Question
For a catchment with an area of 400 $km^2$, the equivalent discharge of the S-curve obtained by summation of 4-h unit hydrograph in $m^3 s^{-1}$ is
The correct answer is
278
Understanding S-Curve Discharge Calculation
The S-curve hydrograph represents the cumulative runoff from a storm with continuous rainfall excess. Its peak discharge rate signifies the steady discharge achieved when the rate of rainfall excess is constant over the catchment.
Key Parameters
- Catchment Area ($A$): 400 $km^2$
- Unit Hydrograph Duration ($t_r$): 4 hours
- Unit Depth ($D$): Assumed 1 cm = 0.01 m
Hydrological Calculation Steps
- Convert the catchment area from $km^2$ to $m^2$: $A = 400 \, km^2 = 400 \times (1000 \, m)^2 = 400 \times 10^6 \, m^2$
- Convert the unit hydrograph duration from hours to seconds: $t_r = 4 \, \text{hours} = 4 \times 3600 \, s = 14400 \, s$
- Calculate the total volume of runoff corresponding to the unit depth (1 cm) over the catchment area: $V = A \times D = (400 \times 10^6 \, m^2) \times (0.01 \, m) = 4 \times 10^6 \, m^3$
- Determine the equivalent S-curve discharge rate ($Q_S$). This is the rate at which the calculated volume ($V$) would be discharged over the unit duration ($t_r$): $Q_S = \frac{V}{t_r} = \frac{4 \times 10^6 \, m^3}{14400 \, s}$
- Simplify the calculation: $Q_S = \frac{40000}{144} \, m^3 s^{-1} = \frac{2500}{9} \, m^3 s^{-1}$
- Compute the final discharge value: $Q_S \approx 277.78 \, m^3 s^{-1}$
Result
Rounding the computed discharge to the nearest whole number yields 278 $m^3 s^{-1}$.
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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 ________ $m^3$ $s^{-1}$.
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