Question
A soil conservation structure has an expected life of 10 years and is designed for a flood magnitude of return period 50 years. The risk of this hydrologic design in percentage (rounded off to two decimal places) is ____________.
Risk Calculation for Hydrologic Design
The risk associated with a hydrologic design represents the probability that the design event (or a more severe event) will occur at least once during the structure's intended lifespan. It is calculated based on the structure's expected life and the return period of the design flood.
Hydrologic Risk Formula
The risk ($R$) is calculated using the following formula:
$ R = 1 - \left(1 - \frac{1}{T}\right)^n $
Where:
- $R$ is the risk (expressed as a decimal).
- $T$ is the return period of the design event (in years).
- $n$ is the expected life of the structure (in years).
Applying the Formula
Given:
- Expected life, $n = 10$ years
- Return period, $T = 50$ years
Substitute the values into the formula:
- Calculate the probability of the event NOT occurring in a single year (non-exceedance probability): $ P_{NE} = 1 - \frac{1}{T} = 1 - \frac{1}{50} = 1 - 0.02 = 0.98 $
- Calculate the probability that the event does NOT occur during the entire lifespan ($n$ years): $ P_{NE\_life} = (P_{NE})^n = (0.98)^{10} $
- Calculate the risk ($R$) by subtracting the probability of non-occurrence over the lifespan from 1: $ R = 1 - P_{NE\_life} = 1 - (0.98)^{10} $
- Compute the numerical value: $ R \approx 1 - 0.8170728 $ $ R \approx 0.1829272 $
- Convert the risk to a percentage and round to two decimal places: $ \text{Risk Percentage} = R \times 100\% \approx 0.1829272 \times 100\% \approx 18.29\% $
The calculated risk is approximately 18.29%.
Was this answer helpful?
Important Questions from Precipitation Analysis
- By depth-area-duration analysis, the maximum average depth of rainfall over a $10^3\text{ km}^2$ catchment due to a 6-hour storm is $80\text{ mm}$. For the same storm, the maximum average depth of rainfall for a $10^4\text{ km}^2$ catchment, in mm, is
- In a basin, rainfall is recorded by five automatic weather stations A, B, C, D and E with respective average annual rainfall of 1020, 810, 675, 940 and 780 mm. In a particular year, the station A was non-operational and the remaining stations B, C, D and E recorded annual rainfall of 890, 725, 980 and 850 mm, respectively. The estimated rainfall at the station A in that particular year in mm is
- A catchment has eight raingauge stations. In a year, the annual rainfall recorded by the gauges (in cm) are 93.8, 106.5, 170.6, 138.7, 87.8, 156.2, 180.9 and 110.3. For a 10% error in the estimation of the mean rainfall, the optimum number of stations in the catchment is
- The normal annual rainfall for 5 rain guage stations A, B, C, D and E in a watershed were 112.7, 120.4, 118.3, 125.2 and 110.6 cm, respectively. In a particular year, the rain gauge installed at station C failed to record rainfall. In the same year the rain gauges at stations A, B, D and E recorded annual rainfall of 114.9, 118.3, 122.6 and 114.5 cm, respectively. The estimated rainfall at station C in that particular year in cm was __________
Need Expert Advice?
