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.

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:

For a trapezoidal channel, the area of flow is given by:

\( A = (b + zd) \cdot d \)

And the velocity of flow:

\( V = \frac{Q}{A} \)

Given:

• Base width b=6m" id="MathJax-Element-17-Frame" role="presentation" style="position: relative;" tabindex="0">b=6mb=6m" id="MathJax-Element-77-Frame" role="presentation" style="position: relative;" tabindex="0">b=6mb=6m" id="MathJax-Element-123-Frame" role="presentation" style="position: relative;" tabindex="0">b=6mb=6m b=6m b=6m $b = 6 \, \text{m}$
• Side slope z=2" id="MathJax-Element-18-Frame" role="presentation" style="position: relative;" tabindex="0">z=2z=2" id="MathJax-Element-78-Frame" role="presentation" style="position: relative;" tabindex="0">z=2z=2" id="MathJax-Element-124-Frame" role="presentation" style="position: relative;" tabindex="0">z=2z=2 z=2 z=2 $z = 2$
• Depth d=1.5m" id="MathJax-Element-19-Frame" role="presentation" style="position: relative;" tabindex="0">d=1.5md=1.5m" id="MathJax-Element-79-Frame" role="presentation" style="position: relative;" tabindex="0">d=1.5md=1.5m" id="MathJax-Element-125-Frame" role="presentation" style="position: relative;" tabindex="0">d=1.5md=1.5m d=1.5m d=1.5m $d = 1.5 \, \text{m}$
• Discharge Q=27m3/s" id="MathJax-Element-20-Frame" role="presentation" style="position: relative;" tabindex="0">Q=27m3/sQ=27m3/s" id="MathJax-Element-80-Frame" role="presentation" style="position: relative;" tabindex="0">Q=27m3/sQ=27m3/s" id="MathJax-Element-126-Frame" role="presentation" style="position: relative;" tabindex="0">Q=27m3/sQ=27m3/s Q=27m3/s Q=27m3/s $Q = 27 \, \text{m}^3/\text{s}$

Substituting values:

\( A = (6 + \cdot 2 \cdot 1.5) \cdot 1.5 = 9 \cdot 1.5 = 13.5 \, \text{m}^2 \)

\( V = \frac{27}{13.5} = 2 \, \text{m/s} \)

Concept:

The gradually varied flow (GVF) equation derived from energy slope and channel bed slope is:

\(\frac{dy}{dx} = \frac{S_0 - S_f}{1 - F_r^2} \)

Given:

Width of stream (b) = 15 m

Depth of stream (y) = 3 m

Discharge (Q) = 29 m³/s

Bed slope S0=15000=0.0002" id="MathJax-Element-7-Frame" role="presentation" style="position: relative;" tabindex="0">S0=15000=0.0002S0=15000=0.0002 $S_0 = \frac{1}{5000} = 0.0002$

Energy slope Sf=121700=0.000046" id="MathJax-Element-8-Frame" role="presentation" style="position: relative;" tabindex="0">Sf=121700=0.000046Sf=121700=0.000046 $S_f = \frac{1}{21700} = 0.000046$

Step 1: Area and velocity

\( A = b \cdot y = 15 \cdot 3 = 45 \, m^2 \)

\( V = \frac{Q}{A} = \frac{29}{45} = 0.644 \, m/s \)

Step 2: Froude number

\( F_r = \frac{V}{\sqrt{g y}} = \frac{0.644}{\sqrt{9.81 \cdot 3}} = \frac{0.644}{5.427} \approx 0.1186 \)

Step 3: Apply GVF equation

\(\frac{dy}{dx} = \frac{0.0002 - 0.000046}{1 - (0.1186)^2} = \frac{0.000154}{1 - 0.0141} = \frac{0.000154}{0.9859} \approx 0.0001562\)

Step 4: Find slope ratio

\( \frac{1}{0.0001562} \approx 6400 \)

Hydraulic Radius

Hydraulic radius is defined as the ratio of the wetted area of a channel to its wetted perimeter. It is an important concept in fluid mechanics and is used in the study of open channel flow. The hydraulic radius is given by the formula:

R = A / P

where:

• A is the cross-sectional area of flow.

• P is the wetted perimeter, which is the length of the boundary that is in contact with the fluid.

Analyzing the Given Options

1. Option 1: Hydraulic radius

   • The hydraulic radius is indeed defined as the ratio of the wetted area to its wetted perimeter.

2. Option 2: Hydraulic depth

   • Hydraulic depth is different from hydraulic radius. It is defined as the ratio of the cross-sectional area of flow to the top width of the surface.

3. Option 3: Depth of flow

   • Depth of flow refers to the vertical distance from the bottom of the channel to the free surface of the water.

4. Option 4: Top width

   • Top width is the width of the water surface in an open channel.

Therefore, the correct answer is option 1: Hydraulic radius.

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

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

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

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

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