Open Channel Flow MCQ Quiz in বাংলা - Objective Question with Answer for Open Channel Flow - বিনামূল্যে ডাউনলোড করুন [PDF]

Section factor:

• Section factor (Z) for critical flow computation: It is the product of the wetted area and the square root of the hydraulic depth, D

Z = A × √ D

40 = A × √4

A = 20 m²

Hydraulic depth (D) = A/T

Where A is the area of cross-section and T is the top width of the channel

4 = 20/T

T = 5 m

Additional Information Section factor: Section factor, Z, for uniform flow computation: It is the product of the wetted area and the hydraulic radius raised to the two-thirds power.

Z = A × R^2/3

Concept:

The relation between Froude number, pre-jump, and post-jump is given by,

\(\frac{y_2}{y_1} = \frac{1}{2}[{\sqrt{{1+8{Fr}^2}}-1}]\)

Where y2 is post-jump depth, y1 is pre-jump depth, Fr is Froude number, and y2 and y1 are called sequent depth ratio.

Calculation:

Given:

\(\frac{y_2}{y_1} = 5\)

∴\(\frac{y_2}{y_1} = \frac{1}{2}[{\sqrt{{1+8{Fr}^2}}-1}]\)

⇒ \(5 = \frac{1}{2}[{\sqrt{{1+8{Fr}^2}}-1}]\)

⇒ \(10 =[{\sqrt{{1+8{Fr}^2}}-1}]\)

⇒ \(11 =[\sqrt{{1+8{Fr}^2}}]\)

⇒ \(121 =[1+8{Fr}^2]\)

⇒ 121 - 1 = 8Fr2

⇒ \(Fr =\sqrt{\frac{120}{8}}\)

⇒ Fr = 3.87

From the above table, we can conclude that the jump will be oscillating in nature. 

Velocity in an open channel is given by

\(V = \frac{1}{n} \times {\left( R \right)^{\frac{2}{3}}}{\left( S \right)^{\frac{1}{2}}}\)

where

n = Manning’s Coefficient

R = Hydraulic radius = (Wetted Area/ Wetted Perimeter)

S = Bed Slope

\(R = \frac{{4 \times 2}}{{4 + \left( {2 \times 2} \right)}} = 1\)

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:

As It can be observed from the graph, specific energy is minimum for a given discharge at a critical state.

As it can be observed from the graph, at critical state discharge is maximum for a given specific energy.

At critical state, specific force is minimum for a given discharge

Additional Information

Open channel flow in a channel is said to be critical under the following conditions:

1. Froude Number is unity i.e. Fr= 1.

2. For a given discharge, the specific energy is minimum.

3. For a given discharge, the specific force is minimum.

4. For a given specific force, the discharge is maximum.

5. For a given specific energy, discharge is maximum.

Mistake Points

• Discharge and specific energy discharge both are different terms.
• Discharge is maximum for a given specific energy.
• The specific force is minimum for a given discharge

Explanation:

Hydraulic jump: Whenever supercritical flow merges into subcritical flow then to reduce the energy a jump is formed, called a hydraulic jump.

Classification of hydraulic jumps:

Answer:

Statement 4: False Ventilation of triangular weir is not necessary.

Weir:

A weir is a concrete or masonary structure, placed in an open channel over which the flow occurs. It is generally in the form of vertical wall, with a sharp edge at the top, running all the way across the open channel. The notch is of small size while the weir is of a bigger size. The notch is generally made of metallic plate while weir is made of concrete or masonary structure.

Triangular weir:

The expression for the discharge over a triangular weir or notch is

Q = \(\frac{8}{{15}} \times {C_d} \times tan\frac{θ }{2} \times \sqrt {2g} \times {H^{\frac{5}{2}}}\)

From this expression flow through the weir depend upon the vertex angle (θ) and coefficient of discharge (C_d) is constant for all the heads, the value of C_d is nearly equal to 0.6.

Rectangular weir:

The expression for the discharge over a rectangular weir or notch is

Q = \(\frac{2}{3} \times {C_d}\; \times L \times \sqrt {2g} \times {H^{\frac{3}{2}}}\)

Advantage of triangular weir over rectangular weir:

A triangular weir is preferred to a rectangular weir due to following reason.

(1) The expression for discharge for a right angled V- notch or weir is very simple.

(2) For measuring low discharge, a triangular weir gives more accurate results than a rectangular weir.

(3) In case of triangular weir, only one reading, (H) is required for the computation of discharge.

(4) Ventilation of triangular weir is not necessary.

Concept:

Weir is defined as a barrier over which the water flows in an open channel. Crest Is the edge or surface over which the water flows. Nappe Is the overflowing sheet of water.

Explanation:

The hydraulic jumps in horizontal rectangular channels are classified into five categories based on Froude number F1 of the supercritical flow as:

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

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

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