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Fy = Flow from other sources <br />• <br />Solve for F„ <br />F, = 100 - (85.10 + 0.30 + 2.30) <br />F, = 12.30 (~ of flow at TR-b) <br />or 7.10 cfs <br />The 12.308 of flow from other sources appears reasonable. The <br />drainage area between TR-a and TR-b contains approximately 208 of the total <br />Trout Creek watershed above TR-b (see Hydrologic Units T-7 and T-8, Exhibit <br />2.5-1 and Table 2.5-1, Section 2.5, Hydrology). Additionally, less than <br />50$ of the drainage area along the east side of the creek is released via <br />the CDPS outfalls. <br />Using the above equation and incorporating the TDS data, a TDS mass <br />balance can be expressed as: <br /> S~ _ (F1 x S1 + Fz) + (Fz x SZ - F~) + (F3 x S3 - F~) + (Fq x S, + F~) <br /> <br />where: ii <br />~u <br />• ~ 6" <br /> - <br />S,, - Annual mean TDS concentration at TR-D (942 mg/1) ~~~ <br /> S~ \e~ <br /> S1 = Annual mean TDS concentration at TR-A (122 mg/1) <br /> S2 = Annual mean TDS concentration at CDPS Outfall 009 (9,285 mg/1) <br /> S3 = Annual mean TDS concentration at CDPS Outfall 007 (3,297 mg/1) <br /> S, = Annual mean TDS concentration from other sources <br />Solve for S„ <br />S, _ ( (Fc x Sz) - (Fi x Si) - (Fz x Sz1 - (F3 x S3) - F~ <br />S6 = [(100 x 942) - (85.10 x 122) - (0.30 x 9,285) - (2.3 x 3,297) <br />= 12.30 <br />S6 = 2,038 mg/1 <br />In order to achieve the TDS levels measured at TR-b/TR-D using the <br />data base available, flow from other sources equivalent to 12.30$ (7.10 <br />cfs) of the annual mean flow measured at TR-b is required containing TDS <br />levels which average 2,038 mg/1. The additional flow containing TDS at <br />this concentration also appears to be reasonable. As presented in Section <br />• 2.5.3.5, the majority of flow in Trout Creek is derived from ground water <br />sources. Springs located along the east side of Trout Creek exhibited TDS <br />Renewal 3 2.SA-5 June 13,1997 <br />