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n�1t��.Jrd� I�ESIGN �VOl �S E�T 5 <br /> (For F�ows up to 120�gp ) <br /> A. FLOW Fstimatcd Sewage Flows in Galloas per day <br /> l d) <br /> Estimated �C' gpd Numbcr r�� T�a r�m -r�� <br /> or measured x 1.5 = gpd. ,a�� <br /> 2 3C0 2?5 lE0 �, <br /> B. SEPTIC TANK LIQUID VQLUMES 4 � 3300 ;� of� <br /> s �so aso i�a "',°� <br /> ����U gallons e 90o szs 3sz `" <br /> � � ioso 600 3�o T mi. <br /> 8 I200 675 I 308 j � <br /> C. SOILS (refer to site evaluation) �°'� <br /> �, <br /> .ZG ' s< <�<T���� oa�a��� y���� <br /> l. Depth to restricting layer = __ inches / � feet �qw,�,��;,Y <br /> « Numher�i! MinimumLyuid I�yuid�aarywiN wichdisposa!& <br /> ?_. Depth of percolation tests = _ inch s Bmnx,� �,��;�Y s,�,s��,=,,,�„ �;h;�;,� <br /> �.�� 2 irc Icts 750 ll25 1500 <br /> 3. Texture � Percolation rate � � mpi ��„< ,� ,5� Z� <br /> Q 5 ar E 1500 2250 3000 <br /> 4. Land slope � /o �.R"�y z� 3� ,�«, <br /> D_ ROCK LAYER DIMEI�SIONS <br /> 1. Multiply flow rate by 0_83 to c�btain required rea of rock layer: A x 0.83 = <br /> �E%�-� gpd x 0.83 sq. ft./gpd = ��'sq. ft. <br /> 2. Select width of rock layer (max 10' if<120 m�� max 5') = �� ft. <br /> 1 <br /> 3. Length of rock layer = area -width = ,, � Q a<.o� �a.4o eo �,�Qo <br /> '=o -� � eb, �._� <br /> �G��"} sq. ft. - i� tt. = S V ft. � a aD,�pPop o�oa a <br /> Q¢v 7p�Qa,�o oe.eo�o� <br /> ��v��o'?oaa n� �. : <br /> 9b� , <br /> -d� f� ft av�.Q �0000000o D ooa <br /> < 20m i <10' Length ��� ftP <br /> P <br /> E. ROCK VOLUME > 20mpi <5' <br /> I 1. Multiply rock area by rock depth to get cubic feet of rock; �'sq. ft. x f <br /> ft. =5��' cu. ft. <br /> 2. Divide cu. ft.by 27 cu. ft./cu. yd. to get cubi yards; <br /> S�`} cu. ft. =27 - �`� cu yd. <br /> 3. Multiply cubic yards by �.4 to get weight of r ck in tons;�cu. yd. x 1.4 <br /> ton/cu. yd. �_� tons <br /> F. ABSOIZPTION WIDTH Absorption Width Sizing Table <br /> 1. Percalation rate in top 12 inches of soil is�� �71 Pcccolarioo Raee in Gallons Etauo ofAbsorpuoa <br /> ���.�^ Miautes pa Inch Soil Teznue per day per width ro Rock <br /> T�xture -� (MP[) squace foot Laycr wdth <br /> Faster than 0.1 Coazx Sand 110 1.00 <br /> 0.1 to 5 Sand 1.20 1.00 <br /> 2. Select allowable soil loading rate from table; o.<<0 5 Fnc Sand o.6o z o0 <br /> 6 to 15 Saady Laam 0.79 1.52 <br /> , �o gpd/ft2 � � o ,��, a <br /> 31 to 45 Siit Loam 0.50 �,yp <br /> 46 to 60 Clay Loam OAS 2.67 <br /> 60 ro 120 Gay 0.24 S.pp <br /> 3. Calculate adsorption width ratio by dividing roc layer s�a��,�,�zo c�ay o.zo 6.00 <br /> loading rate of 1.20 gpd/ft2 by allowable soil loa ' g rate; <br /> 1.20 gpd/ftz= ; �v0 gpd/ft2 = .G� <br /> 4. Multiply adsorption width ratio by rock layer wi th to get <br /> required adsorption width; <br /> �_X� ft = � ft <br /> �I <br />