Presented by: Civil Engineering Academy

Presented by: Civil Engineering Academy

Open-Channel Flow

Uniform Flow (See CERM Ch. 19) Characterized by constant depth volume, and cross section. It can be steady or unsteady Non-uniform Flow *Not on the test

A1V1 = A2V2 (eq. 19.8) Example: Consider an open channel shown in figure. Section 1 has an area of 2 ft 2 and a flow of 15 ft/sec. What must be the area of section 2 if velocity was increased 5 units higher than section 1? Section 1 Solution: Given: S1: Area = 2 ft 2 ; Velocity = 15 ft/sec S2: Area = Unknown; Velocity = 5 units higher that S1 Section 2 Thus (15 + 5) = 20 ft/sec Using continuity equation: A1V1 = A2V2 (2 ft 2 ) x (15 ft/sec) = A2 x (20 ft/sec) A2 = 1.5 ft 2 Answer

Chezy equation: (Eq. 19.9) Manning s equation for C value: Where: Use for small, smooth channels. C is constant (also known as C-Value) R is hydraulic radius (See table 19.2 for details, R=A/P) S is slope of channel (Eq. 19.11(b)) Combining Eq. 19.9 and 19.11 (Eq. 19.12 (b)) Where: Use for large channels. n is roughness coefficient (See App. 19.1 A CERM) R is hydraulic radius (See table 19.2 for details) For flow rate (Q) us Manning s Equation: Q = Av (Eq. 19.13 (b))

Example: Open Channel - Rectangular Section Determine the flow rate of a rectangular channel on a 0.003 slope constructed using rubble masonry. The cross sectional dimension of the channel is 7 ft wide and 4 ft deep. Assume the flow is uniform and steady. Solution: Find the value of important parameters 1. Hydraulic Radius R = A/P: A = 7 ft x 4 ft = 28 ft 2 P = 7 ft + 4 ft + 4 ft = 15 ft R = (28 ft 2 )/(15 ft) = 1.87 ft 2. C Value using Manning s Equation From App. 19.A, the roughness coefficient of rubble masonry is 0.017 3. The flow rate (Q) Q =va = (C(RS)^(1/2))A = (97.29(1.87x.003)^(0.5))28 = = 204.03 ft 3 /s Answer Note: wetted perimeter is the length which is touching water. C = (1.49/0.017) x (1.87)^ (1/6) = 97.29 4 ft 4 ft 7 ft

Example: Open Channel Semi Circle An open channel is to be designed to carry 35.3 ft³/s at a slope of 0.0065. The channel material has n- value of 0.011. Determine the diameter of the most efficient semi-circular section. Solution: 1. For a semi circular section, R= A πd² = 8 P πd 2 = D 4 (Eq. 19.2 read) 2. Manning Equation : Q = Av (Eq. 19.13b) Q= va = ( 1.49 n )AR2/3 S = ( 1.49 0.011 )(πd2 8 )(D 4 )2/3 ( 0.0065) = D 8/3 = 20.74 D = 3.12 ft

Example: Open Channel Slope of bed The bottom of trapezoidal canal is 8.2 ft and its side are both inclined at 60⁰ with the horizontal. Determine the slope of the channel bed if Q = 495 ft 3 /s, v = 5 ft/s and C = 35. Solution: Working Formulas: Solve for d (for trapezoid): See Table 19.2 Solve for the slope: S = 0.0054 S = 0.0054 (answer)

A modified Bernoulli equation can be used, taking the horizontal datum as the channel bottom, z = 0, so the equation becomes But the first term, (p/ ) gives the depth of the water (it is the HGL), so the equation is further simplified and renamed specific energy, E. Setting the first derivative equal to zero and solving for d gives the critical depth, which occurs when the specific energy is at a minimum. And for rectangular channels, this reduces to several forms. Q 2 g A 3 surface width

Specific Energy Diagram is a plot of E versus d for a unique Q. See Ch. 19 of the CERM. d 1 and d 2 are conjugate depths because they share the same value of specific energy and are calculated only when there has been an abrupt energy loss, such as a hydraulic jump.

If the normal depth is calculated to be: less than the theoretical critical depth, then the flow regime is supercritical. greater than the theoretical critical depth, then the flow regime is subcritical. equal to the theoretical critical depth, then the flow regime is critical.

A hydraulic jump occurs only when the upstream normal depth is supercritical and the downstream normal depth is subcritical. For rectangular channels: gd V 1 ²=( 2 )(d 1 +d 2 ) (Eq. 19.94) 2d 1 More variations of this equation are found in the same section of the CERM.

Froude number is another way of calculating flow regime: Fr v gl Q2 b ga 3 (Eq. 19.79) v velocity Q flow g acceleration L length b surface width A cross-sectional area Its is equal to unity at the critical flow regime. Some problems you have to guess and later verify what flow you have.

SXCWE Prob. 37, Depth Derived from the velocity head equation, 2 h v 2g Derived from the continuity equation, Q Av

SXCWE Prob. 37, Depth (continued)

SXCWE Prob. 10, Breadth Where b = 2 m

SXCWE Prob. 10, Breadth (continued) CERM Eq. 19.30(a)

SXCWE Prob. 14, Breadth

SXCWE Prob. 14, Breadth (continued)

SXCWE Prob. 14, Breadth (continued)

Example problems dealing with open channel flow. Next topic: Stormwater Collection & Drainage