Engr Help

Rapidly Varying Flow - Hydraulic Jumps (continued)

Conservation of Energy: P₁/γ + V₁²/2g + z₁ = P₂/γ + V₂²/2g + z₂ + h_L

Now z₁ = z₂ = 0 and P₁ = γ y₁ , P₂ = γ y₂ So the energy equation becomes

y₁ + V₁²/2g = y₂ + V₂²/2g + h_L

Or in terms of specific energy: E₁ = E₂ + h_L

So the flow leaving the hydraulic jump has a lower specific energy than that entering.

The head loss across a hydraulic jump is then h_L = E₁ - E₂ = y₁ + V₁²/2g - y₂ - V₂²/2g

Again from conservation of mass: y₁V₁ = y₂V₂ So V₂ = y₁V₁/y₂

h_L = y₁ + V₁²/2g - y₂ – (y₁V₁/y₂²)/2g = V₁²/2g [ 1 – (y₁/y₂)²] + y₁ – y₂

Recall F_r = V/√gy So for the entering flow F_r1 = V₁/√gy₁ , V₁² = (F_r1)² gy₁

h_L / y₁ = F_r1²/2 [ 1 – (y₁/y₂)²] + [1– (y₂/ y₁)] (expression for head loss across jump)