Understanding Unit Hydrograph Relationships
A unit hydrograph (UH) represents a catchment's response to a unit of effective rainfall. For triangular unit hydrographs of the same duration and base width, the peak discharge is directly proportional to the catchment area.
The relationship can be expressed as:
$ \frac{Q_{p2}}{Q_{p1}} = \frac{A_2}{A_1} $
Where:
- $Q_{p1}$ is the peak discharge of the first unit hydrograph ($UH_1$).
- $Q_{p2}$ is the peak discharge of the second unit hydrograph ($UH_2$).
- $A_1$ is the catchment area corresponding to $UH_1$.
- $A_2$ is the catchment area corresponding to $UH_2$.
Calculating the Second Catchment Area ($A_2$)
We are given:
- $A_1 = 400 \, km^2$
- $Q_{p1} = 60 \, m^3 s^{-1}$
- $Q_{p2} = 90 \, m^3 s^{-1}$
Substitute these values into the proportion:
$ \frac{90 \, m^3 s^{-1}}{60 \, m^3 s^{-1}} = \frac{A_2}{400 \, km^2} $
Simplify the ratio of peak discharges:
$ 1.5 = \frac{A_2}{400 \, km^2} $
Solve for $A_2$:
$ A_2 = 1.5 \times 400 \, km^2 $
$ A_2 = 600 \, km^2 $
Therefore, the catchment area corresponding to $UH_2$ is 600 $km^2$.
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 ________ $m^3$ $s^{-1}$.
