CIV E 530 - OPEN-CHANNEL HYDRAULICS SPRING 2004 - MIDTERM 1 - SOLUTION

PROBLEM 2

y₁ = q/V₁ = 1.0/1.5 = 0.6667 m

y₂ = 0.85 × 0.6667 = 0.5667 m

The specific force principle:

F_o = F₁ - F₂

F = Q²/(gA) + z_barA

In a rectangular channel:

Q = qb

A= yb

z_bar = y/2

Therefore, the force per unit weight (specific force) is:

F = q²b/(gy) + (y²b)/2           [m³]

The force per unit weight per unit width is:

F/b = q²/(gy) + y²/2           [m²]

The force (force per unit weight per unit width) exerted by the obstruction on the flow is:

F_o/b = (F₁/b) - (F₂/b)

F_o/b = [q²/(gy₁) + y₁²/2] - [q²/(gy₂) + y₂²/2]

F_o/b = [0.1529 + 0.2222] - [0.1799 + 0.1606]           [m²]

F_o/b = 0.0346 m²

The unit weight of water is:

γ = 9.81 kN/m³

The force exerted by the obstruction on the flow is:

γF_o/b = 0.3394 kN/m   ANSWER.

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