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, • . MOUND DESIGN WORKSHEET <br /> � ' (For Flows up to 1200 gpd) <br /> A. �.�w Estimaud Sewage Flow i�t Gallons pa Day(gpQ) <br /> Estimated y�D gpd x�,b« �:� <br /> ot 7ype I 'Iype II 7Ype III 'lype JV <br /> or measured — x 1.5 = — gpd. a��� �' <br /> 2 300 21S 180 cos <br /> 3 450 300 218 � <br /> B. SEPTIC TANK LIQUID VOLUMES � a 60o s�s as6 .,,.p. <br /> s �So aso zsa m <br /> � --1 f� O C� gdllOILS 6 900 52S 332 � <br /> 7 1n50 600 370 �� <br /> 8 1200 675 408 <br /> C. SOILS(refer to site evaluation) <br /> „ <br /> � x�� ,� ,.� <br /> 1. Depth to restricting layer= /�'',�y- a t� inches S °� �� �,,� <br /> 2 Depth of permlation tests = 1 a'' inches `A°"' `"""' <br /> 3. Percolation rate i a.•y mpi 23 a'4` ,'o�oo ;•,s o <br /> 4. Land slope 5 % ;Q$ � � <br /> �Q 9 sa�.c.s c:is� <br /> D. ROCK LAYER DIl�iIIVSIONS <br /> 1. Multiply flow rate by 0.83 to obtain required area of rock <br /> layer:A x 0.83 = <br /> . y v . gpd x 0.83 sq.ft./gpd= 3�_sq. ft. <br /> 2 Select width of rock layer(10 feet or less) _ /� ft. <br /> 3. Length of rock layer= area=width= R�� <br /> Z� Sq. ft._ �ft. = 3� ft. ,.,.�.ti.�.ti ti.�.�.�...ti.�.,.�.� <br /> 1•t•1K•!•�•t•l•t•f•r•t•t•l•t•r• <br /> ti.tiK•ti•�K•ti•ti•tiK•Yti•1K•ti.1•ti <br /> tift�tiK:ti:{:�f�f�r f���f���r�f��� idth S10 fk <br /> ♦.ti•S•ti•t.ti.�•tiK•t•ti <br /> J•f•t•t•t•t•t•�•t�t•f•�K•t•r•1• <br /> 1•ti•�K•�•ti•t•ti•S•t•\��•ti•�•ti•ti•'L <br /> .r.f•r.j•r•f•�.�.r•f•f•f•f•t•f•r. <br /> E. ROCK VOLUME ~— �'�' —� :��\�:�, <br /> 1. Multiply rock area by rock depth to get cubic feet of rock; �'T <br /> 3�sq.ft. x ?�a� ft. _��cu.ft. <br /> 2. Divide ru.ft.by 27 cu. ft./cu. yd. to get cubic yards; <br /> �2 cu.ft. s 27=1�cu.yd. � <br /> . 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> ��� cu.yd.x I.4 ton/cu.yd. _�tons. <br /> F. ADSORPTTON WIDTH G�-��t� ►�I .on VYdth Saa� Tibk <br /> 1. Permlation rate in top 12 inches of soil is la y mpi p�,,,�,R,� �-- �� <br /> �,�;,,u��;,,�, Soil Texture �,� m,� <br /> (r ah.wera� <br /> rndih <br /> 2. Select allowable soil loading rate from table; Faster than 0.1 0�sana 1.20 l.00 <br /> •� gpd/ft� O.ito� Sand 120 1.00 <br /> 0.1 to 5 Fne Sand" �•� 2•� <br /> 6 to 15 ndy Loam 0.79 1 S2 <br /> 3. Calculate adso tion width ratio b dividin rock la er 16 to 30 1..oam o.bo 2.00 <br /> rP Y g Y si co as s�ic�m o.� 2•�0 <br /> loading rate of 1.20 gpd/ft2 by allowable soil loading rate; �to bo ct� L,oam 0.45 2.6� <br /> 1.20 gpd/ft2�- ,4 gpd/ft�= a.l.� Slowerthan 120 Clay 0.24 5_00 <br /> "Shc having 50�or mQe d fine or vay fine aand. <br /> 4. Multiply adsorption width ratio by rock layer�vidth to get <br /> required adsorption width; <br /> �2 x_l�ft= ��.� ft `-�;, <br /> :: � <br />