Hydrology Test 2

Concept:

S curve is a hydrograph produced due to continuous effective rainfall at a constant rate for an infinite period .

The maximum ordinate of S-curve is the equilibrium discharge. It has an initial steep portion and reaches a maximum equilibrium discharge at a time equal to the time base of the first unit hydrograph. The average intensity of ER producing the S-curve is 1/D cm/h and the equilibrium discharge,

\({Q_e} = 2.778\frac{A}{D}{m^3}/s\) 

Where,

A = Area of the catchment in km²

D = Duration of rainfall in hour.

Calculation:

A = 600 km², D = 4 hour

\(\therefore {Q_e} = 2.778 \times \frac{{600}}{4} = 416.7\;{m^3}/s\) 

Concept:

→ Estimation of flood of a particular Return period is done using :

(a) Empirical Probability Distribution:

(b) Theoretical Probability distribution:

(i) Log – Pearson’s type III Method

(ii) Gumbel’s Method: Flood discharge X_T for a return period T years is statistically calculated as below,

X_T = X̅ + K_T × σ_n-1

Where,

X_T = Flood discharge for T year return period.

X̅ = Mean of annual data series.

σ_n-1 = Standard Deviation of Annual data series.

K_T = Frequency factor.

Calculation:

x̅ = 30000 m³/s, k_T = 0.2, σ = 300 m³/s

∴ X_T = X̅ + K_T σ

⇒ X₁₀₀ = 30000 + 0.2 × 300 = 30060 m³/s

Runoff depth = \({5\; hr\;\times\;Sum \;of \;DRH}\over{Area\;of\;the\;catchment}\)

\(={{5\times3600\times2625}\over{56700\times10^4}}\) = 0.0833 m = 8.33 cm

Concept:

A hydrograph is a plot between flood discharge and the duration.

If the hydrograph contains only direct runoff then it is called direct runoff hydrograph (DRH). But if it also contains the base flow then it is called flood hydrograph.

Calculation:

∴ Peak ordinate of 4 hour DRH with excess rainfall of 2 cm is 5.25 m³/s.

Concept:

If the probability of an event occurring is P, the probability of the event not occurring in a given year is q = (1 - P).

The binomial distribution can be used to find the probability of occurrence of the event r times in n successive years.

\({{\rm{P}}_{\left( {{\rm{r}},{\rm{n}}} \right)}} = {}_{}^{\rm{n}}{{\rm{C}}_{\rm{r}}}{{\rm{P}}_{\rm{r}}}{{\rm{q}}^{{\rm{n}} - {\rm{r}}}} = \frac{{{\rm{n}}!}}{{\left( {{\rm{n}} - {\rm{r}}} \right)!{\rm{r}}!}}{{\rm{P}}^{\rm{r}}}{{\rm{q}}^{{\rm{n}} - {\rm{r}}}}\)

Where

P_(r,n) = Probability of a random hydrologic event (rainfall) of given magnitude and exceedance probability P occurring r times in n successive years.

E.g.

a) The probability of an event of exceedance probability P occurring 2 times in n successive years is

\({{\rm{P}}_{\left( {2,{\rm{n}}} \right)}} = \frac{{{\rm{n}}!}}{{\left( {{\rm{n}} - 2} \right)!2!}}{{\rm{P}}^2}{{\rm{q}}^{{\rm{n}} - 2}}\)

b) The probability of the event not occurring at all in n successive years is

P_(0,n) = qⁿ = (1 - P)ⁿ

c) The probability of the event occurring at least once in n successive years

P₁ = 1 – qⁿ = 1 – (1 - P)ⁿ

Calculation:

Given: P = 1/50 = 0.02

For rainfall occurring two times in 15 successive years.

n = 15, r = 2

\({{\rm{P}}_{\left( {2,15} \right)}} = \frac{{15!}}{{13!2!}} \times {\left( {0.02} \right)^2} \times {\left( {0.98} \right)^{13}} = 15 \times \frac{{14}}{2} \times 0.0004 \times 0.769 = 0.0323 = 3.23\% \)

Concept:

Unit hydrograph (UH):

A unit hydrograph can be defined as the hydrograph of direct runoff resulting from one unit depth (1 cm) of rainfall excess occurring uniformly over the basin for a specified duration.The unit hydrograph is the unit pulse response function of a linear hydrologic system.

The unit hydrograph is a simple linear model that can be used to derive the hydrograph of any duration resulting from any amount of excess rainfall for the given area.

Peak of UH = \( \frac{{Peak\;of\;DRH}}{{Rainfall\;Excess}}\)

Calculation:

Peak of flood hydrograph = 180 m³/s

Base flow = 30 m³/s

∴ Peak of DRH = 180 – 30 = 150 m³/s

Total rainfall = 2.7 cm

Infiltration loss = 0.3 cm/hour

∴ Rainfall excess = 2.7 – 0.3 × 3 = 1.8 cm

∴ Peak of 3 – h unit hydrograph \(= \frac{{Peak\;of\;DRH}}{{Rainfall\;Excess}} = \frac{{150}}{{1.8}} = 83.33\;{m^3}/s\)