BOD entering T.F = 0.67 x 400 = 268 mg/L
{"Image DocumentName": "BOD entering T.F = 0.67 x 400 = 268 mg/L", "Image ImageDescription": "BOD load to the trickling filter = 10,000 x 268 x 10-6 x103 = 2680 kg BOD/d. The required efficiency of the T.F is: E = ( ) / 268 = = 81.3% Go to figure (), the 81.3% efficiency can be achieved by following combinations: R = 1.0 BOD load = 500 g BOD/m3.d. R = 2.0 BOD load = 620 g BOD/m3.d. R = 3.0 BOD load = 700 g BOD/m3.d. Note: any recycle (R) less than 1.0 is not accepted to achieve 81.3% efficiency because it will give a BOD load less than 500 gBOD/m3.d while the accepted range for one stage high rate trickling filter is 500 \u2013 1500 g BOD/m3.d. So use R = 2.0 BOD load = 620 g BOD/m3.d. Volume of the filter media = Daily BOD load / volumetric BOD load. = 2680 x103 / 620 = 4323 m3. The surface loading rate (Typical range m3/m2.d). Take it as 20 m3/m2.d.", "Image XResolution": "37", "Image YResolution": "37", "Image ResolutionUnit": "Pixels/Centimeter", "Image Software": "SlidePlayer", "Image YCbCrPositioning": "Centered"}