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

Example:

(eq. 19.8)

Consider an open channel shown in figure. Section 1 has an area of 2 ft2 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

Section 2

Solution:Given: S1: Area = 2 ft2; Velocity = 15 ft/secS2: Area = Unknown;Velocity = 5 units higher that S1

Thus (15 + 5) = 20 ft/sec

Using continuity equation: A1V1 = A2V2

(2 ft2) x (15 ft/sec) = A2 x (20 ft/sec)

A2 = 1.5 ft2 Answer

Chezy equation: (Eq. 19.9)

Manning’s equation for C value:

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)

(Eq. 19.11(b))

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

Combining Eq. 19.9 and 19.11

For flow rate (Q) us Manning’s Equation:

(Eq. 19.12 (b))

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

2. C – Value using Manning’s Equation

R = A/P: A = 7 ft x 4 ft = 28 ft2

P = 7 ft + 4 ft + 4 ft = 15 ft

R = (28 ft2)/(15 ft) = 1.87 ft

From App. 19.A, the roughness coefficient of rubble

masonry is 0.017

C = (1.49/0.017) x (1.87)^ (1/6) = 97.29

3. The flow rate (Q)

Q =vA = (C(RS)^(1/2))A =

(97.29(1.87x.003)^(0.5))28 =

= 204.03 ft3/s Answer

Note: wetted perimeter is the length which is touching

water.

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,

2. Manning Equation : Q = Av (Eq. 19.13b)

R= 𝐴

𝑃=

𝜋𝐷²

8𝜋𝐷

2

=𝐷

4(Eq. 19.2 read)

Q= vA = (1.49

𝑛)𝐴𝑅2/3 𝑆 = (

1.49

0.011)(𝜋𝐷2

8)(𝐷

4)2/3( 0.0065) =

𝐷8/3 = 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 ft3/s, v = 5 ft/s and C = 35.

Solution:

Working Formulas:

See Table 19.2

Solve for the slope:

Solve for d (for trapezoid):

S = 0.0054S = 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.

Q2

g

A3

surface width

Specific Energy Diagram is a plot of E versus d

for a unique Q. See Ch. 19 of the CERM.

d1 and d2 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:

V1²=( )(d1+d2) (Eq. 19.94)

More variations of this equation are

found in the same section of the CERM.

2

12

gd

d

Froude number is another way of calculating flow regime:

Its is equal to unity at the critical flow regime. Some problems you have to guess and later verify what flow you have.

Fr v

gL

Q2b

gA3 (Eq. 19.79)

Q

g b

L A

v velocity flow

acceleration surface width

length cross-sectional area

Derived from the velocity head equation,

2

2h

g

v

Derived from the

continuity equation, Q A v

SXCWE Prob. 37, Depth

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.

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