# Runoff — lecture-adapted and authored examples

Subject: Disaster Management

Exam: TEE

Reader: https://deep-learning-study-room.vercel.app/dm/#tee/practice/calculations/runoff

Canonical: https://deep-learning-study-room.vercel.app/read/dm/tee/practice/calculations--runoff/

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.



$$
Q=0.27778\,C I A\quad[Q:\text{m}^3/\text{s},\ I:\text{mm/h},\ A:\text{km}^2]
$$



| 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



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



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

## Step-by-step solution

#### Step 1. Calculate rainfall volume rate



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



#### 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 application · Two catchments in the same rain

A paved catchment can send a larger fraction of rainfall into rapid runoff than a more permeable one. In a rational-method problem, this appears through the runoff coefficient C; keep the area and rainfall units consistent before calculating discharge.

### 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



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



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

## Step-by-step solution

#### Step 1. Find new discharge



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



#### 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 application · Two catchments in the same rain

A paved catchment can send a larger fraction of rainfall into rapid runoff than a more permeable one. In a rational-method problem, this appears through the runoff coefficient C; keep the area and rainfall units consistent before calculating discharge.

### 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.

#### Illustrative application · Two catchments in the same rain

A paved catchment can send a larger fraction of rainfall into rapid runoff than a more permeable one. In a rational-method problem, this appears through the runoff coefficient C; keep the area and rainfall units consistent before calculating discharge.

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.