Ponds and Sedimentation Basins
The process for undertaking a life cycle costing analysis for ponds and sediment basins is the same as described in constructed+wetlands>Life-Cycle Costing - Constructed Wetlands and bioretention+systemslcc>Life-Cycle Costing - Bioretention Systems.
The origin of all of the ‘expected’ values and algorithms in MUSIC’s costing module, as well as the statistical operations used to generate ‘upper’ and ‘lower’ estimates for ponds and sediment basins are explained in Table 1.
Worked Example -
To Manually Adjust the Estimate of V for Sediment Basins and Ponds
One of the alternative algorithms in Table 7-4. allows users to estimate the typical annual maintenance cost using the size attribute V, where V is the volume of material removed from the basin / pond (in m3/year). Currently, music calculates V by adding the estimated volume of gross pollutants, coarse sediment and total suspended solids (TSS) that are trapped in the basin / pond per year.
A worked example is given below showing how to manually calculate an estimate of V that includes TSS, coarse sediment and/or gross pollutants. For example, an estimate may be required of the volume of only trapped coarse sediment and TSS, as these materials could potentially be reused.
Consider an urban catchment in Melbourne 20 ha in size with 50% impervious area that generates stormwater that is to be treated by a 194 m2 sediment basin (sized to trap 80% of the TSS load).
The load of trapped TSS is calculated by right-clicking on the basin’s treatment node icon and examining the Statistics > Mean Annual Loads section of music. In this example, the inflow load is 11,700 (kg/year) and the outflow load is 2,340 kg/year, so the trapped load is 9,360 kg/year. Using a mass to volume conversion factor of 1,800 kg/m3 for sediment, this equates to a volume of 5.20 m3/year.
Using the same procedure for gross pollutants, the inflow load is 2,550 kg/year and the outflow load is 0 kg/year (as music assumes 100% is captured). Using a mass to volume conversion factor of 260 kg/m3 for gross pollutants, this equates to a volume of 9.81 m3/year.
For coarse sediment, it is known that in gross pollutant traps that capture nearly all coarse sediment and gross pollutants, approximately 29% of the volume is sediment (on average). So the load of coarse sediment (m3/year) = the volume of trapped gross pollutants (i.e. 9.81 m3/year) x 0.4085 = 4.01 m3/year.
Now the three elements of the total trapped volume are known, the user can choose which of these should be added to estimate V.
Table 1 Summary of cost-related relationships for ponds and sediment basins.