Summary for Open Channel Flow (continued)

Type 3 (Rectangular channels only) w = width of channel

For Hydraulic Jumps - apply conservation of mass, of linear momentum, and of energy

Conservation of Mass: w y₁ V₁ = w y₂ V₂ = Q

Conservation of Linear Momentum: F₁ ˗ F₂ = ρ Q (V₂ ˗ V₁ ) , F₁ and F₂ are pressure forces

which can be expressed as: y₁²/2 ˗ y₂²/2 = V₁y₁(V₂ ˗ V₁)/g

Conservation of Energy: y₁ + V₁²/2g = y₂ + V₂²/2g + h_L

The combination of conservation of mass and of linear momentum gives information about

the ratio of the depth following the hydraulic jump (y₂) to that leading into the jump (y₁).

(y₂/y₁) = [√ ( 1 + 8 Fr₁²) ˗ 1]/2

and

(y₁/y₂) = [√ ( 1 + 8 Fr₂²) ˗ 1] /2

The energy equation yields the energy loss (head loss due to turbulent mixing) across the

hydraulic jump, h_L .

h_L / y₁ = F_r1²/2 [ 1 – (y₁/y₂)²] + [1– (y₂/ y₁)]

where Fr₁ = upstream Froude Number = V₁² / 2g y₁ and Fr₂ = V₂² / 2g y₂

Finally, construct a specific energy plot to illustrate the hydraulic jump process.

Specific Energy = E E = y + q² / 2gy² where q = Q/w = Vy Click here for a plot.

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Copyright @ 2016 Richard C. Coddington
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