Sea water has a specific gravity of 1.027, and is usually taken as weighing 64.14 lb per cubic foot, while sewage may be taken as weighing 62.45 lb per cubic foot, which is the weight of fresh water at its maximum density. Now the ratio of weight between sewage and sea water is as 1 to 1.027, so that a column of sea water l2 inches in height requires a column of fresh water 12.324, or say 12-1/3 in, to balance it; therefore, in order to ascertain the effective head producing discharge it will be necessary to add on 1/3 in for every foot in depth of the sea water over the centre of the outlet. The sea outfall should be of such diameter that the contents of the reservoir can be emptied in the specified time--say, three hours--while the pumps are working to their greatest power in pouring sewage into the reservoir during the whole of the period; so that when the valves are closed the reservoir will be empty, and its entire capacity available for storage until the valves are again opened. To take a concrete example, assume that the reservoir and outfall are constructed as shown in Fig. 34, and that it is required to know the diameter of outfall pipe when the reservoir holds 1,000,000 gallons and the whole of the pumps together, including any that may be laid down to cope with any increase of the population in the future, can deliver 600,000 gallons per hour. When the reservoir is full the top water level will be 43.00 O.D., but in order to have a margin for contingencies and to allow for the loss in head due to entry of sewage into the pipe, for friction in passing around bends, and for a slight reduction in discharging capacity of the pipe by reason of incrustation, it will be desirable to take the reservoir as full, but assume that the sewage is at the level 31.00. The head of water in the sea measured above the centre of the pipe will be 21 ft, so that [*Math: $21 \times 1/3$], or 7 in--say, 0.58 ft--must be added to the height of high water, thus reducing the effective head from 31.00 - 10.00 = 21.00 to 20.42 ft The quantity to be discharged will be [*Math: $\frac{1,000,000 + (3 * 600,000)}{3}$] = 933,333 gallons per hour = 15,555 gallons per minute, or, taking 6.23 gallons equal to 1 cubic foot, the quantity equals 2,497 cubic feet per min Assume the required diameter to be 30 in, then, by Hawksley's formula, the head necessary to produce velocity = [*Math: $\frac{Gals. per min^2}{215 \times diameter in inches^4} = \frac{15,555^2}{215 * 30^4}$] = 1.389 ft, and the head to overcome friction = [*Math: $\frac{Gals. per min^2 \times Length in yards}{240 * diameter in inches^5} = \frac{15,555^2 * 2042}{240 * 30^5}] = 84.719. Then 1.389 + 84.719 = 86.108--say, 86.11 ft; but the acutal head is 20.42 ft, and the flow varies approximately as the square root of the head, so that the true flow will be about [*Math: $15,555 * \sqrt{\frac{20.42}{86.11} = 7574.8$]