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

Explanation:

To solve this problem, we need to calculate the critical depth for a rectangular channel. The formula for critical depth
y_c in a rectangular channel is:

\(y_c=(\frac{q^2}{g})^{\frac{1}{3}}\)

where, q = discharge per unit width; g = 9.81 m/s²

Calculation:

Given: Q = \(4.8 \text{m}^3/\text{s}\); b = 2.0 m, 

Discharge per unit width: \(q=\frac{Q}{b}=\frac{4.8}{2}=2.4\)

Critical depth \(y_c=(\frac{2.4^2}{9.81})^{\frac{1}{3}}=0.876m\)

Concept:

For triangular open channels, the discharge at critical flow can be found using the critical flow condition:

\(Q_c=\frac{8}{15}\sqrt {g} zy_c^{2.5}\)

where, z = side slope of channel, y_c = critical depth, g = 9.81 m/s²

Calculation:

Given: For 90 degrees, z = 1

y_c = 0.30 m

\(Q_c=\frac{8}{15}\times \sqrt {9.81}\times 1\times0.3^{2.5}\)

\(Q_c=0.11m^3/s\)

Explanation:

• In open channel flow, alternate depths refer to two different flow depths that correspond to the same specific energy for a given discharge.

• For a rectangular channel and constant discharge, two different depths — one subcritical (deep and slow) and one supercritical (shallow and fast) — can have the same specific energy. These are called alternate depths.

• Alternate depths are two different flow depths that exist for the same specific energy at a given discharge in a channel.

• They occur in a channel when the flow shifts between supercritical and subcritical regimes, but total specific energy remains the same.

• Example:

  • For a given discharge, a shallow and fast flow (supercritical) and a deep and slow flow (subcritical) can have the same specific energy.

  • These two depths are known as alternate depths.

 Additional Information

• A hydraulic jump is a sudden transition from supercritical flow (fast and shallow) to subcritical flow (slow and deep) in an open channel. This occurs to conserve energy and momentum when flow velocity decreases rapidly due to downstream conditions.

• It typically forms when high-velocity water is discharged into a channel with a mild slope, causing the water to rise abruptly and form turbulent eddies. The upstream depth is low, while the downstream depth increases sharply after the jump.

• Hydraulic jumps are characterized by intense turbulence, surface rollers, and energy loss in the form of heat, noise, and air entrainment. Despite these losses, the momentum across the jump is conserved.

• The location and formation of a hydraulic jump are influenced by the Froude number before the jump. A Froude number greater than 1 indicates supercritical flow, which is a necessary condition for a jump to occur.

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:

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

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

Explanation:

1) Sutro or Proportional  weir:

2) Parshall flume

3) Weir and notch

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

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

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

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

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

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