Disaster Management · TEE · practice

Channel discharge — optional lecture-formula practice

Optional calculation practice

Supplementary worked examples · no numerical question found in the available 12 TEE sittings

Source of the calculations. M2 PDF p 100 prints Q = A × V in the desilting discussion. It supplies no drain dimensions or worked numerical. All three numerical scenarios below were authored for understanding, not recovered from exam papers.

Definition. Discharge is the volume of water passing a cross-section per unit time. For a known flowing cross-sectional area and average velocity, Q = Av.

Simple meaning. Cross-section mein kitna area hai × paani kitni speed se jaa raha hai.

Q=A_{flow}\,vLaTeX: Q=A_{flow}\,v

Aflow is the wet cross-sectional area in m²; v is average velocity in m/s. Multiplying m² by m/s gives m³/s. This A is different from the catchment area used in the rational method.

Illustrative calculation (not a PYQ). Water flows through a rectangular drain 2 m wide and 1.5 m deep at an average velocity of 0.8 m/s. Find discharge.

Answer

Given

Width=2 m; flowing depth=1.5 m; v=0.8 m/s.

To find

Q.

Formula and why

A_{flow}=\text{width}\times\text{depth},\quad Q=A_{flow}vLaTeX: A_{flow}=\text{width}\times\text{depth},\quad Q=A_{flow}v

The cross-section is rectangular, so width times flowing depth gives wet area.

Step-by-step solution

Step 1. Find area

A_{flow}=2\times1.5=3\text{ m}^2LaTeX: A_{flow}=2\times1.5=3\text{ m}^2

Step 2. Find discharge

Q=3\times0.8=2.4\text{ m}^3/\text{s}LaTeX: Q=3\times0.8=2.4\text{ m}^3/\text{s}

Final answer: 2.4 m³/s.

Illustrative calculation (not a PYQ). Silt reduces the same drain’s flowing depth from 1.5 m to 1 m. Assume the average velocity remains 0.8 m/s. Find the reduction in discharge.

Answer

Given

Width=2 m; new depth=1 m; v=0.8 m/s; old Q=2.4 m³/s.

To find

New Q and reduction.

Formula and why

Q=A_{flow}vLaTeX: Q=A_{flow}v

The fixed-velocity assumption isolates the effect of reduced cross-sectional area.

Step-by-step solution

Step 1. Find reduced area

A_{flow}=2\times1=2\text{ m}^2LaTeX: A_{flow}=2\times1=2\text{ m}^2

Step 2. Find reduced discharge

Q=2\times0.8=1.6\text{ m}^3/\text{s}LaTeX: Q=2\times0.8=1.6\text{ m}^3/\text{s}

Step 3. Calculate loss

2.4-1.6=0.8\text{ m}^3/\text{s}LaTeX: 2.4-1.6=0.8\text{ m}^3/\text{s}

Step 4. Express as percentage

(0.8/2.4)\times100=33.333\%LaTeX: (0.8/2.4)\times100=33.333\%

Final answer: Reduction =0.8 m³/s, or 33.3% under the assumption.

Illustrative calculation (not a PYQ). A channel must carry 6 m³/s at an average velocity of 1.5 m/s. Find the required flowing cross-sectional area.

Answer

Given

Q=6 m³/s; v=1.5 m/s.

To find

Aflow.

Formula and why

Q=A_{flow}v\quad\Rightarrow\quad A_{flow}=Q/vLaTeX: Q=A_{flow}v\quad\Rightarrow\quad A_{flow}=Q/v

Divide both sides by velocity to isolate area.

Step-by-step solution

Step 1. Substitute the known values

A_{flow}=6/1.5LaTeX: A_{flow}=6/1.5

Step 2. Calculate and write units

A_{flow}=4\text{ m}^2LaTeX: A_{flow}=4\text{ m}^2

Final answer: Required flowing area =4 m², under the stated average velocity.

Desilting may restore lost capacity, but actual velocity, downstream water level and channel stability also change. A calculation holding velocity fixed is a teaching comparison, not proof that desilting alone will prevent floods.

Yaad rakho: Catchment area receives rain; channel area carries flow.