1. Fluid flow in an open channel
…or, a tutorial on simplifying impossible equations…

2. What do we know already? Continuity 𝑄= 𝑉 ℎ𝑤 and 𝑄 1 = 𝑄 2 on a reach
Flow resistance, e.g., 𝑉 = 1 𝑛 ℎ 2/3 𝑆 1/2 or 𝑉 = 8𝑔 𝑓 𝑅𝑆
Often assume steady & uniform flow, but natural channels don’t behave that way, so…

3. 3D equations of motion microscopic balance
𝜕 𝑢 𝑥 𝜕𝑥 + 𝜕 𝑢 𝑦 𝜕𝑦 + 𝜕 𝑢 𝑧 𝜕𝑧 =0 𝜕 𝑢 𝑥 𝜕𝑡 + 𝑢 𝑥 𝜕 𝑢 𝑥 𝜕𝑥 + 𝑢 𝑦 𝜕 𝑢 𝑥 𝜕𝑦 + 𝑢 𝑧 𝜕 𝑢 𝑥 𝜕𝑧 = 𝑔 𝑥 − 1 𝜌 𝑚 𝜕𝑝 𝜕𝑥 + 1 𝜌 𝑚 𝜕 𝜏 𝑥𝑥 𝜕𝑥 + 𝜕 𝜏 𝑥𝑦 𝜕𝑦 + 𝜕 𝜏 𝑥𝑧 𝜕𝑧 𝜕 𝑢 𝑦 𝜕𝑡 + 𝑢 𝑥 𝜕 𝑢 𝑦 𝜕𝑥 + 𝑢 𝑦 𝜕 𝑢 𝑦 𝜕𝑦 + 𝑢 𝑧 𝜕 𝑢 𝑦 𝜕𝑧 = 𝑔 𝑦 − 1 𝜌 𝑚 𝜕𝑝 𝜕𝑦 + 1 𝜌 𝑚 𝜕 𝜏 𝑥𝑦 𝜕𝑥 + 𝜕 𝜏 𝑦𝑦 𝜕𝑦 + 𝜕 𝜏 𝑧𝑦 𝜕𝑧 𝜕 𝑢 𝑧 𝜕𝑡 + 𝑢 𝑥 𝜕 𝑢 𝑧 𝜕𝑥 + 𝑢 𝑦 𝜕 𝑢 𝑧 𝜕𝑦 + 𝑢 𝑧 𝜕 𝑢 𝑧 𝜕𝑧 = 𝑔 𝑧 − 1 𝜌 𝑚 𝜕𝑝 𝜕𝑧 + 1 𝜌 𝑚 𝜕 𝜏 𝑧𝑥 𝜕𝑥 + 𝜕 𝜏 𝑧𝑦 𝜕𝑦 + 𝜕 𝜏 𝑧𝑧 𝜕𝑧

4. Simplifying assumption 1: motion limited to x-direction (defined as downstream)
𝜕 𝑢 𝑥 𝜕𝑥 + 𝜕 𝑢 𝑦 𝜕𝑦 + 𝜕 𝑢 𝑧 𝜕𝑧 =0 𝜕 𝑢 𝑥 𝜕𝑡 + 𝑢 𝑥 𝜕 𝑢 𝑥 𝜕𝑥 + 𝑢 𝑦 𝜕 𝑢 𝑥 𝜕𝑦 + 𝑢 𝑧 𝜕 𝑢 𝑥 𝜕𝑧 = 𝑔 𝑥 − 1 𝜌 𝑚 𝜕𝑝 𝜕𝑥 + 1 𝜌 𝑚 𝜕 𝜏 𝑥𝑥 𝜕𝑥 + 𝜕 𝜏 𝑥𝑦 𝜕𝑦 + 𝜕 𝜏 𝑥𝑧 𝜕𝑧 𝜕 𝑢 𝑦 𝜕𝑡 + 𝑢 𝑥 𝜕 𝑢 𝑦 𝜕𝑥 + 𝑢 𝑦 𝜕 𝑢 𝑦 𝜕𝑦 + 𝑢 𝑧 𝜕 𝑢 𝑦 𝜕𝑧 = 𝑔 𝑦 − 1 𝜌 𝑚 𝜕𝑝 𝜕𝑦 + 1 𝜌 𝑚 𝜕 𝜏 𝑥𝑦 𝜕𝑥 + 𝜕 𝜏 𝑦𝑦 𝜕𝑦 + 𝜕 𝜏 𝑧𝑦 𝜕𝑧 𝜕 𝑢 𝑧 𝜕𝑡 + 𝑢 𝑥 𝜕 𝑢 𝑧 𝜕𝑥 + 𝑢 𝑦 𝜕 𝑢 𝑧 𝜕𝑦 + 𝑢 𝑧 𝜕 𝑢 𝑧 𝜕𝑧 = 𝑔 𝑧 − 1 𝜌 𝑚 𝜕𝑝 𝜕𝑧 + 1 𝜌 𝑚 𝜕 𝜏 𝑧𝑥 𝜕𝑥 + 𝜕 𝜏 𝑧𝑦 𝜕𝑦 + 𝜕 𝜏 𝑧𝑧 𝜕𝑧

5. 1D equations of motion
--Mass conservation considered separately-- 𝜕 𝑢 𝑥 𝜕𝑡 + 𝑢 𝑥 𝜕 𝑢 𝑥 𝜕𝑥 + 𝑢 𝑦 𝜕 𝑢 𝑥 𝜕𝑦 + 𝑢 𝑧 𝜕 𝑢 𝑥 𝜕𝑧 = 𝑔 𝑥 − 1 𝜌 𝑚 𝜕𝑝 𝜕𝑥 + 1 𝜌 𝑚 𝜕 𝜏 𝑥𝑥 𝜕𝑥 + 𝜕 𝜏 𝑥𝑦 𝜕𝑦 + 𝜕 𝜏 𝑥𝑧 𝜕𝑧

6. Simplifying assumption 2: drag from bed >> drag from banks, no stretching
𝜕 𝑢 𝑥 𝜕𝑡 + 𝑢 𝑥 𝜕 𝑢 𝑥 𝜕𝑥 = 𝑔 𝑥 − 1 𝜌 𝑚 𝜕𝑝 𝜕𝑥 + 1 𝜌 𝑚 𝜕 𝜏 𝑥𝑥 𝜕𝑥 + 𝜕 𝜏 𝑥𝑦 𝜕𝑦 + 𝜕 𝜏 𝑥𝑧 𝜕𝑧
Q: when might this assumption not be true?

7. Simplifying assumption 3: x-component of gravity scales with bed slope, = 𝑔 sin 𝜃 ≃𝑔 𝑆 0
𝜕 𝑢 𝑥 𝜕𝑡 + 𝑢 𝑥 𝜕 𝑢 𝑥 𝜕𝑥 =𝑔 𝑆 0 − 1 𝜌 𝑚 𝜕𝑝 𝜕𝑥 + 1 𝜌 𝑚 𝜕 𝜏 𝑥𝑧 𝜕𝑧

8. Simplifying assumption 4: pressure at a point along x is hydrostatic, 𝑝≃ −𝜌 𝑚 𝑔(ℎ−𝑧)
𝜕 𝑢 𝑥 𝜕𝑡 + 𝑢 𝑥 𝜕 𝑢 𝑥 𝜕𝑥 =𝑔 𝑆 0 −𝑔 𝜕ℎ 𝜕𝑥 + 1 𝜌 𝑚 𝜕 𝜏 𝑥𝑧 𝜕𝑧

9. Simplifying assumption 5: bed shear stress is approximately 𝜏 𝑧𝑥 ≃ 𝜏 0 ≃ −𝜌 𝑚 𝑔 ℎ−𝑧 𝑆 𝑓
𝜕 𝑢 𝑥 𝜕𝑡 + 𝑢 𝑥 𝜕 𝑢 𝑥 𝜕𝑥 =𝑔 𝑆 0 −𝑔 𝜕ℎ 𝜕𝑥 −𝑔 𝑆 𝑓
Note: friction slope 𝑆 𝑓 is a new parameter that accounts for deviations from steady, uniform flow!

10. Simplifying assumption 6: replace point velocity 𝑢 𝑥 with mean velocity 𝑉
𝜕𝑉 𝜕𝑡 +𝑉 𝜕𝑉 𝜕𝑥 =𝑔 𝑆 0 −𝑔 𝜕ℎ 𝜕𝑥 − 𝑔𝑆 𝑓
Final Equation!

11. Resulting expression: meaning of terms
𝜕𝑉 𝜕𝑡 +𝑉 𝜕𝑉 𝜕𝑥 =𝑔 𝑆 0 −𝑔 𝜕ℎ 𝜕𝑥 − 𝑔𝑆 𝑓
grav. driving stress
time change in 𝑉
downstream change in 𝑉
downstream pressure grad.
new term from momentum bal.
Achtung! friction slope 𝑆 𝑓 is different from 𝑆 0 if either term on the left-hand side is nonzero.

12. OFFICIALLY: the Saint-Venant equation A 1D, simplified momentum balance for open channel flow
Note: Actual water surface follows the hydraulic grade line (HGL), but the flow momentum distribution follows the ENERGY GRADE LINE (EGL)

13. Saint-Venant equation Interpretation
𝑆 𝑓 ≅ 𝑆 0 − 𝜕ℎ 𝜕𝑥 − 1 𝑔 𝜕𝑉 𝜕𝑡 − 𝑉 𝑔 𝜕𝑉 𝜕𝑥
Interpretation: In cases of uniform, steady flow, 𝑆 𝑓 ≃ 𝑆 0 .
Where flow is non-uniform (i.e., natural streams), the 1DSV contains corrections for downstream changes in depth and velocity

14. Reach-scale flow dynamics
𝑆 𝑓 ≅ 𝑆 0 − 𝜕ℎ 𝜕𝑥 − 1 𝑔 𝜕𝑉 𝜕𝑡 − 𝑉 𝑔 𝜕𝑉 𝜕𝑥 Dependent variables are mean stream-wise velocity 𝑉 and mean (for a cross-section) flow depth ℎ, independent variables are 𝑥 and 𝑡. Finally, a solvable system of equations requires a flow resistance equation (e.g., Manning, Chezy, DW) and mass conservation. In macroscopic form: 𝜕𝑄 𝜕𝑥 + 𝜕𝐴 𝜕𝑡 −𝐼=0, 𝑄=𝑉𝐴=𝑉𝑤ℎ=𝛼𝑤 ℎ 𝑏 Here, 𝐼 is any external input (-ve for outflow) and 𝛼 is the function of the slope ( 𝑆 𝑓 ) and roughness from the flow resistance equation. For Manning’s equation, 𝑄= 1 𝑛 𝑤ℎ 5/3 𝑆 𝑓 1/2 , so 𝛼= 𝑆 𝑓 1/2 /𝑛 and 𝑏=5/3.