Section 4: Soil and Water Conservation Engineering
Hydrology
Subtopic: Hydrograph analysis, unit hydrograph theory and application
Question Statement:
A 4-hour Unit Hydrograph (UH) for a drainage basin has a symmetrical triangular shape with a base duration of $T_b = 24.0\text{ hours}$ and a peak discharge of $Q_p = 35.0\text{ m}^3/\text{s}$. The direct catchment drainage area $A$ in square kilometers ($\text{km}^2$) is ________ (round off to one decimal place).
```
Discharge Q (m³/s)
^ Peak Qp = 35.0 m³/s
| *
| / \
| / \
| / \
+-----------+---------------+----------> Time t (hours)
0 4 12 24
```
Answer Key:
151.2
Step-by-Step Solution & Explanation
**Method 1: Hydrograph Volume Integration**
1. By definition, a Unit Hydrograph represents $1.0\text{ cm} = 0.010\text{ m}$ of direct runoff depth over the entire catchment area $A$.
2. Volume of runoff under the triangular hydrograph:
$V = \frac{1}{2} \times \text{Base} \times \text{Peak}$
$\text{Base } T_b = 24.0\text{ hours} = 24.0 \times 3600\text{ s} = 86,400\text{ s}$
$V = \frac{1}{2} \times 86,400\text{ s} \times 35.0\text{ m}^3/\text{s} = 43,200 \times 35.0 = 1,512,000\text{ m}^3$
3. Equating volume to depth $\times$ area:
$V = A \times 0.010\text{ m} \implies A = \frac{1,512,000}{0.010} = 151,200,000\text{ m}^2$
4. Converting to $\text{km}^2$:
$A = \frac{151,200,000}{10^6} = 151.2\text{ km}^2$
**Method 2: Direct Unit Hydrograph Dimensionless Shortcut**
$V = 0.5 \times 24 \times 35 \times 3600 = 1.512 \times 10^6\text{ m}^3$
$A = \frac{V}{0.01\text{ m}} = 151.2\text{ km}^2$