Open Channel Flow MCQ Quiz - Objective Question with Answer for Open Channel Flow - Download Free PDF

Last updated on Jun 11, 2025

Latest Open Channel Flow MCQ Objective Questions

Open Channel Flow Question 1:

Which of the following conditions is the chief characteristic of critical flow?

  1. Q2T/gA= 1
  2. Q2T2/gA3 = 1
  3. Q2R/gA3 = 1
  4. QT2/gA2 = 1

Answer (Detailed Solution Below)

Option 1 : Q2T/gA= 1

Open Channel Flow Question 1 Detailed Solution

Explanation:

The chief characteristic of critical flow is Q2T/gA= 1

  • This equation represents the condition for critical flow in open channel hydraulics.

  • Here, is the discharge, T" id="MathJax-Element-6-Frame" role="presentation" style="position: relative;" tabindex="0"> T is the top width of the flow, g" id="MathJax-Element-7-Frame" role="presentation" style="position: relative;" tabindex="0"> g is acceleration due to gravity, and is the cross-sectional area.

  • When this condition is satisfied, the flow velocity equals the wave velocity (celerity), meaning the flow is at its critical state.

  • At critical flow, the specific energy is minimum for a given discharge.

  • It marks the boundary between subcritical and supercritical flow regimes.

Additional Information

  • Critical flow occurs when the flow velocity is such that the Froude number equals 1. The Froude number (Fr) is a dimensionless parameter defined as the ratio of flow inertia to gravitational forces. 
  • Specific Energy Concept: At critical flow, the specific energy (energy relative to the channel bottom) reaches a minimum for a given discharge, meaning the flow is at an unstable equilibrium.

  • Critical flow plays a key role in the design of hydraulic structures like weirs, spillways, and channel transitions, since it governs flow control and energy dissipation.

  • The state of flow before critical flow is subcritical (slow, deep flow), and after critical flow is supercritical (fast, shallow flow). Understanding this helps engineers predict flow behavior under different conditions.

Open Channel Flow Question 2:

In trapezoidal weir, sides are inclined outward with a slope of

  1. 1 ∶ 4
  2. 1 ∶ 3
  3. 1 ∶ 6
  4. 1 ∶ 5

Answer (Detailed Solution Below)

Option 1 : 1 ∶ 4

Open Channel Flow Question 2 Detailed Solution

Explanation:

  • In a trapezoidal weir, the sides are inclined outward to provide structural stability and help manage flow efficiently.

  • According to standard hydraulic design practices, the slope of the sides is generally taken as 1 vertical to 4 horizontal (1:4).

  • This slope ensures proper flow distribution and ease of construction.

Assignment 1 Basharat 8Q SSC JE ME 24 Jan 18 Morning Satya 8 July Madhu(Dia) Hindi images q7

Additional Information

  • A trapezoidal weir is a hydraulic structure used to measure flow in open channels and rivers. 
  • It has a trapezoidal cross-section with the sides inclined outward, usually at a slope of 1 vertical to 4 horizontal.

  • The trapezoidal shape helps to control the flow smoothly and reduces turbulence compared to sharp-edged weirs.

  • It is commonly used for larger flow measurements because the shape allows handling a wide range of flow rates efficiently.

  • The discharge over the weir is calculated using standard empirical formulas that consider the shape and dimensions.

  • Trapezoidal weirs are easier to construct and maintain compared to rectangular or triangular weirs, making them practical for many irrigation and drainage applications.

Open Channel Flow Question 3:

If the value of rate of change of specific energy is 7.79 × 10-4 m and Sf = 0.00013, the value of bed slope is

  1. 1 in 1000
  2. 1 in 1300
  3. 1 in 1200
  4. 1 in 1100

Answer (Detailed Solution Below)

Option 4 : 1 in 1100

Open Channel Flow Question 3 Detailed Solution

Concept:

The relationship is given by:

Bed Slope

S = dE/dx + Sf

where, dE/dx is rate of change of specific energy, Sf is friction slope, and S is bed slope.

Given: dE/dx = 7.79 × 10-4 m ;  Sf = 0.00013

So, S = 7.79 × 10-4  + 0.00013 = 0.000909

The value of bed slope

= 1/S = 1/0.000909 = 1100.11

So, the value of bed slope is 1 in 1100

Open Channel Flow Question 4:

Most efficient channel section is

  1. Half hexagon in the form of trapezoid
  2. Semicircular
  3. Rectangular
  4. Triangular

Answer (Detailed Solution Below)

Option 1 : Half hexagon in the form of trapezoid

Open Channel Flow Question 4 Detailed Solution

Explanation:

Half hexagon in the form of trapezoid

  • Represents a well-proportioned trapezoidal section commonly used in canal design.

  • Provides an optimal balance between flow efficiency, stability of slopes, and ease of construction.

  • Considered the most efficient channel shape in practical applications.

Additional InformationSemicircular

  • Theoretically most hydraulically efficient due to minimum wetted perimeter.

  • Difficult to construct and maintain, especially in large-scale field works.

  • Rarely used except in small-scale or prefabricated systems.

Rectangular

  • Simple and commonly used in lined drainage and urban water systems.

  • Not hydraulically efficient because of high wetted perimeter and energy loss at corners.

  • Preferred where formwork and modular construction is involved.

Triangular

  • Least efficient hydraulically due to highest wetted perimeter for a given area.

  • Used in specific cases such as roadside drains or where space is limited.

  • Not suitable for large-scale or high-discharge channels.

Open Channel Flow Question 5:

In deriving the equation for the hydraulic jump in a rectangular channel in terms of conjugate depths and initial Froude number

  1. Energy and continuity equations are used
  2. Energy, momentum and continuity equations are used
  3. Continuity and momentum equations are used
  4. Only continuity equation is used

Answer (Detailed Solution Below)

Option 3 : Continuity and momentum equations are used

Open Channel Flow Question 5 Detailed Solution

Explanation:

  • The hydraulic jump is analyzed using the continuity equation to ensure flow rate is constant before and after the jump.

  • The momentum equation is applied to account for the change in momentum due to the sudden change in flow depth.

  • The energy equation is generally not used because energy is not conserved across a hydraulic jump (energy loss occurs due to turbulence).

  • Therefore, the derivation relies primarily on continuity and momentum principles.

 Additional Information

  • A hydraulic jump is a sudden rise in water surface occurring when fast, shallow supercritical flow transitions to slower, deeper subcritical flow.

  • This phenomenon is common downstream of spillways, sluice gates, and steep channels, and it helps dissipate excess energy in the flow.

  • The continuity equation ensures that the flow discharge (volume per second) remains constant before and after the jump.

  • The energy equation relates the specific energy (sum of potential and kinetic energy per unit weight) upstream and downstream, accounting for energy losses due to turbulence and mixing in the jump.

  • By analyzing the depths before and after the jump (called conjugate depths), engineers can calculate the jump characteristics, such as the height of the water surface and energy dissipation.

  • The Froude number, a dimensionless parameter comparing inertial to gravitational forces, determines the flow regime and the occurrence of the hydraulic jump (it occurs when flow transitions from supercritical, Froude number > 1, to subcritical, Froude number < 1).

Top Open Channel Flow MCQ Objective Questions

A rectangular channel of bed width 2 m is to be laid at a bed slope of 1 in 1000. Find the hydraulic radius of the canal cross-section for the maximum discharge condition? Take Chezy’s constant as 50

  1. 0.5 m
  2. 2 m
  3. 1 m
  4. 0.25 m

Answer (Detailed Solution Below)

Option 1 : 0.5 m

Open Channel Flow Question 6 Detailed Solution

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Concept:

Most efficient channel: A channel is said to be efficient if it carries the maximum discharge for the given cross-section which is achieved when the wetted perimeter is kept a minimum.

Rectangular Section:

Full Test 2 (31-80) images Q.55

Area of the flow, A = b × d

Wetted Perimeter, P = b + 2 × d  

For the most efficient Rectangular channel, the two important conditions are

  1. b = 2 × d
  2.  \(R = \frac{A}{P} = \frac{{b\times d}}{{b + 2 \times d}} = \frac{{2{d^2}}}{{4d}} = \frac{d}{2}= \frac{b}{4}\)

Calculation

Given: b = 2 m

 \(R = \frac{2}{4}\)

R = 0.5

quesImage111

Rectangular channel section Trapezoidal channel section
  1. R = y / 2
  2. A = 2y
  3. T = 2y
  4. P = 4y
  5. D = y
  1. R = y / 2
  2. \(A = \sqrt 3 {y^2}\)
  3. \(T = \frac{{4y}}{{\sqrt 3 }}\)
  4. \(P = {{2y\sqrt 3}}{{ }}\)
  5.  D = 3y / 4

Where R = hydraulic radius, A = Area of flow, P = wetted perimeter, y = depth of flow, T = Top width

For obtaining the most economical trapezoidal channel section with depth of flow = 3 m, what is the hydraulic mean radius ?

  1. 1.5 m
  2. 3.0 m
  3. 2.0 m
  4. 1.0 m

Answer (Detailed Solution Below)

Option 1 : 1.5 m

Open Channel Flow Question 7 Detailed Solution

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Concept:

Most economical section is the one whose wetted perimeter is minimum for the given value of discharge.

GATE CE FT 5 (SLOT 1) images Q14b

Hydraulic mean radius,

\({y_m} = \frac{y_0}{2} = \frac{3}{2} = 1.5\; m\)

Important Points

For most economical Channel

S.No.

Shape

Hydraulic Radius

1

GATE CE FT 5 (SLOT 1) images Q14

Rectangular Channel

\({y_m} = \frac{y_0}{2}\)

2

GATE CE FT 5 (SLOT 1) images Q14b

Trapezoidal channel

\({y_m} = \frac{y_0}{2}\)

3

Triangular Channel

\({y_m} = \frac{y_0}{{2\sqrt 2 }}\)

4

896

Circular channel

\({y_m} = 0.29d\)

In a rectangular channel section, if the critical depth is 2.0 m, the specific energy at critical depth is

  1.  3.0 m

  2. 1.33 m
  3. 2.5 m
  4. 1.5 m

Answer (Detailed Solution Below)

Option 1 :

 3.0 m

Open Channel Flow Question 8 Detailed Solution

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Concept:

Critical Specific Energy:

  • The specific energy corresponds to the critical depth of the flow is known as critical specific energy.
  • For rectangular channel, it is equal to -

\(E_c = \frac{3}{2}y_c\)

Here,

yc - critical depth of the flow = \(\left [ \frac{q^2}{g} \right]^{\frac{1}{3}}\)

Here, q - discharge per unit width (m3/s/m)

g - acceleration due to gravity (m/s2)

Calculation:

Given,

Critical depth, yc = 2.0 m

Hence,

Critical specific energy,

Ec = \(\frac{3}{2}\) × yc = \(\frac{3}{2}\) × 2 = 3.0 m

Important Points

  • The following table shows the relationship between different types of sections and critical specific energy -
Type of section Critical specific energy (m)
1. Rectangular \(\frac{3}{2}y_c\)
2. Triangular \(\frac{5}{4}y_c\)
3. Parabolic \(\frac{4}{3}y_c\)

Here,

yc - critical depth (m)

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

  1. S3
  2. S1
  3. S2
  4. either S1 or S2, depending on the magnitude of the slopes

Answer (Detailed Solution Below)

Option 1 : S3

Open Channel Flow Question 9 Detailed Solution

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Explanation-

  • When two-channel sections have different bed slopes the condition is called a break in grade.
  • Under this situation following conditions must be remembered for drawing the flow profile.
    • CDL is independent of the bed slope.
    • The steeper the slope lesser is the normal depth of flow.
    • Flow always starts from NDL and tries to meet NDL.
    • Subcritical flow has downstream control and supercritical flow has upstream control.
  • Steep slope-
    • F7 Savita Engineering 06-4-22 D21

 

Given data and Analysis-

  • Flow is changed from steeper to sleep.
  • So the Normal depth of flow will increase in a steep slope, as the slope is decreased.
  • CDL will remain the same for both the slope.
  • So from the figure shown below it can be concluded that flow will be S3 profile.

F7 Savita Engineering 06-4-22 D22

The head over a rectangular sharp crested notch at the end of a channel is 0.75 m. If an error of 1.5 mm is possible in the measurement of the head, then the percentage error in computing the discharge will be:

  1. 0.5
  2. 0.3
  3. 1.0
  4. 1.5

Answer (Detailed Solution Below)

Option 2 : 0.3

Open Channel Flow Question 10 Detailed Solution

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Concept:

Discharge through rectangular notch

\(Q = \frac{2}{3}{c_d}.b\sqrt {2g} {\left( H \right)^{3/2}}\)

Where,

H = Height of water above will of notch

b = width of the notch

Cd = coefficient of discharge

\(\therefore dQ = \frac{2}{3}{C_d}b\sqrt {2g} \times \frac{3}{2}{\left( H \right)^{1/2}}dH\)

\(dQ = \left( {\frac{2}{3}{C_d}b\sqrt {2g} \times {H^{\frac{1}{2}}} \times H} \right) \times \frac{3}{2}\frac{{dH}}{H}\)

\(dQ = Q \times \frac{3}{2}\frac{{dH}}{H}\)

\(\frac{{dQ}}{Q} = \frac{3}{2}\frac{{dH}}{H}\)

Calculation:

\(\frac{{dQ}}{Q} = \frac{3}{2}\times\frac{{1.5}}{750} \times 100= 0.3\)%

Specific energy of flowing water through a rectangular channel of width 5 m when discharge is 10 m/ s and depth of water is 2 m is:

  1. 1.06 m
  2. 1.02 m
  3. 2.05 m
  4. 2.60 m

Answer (Detailed Solution Below)

Option 3 : 2.05 m

Open Channel Flow Question 11 Detailed Solution

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Concept:

The specific energy may be given as

\(E = y + \frac{{V_1^2}}{{2g}}\)

Calculation:

We know that

Q = AV

⇒ 10 = 5 × 2 × V

⇒ V = 1 m/s

\(\begin{array}{l} E = y + \frac{{V_1^2}}{{2g}}\\ E= 2 + \frac{{{{\left( {1} \right)}^2}}}{{2 \times 9.81}} = 2.05\;m \end{array}\)

In an open channel flow, for best efficiency of a rectangular section channel, ratio of bottom width to depth shall be:

  1. 1
  2. \(\dfrac{1}{2}\)
  3. \(\dfrac{1}{4}\)
  4. 2

Answer (Detailed Solution Below)

Option 4 : 2

Open Channel Flow Question 12 Detailed Solution

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Concept:

For hydraulic efficient rectangular channel for a given area, perimeter(P) should be minimum.

Calculation:

Width of rectangular channel is b and depth of rectangular channel is y

Than Area, A = by

Perimeter, P = b + 2y

P = (A/y) + 2y

For P to be minimum,

dP/dy = 0

(-A/y2) + 2 = 0

A = 2y2

by = 2y2

b/y = 2

Important Points

Rectangular channel section Trapezoidal channel section
  1. R = y / 2
  2. A = 2y2
  3. T = B = 2y
  4. P = 4y
  5. D = y
  1. R = y / 2
  2. \(A = \sqrt 3 {y^2}\)
  3. \(T = \frac{{4y}}{{\sqrt 3 }}\)
  4. \(P = 2\sqrt 3y\)
  5. D = 3y / 4

Identify the flow control device shown in image:

F1 Savita Engineering 30-3-22  D6

  1. Parshall flume
  2. Notch
  3. Sutro weir
  4. None of the above

Answer (Detailed Solution Below)

Option 3 : Sutro weir

Open Channel Flow Question 13 Detailed Solution

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Explanation:

1) Sutro or Proportional  weir:

F1 Savita Engineering 30-3-22  D6

2) Parshall flume

F1 Savita Engineering 30-3-22  D7

3) Weir and notch

F1 Savita Engineering 30-3-22  D9

8 m3/s discharge flows through 4 m wide rectangular channel at a velocity of 2 m/s. The hydraulic mean radius of the channel is

  1. 3/2 m
  2. 2/3 m
  3. 1 m
  4. 4 m

Answer (Detailed Solution Below)

Option 2 : 2/3 m

Open Channel Flow Question 14 Detailed Solution

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Concept:

The wetted area of the channel,

A = \(\frac{Q}{V}\)

The wetted perimeter,

P = (B + 2d)

The hydraulic mean radius of the channel is

\(R = \frac{{\left( {Wetted\;Area,\;A} \right)}}{{\left( {Wetted\;Perimeter,\;P} \right)}}\)

Calculation:

Given data,

Q = 8 m3/s

B = 4 m

V = 2 m/s

The wetted area of the channel,

A = \(\frac{Q}{V}\)

\(A = \frac{8}{2} = 4\;{m^2}\)

A = B × d

⇒ B × d = 4 m2

⇒ 4 × d = 4

d = 1 m

Therefore, wetted perimeter,

P = (B + 2d)

P = (4 + (2 × 1)) = 6 m

The hydraulic mean depth of the channel is

\(R = \frac{{\left( {Wetted\;Area,\;A} \right)}}{{\left( {Wetted\;Perimeter,\;P} \right)}}\)

\(R = \frac{4}{6} = \frac{2}{3} \;m\)

If the channel cut shown in the figure is an economical cut, then what will be its area?

F1  Ram S 20-08-21 Savita D1

  1. A = 1.414 y2
  2. A = 0.5 y2
  3. A = 2 y2
  4. A = 1.914 y2

Answer (Detailed Solution Below)

Option 4 : A = 1.914 y2

Open Channel Flow Question 15 Detailed Solution

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Explanation:

For the economical section, its perimeter should be minimum.

\(A = b \times y + \frac{1}{2} \times y \times y\)

\(A = by + \frac{{{y^2}}}{2}\)

\(b = \frac{A}{y} - \frac{y}{2}\)

Perimeter

\(P = y + b + \sqrt {{y^2} + {y^2}}\)

\(P = y + y\sqrt 2 + \frac{A}{y} - \frac{y}{2}\)

\(\frac{{dP}}{{dy}} = 0\)

\(\frac{{dP}}{{dy}} = 0 = 1 + \sqrt 2 - \frac{1}{2} - \frac{A}{{{y^2}}}\)

A = 1.914 y2
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