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 ________________.
Hydrology Problem: Unit Hydrograph Peak Discharge Calculation
This problem involves calculating the peak discharge of a unit hydrograph given information about a storm event and the resulting flood hydrograph.
Solution Steps
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Calculate Total Infiltration Loss
First, find the total amount of water lost due to infiltration during the storm.
Total Loss = Infiltration Rate $\times$ Storm Duration
Total Loss = 0.2 cm $h^{-1}$ $\times$ 5 h = 1.0 cm
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Determine Effective Rainfall Depth
Effective rainfall is the portion of the total rainfall that contributes to direct runoff.
Effective Rainfall ($P_{eff}$) = Total Rainfall Depth - Total Infiltration Loss
$P_{eff}$ = 9 cm - 1.0 cm = 8.0 cm
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Calculate Peak Direct Runoff
The peak direct runoff is the peak flow of the flood hydrograph minus the constant baseflow.
Peak Direct Runoff ($Q_{peak, direct}$) = Peak Flood Hydrograph Discharge - Baseflow
$Q_{peak, direct}$ = 670 $m^3 s^{-1}$ - 30 $m^3 s^{-1}$ = 640 $m^3 s^{-1}$
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Calculate Peak Discharge of the Unit Hydrograph
A unit hydrograph represents the direct runoff from 1 cm of effective rainfall. To find its peak discharge ($Q_{peak, UH}$), divide the peak direct runoff by the effective rainfall depth in centimeters.
$Q_{peak, UH}$ = Peak Direct Runoff / Effective Rainfall Depth (in cm)
$Q_{peak, UH}$ = $Q_{peak, direct}$ / $P_{eff}$
$Q_{peak, UH}$ = 640 $m^3 s^{-1}$ / 8.0 cm
$Q_{peak, UH}$ = 80 $m^3 s^{-1}$
Thus, the peak discharge of the 5-hour unit hydrograph for this catchment is 80 $m^3 s^{-1}$.
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 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 ________ $m^3$ $s^{-1}$.
