Question 1 |
Identify the cross-drainage work in the figure.
Super passage | |
Aqueduct | |
Siphon aqueduct | |
Level crossing |
Question 1 Explanation:
In given CDW, river is above the canal with underside of trough sufficiently above the full supply level of canal. Hence it is super passage.
Question 2 |
In the context of cross-drainage structures, the correct statement(s) regarding
the relative positions of a natural drain (stream/river) and an irrigation canal,
is/are
In an aqueduct, natural drain water goes under the irrigation canal, whereas in a super-passage, natural drain water goes over the irrigation canal. | |
In a level crossing, natural drain water goes through the irrigation canal. | |
In an aqueduct, natural drain water goes over the irrigation canal, whereas in a super-passage, natural drain water goes under the irrigation canal. | |
In a canal syphon, natural drain water goes through the irrigation canal. |
Question 2 Explanation:
In Aqueduct and Syphon aqueduct under irrigation
canal or canal passes over natural drain.
In Super passage and Canal syphon or simply syphon, natural drain passes over irrigation canal or canal goes under natural drain.
In level crossing, natural drain water goes through the irrigation canal.
In Super passage and Canal syphon or simply syphon, natural drain passes over irrigation canal or canal goes under natural drain.
In level crossing, natural drain water goes through the irrigation canal.
Question 3 |
An unlined canal under regime conditions along with a silt factor of 1 has a width of flow 71.25m. Assuming the unlined canal as a wide channel, the corresponding average depth of flow (in m, round off to two decimal places) in the canal will be ______________
1.25 | |
4.58 | |
3.82 | |
2.94 |
Question 3 Explanation:
R = D (for wide rectangular channel)
\begin{aligned} &\begin{aligned} A f^{2} &=140\left(\frac{2}{5} f R\right)^{5 / 2} \\ (B D) f^{2} &=140\left(\frac{2}{5} f \times D\right)^{5 / 2} \\ (71.25 \times D) \times 1 &=140\left(\frac{2}{5} \times 1 \times D\right)^{5 / 2} \\ D \times 0.5089 &=\left(\frac{2}{5}\right)^{5 / 2} \times(D)^{5 / 2} \\ D^{3 / 2} &=5.029 \\ \text { or } \qquad \qquad \qquad \qquad D &=2.94 \mathrm{~m} \end{aligned} \end{aligned}
Question 4 |
The depth of flow in an alluvial channel is 1.5 m. If critical velocity ratio is 1.1 and Manning's n is 0.018, the critical velocity of the channel as per Kenedy's method is
0.713 m/s | |
0.784 m/s | |
0.879 m/s | |
1.108 m/s |
Question 4 Explanation:
The critical velocity as per Kennedy's method is
given by,
\begin{aligned} V_{0} &=0.55 \mathrm{my}^{0.64}=0.55 \times 1.1 \times(1.5)^{0.64} \\ &=0.784 \mathrm{m} / \mathrm{s} \end{aligned}
given by,
\begin{aligned} V_{0} &=0.55 \mathrm{my}^{0.64}=0.55 \times 1.1 \times(1.5)^{0.64} \\ &=0.784 \mathrm{m} / \mathrm{s} \end{aligned}
Question 5 |
A stable channel is to be designed for a discharge of Q m^{3}/s with silt factor f as per Lacey's method. The mean flow velocity (m/s) in the channel is obtained by
[\frac{Qf^{2}}{140}]^{1/6} | |
[\frac{Qf}{140}]^{1/3} | |
[\frac{Q^{2}f^{2}}{140}]^{1/6} | |
0.48*[\frac{Q}{f}]^{1/3} |
There are 5 questions to complete.