Flood-frequency computations for the river Chambal at Gandhisagar dam, by using Gumbel’s method, yielded the following results:
Return period T (years) |
Peak flood (m^{3}/s) |
50 |
40,809 |
100 |
46,300 |
Estimate the flood magnitude in this river with a return period of 500 years.
(AMIE Summer 2021)
Solution
Using Gumbel's equation
\({x_T} = \bar x + K{\sigma _{n - 1}}\)
\(\begin{array}{l}{x_{100}} = \bar x + {K_{100}}{\sigma _{n - 1}}\\{x_{50}} = \bar x + {K_{50}}{\sigma _{n - 1}}\\({K_{100}} - {K_{50}}){\sigma _{n - 1}} = {x_{100}}{x_{50}}\\ = \,46300 - 40809 = 5491\end{array}\)
But,
\(\begin{array}{l}{K_T} = \frac{{{y_T}}}{{{S_n}}} - \frac{{{{\bar y}_n}}}{{{S_n}}}\\{\rm{where }}{{\rm{S}}_n}\,and\,{{\bar y}_n}\,{\rm{are constants}}{\rm{.}}\end{array}\)
\(\begin{array}{l}\therefore \,({y_{100}} - {y_{50}})\frac{{{\sigma _{n - 1}}}}{{{S_n}}} = 5491\\{\rm{Using equation}}\\{{\rm{y}}_T} = - \left[ {\ln .\ln \frac{T}{{T - 1}}} \right]\\{y_{100}} = - [\ln x\ln (100/99)] = 4.60015\\{y_{50}} = - [\ln x\ln (50/99)] = 3.90194\end{array}\)
\(\frac{{{\sigma _{n - 1}}}}{{{S_n}}} = \frac{{5491}}{{(4.60015 - 3.90194)}} = 7864\)
For T = 500 years
\(\begin{array}{l}{y_{500}} = - [\ln x\ln (500/499)] = 6.21361\\({y_{500}} - {y_{100}})\frac{{{\sigma _{n - 1}}}}{{{S_n}}} = {x_{500}} - {x_{100}}\\(6.21361 - 4.60015)x7864 = {x_{500}} - 46300\\{x_{500}} = 58988\,say\,59000\,{m^3}/s\end{array}\)
A hydraulic structure has a design life of 100 years. What is the risk involved if it is designed for a 50-year return period flood?
(AMIE Winter 2021)
Solution
Here, n = 100 years
T = 50 years and substituting these values of n and T in Eq. (8.17):
\(R = 1 - {(1 - 1/T)^n}\)
\(R = 1 - {(1 - 1/50)^{100}} = 0.867 = 86.7\% \)
A project costing Rs. 2.0 million is expected to produce the benefits of Rs. 3.0 million. Before starting the project, it is observed that the adoption of some advanced technology can increase the benefits to Rs. 3.2 million. If the consultation charge for the advanced technology is Rs. 0.5 million, state whether the adoption of advanced technology is justified?
(AMIE Winter 2020)
Solution
BC ratio without advanced technology = 3.0/2.0 = 1.5.
Incremental cost of advanced technology = 0.5 million rupees.
Incremental benefit from advanced technology = 3.2 - 3.0 = 0.2 million rupees.
B/C ratio with advanced technology = 3.27(2 4 - 0.5) = 3.272.5 = 1.28.
As the incremental cost of advanced technology is more than the incremental benefit, the adoption of advanced technology is not justified.
Further, the BC ratio without the advanced technology is higher than with advanced technology. Hence, it is advisable to continue with the original plan.
Route the following flood hydrograph through a river reach for which K = 12.0 h and x = 0.20. At the start of the inflow flood, the outflow discharge is \(10\,{m^3}/s\).
(AMIE Summer 2021)Solution
Since K = 12 h and 2 Kx = 2 x 12 x 0.2 = 4.8 h,
Δt should be such that 12 h > Δt > 4.8 h.
In the present case, Δt = 6 h is selected to suit the given inflow hydrograph ordinate interval.
Using equations
\(\begin{array}{l}{C_0} = \frac{{ - Kx + 0.5\Delta t}}{{K - Kx + 0.5\Delta t}}\\{C_1} = \frac{{Kx + 0.5\Delta t}}{{K - Kx + 0.5\Delta t}}\\{C_2} = \frac{{K - Kx - 0.5\Delta t}}{{K - Kx + 0.5\Delta }}\end{array}\)
\(\begin{array}{l}{C_0} = \frac{{ - 12x0.20 + 0.5x6}}{{12 - 12x0.2 + 0.5x6}} = 0.048\\{C_1} = \frac{{12x0.20 + 0.5x6}}{{12 - 12x0.2 + 0.5x6}} = 0.0429\\{C_2} = \frac{{12 - 12x0.2 - 0.5x6}}{{12 - 12x0.2 + 0.5x6}} = 0.523\end{array}\)
For the first time interval, 0 to 6 h
\(\begin{array}{l}{I_1} = 10.0;{C_1}{I_1} = 4.29\\{I_2} = 20.0;{C_0}{I_2} = 0.96\\{Q_1} = 10.0;{C_2}{Q_1} = 5.23\end{array}\)
Now, \({Q_2} = {C_0}{I_2} + {C_1}{I_1} + {C_2}{Q_1}\)
Putting values, we get
\({Q_2} = 10.48\,{m^3}/s\)
For the next time step, 6 to 12 h, \({Q_1} = 10.48\,{m^3}/s\). The procedure is repeated for the entire duration of the inflow hydrograph.
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