CE 344 - Topic 2.5 - 2003 - February 16, 2003 9:07 pm

2.5 MOST EFFICIENT CROSS SECTION • For steady uniform flow down a we developed the following relationship:

κ 2 ⁄ 3 1 ⁄ 2 u˜ = --R S (2.5.1) n H 0

where κ = unit coefficient = 1.486 for USCS units/1.000 for SI units (2.5.2)

n = Manning coefficient (2.5.3)

RH = Hydraulic radius (2.5.4)

S0 = bottom slope of the channel (2.5.5) • Total flow down the channel is

κ 2 ⁄ 3 1 ⁄ 2 Q = --AR S (2.5.6) n H 0

where A = cross-sectional area of water (2.5.7)

p. 2.5.1 CE 344 - Topic 2.5 - Spring 2003 - February 16, 2003 9:07 pm

• For a given area of water cross section, Q will be maximum when RH is maximum.

A • Since RH = ------Pw

5 ⁄ 3 κA 1 ⁄ 2 Q = ------S (2.5.8) n 2 ⁄ 3 0 Pw

• Thus for a given n, A and S0 , Q will be a maximum when Pw is a minimum. The corre- sponding cross section will be the most efficient cross section. • A circle has the least perimeter for a given area of any geometric shape.

πr2 r R = ------= -- (2.5.9) H 2πr 2

• A semi-circle has πr2 ≈------’ « 2 ◊ r R = ------= -- (2.5.10) H πr 2

p. 2.5.2 CE 344 - Topic 2.5 - Spring 2003 - February 16, 2003 9:07 pm

• Semi-circular open channel will more water than any other shape (assuming that the area, slope and surface roughness are the same). • Semi-circular shape/circular shape are practical for concrete and steel pipes. • excavated in earth must have a trapezoidal shape with slopes less than the angle of repose of the saturated material. • Thus design is based partly on bank stability considerations.

p. 2.5.3 CE 344 - Topic 2.5 - Spring 2003 - February 16, 2003 9:07 pm

Circular Section Not Flowing Full • Consider the following circular cross section under steady uniform flow conditions:

• Area of water can be computed as:

D2 A = ------(θ – sinθcosθ) (2.5.11) 4

2 D ≈ 1 ’ A = ------θ – -- sin2θ (2.5.12) 4 « 2 ◊

where D = pipe diameter (2.5.13)

θ = angle in radians (2.5.14)

p. 2.5.4 CE 344 - Topic 2.5 - Spring 2003 - February 16, 2003 9:07 pm

• Wetted perimeter is

θ Pw = D (2.5.15)

• The hydraulic radius is computed as A RH = ------(2.5.16) Pw

D≈ sin2θ’ R = --- 1 – ------(2.5.17) H 4 « 2θ ◊

• Recall that for steady uniform flow

κ 2 ⁄ 3 1 ⁄ 2 Q = --AR S (2.5.18) n H 0

2 ⁄ 3 • In order for Q to be maximum, ARH must be a maximum.

p. 2.5.5 CE 344 - Topic 2.5 - Spring 2003 - February 16, 2003 9:07 pm

• Define F as 2 ⁄ 3 F = ARH (2.5.19)

2 ⁄ D ≈ 1 ’ D≈ sin2θ’ 2 3 F = ------θ – -- sin2θ --- 1 – ------(2.5.20) 4 « 2 ◊ 4 « 2θ ◊

• Maximum value of F is found as follows: dF ------= 0 (2.5.21) dθ

• Solving for θ we find: 0 θ = 151.2 (2.5.22)

• This corresponds to

d0 = 0.938D (2.5.23)

• Thus the maximum discharge Q occurs at partial depth. • Note that for a given area A, the semi-circle would still allow for more flow.

p. 2.5.6