To find the peak of the flood hydrograph, we need to determine the peak direct runoff (DR) and add the baseflow.
Step 1: Analyze Unit Hydrograph (UH) Characteristics
- The 6-hour unit hydrograph is represented as an isosceles triangle.
- Peak Discharge of UH ($Q_{p,UH}$): 180 $m^3 s^{-1}$.
- Time to Peak of UH ($T_{p,UH}$): 18 hours.
- For an isosceles triangle, the base duration ($D_{UH}$) is twice the time to peak: $D_{UH} = 2 \times T_{p,UH} = 2 \times 18 \, \text{h} = 36 \, \text{h}$.
- The area under the UH represents the total volume of direct runoff from 1 cm (or 10 mm) of excess rainfall over the watershed.
Step 2: Calculate Rainfall Excess using Phi-Index Method
- Given Phi-index ($\emptyset$) = 3.0 mm $h^{-1}$.
- Period 1 (First 6 hours):
- Total Rainfall = 38 mm.
- Average Rainfall Rate = $\frac{38 \, \text{mm}}{6 \, \text{h}} \approx 6.33 \, \text{mm/h}$.
- Rainfall Excess ($P_{ex1}$) = Total Rainfall - ($\emptyset \times$ Duration) $P_{ex1} = 38 \, \text{mm} - (3.0 \, \text{mm/h} \times 6 \, \text{h}) = 38 \, \text{mm} - 18 \, \text{mm} = 20 \, \text{mm}$.
- Period 2 (Next 6 hours, from h=6 to h=12):
- Rainfall in this period = 106 mm - 38 mm = 68 mm.
- Average Rainfall Rate = $\frac{68 \, \text{mm}}{6 \, \text{h}} \approx 11.33 \, \text{mm/h}$.
- Rainfall Excess ($P_{ex2}$) = Rainfall in period - ($\emptyset \times$ Duration) $P_{ex2} = 68 \, \text{mm} - (3.0 \, \text{mm/h} \times 6 \, \text{h}) = 68 \, \text{mm} - 18 \, \text{mm} = 50 \, \text{mm}$.
- Total Rainfall Excess = $P_{ex1} + P_{ex2} = 20 \, \text{mm} + 50 \, \text{mm} = 70 \, \text{mm}$.
Step 3: Estimate Peak Direct Runoff (DRH)
A common approximation for the peak direct runoff discharge is to scale the peak discharge of the unit hydrograph by the ratio of the total rainfall excess depth to the depth represented by the unit hydrograph (10 mm).
- Peak DRH $\approx Q_{p,UH} \times \frac{\text{Total Rainfall Excess}}{\text{UH Rainfall Depth}}$
- Peak DRH $\approx 180 \, m^3 s^{-1} \times \frac{70 \, \text{mm}}{10 \, \text{mm}}$
- Peak DRH $\approx 180 \, m^3 s^{-1} \times 7$
- Peak DRH $\approx 1260 \, m^3 s^{-1}$.
This approximation provides a value close to the actual peak, considering the complexities of convolution with non-uniform rainfall excess.
Step 4: Calculate Peak Flood Hydrograph Discharge
The peak of the total flood hydrograph is the sum of the peak direct runoff discharge and the constant baseflow.
- Constant Baseflow = 20 $m^3 s^{-1}$.
- Peak Flood Hydrograph Discharge = Peak DRH + Baseflow
- Peak Flood Hydrograph Discharge $\approx 1260 \, m^3 s^{-1} + 20 \, m^3 s^{-1}$
- Peak Flood Hydrograph Discharge $\approx 1280 \, m^3 s^{-1}$.
Important Questions from Unit Hydrograph Theory
- Two catchments M and N are meteorologically similar. The basin length (L), length to the centroid (Lca), and the drainage area (A) of catchment M are $36\text{ km}$, $18\text{ km}$, and $250\text{ km}^2$, respectively. The L, Lca and A for catchment N are $50\text{ km}$, $30\text{ km}$, and $400\text{ km}^2$, respectively.
For 6-h unit hydrograph in catchment M, the peak discharge of $50\text{ m}^3\text{.s}^{-1}$ occurs at $12\text{ h}$ from the beginning of the rainfall excess. Using Snyder’s method, the peak of the 6-h unit hydrograph for catchment N (in $\text{m}^3\text{.s}^{-1}$) is ________. (Rounded off to two decimal places) - A catchment of 720 ha area has 25-year mean rainfall intensity of $100 \, mm \, h^{-1}$ occurring for a duration equal to its time of concentration. During a storm event, the catchment received a total of 7.5 cm design rainfall for 6 hours. Assuming $\phi$-index of $0.25 \, cm \, h^{-1}$ and runoff coefficient of 0.6, the peak ordinate of the 6-h unit hydrograph in $m^3 \, s^{-1}$ will be ________.
- A watershed, with an area of $360 \text{ km}^2$, has a triangular shaped 4-h unit hydrograph with a base of $50 \text{ hours}$. The peak discharge of direct runoff hydrograph due to $3 \text{ cm}$ of rainfall-excess in $4 \text{ hours}$ from the watershed in $\text{m}^3 \text{ s}^{-1}$ is
- The ordinates of a 6-hour S-hydrograph of a catchment are given in Table below.
The catchment has phi-index of $0.25 \text{ cm h}^{-1}$ and baseflow of $10.5 \text{ m}^3 \text{ s}^{-1}$. The peak of the flood hydrograph generated from this catchment due to a storm of 45 mm received during the first 6 h in $\text{m}^3 \text{ s}^{-1}$ isTime (h) 0 6 12 18 24 30 36 42 48 Ordinate ($\text{m}^3 \text{ s}^{-1}$) 0 30 90 180 252 306 342 360 360 The peak of a flood hydrograph due to a 5-hour storm is 670 $m^3$ $s^{-1}$. The total depth of rainfall is 9 cm. Assuming an average infiltration loss of 0.2 cm $h^{-1}$ and a constant baseflow of 30 $m^3$ $s^{-1}$, the peak discharge of the 5-hour unit hydrograph for this catchment in $m^3$ $s^{-1}$ is ________________.
