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C- Lake Spillway <br />Although the lake will have enough volume to store the runoff from an event much larger <br />than the 500 year storm, if the Animas River basin experiences the 100 year storm, the <br />lake will be inundated. The expected elevation of water for this event at this location is <br />666.5 feet, while the top of dam vazies from 6668' to 6664'. Therefore, the true top of dam <br />is 6664' elevation. A trapezoidal spillway of 8 feet bottom width will be installed in the <br />dam to allow water to leave the lake without washing over the surface. The spillway will <br />be cut to a depth of 6662', or 2 feet below the top of dam. Spillway sideslopes will be <br />2.OH1.OV. The spillway will be installed at grade of 3% along the top of dam and will be <br />lined with angular riprap of Dso of 1.0 feet. The downstream portion will be installed at a <br />slope of 30% and will have riprap of DSO of 2.5 feet. The following calculations demon- <br />strate that the riprap is stable for these conditions. <br />Project Name: THOMAS PfT <br />Date: ~ 01(19!02 By: GL <br />Description: DURABLE RIPRAP ROCK FOR 100 YR EVENi <br />SPILLWAY SLOPE <br /> TEST 1 RIPRAP STREAMBED LINING CSU PROCEDURE <br />INPUT D= 0.25 Depth of flow Teet <br />INPUT Y= 62.4 Density of water Ib/cu.R <br />INPUT S= 0.3 slope of channel %)100 <br /> T= 7ractlve force Ib/a.it <br /> T= 4.88 =(D=riS) <br />N= Stability _ = 21T no units <br /> Parameter Y(SG-1)Dso for turbulem flow <br />INPUT Dsa 2.5 ~ Awsrage diameter of riprap in feet <br /> SG= 2.65 Specific Gravity of the rock used <br />iron above N= 0.381816 <br />INPUT O= 42 Angle al repose of riprap Degrees <br /> Z= 16.69924 Bedslope angle Degrees <br /> SF(BOTTOM)= 1.3664&8 xos(Z)'ten(O SAFETY FACTOR <br /> sm + (O) <br /> BOTTOM RIPRAP IS STABLE <br /> TEST 2 RIPRAP BANK LINING <br /> A= 16.69824 Angle of streamlirre (bed assuming uniform flow <br /> (same as bedslope angle Z) DEGREES <br />INPt1T SS= 28.5 Sideslope angle in degrees <br /> B= 18.37756 ten-r (cos(A)/(2'sin(A)/N"tan(O)+sin(A)) <br /> 0.332221 <br /> N'= 0.30082 N• (f+SIN(A+B)/2) BANK STABILITY PARAMETEF <br /> SF(BANK)= 1.235469 =COS(A) •tan(O) 0.862431 <br /> N"fen(O)+sin(SS)'cos(B) 0.698059 <br /> RANK RIPRAP I.S RTARI F <br />Thomas Pit 2/02 21 <br />