or water loss through porous soils. Some irrigation effi- ciencies are given in Table 6.7. The result is the total daily water requirement in gallons per day per square foot. Once this value is known, divide the total actual wastewater flow by this water requirement to find out how big the plot is to be. For example, if a three-bedroom house is expected to produce 150 gal per day, the soil is loam, and fruit trees are to be irrigated, then the plot area is calculated as follows: Rainfall = 13.1 in. per season Total crop consumption = 30.4 in. per season Additional water needed = 30.4 - 13.1 = 17.3 in. per season Growing season = 120 days Crop consumption = 17.3 in. 120 days = 0.14 in./day 0.14 in./day X 0.63 = 0.091 gpd/ft2 0.091 gpd/ft2 0.85 efficiency = 0.11 gpd/ft2 150 gpd 0.11 gpd/ft2 = 1404 ft2 So a plot 35 ft by 40 ft should serve to dispose of the effluent generated in a day and provide a good crop of fruit in the process. It is normal practice to divide the plot into equally sized segments and irrigate one segment per day. This allows resting between waterings, promotes a more even distribution throughout the plot, and you can use a smaller dosing chamber. For sandy soils, the plot is divided into 7 segments; for loamy soils there should be 10 seg- ments; and for clay soils, 12 segments. The smaller number of seg- ments in the more porous soils provides for more frequent dosing. CONVENTIONAL SYSTEM MODIFICATIONS The example plot could be divided into two strips each 20 ft wide, and each of those strips could be segmented into five pieces, each 20 ft by 7 ft (140 ft2. The distribution system can be as simple as a garden hose or as expensive as a series of electronic valves on timers. Manually oper- ated valves would be another choice. Flat rocks should be used as splash pads to prevent erosion where the effluent is discharged. The dose volume is, of course, the actual daily flow from the house, and the rate of delivery should be sufficient to flood the entire segment rather than allow the majority of efflu