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<br />AUTOMATIC COMPUTATION OF STAGE-DISCHARGE RELATIONS AT CULVERTS 5 <br /> <br />2 <br />"IQ Q2 <br />h +-=d +-----;j+ <br />1 2gAl2 c 2g^3 <br /> <br />Q2 <br />1,- <br />w KIK2 <br /> <br />, 1 <br />+\? - <br /> <br />? ') <br />) Q- Q- <br />1 -----;j + I. KK' <br />2gA3 2 3 <br /> <br />in which d is critical depth at the outlet, <br />c <br />The equation is solved by iteration between <br />the two extremes, hI == d2 + z._ and hI ::: <br />1. 5 D + z. <br /> <br />If no solution results because the piezo- <br />metric head at section 1 is greater than <br />1. 5 D + z, solutions to flow equations foJ:" <br />types 5, 6, and 4 are attempted. <br /> <br />If no solution results because of super-- <br />critical flow in the approach section, a mes- <br />sage, "NO SOLUTION TYPE TWO FLOW-- <br />SUPERCRITICAL FLOW AT APPROACH <br />SECTION," is printed. Then solutions for <br />successive type 3 conditions and the type 4 <br />flow equation are tried. <br /> <br />After a successful type 2 solution, flow <br />equations for types 3 and 4 are solved if that <br />option is chosen. <br /> <br />Type 3 Flow <br /> <br />Critical depth does not occur; therefore, <br />d3 is replaced by h4, the depth corresponding <br /> <br />to the- tailwater elevation. Equations and pro- <br />cedures are very similar to those used for <br />type 2 flow. The depth at the entrance, d2, <br /> <br />is obtained from the equation of flow between <br />the entrance and the tail water location, and <br />then it is used to obtain the headwater ele- <br />vation in the equation of flow between the ap- <br />proach cross sectiqn and the tail water location, <br /> <br />The first type 3 flow equation is <br /> <br />d2 + <br /> <br />Q2 Q2 <br />z+---:i=h +-+L <br />2gA2 4 2gA32 <br /> <br />Q2 <br />K2K3 ' <br /> <br />in which h4 is equal to the tailwater elevation <br /> <br />minus the base elevation. The root between <br />the extremes, d2 = de and d2 = D, is obtained. <br /> <br />The initial h4 is computed from the first se- <br /> <br />lected tailwater elevation (card 6) for which <br /> <br />h4 is either greater than de + Z if type 1 flow <br />occurred, or g-reater than d if type 2 flow <br />o('('urr'('(l. (' <br /> <br />'I'h(' HPl'ond t.vpe :l now <:"quatiotl is <br /> <br />2 <br />"IQ <br />hi + . 2 <br />2gAl <br /> <br />2 2 <br />h4+~+L ~ <br />2gA32 w KIK2 <br /> <br />+ (-'- - 1)~ +LL <br />C2 2 A 2 K2K3 ' <br />g 3 <br /> <br />and it is solved for the root between the ex- <br />tremes, ~I =- d2 + z and hI ::: 1. 5 D + z. <br /> <br />If no solution results because of super- <br />critical flow in either the approach section or <br />the culvert barrel entrance, nothing is printed. <br />Then computations for type 3 flow at successi... <br />vely higher tailwater elevations are tried, <br /> <br />Type 4 Flow <br /> <br />Type 4 flow occurs when the culvert is <br />submerged, for which the flow equation is <br /> <br />Q2 (1 ) Q2 2 <br />hi = h4 + -----;j + ""2 - 1 - + L ~, <br />2gA C 2 A 2 K 2 <br />o goo <br /> <br />where A is the cross- section area and K <br />o 0 <br />is th.e conveyance for the full culvert barrel. <br />The equation is solved directly. The value <br />of h4 i::> for the last tailwater elevation listed <br />and should al ways he greater than D. For <br />very steep culverts, the tail water should be <br />high enough to submerge the inlet also. <br /> <br />Type 5 Flow. <br /> <br />Type 5 flow occurs when a culvert -flows <br />part full under a high head. The type 5 flow <br />equation is <br /> <br />h = z+~+(-'-- 1) Q2 <br />1 2 A 2 C2 2 A 2 <br />g 0 g 0 <br /> <br />The root is obtained between the limits <br />hi = 1.5 D + Z and hI = 5.0 D + z. the latter <br /> <br />of which is the maximum elevation given in <br />card 8. <br />