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