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<br />Method 4 Example <br />100 <br /> <br />50- <br /> <br />-- .J.~'ft. <br />n=O,05 :-15ft, n=0.05 <br />~ n=0.03 <br />I <br /> <br />o <br /> <br />50 <br /> <br />100 <br /> <br />200 <br /> <br />25 <br /> <br />Determine location of encroachments so that conveyance with a 1 ft. <br />higher water surface elevation is equal to conveyance for the lower <br />water surface elevation (without encroachments). <br /> <br />Conveyances for natural water surface elevation <br /> <br />K = 1.486 A R2/3/n <br />KLDB = 1.486 (15X50)(15X50/65)2/3/.05 = 113,835 cfs <br />KcH = 1.486 (40X50)(40x50/100)2/3/.03 = 729,923 <br />KROB = 1.486 (15Xl00)(15X100/1l5)2/3/.05 = 246,973 <br />~OTAL = 1,090,731 cfs <br /> <br />Conveyances for higher water surface elevation <br /> <br />KLOB = 1.486 (16X50)(16X50/66)2/3/.05 = 125,442 cfs <br />KCH = 1.486 (41X50)(41X50/100)2/3/.03 = 760,559 <br />KROB = 1.486 (16Xl00)(16X100/116)2/3.05 = 273,472 <br /> <br />KTOTAL = 1,159,473 cfs <br /> <br />1-3 APPENDIX I <br />