Disaster Management · TEE · practice

Runoff — lecture-adapted and authored examples

Optional calculation practice

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

Source of the calculations. The first two examples adapt the barren-land and township values in M2 PDF p 88, with the unit error corrected. The percentage-increase request, time conversion and 20-hectare example add practice. These are lecture examples or authored extensions, not PYQs.

Definition. Runoff is water that flows over the surface rather than being retained or infiltrating. The rational method is a simplified peak-flow estimate for suitable small catchments under stated rainfall assumptions.

Simple meaning. C zyada matlab same rain se zyada runoff. Area ka unit change hua toh factor bhi change hoga.

Q=0.27778\,C I A\quad[Q:\text{m}^3/\text{s},\ I:\text{mm/h},\ A:\text{km}^2]LaTeX: Q=0.27778\,C I A\quad[Q:\text{m}^3/\text{s},\ I:\text{mm/h},\ A:\text{km}^2]
SymbolMeaning
CDimensionless runoff coefficient; a higher value represents a greater runoff contribution.
IRainfall intensity: depth of rain per hour, not total storm depth.
ACatchment area draining to the point of interest, not the channel cross-section.
QEstimated peak discharge: volume per unit time.

Where the conversion factor comes from

1 mm = 0.001 m; 1 km² = 1,000,000 m². So 1 mm falling on 1 km² produces 1,000 m³ of rainfall volume. Dividing one hour by 3,600 seconds gives 1,000/3,600 = 0.27778. The runoff coefficient multiplies this rainfall input.

Q=1000CIA\ (\text{m}^3/\text{h},\ A\text{ in km}^2)LaTeX: Q=1000CIA\ (\text{m}^3/\text{h},\ A\text{ in km}^2)
Q=10CIA\ (\text{m}^3/\text{h},\ A\text{ in hectares})LaTeX: Q=10CIA\ (\text{m}^3/\text{h},\ A\text{ in hectares})

Unit correction: “10 CIA” is compatible with area in hectares and output in m³/h. It is not compatible with area in km² and the same output unit. The lecture mixes these labels; always write the units in your working. 1 km² = 100 hectares.

Illustrative calculation (not a PYQ). A 1 km² catchment receives rainfall of intensity 75 mm/h. Using C = 0.5, estimate runoff in m³/h and m³/s.

Answer

Given

A=1 km²; I=75 mm/h; C=0.5.

To find

Runoff discharge in both units.

Formula and why

Q_{\rm hour}=1000CIA,\quad Q_{\rm second}=Q_{\rm hour}/3600LaTeX: Q_{\rm hour}=1000CIA,\quad Q_{\rm second}=Q_{\rm hour}/3600

The first formula keeps time in hours; dividing by 3,600 converts to seconds.

Step-by-step solution

Step 1. Calculate rainfall volume rate

1000\times75\times1=75000\text{ m}^3/\text{h}LaTeX: 1000\times75\times1=75000\text{ m}^3/\text{h}

Step 2. Apply runoff coefficient

Q=0.5\times75000=37500\text{ m}^3/\text{h}LaTeX: Q=0.5\times75000=37500\text{ m}^3/\text{h}

Step 3. Convert time unit

Q=37500/3600=10.416667\text{ m}^3/\text{s}LaTeX: Q=37500/3600=10.416667\text{ m}^3/\text{s}

Final answer: 37,500 m³/h, or about 10.417 m³/s.

Illustrative calculation (not a PYQ). The same catchment becomes a township with C = 0.9. Find its new runoff and the percentage increase.

Answer

Given

A=1 km²; I=75 mm/h; C changes from 0.5 to 0.9; old Q=37,500 m³/h.

To find

New Q and percentage increase.

Formula and why

Q=1000CIA;\quad \%\text{increase}=\frac{Q_{new}-Q_{old}}{Q_{old}}\times100LaTeX: Q=1000CIA;\quad \%\text{increase}=\frac{Q_{new}-Q_{old}}{Q_{old}}\times100

Only C changes, so the increase reflects land-cover change in this simplified comparison.

Step-by-step solution

Step 1. Find new discharge

Q_{new}=1000(0.9)(75)(1)=67500\text{ m}^3/\text{h}LaTeX: Q_{new}=1000(0.9)(75)(1)=67500\text{ m}^3/\text{h}

Step 2. Subtract old discharge

67500-37500=30000\text{ m}^3/\text{h}LaTeX: 67500-37500=30000\text{ m}^3/\text{h}

Step 3. Divide by the old value

(30000/37500)\times100=80\%LaTeX: (30000/37500)\times100=80\%

Step 4. Convert new discharge

67500/3600=18.75\text{ m}^3/\text{s}LaTeX: 67500/3600=18.75\text{ m}^3/\text{s}

Final answer: 67,500 m³/h =18.75 m³/s; an 80% increase.

Illustrative calculation (not a PYQ). For a 20-hectare catchment with C = 0.6 and rainfall intensity 50 mm/h, estimate discharge in m³/s.

Answer

Given

A=20 ha; C=0.6; I=50 mm/h.

To find

Q in m³/s.

Formula and why

Q=10CIA/3600\quad(A\text{ in ha})LaTeX: Q=10CIA/3600\quad(A\text{ in ha})

The area is in hectares, so use the hectare conversion.

Step-by-step solution

Step 1. Calculate hourly discharge

Q=10(0.6)(50)(20)=6000\text{ m}^3/\text{h}LaTeX: Q=10(0.6)(50)(20)=6000\text{ m}^3/\text{h}

Step 2. Convert to seconds

Q=6000/3600=1.666667\text{ m}^3/\text{s}LaTeX: Q=6000/3600=1.666667\text{ m}^3/\text{s}

Final answer: Q ≈1.667 m³/s.

These are lecture-formula practice examples, not verified past-paper questions. Real drainage design needs rainfall duration, time of concentration, catchment suitability and local design standards; Q alone does not give the total flood volume.

Yaad rakho: Write units before numbers.