HomeMy WebLinkAboutSeptic design - 1999 � � vNo � Q,�
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SEPTIC SEPTIC SYSTEM ESIGN
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Date ,-- ``�
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Owner/Builder �: � '� J :� ��� �.
Address -S '"� S � d �G� . ��r'����s� ��,-1!n�
Site Address � ��� �-'
Home Phone �I 7w"' �� � ! Work Phone Pager/Cell
The following information has been compiled for a sing/e fa ily home:
Bedrooms � GPD --�G� Garbage Disposal N c Lift Pump in Basement ti'0
Septic Tank Capacity ��-�a Pump Tank Capacit ;Z��
System Type: Mound x Trench
Distribution: Gravity Pressure i� Land Slope ��v
a �
Depth to Restricted Layer ?�% Soil Sizing Factor � Perc Rate /;� �y���
Trench System: Drainfield Size/Sq.Ft. Lin I Ft.
S62 Number of Laterals Roc (Tons)
Rock Width Ma Trench Depth Width
, , ,
r
Mound System: Rock Bed `G"X � Sand L er t v�'
� , ,.
Upslope >Z Downsl pe /7 Sideslope /�
Sand Depth 1/�J ?�� Topsoil n Site �"�� Trucked in � ..�
Sand (Tons) ��'—� Rock (T ns) ��_ Topsoil (Tons) i'-i.�'
�
Pump Manufacturer: � �/� �t� :'�!s � ! �
Requirements: GPM ' �-> �� � �. Head �,�I 1�-
�
Force Main Length _/.�^'C � Diameter .C' �
,
Number of Laterals � Length C�^'_-'
Swediund Services • 9520 Laketown Road • haska, MN 55318 • (612) 442-5855
STATE CERT IED
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' N10UND DESIGi�1 WORKSHEET
(For Flows up to 1200 gpd)
A. FLOW D-7 .
Estimated ��� gpd (see pages D-7 or I-3, 4,5) ES��W1[p SCM�G[RO'MS�M G41Dy1�OI�.=
1ar1EEA lrrf pf IKS�OEMCE�
�f
or measured gpd. i °`°"'°'6 ` � � s
: �oo ae ��o son
3 •!O 700 i�� or
• f00 37] t3i
B. SEPTIC TANK LIQCTID VOLUMES � � �n � `'�`
T i 050 .t� 3�0
Z_� �C`.� gallons (see pages C-3 or C-5) ° '_°° °„ '°' ��
C-3
C. SOILS (refer to site evaluation) sflanc T�r+K c�v�cmes, iN GALlONS
/ „�o�.,.�m
1. Depth to restricting layer = 7C9 in es ""`"" "'�"'" '""°.".°.
�.._ �FAOOd LIOUA W�QT� 0�]IOl�I
2. Depth of percolarion tests = %'L in es ,a,,,. ,�� ,,,,
3. Percolation rate i .-�' mpi '°"' ,••° ,�•°
��� ��as »�o
4. Land slope ..� °'o �.�a�
,.o. ..eo
D. ROCK LAYER DIMENSIONS
1. Multiply flow rate by 0.83 to obtain required area of rock
layer: A x 0.83 = _
-;���%� gpd x 0.83 sq. ft./gpd =/r�s . ft.
?„----
2. Select width of rock layer (10 feet or less) = l0 ft.
3. Length of rock layer = area y width = '.:-,�
� �
,, ��e_ sq. ft. i _l!�' ft. _ (:-:%? ft Rock Bed
_ f'f•f.f.f.f.f•f.f.,.f•f•r.f�f T I
1•�•ti•�•�•ti•ti•ti•ti•ti•ti•ti•�•ti•`.• �
f.l.J•t•f•I•f•t l•l.r•l.r•f•f �
ti.ti.�.ti.ti.ti.ti•`•ti•�•ti•ti.ti•�•ti• idth <_1('
� � 1•f•t•f•f.l•f•l l•I./.f•r•1•f
ti�ti•t•ti•ti•ti•ti•�•�•ti�ti•ti•ti�ti•ti•
.f•f.�•f•r•I•r•J•r•f•f•!•,r•f•f
E. ROCK VOLUME �- Length -�
, 1. Multiply rock area by rock depth to get cu 'c feet of rock;
;!-�,? sq. ft. x � ft. =�'��cu. ft.
2. Divide cu. ft. by 27 cu. ft./cu. yd. to get cub c yards;
;1-?� cu. ft. 127 = �,�cu. yd.
3. Multiply cubic yards by 1.4 to get weight o rock in tons; I
�=<cu. yd. x 1.4 ton/cu. yd. _ �to s.
F. ADSORPTION WIDTH I
1. Percolation rate in top 12 inches of soil is mpi E-16
2. Select allowable soi! loading rate from table o page E-16; �-�����d����u��•�� I
�. �'"-' gpd/ft2 _ ��� ' �... I
3. Calculate adsorption width ratio by dividing ro layer �~"" "M� �"� �� �� � '
.,. . ,,� .., ,.a ,�o
laadin� rate of 1.20 �pd/ft2 by allowable soil lo ding rate; ' " "' "' '" "'
�� ��o a w a.•� t... t.eo I
..1�--r�___ ."p�_._.QN".—�.0]__"—.T.O
1.2� gpd/ft2�- �y,���Pd/ft2 = � �- �� .. ..e e.. o.,, ,.� ,.,
.. .,:< a t. ,,, ... ,:a �
Che�K this t�a(ue on page E-16. I
4. Multiply adsorption width ratio by rock layer idth to get ,
required adsorprion width; �
� !� x <=% ft = ��ft
. n _
. . . , -
. . .. . ,
�. � �
DOWNSLOPE DIKE WIDTH
i. If landslope is 3% or more, subtract rock layer wi�lth om
adsorption width to obtain minimum downslope dik toe .
� � ?C� ft- �� ft = � �% feet
2 Calculate Minimum mound size based on geomerery:
a. Deternune depth of cl�e�sand fill at upslope e ge of rock
layer: Separation � 7' feet '
b. Multiply rock layer width by landslope � roo� co�.�
to deternune drop in elevarion; , ' '°°` R° .a
SIOj7E Dlfference S�Der�tlon i t��t
I(,i X � %�` 1�� _ � .� feet Slop• Dltt�r�ne• t
. u �leoe w�otn
c. Add depth of dean sand for separarion (2a) '�" ���� aoe • w�atn
at upslope edge, depth of rock layer (1 foot) to pth of r��.�� oow?�f.w�ac�
cover (1 foot) to find the mound height at the up lope edge �-�-r••`
of rock layer; /
�f �.1-ft+ lft + lft= 'r� feet
d. Enter ta b l e w it h lan dslope and u slope dike ra 'o.
�Select dike multiplier of � ��,J
e. Multiply dike multiplier by u,ps�ope mound he' ht
to find upslope dike widthw5, �� x�..-� 7 / eet
f. Add depth of clean sand for slope difference (2 ) at
downslope edge, to the mound height at the up lope edge
of rock layer c) to fin�he downslope height;
:�e ft+ � ft =.��� feet
g. Enter table with landslope and do�nslope dik ratio.
Select dike multiplier of u'� �
h. Multiply dike mulkiplier by downslop��c�� hei ht
to get downslope dike width: ���x �.`�� r feet
i. Compare the values of step G.1 and Step Zh lect the �-
greater of the two values as the downslope dik width; �?'�
� feet
.-UDf�Op�M�OI�.�.
j. Total mound width is the sum of � ���•�
upslope dike (G.2e) width plus rock ROCC ,a w�a��
layer width (D.2) plus a �oL�o ��a�� `�"" �o �Q .��a��
I x 7C.��..� �-r..�
downslope dike width(G.Zi); � „ o I ��
� �ft + fC� ft + � ft = � feet ��
k. Total mound length is the sum of °o""�•"'°`"
�..�
upslope dike width (G.2e) plus rock layer
len�h(D.3)plus upsl¢.pe dike w�dt (G.2e); � � .
.� ...� ft + �`r. � ft +J� ft = �feet
/ �r
/ TotU l�nqtn
owns ope ' ps upe
3:i �� s:i 6i ri �� i s:i s:i �:i e:i
s.�
o �o �o s.o 60 �.o �o o s.o �o �.o aa
1 )A9 ll7 51b i3a 7S7 2V1 45 �.76 5.66 6 i� 7.�1
2 ].19 l35 556 6.62 6.11 2.57 .757 �51 536 i1� 6.40
� 7 330 t54 S.BB 7J2 l.66 2.75 �� 1.35 S.OA 5.7v 6.15
. � 3A1 l76 6.25 ).D9 9.72 26a .l5 �.17 �.6{ S.Ib 6.06
5 JSJ 5.00 667 157 10.77 261 -JJ 1.00 1.62 5.19 571
� 6 ]66 51b 7.1{ 93E t209 2.5/ 27 7.65 �.0 {.97 S.�l �
7 3d0 55G 7.69 IOJ� I].7] 2.�8 .12 3.70 �?J l70 5.13
S ].95 5.6! !17 I15� 15.91 2.Q ' .m 7S7 I.QS 1.�9 tAa
� 9 �.11 i25 9.W 1).61 1l92 2�6 W ).l5 ).90 IJO �65
10 �29 d67 10.0 I5.00 1JJJ 2J1 !6 311 3.75 {.12 �.k
11 �AE 7.!{ ll.11 1�.A5 JO.0 lI6 '1S 7.b 7.61 3.95 �16
u ��+ ;.�v �sso n.0 ur iz: �o �.�� �.�9 3eo �a 64
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PRESSURE DISTRIBUTIOIV SYSTEM
1. Select number of perforated laterals �v
2. Select perforation spacing = 3 ft.
3. Since perforations siloul��rnot be placed closer tha 1 ft. to
the edge of the rock layer (see p. F-14), subtract 2 f . from the
rock layer length.
;
(�, ,.__. t f t
Ra k laycr lenRth - 2 1 t. — (�i t.
4. Determine the number of spaces between perforat'ons.
Divide the length above by perforation spacing an round E-17a
do�vn to nearest �hole number. fj TABLEOFPERFORATION DfSCHARGESf�lCf".'
Head Perforation ciiune!e:hnctes)
Length perf. spacing =�e�> ft. �'= � ft. _ •��s aces �i32 '!a
�3) C�� 1.Oa 0.�6 0.;�
1.5 0.69 0.90
5. Number of perforations is equal to,q'ne plus the n mber of Z.ob a.�� i.oa
zs o�� �.�;
perforation spaces . 3.o G.v3 t_9
4.0 1.13 1.4i
5.0 1.26 1.5�
r'�'; spaces + 1 = �.'_ � perforarions/lateral aUse 1.0 foot of head for residential sysrems
bUse 2A feet of head Ior other establis�ments
6. Multiply perforations per lateral by number oE lat rals to
_� get total number of perforations. E-17b
.__, --. /
.....__.u..�w.s�ae�.a� �..1e m�e our.,�..s,,Q�,..e.:�e
�' ,G. i ; /n r�.,v�<�os�e.�..�.,m
la�erals X perts�:aterat— �� � I'erFnrat101iS. ��r'p'OO� 1.25 inch 1.5 incr. � 2.0 ir_r:
2.5 14 18 � 28
7. De�e�ine require� flcr: rate by :nultiplyi g 3.3 iz i6 � �s
number of perforations by flow per perforation a.o i� ts � z�
(see page E -lii ; s.o io ia f r_
//` ,1 f� ��/ r,r�-�f:�
perfs X �+r:/,c••rf �—j��' :h[' � �i �.._1 �
Wr,OI]lL[�'ID�i PO�J'�[]1K O'f�1i�Rqw ri•�r �
8. If laterals are connected to header pipe as shown n page E- �
�y l
15, select minimum required lateral diameter fro table on � �`;,._��• �
page E-17; enter table with perforarinn spacing ar. number �_.�r,�•-� d��J'
of perforations per lat��al. Seleci �ninimum diam ter for �,j
perforated lateral = r :�� inche�.
' F.-12
,..,..
�?. if perfc�r���ed later�: ;ystem is attached to manifol pipe near �„�„�,,:__. �r-
the center, a�, on page E-12, perforated lateral len h and �``^� -�
num`�er oE ��erforat�ons per lateral will be approxi ately one �'��'�
` "�'AL•�
halE oE that ir� step 8. Using these values, select m nimum _� . ,_
diameter for �erf��rated lateral from page E-17 as �..�''` —'�—
inches.
�, . • � � •
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, 9
PUMP SELECTION PROCEDLIRE I
A. Determine pump capacity:
Gravity Distribution
1. Minunum suggested is 20 gpm
2. I�.1dXimum SuggeSted i5 45 gPrr��� Perforation Discharges in GPM
Head Perforation diameter
Pressure Distibution (feet) (inches
3. a. Select number of perforated laterals 7/3z 1/4
b. Select perforation spacing= feet. 1.oa o..�6 o.7a
c. Subtract 2 ft. from the rock layer length. t.5 0.69 0.90
Rock Iryer IengN
-2 ft. = feet. �.- 2.ob 0.80 1.04
d. Determine the number of spaces between perforations. a Use t.o foot single homes.
Length perf.spacing= ft.= ft. - spaces b Use 2.0 feet for anything else.
e. spaces+1 = perforations/lateral
f. Multiply perforations;per lateral by number;qf laterals to
get total number of perforations. A�a�s x �,T�= p rforations.
g• � X �m,��- gPm• .
i
"= SELECTED PUMP CAPACITY��pm
, �
B.Determine head requirements:
1. Elevation difference between pump and point of discharg .
'�'�d feet
2. If pumping to a pressure distribution system,five feet for ressure SoJ Ceatment systvn
required at manifold if gravity system,zero. °:°�"`-"°: �/ O
�_feet Total p�pe I�q�t,
3. Fricrion loss
-� a. Enter friction loss table with gpm and pipe diameter. i,,,e � gJevanon Dlffernnce
Read friction loss in feet per 100 feet from table(F-14). P'�`
� ------ -- -
F.L. _ �� � ft./100 ft of pipe -�f �
b. Determine total pipe length from pump to discharge --'�°""--�---�� �-- �- �� ��
point. Estimate by adding 25 percent to pipe length fo fitting
loss,or use a fitting loss chart(F-15 feet).
Equivalent pipe length- 1.25 times pi�e length= I
l�o x 1.25=��feet Friction Loss in Plastic Pipe
c. C a lcu la te to t a l f r i c ti o n l o s s b y m u l ti p l y i n g
friction loss in ft/100 ft by equivalent pipe length.l �,. Nominal
pipe dia.
Total friction loss= 3��_x l t� 3 =100= � � feet
4. Total head required is the sum of elevation differenc�, ���te 1.5" 2" 3"
special head requirements,and total friction loss.
�� + '� � ` 20 2.47 0.73 0.11
+ � 25 3.73 1.11 0.16
(1) (2) (3c) 30 523 1.55 0.23
� 35 6.96 2.06 0.30
TOTAL HEAD � 40 8.91 2.64 0.39
Z feet 11.07 0.48
50 13.46 3.99 0.58
55 4.76 0.70
C. Pump selection bo 5.60 o.s2
65 6.48 0.95
70 7.44 1.09
��z 1. A pump must be selected�to deliver at le st
gpm (Step A) with at least �zfeet of total ead (Step B).
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•
�izin$ of Pump Station
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1. Dctcrminc Surfacc Arca I
Rcctanglc= Arca = L x W " w�dih
L
x = square feet 1.�„6�f�
�
Circic= Arca =n x(Radius)Z
3.14 x x_ = square feet K;,�;,,�
Other=Cet Surface Area from Manufacturer ' n=3.t4
square feet I
2. Calculate Gallons Pcr lnch
Thcre arc 7.5 gallons per cubic foot of volumc,thcreforc you ust muitiply thc arca
times the conversion factor and divide by 12 inches per foot t calculate gallons per inch
Arca x 7.5 gpft'+12 inchs per foot
- x 7.5+12 = /._�<% gallons/inch • I - -
3. Calculate Gallons to Cover Pump(with 2 inches of H�ater cov ring pump) Estimala!Scwagc Flows in Gallons per day
(Height(in)+ 2 inchcs) x gLallons/i�uh(#i2) �g��
( i"� + . )x = _.<-.� �,:.'gallons of r Type I Typc I[ Typc III TYpc
�d..
Iicdrooms 1 V
4. Calculate Total Pumpout Volume 2 300 225 180
a. To maximize pump life select sur�p sizc for 4 to 5 pump oper� tions per day. 3 450 300 218 �
"?:.': y`;> $Pd+4= �� l galions Exr dosc 4- 600._ 375 256 �.i��
. b. Calculate drainback C.S� ' 750 45U 294 ;�
6 900� 525 332 �Yr��,
l. Determine total pipe length, %;�'"�.�feett� 7 1050 600 370 ""`
2. Detcrminc liquid volume of pipe,i ' �':gallons per 1(H)fcYt. 8 1200 675 408 «,i��,,,;,,,:
3. Multi�ly Icngth by,valume: Drainback quantity=
'-`' '; fcet x :'-' ',�.' allons/]00 ft. _ .�_f allons.
i g �
� Pi diamelet inches G�Iiexu r 100(ccl
c. Total pump out .volume equals dose volum�e+drainback � � , 1 4.4
�%� Qallons per dosc+ �'�__ gallons= -- f ' gallons 1.25 7.77
l 5, �.10.58--
5. Calculate Volume for Alarm(typically 2 t�3 inches} ' 2 -r�'I�43 .'
Depth(in)x gallons/inch(#2)= 2.5 ��-24.87
�r� x �--> _,-=� .. gallons 3 38.4
4 66.1
6. Calculate Reserve Capacity(75% the daily flow?
Daily flow(see page D-7)x.75= : .
`i�_">�� x.75=.'-�':- c'_ gallons I
Reserve Capacity
7. Calculate total gallons
gallons over pump+gallons pumpout+i;aiic�ns alarm+gall ns reserve capcity
#3+ #f4 c+#t 5 +# 6
-�_'` + - �/ + '��, + � �•�= f'`, " gallonS Alarm
Pump On
S. T�tal Depth (Total gallon dividcd by gallc�n per inch) -
Total Gallon (�7)+�allon/i ch(#2)
� ' : + - - -��' -�' inches To 1 Pumpout Volumc
— Pump OfF
Pump Hcight
9. Float Scparation Distancc(cqual total pumpout volumc)
Total pumpout volumc(#4c)+�allons/inch (#2) '
� /% % �i - '�inclies