he under- drain is the same too, except that the effluent is fed into a recircula- tion tank instead of being pumped to a disposal area. As is shown in Figure 6.27, this collected effluent is periodically pumped back to the sand filter surface. The pumping frequency is controlled by a time clock mechanism. If the liquid level in the recirculation tank is low, the float valve will be open, and the effluent pumped from beneath the sand will flow back into the recirculation tank. But if the level in the recirculation tank is high, the float valve will be closed and the effluent will bypass the recirculation tank and be pumped directly to the disposal area. Mechanisms other than the float valve CONVENTIONAL SYSTEM MODIFICATIONS Discharge Filter Floating Ball Wire Basket FLOAT VALVE Discharge Filter SPLITTER BOX Figure 6.28 Proportional flow controls. can be use to portion out the amount of effluent to be recirculated (Figure 6.28). Each dose to the sand filter surface should flood the filter to a depth of about 2 to 3 in. (0.17 to 0.25 ft). The volume necessary to do this (the dosing volume) is calculated by multiplying the sand filter area by the depth of flooding. For example, if the sand chosen has an effective size of 0.80 mm, the loading rate is 3.8 gpd/ft2 (from Table 6.8). If the discharge is 300 gpd, then the filter area must be 300 gpd 3.8 gpd/ft2 = 79 ft2 Then multiplying 79 ft2 X 0.17 ft = 13.4 f ft3, and converting ft3 to gal, 13.4 ft3 X 7.5 gal/ft3 = 100.7 gal ONSITE WASTEWATER DISPOSAL The length of time the clock must be set to run the pump in each cycle is found by dividing the effluent volume per cycle by the flow rate that the pump provides. Since the pump must be able to empty the recirculation tank in 20 min or less, and since the recirculation volume is normally about half the volume of the septic tank (or a minimum of one day's flow from the house), we can calculate a flow rate for the pump. For example, if we assume a 1000-gal septic tank, then the recirculation tank would be about 500 gal and 