Changes

===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>

{{plainlist|Where:

{{plainlist|

*''Y_0.05'' is five percent of the average annual yield (L)

*''Y_0.05'' &#61;  Five percent of the average annual yield (L)

*''A_c'' is the catchment area (m²)

*''A'' &#61;  The catchment area (m²)

*''C_{vol, A}'' is the annual runoff coefficient for the catchment

*''C_{vol, A}'' &#61;  The annual runoff coefficient for the catchment

*''R_a'' is the average annual rainfall depth (mm)

*''R_a'' &#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_{vol, A}'' 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_{vol, A}'' 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>

{{plainlist|Where:

{{plainlist|

*''D_0.05'' is five percent of the average annual demand (L)

*''D_0.05'' &#61; Five percent of the average annual demand (L)

*''P_d'' is the daily demand per person (L)

*''P_d'' &#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:

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>

{{plainlist|Where:

{{plainlist|

*''V_S'' &#61; Storage volume required (L)

*''V_S'' &#61; Storage volume required (L)

*''A'' &#61; The catchment area (m²)

*''A'' &#61; The catchment area (m²)