Engr Help

Open Channel Flow Gradually Varied Flow

Example: Water flows in a 5 ft wide rectangular channel with a flowrate of Q = 30 ft³/sec. The

upstream depth of y₁ = 2.5 ft as shown in the figure below. Determine the flow depth and

surface elevation at section 2 with e = 0.2 ft.

Strategy: Apply principles of conservation of mass, conservation of energy, and use the specific

energy diagram for interpretation of the flow results.

From conservation of mass (steady, incompressible, uniform flow) ρ₁A₁V₁ = ρ₂A₂V₂

and A₁V₁ = A₂V₂ = Q since ρ = constant for incompressible fluid

A₁ = (2.5)(5.0) = 12.5 ft² , V₁ = Q/ A₁ = 30/12.5 = 2.4 ft/sec

A₁ = 5y₂ ft² , V₂ = Q/ A₂ = 30/5y₂ = 6/ y₂

From conservation of energy (steady, incompressible, uniform flow neglecting head loss due to

friction, h_L)

y₁ + V₁²/2g + z₁ = y₂ + V₂²/2g + z₂

For the datum shown z₁ = 0 and z₂ = e = 0.2 ft , y₁ = 2.5 ft

So 2.5 + (2.4)²/64.4 = y₂ + (6/ y₂)²/64.4 + 0.2

or after multiplication the cubic equation becomes: y₂³ – 2.38944 y₂² + 0.559006 = 0

Solution by trial and error results in two positive roots: y₂ = 2.28 ft and y₂ = 0.55 ft

Next use the specific energy diagram to determine the depth and surface elevation downstream.