250+ TOP MCQs on Canal Falls – Design of Syphon Well Drop and Answers

Irrigation Engineering Multiple Choice Questions on “Canal Falls – Design of Syphon Well Drop”.

1. ______________ fall is adopted for smaller discharges and larger drops.
a) Ogee fall
b) Sarda fall
c) Glacis fall
d) Well-type fall
Answer: d
Clarification: A Cylinder fall popularly known as Syphon or Well-Type fall is useful for affecting larger drops and for small discharges. They are commonly used where high levelled small drains do outfall into low levelled bigger drains and also as tail escapes for small canals.

2. Consider the following statements.
I. Syphon falls are commonly used where high levelled small drains do out-fall into a low levelled bigger drain.
II. The D/s well is necessary in case of falls greater than 1.8 m and for discharges greater than 0.29 cumecs.
III. The design discharge is determined based on the fact that V-notch is used.
Which of the following statement is correct?
a) I only
b) II only
c) I and II
d) I, II and III
Answer: c
Clarification: The design discharge of a notch fall is obtained by adding the discharge of rectangular and V-notch i.e. trapezoidal notch taking the value of the coefficient of the discharge as 0.75. The notches are designed by taking into consideration full supply discharge and half supply discharge w.r.t normal water depths in the channel in both cases.

3. Calculate the velocity over the notch (V1) and the velocity through the pipe (V3) if the full supply discharge is 2 cumecs. The area of flow over the trapezoidal notch is 0.58 m2. Assume the value of diameter of the pipe used as 1 m.
a) V1 = 3.44 m/sec and V3 = 2.54 m/sec
b) V1 = 2.54 m/sec and V3 = 3.44 m/sec
c) V1 = 3.44 m/sec and V3 = 3.44 m/sec
d) V1 = 0.44 m/sec and V3 = 2.04 m/sec
Answer: a
Clarification: The velocity over the notch (V1) = full supply discharge / Area over the notch and the velocity through the pipe (V3) = Full supply discharge / Area of pipe.
V1 = 2/0.58 = 3.44 m/sec
V3 = 2/0.785 = 2.54 m/sec (Area of pipe = 0.785 m2).

4. What is the height of the centre of pressure above the water level in the inlet well if the R.L of the centre of pressure is 150.83 m and the R.L of water level in inlet well is 148.30 m?
a) 2.53 m
b) 1.78 m
c) 3 m
d) 1.99 m
Answer: a
Clarification: The height of the centre of pressure (Y) above the water level in the inlet well is given by –
Y = R.L of C.P – R.L of water level in inlet well
Y = 150.83 – 148.30
Y = 2.53 m.

5. Calculate the diameter of the inlet well if the X and Y coordinate of the jet (issuing from the centre of pressure) w.r.t the water surface in the inlet well is 1.05 m and 1.88 m respectively.
a) 1.575 m
b) 2.82 m
c) 1.79 m
d) 0.83 m
Answer: a
Clarification: The diameter of the inlet well is kept as 1.5 times the value of X.
Given, X = 1.05 and Y = 1.88
Diameter = 1.5 x 1.05 = 1.575 m.

6. Which of the following loss is not considered while calculating the total head loss between the inlet well and D/s FSL?
a) Entry and exit losses
b) Loss due to friction
c) Loss due to enlargement of the section
d) Loss due to contraction of section
Answer: d
Clarification: The head loss between the inlet well and the D/s FSL is given by adding the losses due to entry, exit, loss due to sudden enlargement and loss due to friction in the pipe.
HL = 0.5 V22/2g (Entry loss) + (V2 – V3)2/2g (sudden enlargement loss) + fLV32/2gd (Friction loss) + V32/2g (Exit loss)
Where, V2 = Velocity of entry in pipe, V3 = Velocity through the pipe, L = assumed pipe length, d = diameter of the pipe, f = Darcy’s coefficient of friction = 0.012 and g = acceleration due to gravity.

7. The discharge formula used in the design of the syphon well drop is ____________
a) 8/15 Cd. (2g)1/2. tan (Q/2). H5/2
b) 2/3 Cd. (2g)1/2. L. H3/2
c) 8/15 Cd. (2g)1/2. tan (Q/2). H3/2
d) 2.22H3/2 [L + 0.4nH]
Answer: d
Clarification: The discharge passing through a trapezoidal notch is given by adding the discharge of a rectangular notch and V-notch which is given by-
Q = 2/3 Cd. (2g)1/2. L. H3/2 + 8/15 Cd. (2g)1/2. tan (Q/2). H5/2
Q = 2.22H3/2 [L + 0.4nH]
(Where n = 2 tan (Q/2) and coefficient of discharge = 0.75).

8. Which of the following statement is wrong?
a) In a trapezoidal notch fall, the top width of the notch is kept between 3/4th to full water depth above the sill of the notch
b) The syphon fall is designed to maintain a fixed supply level in the channel
c) A cylinder fall is popularly known as syphon well drop
d) Energy dissipators are provided for small discharges
Answer: d
Clarification: In syphon well drops, the downstream well is provided for discharges greater than 0.29 cumecs. A trapezoidal notch is constructed in the steining of the well, the water falls into the inlet well from where it emerges near the bottom dissipating its energy in turbulence inside the well.

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