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: $15,555 * \sqrt{\frac{20.42}{86.11} = 7574.8$]
[Illustration: FIG 34 DIAGRAM ILLUSTRATING CALCULATIONS FOR THE
DISCHARGE OF SEA OUTFALLS]
—say 7,575 gallons. But a flow of 15,555 gallons per minute is required, as it varies approximately as the fifth power of the diameter, the requisite diameter will be about