Time (h) 0 1 2 3 4 5 6 Discharge ($m^3$ $s^{-1}$) 1 7 26 37 27 13 1
Hydrology: Unit Hydrograph Peak Calculation
This solution determines the peak discharge and its time of occurrence for a 1-hour unit hydrograph based on provided discharge data, excess rainfall, and base flow.
Given Data
- Base Flow: 1 $m^3s^{-1}$
- Excess Rainfall (0-1 h): 2 cm
- Excess Rainfall (1-2 h): 3 cm
- Observed Discharge ($m^3s^{-1}$): [1, 7, 26, 37, 27, 13, 1] at times [0, 1, 2, 3, 4, 5, 6] hours.
Step 1: Calculate Direct Runoff (DR)
Subtract the constant base flow from the observed discharge for each time step.
| Time (h) | Observed Discharge ($m^3s^{-1}$) | Base Flow ($m^3s^{-1}$) | Direct Runoff (DR) ($m^3s^{-1}$) |
|---|---|---|---|
| 0 | 1 | 1 | 0 |
| 1 | 7 | 1 | 6 |
| 2 | 26 | 1 | 25 |
| 3 | 37 | 1 | 36 |
| 4 | 27 | 1 | 26 |
| 5 | 13 | 1 | 12 |
| 6 | 1 | 1 | 0 |
Step 2: Apply Unit Hydrograph Convolution
The direct runoff ($DR$) is generated by two successive rainfall bursts. Using the 1-hour unit hydrograph ($UH_1$), the relationship is:
$DR(t) = R_1 \times UH_1(t) + R_2 \times UH_1(t-1)$
Where $R_1 = 2$ cm and $R_2 = 3$ cm are the excess rainfall amounts in the first and second hours, respectively.
Step 3: Solve for Unit Hydrograph Ordinates
Calculate the $UH_1$ values step-by-step:
- At t=1 h: $DR(1)=6$. Since $UH_1(0)=0$, $6 = 2 \times UH_1(1)$. Thus, $UH_1(1) = 3 \; m^3s^{-1}$.
- At t=2 h: $DR(2)=25$. $25 = 2 \times UH_1(2) + 3 \times UH_1(1) = 2 \times UH_1(2) + 3 \times 3$. Solving yields $2 \times UH_1(2) = 16$, so $UH_1(2) = 8 \; m^3s^{-1}$.
- At t=3 h: $DR(3)=36$. $36 = 2 \times UH_1(3) + 3 \times UH_1(2) = 2 \times UH_1(3) + 3 \times 8$. Solving yields $2 \times UH_1(3) = 12$, so $UH_1(3) = 6 \; m^3s^{-1}$.
- At t=4 h: $DR(4)=26$. $26 = 2 \times UH_1(4) + 3 \times UH_1(3) = 2 \times UH_1(4) + 3 \times 6$. Solving yields $2 \times UH_1(4) = 8$, so $UH_1(4) = 4 \; m^3s^{-1}$.
- At t=5 h: $DR(5)=12$. $12 = 2 \times UH_1(5) + 3 \times UH_1(4) = 2 \times UH_1(5) + 3 \times 4$. Solving yields $2 \times UH_1(5) = 0$, so $UH_1(5) = 0 \; m^3s^{-1}$.
The calculated 1-hour unit hydrograph ordinates ($m^3s^{-1}$) are approximately: [0, 3, 8, 6, 4, 0, ...].
Step 4: Identify Peak Discharge and Time
From the calculated unit hydrograph ordinates:
- The peak discharge is the maximum ordinate, which is 8 $m^3s^{-1}$.
- The time of occurrence for this peak is 2 hours.
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 6-hour unit hydrograph of a watershed is represented by an isosceles triangle with the peak of 180 $m^3$ $s^{-1}$ and time to peak of 18 hours. The phi-index ($\emptyset$) of this watershed is 3.0 mm $h^{-1}$ and the constant baseflow is 20 $m^3$ $s^{-1}$. The accumulated rainfall received in this watershed at 6 h and 12 h from the start of the storm are 38 mm and 106 mm, respectively. The resulting peak of the flood hydrograph due to this storm event is ________ $m^3$ $s^{-1}$.
