation step of Single Step Euler Method is ______
0.247 | |
0.625 | |
0.793 | |
1.265 |
Question 10 Explanation:
\begin{aligned} \frac{d y}{d x}&=\frac{1-e^{-\frac{10}{3} \ln y}}{250-45 e^{-3 \ln y}}\\ y_{1} &=y_{0}+h f\left(x_{0}, y_{0}\right) \\ y_{1} &=0.8+1\left(\frac{1-e^{-\frac{10}{3} \ln 0.8}}{250-45 e^{-3 \ln 0.8}}\right] \\ &=0.8+1\left(\frac{-1.1039}{162.109}\right) \\ &=0.793 \mathrm{m} \end{aligned}
There are 10 questions to complete.