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GENERAL32315
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Last modified
8/24/2016 7:54:56 PM
Creation date
11/23/2007 7:15:55 AM
Metadata
Fields
Template:
DRMS Permit Index
Permit No
C1981033
IBM Index Class Name
General Documents
Doc Date
7/8/1982
Doc Name
CLARIFICATION OF BEAR 3 STIPULATION RESPONSE
From
ACZ INC
To
MLRD
Permit Index Doc Type
STIPULATIONS
Media Type
D
Archive
No
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' Revised 7/8/82 <br />' Calculation of Inflow to Sediment Control Structure <br /> Time Parameters <br />' <br /> Time of Concentration <br />' <br /> From Figure 2.34, p. 82 <br />' s = 40% nearly bare, untilled and alluvial <br />~ fans, western mountains <br /> v = 64 fps <br /> t <br />= L/V = (1990 ft) (64 ft/s) = 31.1 s = 0.52 min <br /> c <br />' or from Equation 2.58, p. 83 <br /> = 0.0078L0.77 (L/H)0.385 = 0.0078 (1990) <br />t 0.77 (1990/759)0.385 <br /> c <br />' t <br />= 3.92 minutes <br /> c <br />' Lag Time <br /> Equation 2.59, p. 84 ~ ~~),~ <br />' tL = 0.6 tc = 0.60(3.92) = 2.35 min <br /> <br />' Equation 2.60, p. 84 SCS method <br /> tL = [L0.8(S+1)0.7 + [1900 Y0.51 <br /> <br /> L = 1990 ft <br /> Y=404~i <br /> S = 1000/CN - 10 = 1000/79 - 10 = 2.66 <br />' tL = [(1990)D'8(2.66+1)0'7] e [1900 (40.4)0 '5] = 0.09 hrs = 5.37 min ;~ <br />' Time to Peak <br /> Equation 2.40, p. 73 <br /> <br /> = tL + D/2 <br />t <br /> p <br /> assume D = 1/3 t <br />' p <br /> tp = tL + 1/6 tp <br /> <br /> = 6/5 tL = 6/5 (5.37) = 6.44 min <br />t <br />' p <br /> <br />1 <br />
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