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� ' <br /> � .�_ OS�P Mound Desi�n Work heet <br /> ; r�, , <br /> UNIYER5ITY <br /> Msnnesota Poflution �'�� S�� e OF MINNESOTA � ���� <br /> Contro!Agency p ���`. <br /> 1. SY57EM StZIMG: P�oJectiD: v 11.09.22 <br /> A.Oesfgn F(ow(Ftow&Sol(• f.A}: 750 GPD `f'AB�E ��(a <br /> B. Sofl Loadfns Rate(Ftow&Soil-3.Cj: O.6Q GPd/1t2 �-p��� �S FOR aETERMINING BOTTOM ABSORPTlON AREA <br /> AND BSOR�TION RAT}OS USING PERGOLATION iESTS <br /> C.Depth to Lfmlt/ng tonditlon: 1.0 ft Trcetmcnt Level C rrea�cM ie,��i�,e-z,a, <br /> D.Percent[.�xrd 5lope: 6.0 ;� tercot� Rate u� M,,,,a M�^�O" M,,,,,� <br /> E.Ueslgn Medla Looding kate: 1.2 GPD/ft2 ` ��am�)3 ��'°O (�rt'� �O�O <br /> F.Mound A6sorption Rotfo(Table IXa): 2.00 m i _ t . � <br /> G.Design Contvur Locdlnq Race: 92.0 GPD/ft 'to5 �•Z t �.s � <br /> ' iwsn.,assnd o.e 2 � t.e <br /> � � � Tablel . an9foa t etano <br /> .A'AUN�CONiOEtR lDAOWG RA7ES: elo t5 0.78 1.6 1 1.6 <br /> lz,naswM ' 79xture•do�t:pd �� ���f III l0 34 0.6 2 0.T8 2 <br /> Verc Wto �R mcemd abyxptlon rat lo�'� 3f to45 O.b 2A 0.78 2 <br /> Rata: <br /> 4fi to 6YJ -.. 0_66 2.8 0.6 26 <br /> =GOCiy�i lA.1.3.2.0.2.a.1.6 eR <br /> , 61to 120 . 6 0.� 6,3 <br /> ,t-)20 n�pi PR 4.0 ;12 �t20 _ . . _ <br /> �t2o ma� .S.G' cb' 'Systems wilh these lues are rwt Type 1 systertu. Contaur loading Rate(linear <br /> oading rate)ts a recarn�ended vatue. <br /> 2. D15PERSAi MEDIA S1ZtNG <br /> A. Catculate Requtred Rlspeiwt BedArea:Deslqn flow(i.A)�Desfgn/rledia L ing Rale (1.E��ttZ <br /> !t a larger dispersal medla area 750 GFD� 1.20 GPD/it2 = 625 (�2 <br /> ts desiad,enter stre: �Ft� <br /> 8. Calculate Dlsperw!Bed Wfdth:Contour Loadfng Rate (S.Gj:Design Medlo lr+g Rate (1.E)=eed Width <br /> 12.0 fc : 1.2 gpd/ftz = 10 tt <br /> C.Calculate Dfsperwl 8ed Length: Oispersaf Bed Area (2.A}�8ed Wldth (2,B) 8ed�ength <br /> b25 Rx : 10 tc - 63 it <br /> D.Selett Olspersa(Medla: <br /> E. 1F using a registered product,enter the Component Lenyth: in s 12 = �(� <br /> F. if using a regtstered product,enter the Comporant lY(dth; in: f2 e �ft <br /> G.Number of Components per Row=Bed Length (2.C)divided by Component L ngth {4.J}(Round up) <br /> � ft = � ft= �co ents/row <br /> H.Number of Rows �8ed Wtdth (2.6)dfvided by Component Width (4.K)(Roun up) Note:CLR of 10.3 <br /> �- ���� goUft resuks fn 4 foot <br /> lu laadiRg Rate an Oes43n`�+�+MY P��t51 it�is namber is a e rn�mber wlde bed. <br /> �� ft+ � ft� �� � <br /> j, Total Num6er of Components =Nvmber of[on�onen[s per Row X Number Rows <br /> � X �� _ ��co enu <br /> I <br />