then use a smaller slope.
This
and is too small.
The slope must be
elevates the end of the sewer above
about doubled to obtain acceptable
that obtained by the first method.
values,
so
the
change
in
invert
Since in this case the second method
elevations of manholes 4 and 3 is 2
necessitates less excavation, it will
feet. This design is tabulated in row
be used.
The best way to determine
12. The sewer at manhole 3 is 1 foot
the amount of drop to be used is to
lower
than
is
needed
for
cover
work backwards using the profile and
considerations.
charts. The first step is to draw in
a tentative invert.
This invert
f. MH 3 to MH 2. The ground drops
should be drawn so that it gives
2 feet in 400 between MH 3 and 2 MH
minimum cover at the lower manhole and
which is close to the minimum slope
at the greatest slope consistent with
and so is used to design this section.
cover requirements between manholes.
It is tabulated in row 13. All along
The second step is to find the minimum
this section the sewer is 1 foot lower
slope which will give a full flow
than is needed for cover, but this
equal to the actual flow and a
foot cannot be gained back because the
velocity greater than 2 feet per
slope of the sewer would then be too
second.
In this case a flow of 329
shallow.
gallons per minute in an 8-inch pipe,
a slope of 0.45 percent is necessary
g. MH 2 to MH 1. Between MH 2 and
and a velocity of 2.25 is obtained
MH 1 the ground drops 5 feet in 400.
from figure 14.
Whichever slope is
This is a much greater slope than is
greater should be used.
In this case
needed by the sewer, so instead of
the first condition gives a greater
dropping the sewer 5 feet, it is
slope.
The design is tabulated in
dropped only 4 feet.
This gives back
rows 7 and 8. In row 7 a 6-inch pipe
the foot of elevation lost between
was tried and the minimum slope that
manholes 4 and 3. The drop may not be
could be used was 1.67.
This is too
less than 4 feet because then the
small for 6-inch pipe.
With 6-inch
sewer would be above minimum depth for
pipe, the minimum slope to get a flow
cover.
This
sewer
is
therefore
of 329 gallons per minute was 2.2
designed with a 4 foot drop in 400
percent. This might be acceptable for
feet
(1
percent
slope)
and
is
this section but certainly is not for
tabulated in row 14.
the rest of the sewer. Note that the
slope of the ground is about 1 percent
h. MH 1 to the lagoon.
The line
from MH 7 to the lagoon. Since 6-inch
from MH 1 to the lagoon is tabulated
pipe will need at least 2.2 percent
in row 15.
slope to handle the flow, the 6-inch
pipe would have to go deeper and
i. Building D to MH 7 and MH 6.
deeper.
Eight-inch pipe can handle
This house connection could have been
the flow with a slope of less than 1/2
designed as soon as the main between
percent and thus can follow roughly
manholes 7 and 6 was designed.
The
the ground slope.
For this reason a
elevation of the connection with the
change to 8-inch pipe was made at
main can now be computed. It will be
manhole 7. The design from MH 6 to MH
the elevation of the outlet invert at
5 is straight forward and is tabulated
manhole 7 less the slope of the main
in row 9, figure 17. The design from
multiplied by the distance to the
MH 5 to MH 4 is also straight forward
connection:
and is tabulated in row 10.
Notice
that a lamphole can be used instead of
Connection elevation =290.3-(1.67%x200')
manhole 5 as all the conditions for a
=290.3-(.0167x200)
lamphole are fulfilled.
=290.3-3.3
=287.0 feet
e. MH 4 to MH 3. The ground slope
from MH 4 to MH 3 is used in row 11
4-24
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