The Sewerage of Sea Coast Towns Part 9

[*Math: $15,555 * sqrt{frac{20.42}{86.11} = 7574.8$]

[Ill.u.s.tration: FIG 34 DIAGRAM ILl.u.s.tRATING 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 a.s.sume 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 a.s.sumed, a diameter of 34.64 in is shown to be required, and when 40 in is a.s.sumed 39.94 in is indicated.

Let _a_ = difference between the two a.s.sumed diameters. _b_ = increase found over lower diameter. _c_ = decrease found under greater diameter. _d_ = lower a.s.sumed 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.

This result differs somewhat from the one previously obtained, but there remains a third method, which we can now make trial of--namely, Saph and Schoder's formula for the discharge of water mains, V = 174 3rd_root(R^2) x S^.51*. Subst.i.tuting values similar to those taken previously, this formula can be written

2497/(15 pi d^2) = 174 x 3rd_root(d_2)/3rd_root(4^2) x 1^.51/300^.51

or d^2 x 3rd_root(d^2) = 3rd_root(d^6) = (2497 x 3rd_root(16) x 300^.51) / (174 x 15 x 3.14159)

or* log d = 3/8 (3.397419 + 0.401373 + (54 x 2.477121) - 2.240549 - 1.176091 - 0.497150) = 3/8 (1.222647) = 0.458493

* d = 2.874* feet = 34.388 say 34 1/2 inches.

By Neville's general formula the velocity in feet per second = 140 SQRT(RS)-11(RS)^(1/3) or, a.s.suming a diameter of 37 inches,

V = 140 X SQRT(37/(12 x 4) x 1/300) - 11 (37/(12x4x300))^(1/3)

= 140 x SQRT(37/14400) - 11 (37/1440)^(1/3)

= 7.09660 - 1.50656 = 5.59 feet per second.

Discharge = area x velocity; therefore, the discharge in cubic feet per minute

= 5.59 x 60 x (3.14159 x 37^2)/(4*12^2) = 2504 compared with

2,497 c.f.m, required, showing that if this formula is used the pipe should be 37 in diameter.

The four formulae, therefore, give different results, as follows:--

Hawksley = 40 in Neville = 37 in Santo Crimp = 36 in Saph and Schoder = 34-1/2 in

The circ.u.mstances of the case would probably be met by constructing the outfall 36 in in diameter.

It is very rarely desirable to fix a flap-valve at the end of a sea outfall pipe, as it forms a serious obstruction to the flow of the sewage, amounting, in one case the writer investigated, to a loss of eight-ninths of the available head; the head was exceptionally small, and the flap valve practically absorbed it all. The only advantage in using a flap valve occurs when the pipe is directly connected with a tank sewer below the level of high water, in which case, if the sea water were allowed to enter, it would not only occupy s.p.a.ce required for storing sewage, but it would act on the sewage and speedily start decomposition, with the consequent emission of objectionable odours. If there is any probability of sand drifting over the mouth of the outfall pipe, the latter will keep free much better if there is no valve. Schemes have been suggested in which it was proposed to utilise a flap valve on the outlet so as to render the discharge of the sewage automatic. That is to say, the sewage was proposed to be collected in a reservoir at the head of, and directly connected to, the outfall pipe, at the outlet end of which a flap valve was to be fixed. During high water the mouth of the outfall would be closed, so that sewage would collect in the pipes, and in the reservoir beyond; then when the tide had fallen such a distance that its level was below the level of the sewage, the flap valve would open, and the sewage flow out until the tide rose and closed the valve. There are several objections to this arrangement. First of all, a flap valve under such conditions would not remain watertight, unless it were attended to almost every day, which is, of course, impracticable when the outlet is below water. As the valve would open when the sea fell to a certain level and remain open during the time it was below that level, the period of discharge would vary from, say, two hours at neap tides to about four hours at springs; and if the two hours were sufficient, the four hours would be unnecessary. Then the sewage would not only be running out and hanging about during dead water at low tide, but before that time it would be carried in one direction, and after that time in the other direction; so that it would be spread out in all quarters around the outfall, instead of being carried direct out to sea beyond chance of return, as would be the case in a well- designed scheme.

When opening the valve in the reservoir, or other chamber, to allow the sewage to flow through the outfall pipe, care should be taken to open it at a slow rate so as to prevent damage by concussion when the escaping sewage meets the sea water standing in the lower portion of the pipes. When there is considerable difference of level between the reservoir and the sea, and the valve is opened somewhat quickly, the sewage as it enters the sea will create a "water-spout," which may reach to a considerable height, and which draws undesirable attention to the fact that the sewage is then being turned into the sea.