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]