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

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:

• For a given specific energy E, the critical depth yc for a rectangular channel is given by yc = 2/3E.
• This relation comes from the condition of minimum specific energy for a given discharge, which occurs at the critical depth. It is derived from the energy equation in open channel flow theory.

 Additional Information

• Specific energy (E) is the total energy relative to the channel bottom, measured as the sum of the flow depth (y) and the velocity head (v²/2g). 

• For a given discharge, the specific energy varies with flow depth, and the flow can be subcritical (deep and slow), critical, or supercritical (shallow and fast).

• The critical depth " id="MathJax-Element-20-Frame" role="presentation" style="position: relative;" tabindex="0">
  y_cy_c​ is the depth at which the specific energy is minimum for a given discharge.

• At critical depth, the flow velocity equals the wave velocity (celerity), and the Froude number equals 1.

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 \( n \):

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

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,