Part 9 (1/2)

[Math: 15,555sqrt{frac{2042}{8611} = 75748]

[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{305 times 15,555}{7575}] = 3464 inches

Now assume a diameter of 40 in, and repeat the calculations

Then head necessary to produce velocity

[Math: = frac{15,5552}{215 times 404}] = 0044 ft, and head to overcome friction =

[Math: frac{15,5552 times 2042}{240 times 405}]

= 20104 ft Then 0044 + 20104 = 20148, say 2015 ft, and the true floill therefore be about

[Math: 15,555sqrt{frac{2042}{2015}}]

= 15,659 gallons, and the requisite diameter about

[Math: sqrt[5]{frac{40515,555}{15,659}}]

= 3994 inches

When, therefore, a 30 in diameter pipe is assumed, a diameter of 3464 in is shown to be required, and when 40 in is assumed 3994 in is indicated

Let _a_ = difference between the two assumed diameters _b_ = increase found over lower diareater diameter _d_ = lower assumed diameter

Then true diameter =

[Math: d + frac{ab}{b+c} = 30 + frac{10 times 464}{464+006} = 30 + frac{464}{47} = 39872],

or, say, 40 in, which equals the required dia at the size would be to calculate it by Santo Crie, namely, velocity in feet per second =

[Math: 124 sqrt[3]{R2} sqrt{S}],

where R equals hydraulic th; the fall being taken as the difference in level between the sewage and the sea after allowance has beendensities In this case the fall is 2042 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,