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The maximum flow rate through a bed of filer media (''Q<sub>max</sub>'') may be calculated:
The maximum flow rate through a bed of filer media (''Q<sub>max</sub>'') may be calculated:
<math>Q_{max}=K_{m}\times A_{p}\times \left (\frac{\sum d}{d_{m}}  \right )\times 3.6 \times 10^{-3}</math>
<math>Q_{max}=\frac{K_{m}\times A_{p}\times \left (\frac{\sum d}{d_{m}}  \right )}{3600}</math>
{{Plainlist|1=Where:
{{Plainlist|1=Where:
*''K<sub>m</sub>'' is the hydraulic conductivity of the filter media (mm/hr),  
*''K<sub>m</sub>'' is the hydraulic conductivity of the filter media (mm/hr),  
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===Example calculation===
===Example calculation===
A [[stormwater planter]] with footprint of 8 x 1.5 m is planned to received runoff from an adjacent rooftop. The initial design for the planters includes 750 mm depth of filter medium, 50 mm rock mulch, and a further ponding of 300 mm. The underdrain pipe will be embedded into high performance bedding or similar, with a strip of geotextile over the top to prevent migration of the filter media into the pipe. The lab test states that the medium has a hydraulic conductivity of 25 mm/hr. The maximum flow through the medium will be calculated and a comparison made with the maximum flow through the pipe to see....  
A [[stormwater planter]] with footprint of 8 x 1.5 m is planned to received runoff from an adjacent rooftop. The initial design for the planters includes 750 mm depth of filter medium, 50 mm rock mulch, and a further ponding of 300 mm. The underdrain pipe will be embedded into high performance bedding or similar, with a strip of geotextile over the top to prevent migration of the filter media into the pipe. The lab test states that the medium has a hydraulic conductivity of 25 mm/hr. The maximum flow through the medium will be calculated and a comparison made with the maximum flow through the pipe to see....  
 
<math>Q_{max}=\frac{25 mm/hr\times 12\ m^{2}\times \left (\frac{1.1\ m}{0.75\ m}  \right )}{3600\ s/hr}</math>
<math>Q_{max}=25 mm/hr\times 12\ m^{2}\times \left (\frac{1.1\ m}{0.75\ m}  \right )\times 3.6 \times 10^{-3}</math>

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