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c Given flow (Q) = 12m3/day Y c Y V Y 0 ^
A hrs X
Y mrea of the reactor (m) =
X = 1m2 Shape of the reactor : Rectangular unit of net dimensions 1x1 m , free body to be provided can be 0.2 m
Design of GLS Y Ênclination angle A 0 Y Èolume of the GLS = of the total volume of the reactor Y ffective volume of the reactor = area × depth = 1m2 × 2.2 m = 2.2 m2 Y Èolume of GLS = × 2.2 m3 =0.11 m3 (×A sides ) =0.AA m3 Depth to compensate volume of GLS
^
= 0.AA m3=
0.AAm
1 m2 Overall depth of the reactor = 2m + 0.2 m +0.A m = 2.6A m
