 in the mound, (2) overcome the friction of the water on the sides of the pipes, and (3) provide sufficient head to squirt the effluent out of the holes in the distribution pipe at the desired pressure. The total of all of this work is the minimum head required for the pump. This calculation is completed in four easy steps: Calculate the total elevation difference (in ft) between the pump intake and the distribution system in the mound; 2. Determine the lengths of the supply pipe, manifold, and laterals, and using Table 4.4, add up the head loss due to friction (this should always be less than about 25 ft of head; if it is more, too much pump capacity is being wasted overcoming friction and larger pipe should be used throughout); 3. Establish the head needed at the distribution laterals; 4. Add these together. For example, assume there is a 9-ft rise from the pump in the dosing chamber to the laterals in the mound. As in the last example, there will be 150 ft of 1-in. lateral pipe in six 25-ft "legs." The mani- fold will be 6 ft of 2-in. PVC pipe and the supply line will be 75 ft of 2-in. PVC pipe. We have calculated the desired total flow to the distribution system as 27 gal per min. The flow entering each leg of the lateral is 27 divided by six, or 4.5 gal per min. Checking with Table 4.4 for the head loss due to friction, we find that for 4 gal per min in 1-in. pipe the head loss is 1.01 ft per 100-ft length. For a flow of 5 gal per min, that loss is 1.52 ft. Since 4.5 gal per min is about 122 ONSITE WASTEWATER DISPOSAL halfway between, estimate a head loss about halfway between the respective losses given in the table. So the head loss should be about 1.25 ft for a low of 4.5 gal per min through 100 ft of 1-in. diameter PVC. Since we've got 150 ft, the total loss would be 150 ft 1.25 ft of head = 1.9 ft of head 100 ft At 27 gal per min, 16 ft of 2-in. PVC would lose 16 ft 1.28 ft of head = 0.1 ft of head 100 ft At 27 gal per min, 75 ft of 2-in. PVC would lose 75 ft X 1.28 ft of head = 1.0 ft of head 100 ft To