ENGINEERING
TECHNICAL
FIELD
NOTES
DATA
INFORMATION
TECHNICAL
RETRIEVAL
REPORTS
CURRENT
TEXTS
AWARENESS
SYSTEM
Notes
Field
Volume
5 Numbers
March-April
3
and 4
1973
Design Aids for Lagoon Wastewater
Treatment Systems
William C. Kolzow
An Economic
Analysis For Flood
Risk Costs
John Jakel
JýS
FOREST
SERVICE
U.S.
DEPARTMENT
OF AGRICULTURE
ENGINEERING
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FIELD NOTES
DESIGN AIDS FOR LAGOON WASTEWATER TREATMENT
By
William
C.
Sanitary Engineer
SYSTEMS
Kolzow
Office
Washington
INTRODUCTION
The
Forest
Corrective action
water pollution
to abate
Order 11507.
Year 1974 we
In Fiscal
FY
level
of the remaining
in later
years with
pollution abatement
for water
approximately equal to that
presently underway
on some
action
sources.
pollution
Program under
Abatement
will initiate
is
in
1973.
Many
in
program
possible water
$5 million program to be followed
1600 projects through an anticipated
a requested
are
on over 200 of these projects
the current Water Pollution
or has been completed through
Executive
1800 projects that
Service has identified over
1600 plus projects involve
of the
sewage lagoons
the vernacular
the design of possible waste
unaerated
either
or aerated.
These lagoons particularly
those that are not artificially
aerated require the use of much land that could
used for multi-use purposes.
Therefore
it
is
to the design engineer
helpful
or
ponds
stabilization
be
otherwise
and/or
planner
to
in-dividual
time-savers
rapidly estimate
Figures
1
shown
in
such
through
land
6 presented
for facultative
are intended
these figures
to be able to vary
area impacts
as
lagoons
design from original
should
design and estimating aids
proposals.
estimates. The
aid in such
- not
charts
solutions to
as
situations.
The
charts
portion
presented
aerated
system design
of treatment
equations
for surface
solutions.
must apply
the lagoon designer
lagoons
figs.
The
in his
7
charts
and
8
be applied
can
graphically
work. Used
to a
one of many complex
solve
wisely they
directly
should
prove
to the designer.
The
field
will
following
and accompanying
figures
situation presents
use
the charts
as
an almost
time-saving
totally
aids
intended
write-ups are
as
only and
not
as
only. Each
design aids
different and complex problem.
The prudent
engineer
substitutes for developed competence
or initiative.
in-cluded
Editors Note
near future.
is
treatment
We
to the
solicit
WO
need for more material
a pressing
in his endeavors
field engineer
wastewater
There
systems.
to provide
We
your input
Division
intend
potable
safe sanitary
to publish
for this proposed
water
an Engineering
Report.
of Engineering.
1
in
of the type presented
Please
this
article
supplies and effective
Technical
Report on
send material
that
this
to
assist
the
efficient
subject
you would
like
in the
ESTIMA TING FA CUL TA TIVE LAGOON RETENTION TIMES
1
lagoons
when
water
depths
for
4
through
Figures
such
lagoons
35
are
expressing
for estimating the retention
presented
are
are square interior
and 9
7
The
feet.
side
charts
loading that is pounds
lagoon
time in unaerated
slopes are constructed
are
at
31
per acre
or people
and working
method used
based on the conventional
B.O.D.5
facultative
population
served per acre.
The volume
can be calculated
of any lagoon
from the equation
VDXZ-SDX-SDZ3S2D2
where
V
volume
D
liquid
V to
the following
depth
in lieu
of
ft.
lagoon
width
Z
lagoon
length
water
at
gallons
surface
water
at
embankment slope
and recognizing
ft.
surface
ratio
of horizontal
X and
that
ft.
Z
to vertical
are identical
for a square lagoon
yields
expression
V
Substituting
ft.
X
S
Converting
cu.
3 for S
X yields
31
side
7.48DX2
2SDX
-
slopes and substituting
Vgallons
7.48D
3 S2D2
term
a dimensional
43560Ah
-
A
in
surface
acres
2
3.464D
Example Problem
A waste
treatment
GPCD.
Operating
facultative
organic
is
to be designed
experiences
in
are
to serve 250 persons per day. Per capita
determined to be
rate
of 20
pounds
the net daily flow
the retention
time
0.1
lbs.
the general area of the site have
lagoon with a 3-foot operating
loading
Calculate
system
and hydraulic loadings
design organic
depth can
B.O.D.5
Q the
be
per lagoon
minimum lagoon
t.
2
surface
and 40
indicated that a
effectively
surface
B.O.D.5
operated
at
an
acre.
area required
A and
Q 40 GPCD250
A_
tfrom
fig.
1
250
persons0.1
20
lb.
persons
lb.
B.O.D.
B.O.D. per acre
111 days
3
10000 GPD
1.25 acres
Lagoon
is
Square
Working
with
Depth
31
D3
Side Slopes
Feet
100
50
40
30
14o
7
Mr
o
20
44-0
10
C
U
5
n
4
3
V
0
O
ý
2
U
-H10
HII
HHM
0.
H14
0
0.5
0.4
HAi
0.3
4V
0.2
-
0.1
I
2
3
4
5
10
t
Figure
20
30
Time
Retention
1.-Retention
50
100
200 300
Days
Time for Unaerated Facultative Lagoon
3 Feet
Working Depth D
4
1000
100
r
50
40
j
1
r
k
30
TT
1
20
LEH
77-N
FO
V
Q
0
a
w
N
FI
_t
r
4
VTý
a
Fr
Gc
ý
i
1
i
4-4
I
2
4t
rtk
I
_
tit
Q5
u
04
rr
1.
0.3
02
i
F
1
2
3
4
5
2.-Retention
t-t
20
10
t
Figure
y
30
Retention
50
Time
100
200
300
Days
Time for Unaerated Facultative Lagoon
Working Depth
D
5 Feet
5
1000
100
if-
th
50
40
T9XV
.4
4
30
h
- - -
-
-
-
----------4-H-
-
w
-
20
lk
F.
10
O
0
t
Q
O
ZZ
-
t
to
3
_
I
2
it
4
h
_
0.5
a
.
0.4
l
0.3
_r
0.2
r
H4
i
V1.1
wo
.
jjL
To
ol
I
lo
I
i
fit
Fl-ý
t.
0
S
4
-
v
2
3
4
5
IN
w
20
10
t
Figure
3.-Retention
30
50
Time
Retention
100
200
Days
Time for Unaerated Facultative Lagoon
Working Depth
D
7
Feet
6
300
1000
100
w
50
_f
T
ME
40
30
20
H
10
a
0
v
Q
M
-fl-Iff
o
0
1_4
flj
5
4
Q
0
0
c
3
2
U
TIM
w
Q
I
Eff
14
-M
0.5
0.4
0.3
OX
0.2
100
7
0
ED
P
3
4
5
1.
1
t
0.1
1
0
-
4.-Retention
fl-Sr
-HtHffl
W
U.
01
20
10
30
Retention
Figure
11
50
Time
100
200
300
Days
Time for Unaerated Facultative Lagoon
D 9 Feet
Working Depth
7
1000
UNAERATED FACULTATIVE LAGOON DESIGN
For many domestic wastes the following
of lagoon
size
when
the effects
empirical relationship
provide
a rapid estimate
are involved
of temperature
V
will
CNpgSaff035-T
where
V
C
lagoon volume
10.7
X
10-8
acre-feet
where temperaturevariations
designs are based on a depth of 6 feet
depth with one extra foot
NP
population
q
gpcd
Sa
ultimate
For
effect.
sulfide
correction
as
to compensate
for a
estimating purposes one can
120-125 percent
of
B.O.D.5
normally 1.0 for domestic waste
toxicity factor
f
mg/1
influent B.O.D.
assume ultimate B.O.D.
algal
working
for solids storage
served
possible sludge
f
are great and
a 5-foot
for
1
SO4
concentrations
less
than
500 mg/1
O
temperature coefficient
1.072
T
but
f
and
Npq
f
Q
to 1.085
average
V
is
of the coldest
temperature
gallons per day daily
flow
therefore
expressed
in
gallons instead
10.710-8QSa035-T
of acre-feet
8
operating
substituting
yields
V
If
variable from approximately
month oC
the numerical
values for
C
V
10.710-8435607.48QSaO35-T
0.035QSaO35-T
and
Q
t
5 and 6 are presented
Figures
O
Temperature coefficient
shown
in
Example
the
two
T
0.035SaO35
retention
time in days
for rapid
estimating of lagoon size based on retention
and 1.085
values of 1.072
were used for developing
time.
the charts
figures.
Problem
Given the following
information estimate
Ultimate
Quantity
influent B.O.D.
Q
Temperature
50000
mean
figure
Knowing
5
t
total
and surface loading
320 mg/1
air coldest
O
month
5C
1.085
130 days
time and
the retention
depth of 5 feet
pond required
GPD
Temperature coefficient
From
the size of facultative
depth of 6
rate
for a
total
daily
flow and assuming a
feet you can
number
of specific
9
now
facultative
operating
readily determine surface area
lagoon sections
figs.
1
through
4.
t
0.035 Sa
A
35-T
A1.085
1000
500
400
300
E
d
loo
if
i
200
d
CO
50
3
c
0
N
N
X1
1
E
11
10
I
2
3
4
5
20
10
t
Figure
30
Retention
5.-Unaerated
O
Time
Facultative
1.085
10
40 50
100
Days
Lagoons
200
300
500
1000
t
0.035 S.
e
e35-T
1.072
1000
-
-------------500
400
300
200
E
0
l00
-
ao
50
c
v
E
10
4ooooooo
cn
ýz
-1
1
x
fif
III-EI
1
I
2
3
4
5
10
t
Figure
20
30
Retention
40 50
Time
6.- Unaerated Facultative
O
1.072
11
100
Days
Lagoons
200
300
500
IT
1000
AERATION REQUIREMENTS FOR LAGOONS
The oxygen
that is
transfer capacity
20C
converted
of various
and zero dissolved
to any other
aerators
oxygen
by
condition
N
UTILIZING
usually reported
is
The
pure water.
in
SURFACE AERATION
oxygen
conditions
standard
at
transfer
can be
capacity
the equation
No FCs9 17L
1.024T-20
where
N
transferred per horsepower hour at lagoon operating
oxygen
conditions
No
lb/hp-hr.
transferred per horsepower hour in water
oxygen
and zero dissolved
F
altitude
CSW
saturation
mg/
CL
in
factor
the lagoon
see
fig.
contents
7.
T
temperature
at
1.
oxygen
operating
level
in
the lagoon
mg/1.
temperature correction.
T
mean lagoon temperature
a
ratio
of oxygen
rates
in
7 and 8 are presented
Figures
of oxygen
20C
lb/hp-hr.
pressure correction
dissolved
1.024T-20
oxygen
at
C.
transfer rates
in
sewage to oxygen
transfer
water.
for graphical solution
of the preceding
For most
equation.
concen-tration
con-centration
CL
cases
practical
from 0.7 to
in
applied
0.9.
the
Example
The
be
value
1.5
to 2.0
of CSW
mg/1 No
is
design engineers
pure water
is
but such
approximately 3 lb/hp-hr
depends upon the
liquid which must be established
by some
in
will
that CS
usage
in
total
dissolved
the design problem.
Project elevation
8000
will
should
be cautiously applied.
feet F
0.74
12
from fig.
7
vary
T.D.S.
A rule
of
thumb
will be 0.9 to 0.95 times the saturation
Problem
Given
solids
and
CL
desired
No
a
1.5
mg/1
31b/hp-hr
used for domestic
0.9 often
wastes
in
the absence
of actual
data
T
20C
CSW
8.7 mg/1
Calculate
FCsw 0.748.7
FCSW
Enter
T
figure
value of
8 at
FCsw- CL
20C then
to the
-
CL
6.44
4.94 proceed
No value
-
6.44
1.5
to the
4.94
value
0.9 then
3.0 then read the value of
1.46 lb 02/hp-hr.
13
to the temperature
N
12000
i
11000
10000
9000
8000
1
w
w
w
U
w
0
7000
6jQ00
w
5000
4000
3000
2000
1000
0
L
1.0
095
0.9
0.85
0.8
11
Figure
7.-Altitude
Correction
14
0.75
0.7
C
Factor
F
vs.
Altitude
0.65
0.6
100
EXAMPLE
Given
FCsw C
024T-so
PROBLEM
F0.74
oý
.N
-J
0
No
1
9.17
CL
20
v
\3Sý
ý
e
8
-ý
j
Csw8.7mg/I
-
6
Calculate
3
Follow
FCsW-CL4.94
Enter chart
Op
0
Q
a
ý
1.5mq/I
0c0.9
10
1
from chart
desired
No3lb./Hp-Hr.
15
0
8000ft.
Elev.
Proj.
21 ýR
Read
at
flow
N
4.94
path
1.46
lb./Hp-Hr.
-
2
X
E
1.0
-3
-.
E
Ott
.
-11
O
itg
iqt
m
t4
7-77
a
IýýI
s
LFZ
11lýl
III
t
rr
t
r
llil
71
II
it
All
ýo
T
0.1
0.5
LBS.
2
1.0
0/HP-HR.
in
3
II.
4
WASTEWATER
5
N
10
AN ECONOMIC ANALYSIS FOR FLOOD
By John
In
our engineering
all
alternate proposed.
a flood
plain versus other
Order 11296
Executive
1
we need
analysis
of each
Jakel
Engineer Snoqualmie National
Civil
Here
is
RISK COSTS
Forest
up for the decision-maker
to cast
one technique
- the
costing
of
at
every level
a cost
locating of a facility in
open to management
options
requires that
un-necessary
All executive
roads
structures
of
new
agencies
and
facilities
as far as
use of flood
The key words
in
plains
To thoroughly
some
plain
value
be used
as
economic
the degree
what
dollar
of determining
in
risks
hazards and preclude the
plains. The flood hazards of any
of locating
the same
in
is
It
can
planning
also
new
a
and
independently
within
facility
a flood
One method
established.
of
below.
The
two-fold.
any other annual
way
involved
must be
described
cost of flood risk
cost comparison of alternatives.
of economic
is
or
facilities.
uneconomic
cost of flood risks
the dollar
an annual cost
is
planning
uneconomic hazardous
the
plain must be evaluated
the economics
evaluate
for determining
only the
determining
The
basis
a flood
buildings
the location
flood
of flood
use
within
of Federal
when
hazards
preclude
evaluate
are
or unnecessary
particular planned site location
subjectively.
shall
flood
connection with such
the above statement
uneconomic hazardous
evaluate
shall
practicable
in
for the construction
responsible
directly
or other facilities
costs can and should
or recurring
be used by
construction
line
cost
is
used
in
an
managers to determine
of Federal
on flood
facilities
plains.
This
cost
of flood
flood insurance.
risk
The
may be
viewed
cost of flood
as
riskR
the probability of a particular storm i.e.
by
that storm. In this case the damage
.damag
f
occurring
frequency
Algebraically
is
the reciprocal
of flood
for flood
the relation
Comments by
the
20-year
storm
assumed to be
of flood
plain
damage
or expected
e.g.
frequency
5 10
Division
1/20 times the damage
The probability p
f which
of Engineering
16
fair
is
P1/f
charge
for
determined by multiplying
inflicted
of that
being investigated
or 100 year flood p
Pvalue of structurep
Washington Office
is
total.
would be written
R
1
is
an annual amount that would be a
probability
where
1/f.
This
equation
is
true for any particular year however
that P depreciates
and
interest
some
at
with time. By assuming
which
algebraic gymnastics
straight
annual
an average
6 percent
will
the analysis
line
is
depreciation
cost of flood
risk
be outlined subsequently
complicated by the
may be
determined.
the following
fact
economic
20-year
a
life
Thru
relationship
was developed
P.620/f
Rave
where
P
initial
f
flood
i
time
cost
or value
of building
frequency being investigated
of
rate
money
6%
interest
fac-tors
as-sumes
economic
n
Rave
The
factor
could
To
the relative
various
last
section
dependent
cost of flood
flood
interests
of
i
6% and
and economic
n
and applies them to a $10000
formula to include
the data provided
20 years and
similar
lifes.
magnitude of the cost of flood risks Table
frequencies
of the Rave
risk
on the assumptions
be developed for different
illustrate
derivation
is
20 years
structure
annual
average
.62
life
in
was developed.
I
initial
Table
structure
I
is
cost.
It
The
summarized
of this article.
TABLE
I
- Relative
Magnitude of the Cost of Flood Risks
Initial
Flood Frequency
1
year
Structure
Cost
-P
$10000
P/f
10000
5
10000
2000
10
10000
1000
Rave620
$6200
1240
620
15
10000
667
413
20
10000
500
310
207
30
10000
333
40
10000
250
155
50
10000
200
124
100
10000
100
62
17
in
the
Using the
average
figures
cost of flood
risk
risk
of.
P.
it
be
estimate
possible to quickly
is
Simply multiply the
a facility has a P value of
if
frequency would
for a 20-year
I
Table
for any facility value
For instance
by P/10000.
figure
column
in the right-hand
cost of flood
$45000
listed
the
annual
x 310
45000/10000
Rave
.62
Rave
average
the
$1395
per year.
The above
calculations only give
improvements
However
themselves
-
those costs
not include
does
it
P/f and add
it
to the Rave
formulation
its
rate
R
the basic equation
for 20 years.
i.e.
decreasing
RP
For the
amount P/20 each
P
f
first
year. Since
each year
Simply use the
OF THE
1/f
i
P/f described
n
and
year
risks
for 20 years
the
Spreading
R
for the
P/f
equal to P/fn
depreciating
is
following
R
P/f
is
calculated
is
as
Grant
Editors Note
article
PW
Ppwf-6%-20/f
PW
P11.47/f
PW
P7.11/f
-
-
years P
a simple
has
Pgpwf-6%-20/f
encouraged
to
this
of Engineering
of
a straight
product
R
by
must
line
the
also
the present
worth
PW
of
20
Forest
annual
cost of Flood
Risk
gives
P.620/f
Economy Fourth
The procedure stated in the above
the Division
at
decreased
P7.11 .08718/
P7.11 crf-6%-20/f
Ireson Principles
is
those inputs used
P87.23/f 20
back over the 20 years for an average
computerized approach
Rave
follows
Rave
2
of value
or P/f 20.
Rpwf-6%-20 - ggpwf-6%-20
Rave
loss
relationship
original
earlier and retaining
PW
PW
total
FORMULA
lets assume that P
the basic formula
by an amount
Rave
Using the general formula for summing a decreasing gradient2
flood
and
value for the structures.
DERIVATION
in
to or loss of the land.
possible damage
be determined.
with the structures or
associated
risks
by using an appraisal value and assuming no depreciation
a cost for risk to the land can also
Using
of flood
article
Economics
is
to
be a precise solution.
and Marketing Research to develop
problem. We hope to publish
18
Edition Page 69
not considered
the results in the near future.
This
a detailed
be