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Open Channel Flow Gradually Varied Flow (Cont)

Example: (continued) (Use of Specific Energy Diagram)

Recall E = y + q²/2gy² Now q = Q/channel width = 30/5 = 6

So E = y + (36/64.4) y² = y + 0.559/ y² plotted in the figure below.

Starting at the upstream depth, y₁ = 2.5 ft then the downstream depths are either

y_2a = 2.28 ft or y_2b = 0.55 ft. The critical depth is determined from y_c = (q²/g)^1/3 .

So y_c = (q²/g)^1/3 = (6²/32.2)^1/3 = 1.037 ft. Emin = (3/2) y_c = 1.56 ft .

Next examine the Froude numbers for the downstream depths. F_r= V₂/√gy₂

For y₂ = y_2a = 2.28 , V₂ = Q/A₂ = 30/(2.28)(5) = 2.63 ft/sec F_r= 2.63/√(32.2)(2.28) = 0.31

For y₂ = y_2b = 0.55, V₂ = Q/A₂ = 30/(0.55)(5) = 10.91 ft/sec F_r= 10.91/√(32.2)(0.55) = 2.59

If y₂ = y_2b = 0.55 is the correct downstream depth, then F_r> 1. But flow from station 1 to station 2b must stay on the constant q-line of q = 6 ft²/sec. This result cannot occur without passing through the critical depth of y_c = 1.037 ft. So flow must proceed from depth y₁ at station 1 to depth y_2a at station 2a . Result: flow downstream flow depth = 2.28 ft and the

downstream free surface elevation = 2.28 + 0.2 = 2.48 ft.

To analyze conditions for the downstream depth to reach y_2b click here to continue.

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