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

Explanation:

The chief characteristic of critical flow is Q2T/gA3 = 1

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

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

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.

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.

Concept:

The relationship is given by:

Bed Slope

S = dE/dx + S_f

where, dE/dx is rate of change of specific energy, S_f 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

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.

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).

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:

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

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

Concept:

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

Hydraulic mean radius,

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

Important Points

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,

y_c - 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, y_c = 2.0 m

Hence,

Critical specific energy,

E_c = \(\frac{3}{2}\) × y_c = \(\frac{3}{2}\) × 2 = 3.0 m

Important Points

• The following table shows the relationship between different types of sections and critical specific energy -

Here,

y_c - critical depth (m)

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-
  • 

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.

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:

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

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

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

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

Explanation:

1) Sutro or Proportional  weir:

2) Parshall flume

3) Weir and notch

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

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

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

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

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

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

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