By

Assoc Prof Dr. Ir. Lariyah Mohd. Sidek Water Engineering Unit

Hydraulic Engineering CEWB222

HYDRAULICS -

CHAPTER 2 (OPEN CHANNEL

PROPERTIES)

Natural channel

• All watercourses that occurs naturally, for example streams

and rivers

• The Hydraulic properties are generally irregular

• Not regular nor prismatic and their materials of construction

can vary widely.

• The surface roughness will often change with time, distance

and elevation.

Types of open channel

Man-made channel Constructed in a regular cross-section shape throuhout and thus

are PRISMATIC CHANNELS.

Made from concrete, steel or earth with well defined surface roughness.

All watercourses that are constructed by human effort, for example :

Canal : a long and mild-sloped channel built in the ground that may be unlined or lined with stone masonry, concrete ,etc.

Flume : a channel of concrete, masonry, etc that is usually supported on or above the surface of the ground to convey water across a depression.

Chute : a channel with steep slopes.

Culvert : a covered channel of comparatively short length installed to drain water through highway and railroad embankments.

Types of open channel

Canal

Trunk Drain - Wet Period

Rigid Boundary Canal

FLUME

Existing Culvert

Existing Pipe Culvert

Existing Box Culvert

• Prismatic channel : a channel with

unvarying cross section and

constant bottom slope.

• Non-prismatic channel : a channel

with varying cross section and

constant bottom slope.

Channel Geometry

CHANNEL GEOMETRY

GEOMETRIC PROPERTIES • Depth (y) – the vertical distance from the lowest point of the

channel section to the free surface

• Stage (z) – the vertical distance from the free surface to an

arbitrary datum

• Area (A) – the cross-sectional area of flow, normal to the direction

of flow

• Wetted perimeter (P) – the length of the wetted surface measured

to the direction of flow

• Top Surface width (T) – width of the channel section at the free

surface

• Hydraulic Radius (R)- The ratio of area to wetted perimeter (A/P)

• Hydraulic Mean Depth (Dm) – The ratio of area to top surface

width (A/T)

12

CHANNEL GEOMETRY

13

Channel type Area

A

Wetted

Perimeter, P

Hydraulic radius

R

Top width

T

Hydraulic

depth, D

Section factor

Z

by

b+2y

y2b

by

b

y

5.1by

(b+zy)y

212 zyb

212

)(

zyb

yzyb

zyb 2

zyb

yzyb

2

)(

zyb

yzyb

2

5.1

2zy

212 zy

212 z

zy

zy2

y21

5.2

2

2zy

Ty3

2

T

yT

*2

3

8

22

2

83

2

yT

yT

Y

A

2

3

y32

5.169

2 Ty

2

0sin8

1 d

021 d

0

sin1

4

1d

ydy 02

21sin

sin

8

1

5.2

0

21

5.1

sin32

sin2d

y

b

RECTANGULAR CHANNEL

T Area (A) = by

Wetted perimeter (P) = b+2y

Hydraulic Radius (R) = A/P= y2b

by

Top width (T) = b

Hydraulic Depth (D) = y

b

y

TRAPEZOIDAL CHANNEL

Area = (b+zy)y

P = 212 zyb

212

)(

zyb

yzyb

R = A/P =

T = zyb 2

D = A/T = zyb

yzyb

2

)(

y

b

z

1

T

A concrete rectangular channel has a bottom

width of 4 meters. If the channel is on a 1%

slope and is flowing 1 meter deep throughout its

length, what is the cross sectional area, wetted

perimeter and hydraulic radius?

Solution:

If the shape of the channel is rectangular, the

discharge computed as follows:

A = by = (4) (1) = 4m2

P = b+2y = 4 + 2 (1) = 6m

R = 4m2 / 6m = 0.67m

Example 1

T

4

1

A concrete trapezoidal channel has a bottom width of 4

meters and 45 side slopes. If the channel is on a 1%

slope and is flowing 1 meter deep throughout its

length, what is the cross sectional area, wetted

perimeter and hydraulic radius?

Solution:

If the shape of the channel is trapezoidal, the discharge

computed as follows:

A = (b+zy)y = (4+(1)(1))1 = 5.00m2

P = (b+ 2y (1+ z2)0.5 )= 4 + 2 (1) (2)0.5 = 6.83m

R = A/P = 5.00m2/6.83m = 0.73m

Example 2

T

4

1 1

1

Water flows at a depth of 1.83 m in a trapezoidal,

concrete-lined section with a bottom width of 3 m and

side slopes of 2:1 (H:V). The slope of the channel is

0.0005 and the temperature is 20o C. Assuming uniform

flow conditions, estimate the cross sectional area (A),

wetted perimeter (P) and hydraulic radius (R)?

Solution:

Flow area, A = (b+zy)y = (3 + 2 (1.83)) x 1.83

= 12.2 m2

Wetted perimeter, P = 11.2 m

Hydraulic Radius, R = A/P = 12.2/ 11.2 = 1.09 m

Example 3

Example 4

Example 4 (cont’)

TEXTBOOKS REFERENCES

• Lariyah, M.S. & Norazli, O., Hydraulic Notes,

UNITEN, 2006

• Roberson, J. A., Cassidy, J. J. & Chaudry, H.,

Hydraulic Engineering, 1st S.I. Ed., John

Wiley, 1998

• Chin, D., Water Resources Engineering,

Prentice Hall, 2006