Logo

AskSia

Plus

The rainfall hyetograph for a storm over Sydney is shown below in Table 2. Given...
May 1, 2024
The rainfall hyetograph for a storm over Sydney is shown below in Table 2. Given the storm duration was 30 minutes, calculate the approximate Annual Exceedance Probability for the event. Table 3 gives design rainfall Intensity-Frequency-Duration (IFD) data from the 2016 IFDs from the Bureau of Meteorology for Sydney. [2 marks] Table 2. Recorded Rainfall (mm) \begin{tabular}{|c|c|} \hline Time (minutes) & Rainfall (mm/period) \\ \hline 0−50-5 & 16.7 \\ \hline 5−105-10 & 0.5 \\ \hline 10−1510-15 & 1 \\ \hline 15−2015-20 & 7.1 \\ \hline 20−2520-25 & 10 \\ \hline 25−3025-30 & 4 \\ \hline \end{tabular} Table 3. Design Intensity-Frequency-Duration Rainfall Data (depth in mm\mathrm{mm} ) for Sydney \begin{tabular}{|c|c|c|c|c|c|c|c|} \hline \multirow{2}{*}{ Duration } & \multicolumn{7}{|c|}{ Annual Exceedance Probability (AEP) } \\ \hline & 63.2%63.2 \% & 50%\mathbf{5 0 \%} & 20%20 \% & 10%10 \% & 5%5 \% & 2%2 \% & 1%1 \% \\ \hline 5min⁡5 \min & 8.23 & 9.24 & 12.4 & 14.6 & 16.7 & 19.6 & 21.8 \\ \hline 10 min10 \mathrm{~min} & 13.0 & 14.6 & 19.8 & 23.3 & 26.7 & 31.3 & 34.8 \\ \hline 15 min & 16.2 & 18.3 & 24.7 & 29.1 & 33.4 & 39.1 & 43.4 \\ \hline 30 min30 \mathrm{~min} & 22.2 & 25.0 & 33.7 & 39.6 & 45.5 & 53.3 & 59.2 \\ \hline 1 hour & 28.8 & 32.3 & 43.4 & 51.0 & 58.5 & 68.7 & 76.7 \\ \hline 2 hour & 36.7 & 41.1 & 55.0 & 64.8 & 74.6 & 88.0 & 98.6 \\ \hline 3 hour & 42.5 & 47.5 & 63.7 & 75.2 & 86.8 & 103 & 116 \\ \hline 6 hour & 55.5 & 62.1 & 83.8 & 99.5 & 116 & 138 & 156 \\ \hline 12 hour & 73.9 & 83.0 & 114 & 136 & 159 & 191 & 216 \\ \hline 24 hour & 98.6 & 112 & 155 & 187 & 220 & 265 & 300 \\ \hline \end{tabular} a) Estimate the maximum AEP for embedded storms of 5 minutes, 10 minutes and 15 minutes that occurred within the 30-minute storm from part a). [3 marks]
Solution by Steps
step 1
Calculate the total rainfall by adding the rainfall amounts for each period
step 2
Total rainfall =16.7+0.5+1+7.1+10+4= 16.7 + 0.5 + 1 + 7.1 + 10 + 4
step 3
Total rainfall =39.3= 39.3 mm
step 4
Calculate the average rainfall intensity by dividing the total rainfall by the storm duration
step 5
Average rainfall intensity =total rainfallstorm duration=39.3 mm30 minutes= \frac{total \ rainfall}{storm \ duration} = \frac{39.3 \ mm}{30 \ minutes}
step 6
Average rainfall intensity =1.31= 1.31 mm/minute
step 7
Use Table 3 to find the AEP for the average rainfall intensity of 1.31 mm/minute for a 30-minute duration
step 8
The AEP closest to 1.31 mm/minute for a 30-minute duration is between 63.2%63.2\% and 50%50\% since 22.2 mm (for 63.2%63.2\% AEP) is less than 39.3 mm and 25.0 mm (for 50%50\% AEP) is more than 39.3 mm
step 9
Therefore, the approximate AEP for the event is between 63.2%63.2\% and 50%50\%
Answer
The approximate Annual Exceedance Probability (AEP) for the event is between 63.2%63.2\% and 50%50\%.
Key Concept
Average Rainfall Intensity and AEP
Explanation
The average rainfall intensity is calculated by dividing the total rainfall by the duration of the storm. The AEP is then estimated by comparing the average rainfall intensity to standard values given for specific durations and AEPs.
Solution by Steps for Part a)
step 1
Identify the maximum rainfall for each embedded storm duration of 5, 10, and 15 minutes from Table 2
step 2
Maximum rainfall for 5 minutes is 16.7 mm, for 10 minutes is 16.7+0.5=17.216.7 + 0.5 = 17.2 mm, and for 15 minutes is 17.2+1=18.217.2 + 1 = 18.2 mm
step 3
Use Table 3 to find the AEP for each maximum rainfall amount for the corresponding durations
step 4
For 5 minutes, 16.7 mm corresponds to an AEP of 5%5\%
step 5
For 10 minutes, 17.2 mm is between the AEP values for 63.2%63.2\% (13.0 mm) and 50%50\% (14.6 mm)
step 6
For 15 minutes, 18.2 mm is between the AEP values for 63.2%63.2\% (16.2 mm) and 50%50\% (18.3 mm)
step 7
Therefore, the maximum AEP for embedded storms of 5 minutes is 5%5\%, and for 10 and 15 minutes, it is between 63.2%63.2\% and 50%50\%
Answer for Part a)
The maximum AEP for embedded storms of 5 minutes is 5%5\%, and for 10 and 15 minutes, it is between 63.2%63.2\% and 50%50\%.
Key Concept for Part a)
Embedded Storms and AEP
Explanation for Part a)
For embedded storms, the maximum rainfall within the given duration is compared to the IFD data to estimate the AEP for each duration.
If the storm fell over a 100-hectare catchment calculate the total runoff volume from the event assuming a continuing loss of 1.2mm/hr and no initial loss.
Solution by Steps
step 1
Convert the area from hectares to square meters
step 2
1 hectare is equal to 10,000 m210,000 \ m^2. Therefore, 100 hectares=100×10,000 m2=1,000,000 m2100 \ hectares = 100 \times 10,000 \ m^2 = 1,000,000 \ m^2
step 3
Calculate the continuing loss in meters
step 4
Since the continuing loss is 1.2 mm/hr1.2 \ mm/hr, we convert it to meters by dividing by 10001000: 1.2 mm/hr=1.21000 m/hr=0.0012 m/hr1.2 \ mm/hr = \frac{1.2}{1000} \ m/hr = 0.0012 \ m/hr
step 5
Integrate the continuing loss over the time period of 1 hour
step 6
The total loss over 1 hour is 0.0012 m/hr×1 hr=0.0012 m0.0012 \ m/hr \times 1 \ hr = 0.0012 \ m
step 7
Calculate the total runoff volume
step 8
Total runoff volume =(Area−Continuing loss×Time)×Rainfall intensity= (Area - Continuing \ loss \times Time) \times Rainfall \ intensity. Since there is no initial loss and the rainfall intensity is not given, we assume it to be the same as the continuing loss for calculation purposes
step 9
Total runoff volume =(1,000,000 m2−0.0012 m×1 hr)×0.0012 m/hr= (1,000,000 \ m^2 - 0.0012 \ m \times 1 \ hr) \times 0.0012 \ m/hr
step 10
Simplify the expression to find the total runoff volume
step 11
Total runoff volume =(1,000,000 m2−0.0012 m2)×0.0012 m/hr= (1,000,000 \ m^2 - 0.0012 \ m^2) \times 0.0012 \ m/hr
step 12
Total runoff volume =(999,999.9988 m2)×0.0012 m/hr= (999,999.9988 \ m^2) \times 0.0012 \ m/hr
step 13
Total runoff volume =1199.99999856 m3= 1199.99999856 \ m^3
Answer
The total runoff volume from the event is approximately 1200 m31200 \ m^3.
Key Concept
Runoff volume calculation
Explanation
The total runoff volume is calculated by multiplying the catchment area by the rainfall intensity, subtracting any continuing losses over the time period of the storm event.
If the catchment area was 60 km2, would you expect the catchment average rainfall to be larger or smaller than catchment average rainfall over the 100-hectare catchment? Explain your answer.
Solution by Steps
step 1
Convert the catchment areas into the same units to compare them directly
step 2
According to the asksia-ll calculation, 60 km^2 is equivalent to 6000 hectares
step 3
The asksia-ll calculation also shows that 100 hectares is equivalent to 1 km^2
step 4
Compare the two catchment areas: 6000 hectares (60 km^2) and 100 hectares (1 km^2)
step 5
Since 6000 hectares is larger than 100 hectares, the catchment area of 60 km^2 is larger than the catchment area of 100 hectares
Answer
The catchment average rainfall over a 60 km^2 area would be expected to be smaller than the catchment average rainfall over a 100-hectare catchment, assuming uniform rainfall distribution, because the larger area would likely have more variation in rainfall.
Key Concept
Catchment Area and Average Rainfall
Explanation
The average rainfall over a catchment area is not solely dependent on the size of the catchment but also on the distribution of rainfall. However, in a larger catchment, there is a higher chance of variation in rainfall, which could lead to a lower average when compared to a smaller, more uniform area.
© 2023 AskSia.AI all rights reserved