tal head loss due to friction is, therefore, 1.9 ft plus 0.1 ft plus 1.0 ft or 3.0 ft. We will disregard the friction due to bends in the pipes. This loss plus the 9-ft rise from the pump to the laterals makes about 12 ft of head loss. Finally, adding the 3 ft of pressure head needed at the holes in the laterals (for a total of 15 ft), we know that the pump selected must be able to deliver 27 gal per min at a 15-ft head. We have not included an additional 2 gal per min flow through an antisiphon hole inside the dosing chamber. If such a hole were necessary, that flow rate would have to be added to the pump delivery rate capacity needed. These are the specifications we must give the pump dealer. Figure 4.8 shows a pump curve for a typical pump. As long as the intersection of the head we need and the dosing rate we used fall below the line for this particular pump, the pump will do the job. Some pump curve charts will also have an efficiency curve. The optimum pump for our job would be one which would meet our head and delivery rate requirements while working at maximum efficiency. The last consideration is that of dosing volume. Volume per dose should be a minimum of about 10 times the volume of the distribu- tion laterals in order to encourage equal volumes of effluent exiting through each hole. A smaller dosing volume could result in over- loading the part of the system nearest the pump or manifold CONVENTIONAL SYSTEM MODIFICATIONS because of the time it takes for the distribution system to reach optimum operating pressure. For this calculation it is useful to know that the volume of a pipe is equal to the square of its inside diameter in inches times its length in feet, times the constant 0.0408. As an example of calculating the minimum dosing volume, assume: Total lateral length = 25 ft X 6 = 150 ft; Total lateral volume = (1 in.)2 X 0.0408 X 150 ft = 6.1 gal Dosing volume = 6.1 gal X 10 = 61 gal For a two-bedroom house (300 gal per day) this would be a maxi- mum dosing frequency of five times per day (300 gal