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Mound Design Worksheet (For flows up to 1200 gpd) <br /> All boxed rectangles must be entered,the rest will be calculafed. i� -- -� ---�------ -------- �--�-�---�-�---�-----i <br /> A. FLOW { A l: Lshmaied Sevtiage Ilows in t,allo�;p�t�v� ! <br /> Estimated 750 gpd(see figure A-1) ' bedr�ooms � Uass� �n C�ass II � Qoss III f (�oss IU� <br /> or measured � x 1.5(safety factor)= 0 9pa 1 �'JJ � �1� I ?9� � 6fJ�6 <br /> ' I <br /> B. SEPTIC TANK LIQUID VOLUMES ; � r �,�D I �J(1 � 218 � ctthe <br /> ! ,1 WU � 3'];i 25a � voiu�s � <br /> Septic tank capacity 2-1300 gallons(see figure C-1) i �; � � „ ' <br /> : I;i�J � =1�J ��4 � in m� � <br /> C. SOILS(Site evaluation data) j � � y�� � �'1' � '��� � `���• <br /> � I���� � b-JO , 37J �,or ill � <br /> 1. Depth to restricting layer- 1.33 feet � � <br /> ; }} ? I�L�1 � L7;� l 40� cc�u�s� <br /> 2. Depth of percolation tests= inches �--��m�-.--.�.--��-•�mm�--�m��-�-p•�--�•��� <br /> 3. Texture CLAY LOAM <br /> 4. Soil loading rate(see Figure D-33) 0.45 gpol ft2 <br /> Percolation rate 11.to 29 MPI <br /> 5. %Land Slope � % IJ:t�: :�b�orpitun 4�7d11�Sitin�7a61e <br /> �q.�rn;,cinn?lafr� (.nad�ng ILaJo S <br /> �*J'�� �ir�per `� �oill..�vturr � t. drn.ti � .�I;ti�xpi�:^n <br /> ��'�; �l'J��l' Iyltk��3 �cifieti tii�psllanw� �° k i i...,n � �' � � k«nn <br /> . _ . . .__.+.,�I�I�_._...._�. . ..._. .._ .,�l �f-...:.o�.!...i..,.. ........_..... <br /> I_,'_< <'�ili)2i�.?si I�dsr�i-:an5 � �iv�'�a1d � F:.di � t 0(7 <br /> i�11131�iC1'tl; �1'illll�l tFt1B::.'l''i(� � �(ti+� ��C.� 1�111 'dll�l � h} c um5,.nd � <br /> i' ° �k'Ii�l1 41vi1?.�. � 1 <br /> {� � `� � �.��i. ��.. . � lii^�s,L.md , <br /> ._.'l'(�Fif.`I�LS �.�1�;dC`s21 ��i. .,�1(iES'(Sln �I� I,'iili�t�. I n 5n^d � .E <br /> . . �:f� �c. .._.4 S pa+1�'�.,o:i7R �.._.._ �i .._...}_. I..\".... <br /> f <br /> .. �' '!i n.tU d I n.�m � L:7 � .':00 <br /> i.i;I��1:55 �SI� ���: ���i:' .... ;5� 04.. t tiut . . . <br /> ' ...., � °rr.iln ; �>SIS � �4U <br /> �:;I�•� � YJ'J 3 yl��� tiS.��tf� . � .Sll. ` . . . _.. <br /> �.: s{,m GO � inih�.E:e.�I.r.m�� �t a5. . �.. . y G 7 <br /> 5 l;l'ri � ;;i ���.SU .... €�ilc, .:Lat=i cam <br /> ^, � { �..-l1�' � � �� ( r .I�a,�m <br /> �..;'�ill''7 Iut1 .;�����-� � } ) _ . __ _- <br /> !:i tn I?U �{ Slh(:)r ;}..'4 �OU <br /> � � `,7rzde-�.�a� � <br /> (:lati <br /> ._ _._ . _ .._..... ..t. ._.... <br /> ..51c�a�:•rlhr.n..�_fY.� -. . . } . .. <br /> t ? i <br /> .>�cr.i Jcsi�•ac�i fm:tm::c xriix:i:�.�a ti•oU:.v ::�:�cllY <br /> D. ROCK LAYER DIMENSIONS <br /> 1. Multiply average design flow(A}by 0.83 to obtain required area of rock layer:Item A x 0.83= <br /> 750 gpd x 0.83 ft"gpd= 630.0 ft <br /> *WRIGHT COUNTY REQUIRES 450 S.F.OF ROCKBED FOR A THREE BEDROOM HOME <br /> 2. Determine rock layer width =0.83 ft Igpd x Linear Loading Rate(LLR)(see LLR chart) <br /> 0.83 ft Igpd X 12 = 10.0 ft <br /> LLR Chart <br /> PerkRate LLR ����� <br /> <120 MPI <=12 .�,.a�.,��. ��i�,��� <br /> >=120 MPI <=6 �������, <br /> 3. Length of rock layer=area divided by width= <br /> 630 ft I 10 feet= 63.0 feet <br /> E. ROCK VOLUME <br /> 1. Multiply rock area by rock depth to get cubic feet of rock <br /> 630 X 1 ft= 630.0 ft3 <br /> 2. Divide ft3 by 27 ft3(yd3 to get cubic yards <br /> 630.0 ft3 / 27 = 23.3 yd3 <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> 23.3 yd3 X 1.4 ton/yd3 = 32.7 tons <br />