Gradually Varied MCQ Quiz - Objective Question with Answer for Gradually Varied - Download Free PDF

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

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

Explanation:

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

For minimum wetted perimeter,

d = b/2

∴ d/b = 1 : 2

Note:

Explanation:

Given Data:

• Channel Width (B) = 5 m
• Depth (y) = 1 m
• Discharge (Q) = 20 m³/s
• Bed Slope (S) = 1/900

Step 1: Calculate Flow Area (A)

\( A = B \times y = 5 \times 1 = 5 m^2 \)

Step 2: Calculate Wetted Perimeter (P)

\( P = B + 2y = 5 + 2(1) = 7 m \)

Step 3: Calculate Hydraulic Radius (R)

\( R = \frac{A}{P} = \frac{5}{7} m \)

Step 4: Find Velocity (V) using Q = AV

\( V = \frac{Q}{A} = \frac{20}{5} = 4 \text{ m/s} \)

Step 5: Solve for Manning’s Constant (n)

Using Manning’s equation:

\( V = \frac{1}{n} R^{2/3} S^{1/2} \)

Substituting values:

\( 4 = \frac{1}{n} \times \left(\frac{5}{7}\right)^{2/3} \times \left(\frac{1}{900}\right)^{1/2} \)

Solving for \( n \):

\( n = \frac{1}{120} \times \left(\frac{5}{7}\right) \)

Turbulent Flow in Open Channels:

• Flow in an open channel is classified based on the Reynolds number (Re), which is a dimensionless quantity used to predict flow patterns in different fluid flow situations.
• The Reynolds number is given by the formula:  \(R_e = {\rho VD \over \mu}\), where ρ is the fluid density, V is the flow velocity, D is the characteristic length (such as hydraulic diameter), and μ is the dynamic viscosity of the fluid.
• If the Reynolds number of the flow in an open channel is more than 2000, the flow is considered to be turbulent. Turbulent flow is characterized by chaotic changes in pressure and velocity.
• In contrast, flow with a Reynolds number less than 500 is typically classified as laminar flow, where the fluid flows in parallel layers with minimal disruption between them.
• Transitional flow occurs between Reynolds numbers of approximately 500 and 2000, where the flow can shift between laminar and turbulent states.

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

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:

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:

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

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

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

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

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

Concept

For a hydraulic efficient rectangular channel

\(depth\;of\;water\;level = \frac{{width\;of\;channel}}{2}\)

\(y = \frac{B}{2}\;\)

\(Hydraullic\;radius\;\left( R \right) = \frac{{wetted\;area}}{{wetted\;perimeter}} = \frac{{B \times y}}{{B + 2 \times y}}\)

\(R = \frac{{B \times \frac{B}{2}}}{{B + 2 \times \frac{B}{2}}} = \frac{B}{4}\)

Calculation

Given, Width of channel (B) = 5 m

\(y = \frac{B}{2} = \frac{5}{2} = 2.5\;m\)

\(R = \frac{B}{4} = \frac{5}{4} = 1.25\;m\)

∴ Hydraulic radius (R) = 1.25 m 

Concept:

The Chezy equation can be used to calculate the discharge in the open channel.

\(\rm Q= A × C\sqrt {mi} \)

A - Area of the channel

C - Chezy's coefficient

m - Hydraulic radius (or) Hydraulic mean depth = A/P

P- Perimeter of the channel

i - Bed slope

Calculation:

Given:

b = 4 m, d = 2 m,C = 50

\({\rm{i}} = \frac{{\rm{1}}}{{\rm{900}}} = 0.0011\)

A = 4 × 2 = 8 m²

\({\rm{m}} = \frac{{\rm{A}}}{{\rm{P}}} = \frac{{4\; \times\;2}}{{4 \;+ \;2\; + \;2}} = 1\)

\({\rm{Q}} = \rm A \times {\rm{C}}\sqrt {{\rm{mi}}} = 8 \times50 \sqrt {1\times 0.0011} \)

∴ Q = 13.3 m³/s

Concept:

For critical flow, y = y_c

Specific energy at critical flow, E_c 

\({E_c} = {y_c} + \frac{{V_c^2}}{{2g}}\)

\({E_c} = {y_c} + \frac{{Q_c^2}}{{2gA_c^2}}\)

For critical flow:

\(\frac{{{Q^2}T}}{{gA_c^3}} = 1 \Rightarrow \frac{{{Q^2}}}{{gA_c^2}} = \frac{{{A_c}}}{T}\)

Where T is the top width

( T = B) 

Put, \(A = \frac{1}{2} \times Base \times Height\)

\({E_c} = {y_c} + \frac{{{A_c}}}{{2T}} = {y_c} + \frac{{0.5 \times B \times {y_c}}}{{2 \times B}} = {y_c} + \frac{{{y_c}}}{4}\)

∴ \(\frac{{{y_c}}}{{{E_c}}} = \frac{4}{5}\)

Concept:

In an open channel flow control section is a section where for a given discharge the depth of flow is known or it can be controlled.

The Froude number (F_R) is given by

\({F_R} = \frac{V}{{\sqrt {gH} }}, \sqrt {gh} - \rm Celerity\)

V - Velocity of flow

H - Hydraulic depth

Celerity is the velocity of a wave when the flow is disturbed.

Subcritical flow:

In Subcritical flow F_R < 1, 

\(∴ \sqrt {gh} > V\)

Hence velocity of the wave is more than the velocity of flow, hence it can travel upstream. Here the downstream conditions are controlled.

∴ For subcritical flow downstream is the control section.

Supercritical flow:

In Supercritical flow FR > 1, 

\(∴ \sqrt {gh} < V\)

Hence velocity of the wave is less than the velocity of flow, hence it cannot travel upstream. Here the upstream conditions are controlled.

∴ For supercritical flow upstream is the control section

Key Points

If the wave can travel upstream, downstream is the control section.

If the wave can travel downstream, upstream is the control section.

Concept:

When the width of the channel is reduced the term q (Discharge per unit width) increases. The no choke condition means critical depth of flow is not reached.

Case 1: Width of channel is B

The specific energy curve is drawn for the channel

Case 2: Width of channel is B₁ (B₁ ≤ B)

The specific energy curve corresponding to the width B₁ is drawn.

When we consider the subcritical portion the graph,