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===Simple===
===Simple===
Five percent of the average annual yield can be estimated:
Five percent of the average annual yield can be estimated:
$$Y_{0.05} = A \times C_{vol,A}\times R_{a} \times e \times 0.05$$
<math>Y_{0.05} = A_{c} \times C_{vol,A}\times R_{a} \times e \times 0.05</math>
where:
{{plainlist|Where:
{{plainlist|
*''Y<sub>0.05</sub>'' is five percent of the average annual yield (L)
*''Y<sub>0.05</sub>'' &#61;  Five percent of the average annual yield (L)
*''A<sub>c</sub>'' is the catchment area (m<sup>2</sup>)
*''A'' &#61;  The catchment area (m<sup>2</sup>)
*''C<sub>vol, A</sub>'' is the annual runoff coefficient for the catchment
*''C<sub>vol, A</sub>'' &#61;  The annual runoff coefficient for the catchment
*''R<sub>a</sub>'' is the average annual rainfall depth (mm)
*''R<sub>a</sub>'' &#61;  The average annual rainfall depth (mm)
*''e'' is the efficiency of the pre-storage filter}}
*''e'' &#61; The efficiency of the pre-storage filter}}
*Filter efficiency (''e'') can be reasonably estimated as 0.9 pending manufacturer’s information.<br>
*Filter efficiency (''e'') can be reasonably estimated as 0.9 pending manufacturer’s information.<br>
*In a study of three sites in Ontario, STEP found the annual ''C<sub>vol, A</sub>'' of the rooftops to be around 0.8 [http://www.sustainabletechnologies.ca/wp/home/urban-runoff-green-infrastructure/low-impact-development/rainwater-harvesting/performance-evaluation-of-rainwater-harvesting-systems-toronto-ontario/]. This figure includes losses to evaporation, snow being blown off the roof, and a number of overflow events.
*In a study of three sites in Ontario, STEP found the annual ''C<sub>vol, A</sub>'' of the rooftops to be around 0.8 [http://www.sustainabletechnologies.ca/wp/home/urban-runoff-green-infrastructure/low-impact-development/rainwater-harvesting/performance-evaluation-of-rainwater-harvesting-systems-toronto-ontario/]. This figure includes losses to evaporation, snow being blown off the roof, and a number of overflow events.


Five percent of the average annual demand can be estimated:
Five percent of the average annual demand can be estimated:
$$D_{0.05} = P_{d} \times n\times 18.25$$
<math>D_{0.05} = P_{d} \times #\times 18.25</math>
Where:
{{plainlist|Where:
{{plainlist|
*''D<sub>0.05</sub>'' is five percent of the average annual demand (L)
*''D<sub>0.05</sub>'' &#61; Five percent of the average annual demand (L)
*''P<sub>d</sub>'' is the daily demand per person (L)
*''P<sub>d</sub>'' &#61; The daily demand per person (L)
*''#'' is the number of occupants}}
*''n'' &#61; The number of occupants}}
 


Then the following calculations are based upon two criteria:
Then the following calculations are based upon two criteria:
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When \(Y_{0.05}/D_{0.05}<0.33\), the storage volume required can be estimated:
When \(Y_{0.05}/D_{0.05}<0.33\), the storage volume required can be estimated:
<math>V_{S} = A \times C_{vol,E}\times R_{d} \times e</math>
<math>V_{S} = A \times C_{vol,E}\times R_{d} \times e</math>
Where:
{{plainlist|Where:
{{plainlist|
*''V<sub>S</sub>'' &#61; Storage volume required (L)
*''V<sub>S</sub>'' &#61; Storage volume required (L)
*''A'' &#61; The catchment area (m<sup>2</sup>)
*''A'' &#61; The catchment area (m<sup>2</sup>)

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