Laserfiche WebLink
<br />Method 4 Example <br />100 <br /> <br />25 <br /> <br /> -- _ v:::: 1ft, - <br />, n=0,05 :-15ft. <br /> n=0,05 <br />" <br /> ~ n=0.03 <br /> I <br /> <br />50- <br /> <br />o <br /> <br />50 <br /> <br />100 <br /> <br />200 <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 />KLOB = 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)(15Xl00/115)2/3/.05 = 246,973 <br /> <br />KTOTAL = 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 />~OB = 1.486 (16Xl00)(16X10O/116)2/3.05 = 273,472 <br /> KTOTAL = 1,159,473 cfs <br /> <br />1-3 <br /> <br />APPENDIX I <br />