THE DISCHARGE OF SEA OUTFALL SEWERS.
The head which governs the discharge of a sea outfall pipe is
measured from the surface of the sewage in the tank, sewer, or
reservoir at the head of the outfall to the level of the sea.
As the sewage is run off the level of its surface is lowered,
and at the same time the level of the sea is constantly varying
as the tide rises and falls, so that the head is a variable
factor, and consequently the rate of discharge varies. A curve
of discharge may be plotted from calculations according to
these varying conditions, but it is not necessary; and all
requirements will be met if the discharges under certain stated
conditions are ascertained. The most important condition,
because it is the worst, is that when the level of the sea is
at high water of equinoctial spring tides and the reservoir is
practically empty.
Sea water has a specific gravity of 1.027, and is usually taken
as weighing 64.14 lb per cubic foot, while sewage may be taken
as weighing 62.45 lb per cubic foot, which is the weight of
fresh water at its maximum density. Now the ratio of weight
between sewage and sea water is as 1 to 1.027, so that a column
of sea water l2 inches in height requires a column of fresh
water 12.324, or say 12-1/3 in, to balance it; therefore, in
order to ascertain the effective head producing discharge it
will be necessary to add on 1/3 in for every foot in depth of
the sea water over the centre of the outlet.
The sea outfall should be of such diameter that the contents of
the reservoir can be emptied in the specified time--say, three
hours--while the pumps are working to their greatest power in
pouring sewage into the reservoir during the whole of the
period; so that when the valves are closed the reservoir will
be empty, and its entire capacity available for storage until
the valves are again opened.
To take a concrete example, assume that the reservoir and
outfall are constructed as shown in Fig. 34, and that it is
required to know the diameter of outfall pipe when the
reservoir holds 1,000,000 gallons and the whole of the pumps
together, including any that may be laid down to cope with any
increase of the population in the future, can deliver 600,000
gallons per hour. When the reservoir is full the top water
level will be 43.00 O.D., but in order to have a margin for
contingencies and to allow for the loss in head due to entry of
sewage into the pipe, for friction in passing around bends, and
for a slight reduction in discharging capacity of the pipe by
reason of incrustation, it will be desirable to take the
reservoir as full, but assume that the sewage is at the level
31.00. The head of water in the sea measured above the centre
of the pipe will be 21 ft, so that
[*Math: $21 \times 1/3$],
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
[*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.
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*. Substituting
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, assuming 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 formulæ, therefore, give different results, as
follows:--
Hawksley = 40 in
Neville = 37 in
Santo Crimp = 36 in
Saph and Schoder = 34-1/2 in
The circumstances 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 space 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.