This set of Irrigation Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Regulation Modules – Semi Modules Types”.
1. Which of the following is not a type of semi-modules?
a) Free pipe outlet
b) Kennedy’s Gauge outlet
c) Open flume outlet
d) Gibb’s module
View Answer
Explanation: Gibb’s module is the most important and widely used rigid module. It was designed by A. S Gibb, formerly Executive Engineer in Punjab Irrigation Department. The discharge range varies from 0.03 to 0.45 cumecs.
2. Which of the following is the simplest and the oldest type of a flexible outlet?
a) Kennedy’s gauge outlet
b) Open flume outlet
c) Free pipe outlet
d) Adjustable orifice semi-module
View Answer
Explanation: Silt conduction for pipe outlet discharging freely into the air is quite good and has high efficiency. This can be provided only at a few places where there is a sufficient level difference available between the distributary and the watercourse.
3. Calculate full supply discharge for a free pipe outlet for the following data.
Head on the upstream side above the centre line of pipe = 0.55 m
Coefficient of discharge = 0.62
The diameter of the pipe = 25 cm
a) 8 lit/sec
b) 9 lit/sec
c) 0.9 lit/sec
d) 9 cumecs
View Answer
Explanation: The discharge can be computed by –
Q = Cd. A. (2gH)1/2 where H is the head on U/s side measured from FSL of distributary up to the centre of pipe outlet and A is the area of cross-section of the pipe.
Q = 0.62 x [3.14/4 x 0.25 x 0.25] x (2 x 9.81 x 0.55)1/2
Q = 0.09 cumecs or 9 lit/sec.
4. Which module is mostly adopted and is considered to be the best of all the modules?
a) Adjustable orifice semi-module
b) Open flume outlet
c) Venturi-flume outlet
d) Free pipe outlet
View Answer
Explanation: Adjustable Proportional Module (APM) is also called Adjustable Orifice Semi-Module (AOSM). It consists of an orifice provided with a gradually expanding flume on the downstream side of the orifice. This outlet is very popular due to several merits over other types and is used in Punjab and Haryana.
5. Calculate the minimum modular head loss involved in ASOM if the head measured from the upstream water level in the distributary is 1.5 m and the width of the throat as 0.16 m.
a) 1.50 m
b) 1.15 m
c) 1.07 m
d) 1.00 m
View Answer
Explanation: The minimum modular head loss involved in ASOM is given by the formula –
HL = 0.82 HS – W/2 where HS = Head measured from the U/s level in the distributary to the lowest point of the roof block and W = width of the throat
HL = (0.82 x 1.5) – 0.08 = 1.15 m.
6. The proportionality of APM was sacrificed to enable the outlet to carry higher silt charge by increasing the setting to 8/10.
a) True
b) False
View Answer
Explanation: The APM (Adjustable Proportional Module) has a setting of 6/10 which is aimed at exact proportionality. But the channels using such module silted very badly and hence, the setting was increased to 8/10 to enable the outlet to carry higher silt charge. The outlet is then known as A.O.S.M.
7. Calculate the discharge through an open flume outlet for the following data:
Coefficient of discharge = 1.71
Width of the throat = 0.15 m
Head over the crest measured from FSL of distributary = 1.8 m
a) 0.62 cumecs
b) 0.82 cumecs
c) 0.62 lit/sec
d) 6.2 lit/sec
View Answer
Explanation: The discharge through an open flume outlet is given as –
Q = Cd. W H3/2 where W = 0.15 m, Cd = 1.71 and H = 1.8 m
Q = 1.71 x 0.15 x 1.83/2 = 0.62 cumecs.
8. The only irrigation semi-module through which the discharge is not proportional to H1/2 where H is the head causing flow through the module is _____________
a) kennedy’s gauge outlet
b) open flume outlet
c) free pipe outlet
d) adjustable orifice semi-module
View Answer
Explanation: The discharge through free pipe outlet and Kennedy’s flume outlet is given by-
Q = Cd. A. (2gH)1/2 where H is the head on U/s side measured from FSL of distributary up to the centre of pipe outlet, Cd is the coefficient of discharge, and A is the area of cross-section of the pipe.
The discharge through an open flume outlet is given as: Q = Cd. W H3/2 where W is the width of the throat.
The discharge through AOSM is given by- Q = Cd. (W. y0).(2gH)1/2 where ‘y0’ is the height of an orifice opening.
9. An irrigation outlet is said to be proportional, when its ______________
a) Setting = outlet index/ channel index
b) Setting = channel index/ outlet index
c) Setting = channel index x outlet index
d) Setting = channel index + Outlet index
View Answer
Explanation: For a proportional outlet H/y i.e. outlet setting should be made and is equal to the outlet index/channel index. It is the ratio of the depth of the sill of a module below the FSL of the distributary to the FS depth of the distributary.
10. The principal feature of an ASOM is similar to those of a flume regulator.
a) True
b) False
View Answer
Explanation: On the upstream side, a flume regulator with horizontal crest and curved water approach is used and the downstream wings expanding to the width of the watercourse. The provision of cast iron block is done, unlike gates around which masonry is done.
11. Choose the incorrect advantage of AOSM.
a) The adjustment can be made by raising or lowering the roof blocks at low costs by dismantling the masonry in which the roof bolts are fixed
b) Any undue tampering of roof blocks by the cultivators can be easily detected
c) The module is not perfectly rigid
d) It is the most simple and cheaper outlet
View Answer
Explanation: In ASOM, by suitable adjustment of the roof block the opening height can be changed easily after dismantling the masonry around it. But the roof block cannot be readjusted without breaking the masonry around it and thus, the outlet discharge cannot be easily tempered by the cultivators. The adjustment in dimensions has to be done at a slight cost of re-doing the masonry and hence, the module is perfectly rigid.
12. For a wide trapezoidal channel, the channel index is ______
a) 2/3
b) 5/3
c) 4/3
d) 1/3
View Answer
Explanation: The discharge is proportional to the y5/3 for a wide trapezoidal channel. Hence, the channel index is generally 5/3. The discharge through an orifice type outlet is proportional to H1/2 and channel index for such outlet is 1/2.
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