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67 <br />The first orifice will have a eter of 1". a area of this orifice is 0.00545 sq. ft. The release <br />rates through this orifice were calculated at 0.50' intervals between 0.01' and 3.00'. <br />0.01' - Q = .62(.00545) 2(32.2)(.O1) <br />Q = 0.003 c. f. s. <br />0.50'- Q = .62(.00545) 2(32.2x.50) <br />Q=0.019c.f.s. <br />1.00' - Q = .62(.00545) 2(32.2)(1,00) <br />Q = 0.027 c.f.s. <br />1.50' - Q = .62(.00545) 2(32.2x1.50) <br />Q = 0.033 c.f.s. <br />2.00' - Q = .62(.00545) 2(32.2)(2.00) <br />Q = 0.038 c.f.s. <br />2.50' - Q = .62(.00545) 2(32.2x2.50) <br />Q = 0.043 c.f.s. <br />3.00' - Q = .62(.00545) 2(32.2)(3.00) <br />Q = 0.047 c. f. s. <br />Ic <br />The 12" standpipe portion of the control structure will be open on the top. A portion of the flow <br />from the larger storms will pass through this pipe and it will also serve as an overflow device. The <br />pipel2" overflow structure will be placed at "h" of 2.50'. The area of a 12" orifice is 0.785 sq. ft. <br />The release rates through the second orifice were calculated at even 0.50' intervals from 2.5l' to <br />3.00'. <br />2.5l' — 2.50' = 0.01' Q = .62(.785) 2(32.2)(.01) <br />Q = 0.39 c.f.s. <br />3.00' — 2.50' = 0.50' Q = .62(.785) 2(322)(50) <br />Q = 2.76 cTs. <br />9 <br />