CE 344 - Topic 2.5 - Spring 2003 - February 16, 2003 9:07 pm
2.5 MOST EFFICIENT CROSS SECTION
• For steady uniform flow down a channel we developed the following relationship:
κ 2⁄3 1⁄2
ũ = --- R H S 0
n
(2.5.1)
where κ = unit coefficient = 1.486 for USCS units/1.000 for SI units
(2.5.2)
n = Manning coefficient
(2.5.3)
R H = Hydraulic radius
(2.5.4)
S 0 = bottom slope of the channel
(2.5.5)
• Total flow down the channel is
κ 2⁄3 1⁄2
Q = --- AR H S 0
n
(2.5.6)
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 R H is maximum.
A
Pw
• Since R H = -----5⁄3
κA
1⁄2
Q = --- ---------S
n P2 ⁄ 3 0
(2.5.8)
w
• Thus for a given n, A and S 0 , Q will be a maximum when P w is a minimum. The corresponding cross section will be the most efficient cross section.
• A circle has the least perimeter for a given area of any geometric shape.
2
RH
r
πr
= --------- = --2
2πr
(2.5.9)
• A semi-circle has
2
RH
πr
-------r
2
= -------------- = --2
πr
p. 2.5.2
(2.5.10)
CE 344 - Topic 2.5 - Spring 2003 - February 16, 2003 9:07 pm
• Semi-circular open channel will discharge 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.
• Canals excavated in earth must have a trapezoidal shape with slopes less than the angle
of repose of the saturated bank material.
• Thus canal 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:
2
D
A = ------ ( θ – sin θ cos θ )
4
(2.5.11)
2
D
1
A = ------ θ – --- sin 2θ
2
4
(2.5.12)
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
P w = Dθ
(2.5.15)
• The hydraulic radius is computed as
A
R H = -----Pw
(2.5.16)
sin 2θ
D
R H = ---- 1 – -------------2θ
4
(2.5.17)
• Recall that for steady uniform flow
κ 2⁄3 1⁄2
Q = --- AR H S 0
n
• In order for Q to be maximum, AR 2H⁄ 3 must be a maximum.
p. 2.5.5
(2.5.18)
CE 344 - Topic 2.5 - Spring 2003 - February 16, 2003 9:07 pm
• Define F as
2⁄3
F = AR H
2
sin 2θ
D
D
1
F = ------ θ – --- sin 2θ ---- 1 – -------------2θ
4
2
4
(2.5.19)
2⁄3
(2.5.20)
• Maximum value of F is found as follows:
dF
------- = 0
dθ
(2.5.21)
• Solving for θ we find:
θ = 151.2
0
(2.5.22)
• This corresponds to
d 0 = 0.938D
• 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
(2.5.23)