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PERMFILE72878
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PERMFILE72878
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Entry Properties
Last modified
8/24/2016 11:22:25 PM
Creation date
11/21/2007 12:28:08 AM
Metadata
Fields
Template:
DRMS Permit Index
Permit No
C1983059A
IBM Index Class Name
Permit File
Doc Date
12/11/2001
Section_Exhibit Name
EXHIBIT 9 DRAINAGE & SEDIMENT CONTROL PLAN CALCULATIONS
Media Type
D
Archive
Yes
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SED.WK1 <br />• SEDIMENT POND CAPACITY <br />The sediment pond is constructed in the southeast portion of <br />the facilities area. Calculations on the following pages <br />indicate that the pond is to have a capactiy of 0.48 A-F <br />and the emergency spillway is to have a capactiy of 8.62 CFS. <br />The purpose of this calculation is to determine the sediment <br />volume to be expected in the pond, to check the size of the <br />emergency spillway and to size collection ditchs B & C. <br />REQUIRED WATER VOLUME - Acre Feet <br />SEDIMENT VOLUME <br />0.480 <br />The only area on the site that will contribute significant <br />sediment to the pond is the coal pad, Area 7. Deposition <br />from the other areas will occur before run-off from them <br />reaches the sediment pond. <br />USING THE UNIVERSAL SOIL LOSS EQUATION <br />A= R K L S C P <br />R = rainfall factor 30.00 <br />K = soil erodibility factor - Conservative 0.70 <br />LS = combined length slope factor 2$ - 900' 0.39 <br />C = cropping management factor 1.00 <br />P = erosion control practice factor 1.00 <br />n <br />U <br /> <br />A = sediment,tons/acre/year 8.19 <br />THREE YEAR SEDIMENT VOLUME ACRE FEET <br />5.2 Acres @ 115 #/Cubic Feet 0.05 <br />TOTAL REQUIRED SEDIMENT POND CAPACITY -- AF 0.53 <br />EMERGENCY SPILLWAY <br />MANNING'S EQUATION <br />FLOW IN A TRAPIZODIAL CHANNEL <br />DEPTH D 0.500 FEET <br />WIDTH W 10.000 FEET (Bottom Width) <br />SLOPE S 0.020 EXPRESS AS DECIMAL <br />"N" N 0.040 EXPRESS AS DECIMAL <br />SIDE SLOPE X 1.500 SLOPE = X:1 <br />AREA=(W*D)+(X*D^2)= 5.36 FEET^2 <br />WETTED PERIMETER=(W)+2((XD^2+D^2))^.5 11.80 FEET <br />R=AREA/WETTED PERMITER 0.46 <br />V=(1.49/N)*(R^.667)*(S^.5)= 3.12 FT/SEC <br />Q = A X V = 16.76 CFS <br />Q REQUIRED 8.62 CFS <br />-5- 6/03/93 <br />
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