for the example system of 75 ft X 2.05 ft of head = 1.5 ft head 100 ft Friction loss from the laterals would be the friction from six 40-ft, 1-in. diameter laterals, each with a flow rate of 34.6 6 gpm = 5.8 gpm From Table 4.4, we find that at six gal per min, 1-in. pipe will add 2.14 ft of head per 100 ft of pipe. The total friction would then be 40 ft X 2.14 ft of head X 6 laterals = 5.2 ft of head 100 ft The head loss in going from a 2-in. manifold to the smaller 1-in. lateral is about 0.03 ft of head per reduction fitting because of the flow into smaller diameter pipe, plus a few more hundredths of a foot due to the change in direction of flow. If you added up the friction from all the fittings, the total might be significant. But the friction losses are not additive in a system of parallel pipes. When you add another lateral, the water flow rates in all the laterals are decreased, and the lower the flow rate, the lower the friction loss (Table 4.4). What all this means is that we can usually disregard friction from fittings in parallel distribution systems with flow rates in the range we've been discussing. The pump will have to overcome the elevation head, the required pressure head in the laterals, and the friction head. For example, assume that the elevation head is 9 ft. The pressure head is usually set at 3 ft, and we have found, in our example, a friction head of 1.5 ft (supply and manifold pipes) plus 2.1 ft (laterals). Therefore, the total head to be delivered by the pump is 15.6 ft. We would, for our example system, like to have a pump that could deliver 35 gal per min at a pressure head of 16 ft with reasonably high efficiency. There are several ways to get by with a smaller pump: (1) reduce the pressure at the distribution holes from 3 ft to the 2-ft minimum; 150 ONSITE WASTEWATER DISPOSAL (2) use larger diameter pipe; (3) reduce the hole size to the one- eighth in. minimum; and (4) raise the pump in the dosing chamber to reduce elevation head. Theoretically, to lift water 15 ft and deliver 35 gal