[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 [*Math: \sqrt[5]{\frac{30^5 \times 15,555}{7575}] = 34.64 inches. Now assume a diameter of 40 in, and repeat the calculations. Then head necessary to produce velocity [*Math: = \frac{15,555^2}{215 \times 40^4}] = 0.044 ft, and head to overcome friction = [*Math: \frac{15,555^2 \times 2042}{240 \times 40^5}] = 20.104 ft Then 0.044 + 20.104 = 20.148, say 20.15 ft, and the true flow will therefore be about [*Math: 15,555 * \sqrt{\frac{20.42}{20.15}}] = 15,659 gallons, and the requisite diameter about [*Math: \sqrt[5]{\frac{40^5 * 15,555}{15,659}}] = 39.94 inches. When, therefore, a 30 in diameter pipe is assumed, a diameter of 34.64 in is shown to be required, and when 40 in is assumed 39.94 in is indicated. Let _a_ = difference between the two assumed diameters. _b_ = increase found over lower diameter. _c_ = decrease found under greater diameter. _d_ = lower assumed diameter. Then true diameter = [*Math: d + \frac{ab}{b+c} = 30 + \frac{10 \times 4.64}{4.64+0.06} = 30 + \frac{46.4}{4.7} = 39.872], or, say, 40 in, which equals the required diameter. A simpler way of arriving at the size would be to calculate it by Santo Crimp's formula for sewer discharge, namely, velocity in feet per second = [*Math: 124 \sqrt[3]{R^2} \sqrt{S}], where R equals hydraulic mean depth in feet, and S = the ratio of fall to length; the fall being taken as the difference in level between the sewage and the sea after allowance has been made for the differing densities. In this case the fall is 20.42 ft in a length of 6,126 ft, which gives a gradient of 1 in 300. The hydraulic mean depth equals [*Math: \frac{d}{4}]; the required discharge, 2,497 cubic feet per min, equals the area, [*Math: (\frac{\pi d^2}{4})] multiplied by the velocity, therefore the velocity in feet per second = 4/(pi d^2) x 2497/60 = 2497/(15 pi d^2) and the formula then becomes 2497/(15 pi d^2) = 124 x * 3rd_root(d^2)/3rd_root(4^3*) x sqrt(1)/sqrt(300) or d^2 x 3rd_root(d^2) = 3rd_root(d^6) = (2497 x 3rd_root(16) x sqrt(300)) / (124 x 15 x 3.14159*) or (8 x log d)/3 = log 2497 + (1/3 x log 16) + (* x log 300) - log 124 - log 15 - log 3.14159; or log d = 3/8 (3.397419 + 0.401373 + 1.238561 - 2.093422 - 1.176091 - 0.497150) = 3/8 (1.270690) = 0.476509. * d = 2.9958* feet = 35.9496, say 36 inches. As it happens, this could have been obtained direct from the tables where the discharge of a 36 in pipe at a gradient of 1 in 300 = 2,506 cubic feet per minute, as against 2,497 cubic feet required, but the above shows the method of working when the figures in the tables do not agree with those relating to the particular case in hand.