er in.; 58 ONSITE WASTEWATER DISPOSAL then 192 60 = 3.2 gal per min. Naturally, if one day's accumulation of wastewater is delivered to a drain field in one dose, all of it will not be absorbed immediately. Therefore, if there is to be no seepage to the ground surface, ade- quate storage volume must be available in the trench to hold the wastewater until it can filter into the soil. If the drain field is con- structed according to recommendations in this book, there should be sufficient storage volume in the trench available to hold 100% of the total daily wastewater flow. An example of a typical dosing system used to supply an absorp- tion trench follows. Example 3 If we assume: (a) 2-bedroom house, 300 gpd; (b) 60 min per in. percolation rate; (c) 3-ft inside diameter dosing chamber; and (d) 12-in. high pump inlet; then: (a) reserve capacity = 1 day's flow = 300 gal; (b) alarm capacity = 50 gal; (c) 1 dose = 192 gal (as above); and (d) permanent reservoir = (36 in.)2 X 0.0408 X 1 ft = 53 gal. Therefore, the total volume of the chamber should be: 300 gal (1 day's reserve) + 50 gal (alarm capacity) + 192 gal (1 dose) + 53 gal (permanent reservoir) = 595 gal. The length of a 3-ft inside diameter storm drainpipe needed to hold 595 gal would be: 595 gal (36 in.)2 X 0.0408 = 11.25 ft = 11 ft 3 in. WASTE PROCESS A 5-ft diameter culvert would have to be only 4 ft long. The distance between the "off" and "on" positions of the pump switch determines the volume per dose. That distance is calculated as follows: (a) 3-ft inside diameter dosing chamber; (b) 192 gal dose; Distance = 192 gal = 3.6 ft = 3 ft 7 in. (36 in.)2 0.0408 If the dosing chamber is rectangular, simply find volumes by mul- tiplying the inside dimensions: length (ft) X width (ft) X height (ft) = volume (ft³). Then find gallons by multiplying: ft3 X 7.48 = gal. The height for any given volume is found by: volume (gal) height (ft) = length (ft) X width (ft) X 7.48 For example, to find the height of 192 gal of liquid in a 3-ft by 5-ft dosing tank (inside dime