Fluvial Hydraulics: Water Surface Profiles and Computation
This content delves into the intricate topic of fluvial hydraulics, focusing on the computation of water surface profiles in natural water courses. It explores equations, channel variations, and the utilization of Manning's and Chezy's formulas to analyze nonuniform flows. Examples and illustrations further elucidate the concepts presented.
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Presentation Transcript
BAE 6333 Fluvial Hydraulics Nonuniform Flow
Computation of Water Surface Profiles Three forms of this equation: : All Channels 2 2 1 1 Q 2 Q 2 ( ) = ) 1 ( h h S x x + + 1 1 i i f i i 2 2 2 C A R g A A h + 1 i i 2 2 1 1 Q 2 Q 2 ( ) ( ) ( ) + + = ) 2 ( h z h z x x + + 1 1 i i i i 2 2 2 C A R g A A h + 1 i i Prismatic Channels: 2 U ( ) ( ) ( ) = 3 ( ) a H H S x x Chezy + 1 + s s f i i 1 i i 2 C R h 2 2 n U 3 / 4 ( ) ( ) ( ) = 3 ( ) ' b H H S x x Manning s + 1 + s s f i i 1 i i R h
For natural water courses Channel bed varies continuously, so it is common to investigate the position of the water table with respect to a horizontal datum (h+z): 2 2 2 2 1 1 Q Q Q Q ( ) + + + + = h z h z x x K + 1 i i ss 2 2 2 2 2 2 2 2 2 gA gA C A R g A A h + 1 i i + 1 i i 2 2 2 2 2 1 1 Q Q U n Q ( ) + + + + = h z h z x x K + 1 i i ss 3 / 4 h R 2 2 2 2 2 2 2 gA gA g A A + 1 i i + 1 i i ? What is K ss
Example 4.C Graf A small river conveys a discharge of Q = 3 m3/s. The geometry of sections 1 to 6 are reported along with the Manning s n value. The local head-loss coefficients are estimated to be: reach: 1-2 2-3 Kss: 0.10 0.30 0.30 0.15 0.10 Compute the water surface profile in the reach between sections 1 to 6 for h = 401.2 m and h = 402.4 m. 3-4 4-5 5-6
First point Geometry of channel is irregular and geometric characteristics can not be given with simple equations... How do you get A, B, P, and Rh?
First Question What kind of water-surface profile do we have? Need hc and hn How do we do this for a channel that is changing continuously?
We can now compute the water surface profile Graf uses the standard step method 2 2 2 2 2 1 1 Q Q U n 3 / 4 Q 2 ( ) + + + + = h z h z x x K + 1 i i ss 2 2 2 2 2 2 gA gA g R A A + 1 h i i + 1 i i = H H h h + 1 i i r s
