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6. Eerg, Mometum coefficiets for differet velocit distributios Rehbock obtaied ) For Liear Velocit Distributio α + ε Vmax { } Vmax ε β +, i which ε v V o Give: α + ε > ε ( α ) Liear velocit distributio Substitutio for " ε "ithe expressio for " β", v α - + α α + Vmax o β + α+ β (Liear relatio) α.6..8 β..4.6 The plot is show below...9.8.7 Scale X-axis cm.α Y-axis cm.β.6.5.4.......4.5.6.7.8.9.....4.5.6.7.8 Kietic eerg correctio factor α () He obtaied for Logarithmic Velocit Distributio the followig equatios.

Kietic Eerg correctio factor, α + ε ε Mometum correctio factor, β + ε.5 v i which ε * V Give : β + ε > ε ( β ) ( β ) > ε Substitutig for " ε " i the expressio for " α ", α + ( β-) - ( β- ) + ( β - ) - ( β - ) / > α β ( β ) α β...7..4..57.4.694.5.79.6.875.7.99.8.969.9.99...99..97..955.4.887.5.86.6.75.7.667.8.57 /

The plot is show below..8 α+ε ε β+ε.6 α.4. β Relatioship betwee α ad β...4.6.8.

6.. Derivatio of relatioships Assumig a wide chael with the two - dimesioal velocit distributio give b v si ad V v determie " α " ad " β " ( as a fuctio of expoet i secod case). Hece show V α - that (a) For lamiar case.76 ad β - α - ( + )(+) (b) for turbulet case. β - (+) Solutio: v Case ( a ) : si V where v is the velocit at a depth of " " from boudar, is the total depth of flow i wide chael. Let B the width of wide chael. V V si Mea velocit V v da A V V V si B d si d B V - cos -V -V cos { cos cos() } - V { } V V

6.. Kietic Eerg Correctio Factor : B α v da V si B d V A V { } si d 8 si si si cos si cos A - si A si cos si si si si cos si si cos si ( A+B ) +si (A -B ) sia cosb si + + si si cos si + si si si si (- A) -sia si si si cos si si + si 4 4 α si d 8

si d - si d+ si d 8 4 4 cos cos + cos 8 4 4 { ( ) ( ) ( )} + 8 6 8 { } { } o 6 - - + 6 8 6 8 8 6 6 α 6

6.. Mometum correctio factor cos A - si A si cos 4 o β V da V si B d V A V { } B o si d 4 o β si d 4 { ( ( ) } cos d [ ] si 4 - ) 4 4 o d β 8 α 6 6 8.76 β 6 8 8 o

v case (b ): V where v is the velocit at a depth " " from boudar, is the total depth of wide chael. Let B the width of wide chael v V Mea velocit V A V V + ( ) + ( ) + V V ( ) + + + ( ) + V B d d B ( ) V > V + Kietic eerg correctio factor : v. da α v da V B d VA V { } B o + ( +) d ( ) + o ( ) + + + > + + + + ( +) > α +

Mometum correctio factor V V B β da B d VA V { +} ( +) ( ) d + + + + ( ) ( +) > β + + + + + ( +) + + +-- α + (+) β ( +) + +-- + (+) If 7 ( ) { + + } 7 α + * + 7 α ( + )( +) β (+).497 > α.449.485 ( ) { + + } 7 β + * + 7.6 > β.58.857

Example: Obtai α ad β for the velocit distributio give below u.4 +.6, h., h Solutio: u ( ud). 4 +. 6 h h (.4 ) +.6.7 m/s h ( ) ( ) α u d.4+.6 d u h.7 * ( +. +. ) α.64+.6 4 88 d.4.64 +.6.4 +. 4 +. 88 α.8 4 Problems:. The velocit distributio ( i m/s ) i a ope chael m deep ca be represeted b the equatio, / v().6 +.4 ( ) Calculate the eerg correctio factor. Here i is the height above bed ad o m.. I a chael of trapezoidal cross sectio the velocities were measured at mid depth at various sub areas. Compute the average values of α ad β for a give cross sectios. 5 m.8 m/s.9 m/s. m/s. m/s. m/s. m/s.9 m/s.8 m/s : m : 5 m

. For a assumed velocit distributio V 5.75V* log Prove that K α + ad β + i which Vmax, Vmax is the maximum velocit, Vis the mea velocit. V