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Koehler Tunnel bulkhead design Page 7 July 31, 2003 <br />Solution to Bernoulli equation <br />Co +2& +Z1)+hA=\~2 +2gc +Zz)+hg+np <br />62.4 0.18442 0 0.18442 218(0.03401) 0.04(30X1.03401 <br />062.4 + 2(32.2)D4 +~)+~=(62.4 + 2(32.2)Da -F O)+Q+ 2(32.2)D~ + 2D3(32.2) <br />1.0= O.OOIg042 + 0.00063379 <br />D D <br />0 = 1.OD5 - 0.0012042D - 0.00063379 <br />D = 0.2477 ft (2.972 in) for welded and seamless steel pipe <br />A 3-in nominal diameter Schedule 40 standard weight steel pipe <br />[(3.068-in (0.2557-ft) inside diameter] will meet the bypass flow <br />requirement. <br />Calculate °f" (Moody friction factor) based on associated Reynolds <br />number (N,1~) <br />N>zE _ ~`~ ft = kinematic viscosity = 1.405 x 10.5 ,fl-oo <br />Q 63(0.002228) _ 0.18439 <br />N~ = Diu o De(OD1~ 39) = o.Dca~39 (o.2ss xgaos:lo-S) = 5.13253 x 104 <br />e = 0.0002 p 2-20, Table 3.8 <br />E 0.0002 _ 0.00078217 <br />De - 0.2337 - <br />f = 0.0230 p 2-21, Figure 3.13 <br />(0 + q~ +Z 1) t hp = ( p2 + 2~ +Z2) +hg +h1 <br />62.4 0.184392 1 0 0.184392 1 2.28(0.034014) 0.0230(30X1.034014 <br />C 62.4 + 2(32.2)D~ +0/ f 0 = (62.4 + 2(32.2)D~ -FO/+0+ 2(32.2)IN + 2D5(32.2) <br />1.Oft = 0.0012042 + 0.00036443 <br />D4 DS <br />0 = 1.ODe3 - 0.0012042De - 0.00036443 <br />D = 0.2298 ft (2.758 in) <br />A 3-in nominal diameter Schedule 40 standard weight stainless <br />steel pipe [3.068-in (0.2557-ft) inside diameter] will meet the <br />bypass flow requirement. <br />