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

Explanation:

The discharge (Q) through a rectangular notch in an open channel flow is given by the formula:

Q=CLH^3/2

Where:

• Q is the discharge (flow rate),

• C is a constant depending on the shape of the notch,

• L is the length (width) of the notch,

• H is the head (the height of water above the bottom of the notch).

From this formula, you can see that discharge Q is directly proportional to the width (L) of the notch. This means that if the width of the notch is doubled, the discharge will double as well, assuming the head (H) remains the same.

 Additional Information

Effect of Changing the Width (L):

1. The discharge is directly proportional to the width (L) of the rectangular notch. This means that if the width is doubled, the discharge will double, as long as the head (H) remains constant.

Discharge Coefficient (C):

1. The discharge coefficient C depends on factors like the shape of the notch, the flow type (subcritical or supercritical), and the specific characteristics of the flow.

2. For a sharp-crested weir (a type of notch), C is a constant that generally depends on experimental results, and it can be adjusted based on the characteristics of the flow and the notch.

Explanation:

Catchment Area

• The catchment area refers to the entire geographical region that drains into a specific stream, river, or other water body. In the context of a dam, it is the area from which all precipitation (rainfall, snowmelt, etc.) flows into the reservoir behind the dam. The boundaries of a catchment area are determined by natural topographical features, such as ridges and hills, which separate one drainage basin from another.

When a dam is constructed across a river or stream, the catchment area becomes a critical factor in determining the water inflow into the reservoir. This inflow is essential for various purposes, including hydroelectric power generation, irrigation, water supply, flood control, and more. The total volume of water that can be collected in the reservoir is directly proportional to the size and characteristics of the catchment area. Key aspects of the catchment area include:

• Size of the Catchment Area: Larger catchment areas generally contribute more water to the reservoir because they collect precipitation over a broader region. However, the efficiency of water collection also depends on other factors like soil permeability and vegetation cover.
• Topography: The shape and slope of the terrain influence the flow of water. Steeper slopes facilitate quicker runoff, while flatter areas may allow more water to infiltrate into the ground.
• Climate: The amount and pattern of rainfall in the catchment area determine the volume of water that can be collected. Seasonal variations, such as monsoons, play a significant role in regions with distinct wet and dry periods.
• Land Use: Human activities, such as agriculture, deforestation, and urbanization, can impact the runoff characteristics of the catchment area. For instance, deforestation may lead to increased surface runoff and reduced groundwater recharge.
• Hydrology: The natural drainage network, including rivers, streams, and tributaries, influences how efficiently water is channeled into the reservoir.

Importance of the Catchment Area in Dam Construction:

Understanding the catchment area is vital for the planning and operation of a dam. Key reasons include:

1. Reservoir Capacity Estimation: The size and characteristics of the catchment area help engineers calculate the maximum potential water inflow, which is crucial for determining the reservoir's storage capacity.
2. Flood Management: A well-designed dam takes into account the peak runoff from the catchment area during heavy rainfall to prevent overflow and downstream flooding.
3. Water Resource Planning: Accurate knowledge of the catchment area allows for efficient allocation of water for various purposes, such as irrigation, drinking water supply, and industrial use.
4. Environmental Impact Assessment: The catchment area influences the ecological balance of the region, affecting flora, fauna, and local communities. Proper assessment ensures sustainable development.

Explanation:

Open Conduit

• An open conduit is a channel or passage that allows the flow of liquid (usually water) in an open and unpressurized condition. The liquid flows under the influence of gravity, and there is a free surface exposed to the atmosphere. Open conduits are commonly used in hydraulic systems, irrigation, drainage systems, and water supply networks where pressure is not a primary factor in the liquid's movement.

Flume:

• A flume is an artificial channel specifically designed to convey water or other liquids in an open condition. The water flows freely, with its surface exposed to the atmosphere. Flumes are often constructed using materials such as concrete, metal, or wood, and they are frequently used in applications such as hydroelectric power plants, irrigation systems, and water transport for industrial purposes. The key characteristic of a flume is that it allows the liquid to flow under gravity without being confined under pressure, making it a classic example of an open conduit.

Working Principle: A flume operates by channeling water along a specific path, maintaining a controlled flow rate and direction. The flow is governed by gravitational forces, and the design of the flume ensures minimal resistance to the movement of water. Flumes can also include features to measure the flow rate, such as weirs or flow meters, making them useful in hydraulic engineering.

Applications:

• Hydroelectric Power Generation: Flumes are used to channel water to turbines in hydroelectric plants.
• Irrigation Systems: Flumes transport water to agricultural fields for irrigation purposes.
• Industrial Use: Flumes are employed in industries for conveying water or other liquids between processes.
• Drainage: Flumes are used in drainage systems to manage surface runoff and prevent flooding.

Explanation:

Spillways

Definition: A spillway is a structure built into or near a dam to safely discharge excess water from a reservoir when it reaches or exceeds a certain level. The primary purpose of a spillway is to prevent overtopping of the dam, which could lead to structural failure and catastrophic downstream flooding. By providing a controlled pathway for excess water, spillways ensure the safety and integrity of the dam while also protecting downstream areas.

Working Principle: When the water level in the reservoir rises beyond the designed maximum level, it flows into the spillway. The spillway channels the water away from the reservoir and releases it downstream. Spillways are designed to handle specific maximum flow rates based on the dam's capacity and the potential inflow during extreme weather events, such as heavy rainfall or rapid snowmelt.

Types of Spillways:

• Ogee Spillway: This is the most common type of spillway, with a curved profile that allows water to flow smoothly over the crest and down the spillway.
• Chute Spillway: A steep channel that carries water away from the reservoir, designed to handle high flow rates.
• Side Channel Spillway: Positioned parallel to the dam, it diverts water to a channel leading away from the dam.
• Shaft Spillway: Also known as a "morning glory" spillway, it features a vertical shaft that leads to a horizontal tunnel for water discharge.
• Stepped Spillway: Includes steps that dissipate energy as water flows down, reducing erosion and turbulence.

Concept:

The discharge through a Crump’s open flume outlet is given by:

\( Q = C \cdot L_t \cdot H^{3/2} \)

• \( C = 1.60 \)
• \( L_t = 7 \, \text{cm} = 0.07 \, \text{m} \)
• \( H = 0.64 \, \text{m} \)

Substituting values:

\( Q = 1.60 \cdot 0.07 \cdot (0.64)^{3/2} \approx 0.0573 \, \text{cumecs} \)

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

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

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

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

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