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
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]| Symbol | Meaning |
|---|---|
| C | Dimensionless runoff coefficient; a higher value represents a greater runoff contribution. |
| I | Rainfall intensity: depth of rain per hour, not total storm depth. |
| A | Catchment area draining to the point of interest, not the channel cross-section. |
| Q | Estimated 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)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}/3600The 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}Step 2. Apply runoff coefficient
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}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}}\times100Only 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}Step 2. Subtract old discharge
67500-37500=30000\text{ m}^3/\text{h}Step 3. Divide by the old value
(30000/37500)\times100=80\%Step 4. Convert new discharge
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})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}Step 2. Convert to seconds
Q=6000/3600=1.666667\text{ m}^3/\text{s}Final answer: Q ≈1.667 m³/s.
Yaad rakho: Write units before numbers.