Hydraulic Engineering Eng. Osama Dawoud. Lecture 14 Open Channel Flow.

Hydraulic EngineeringHydraulic

Engineering

Eng. Osama Dawoud

Eng. Osama Dawoud

Lecture 14

Open Channel Flow

Open Channel

• Open channel hydraulics, a subject of great importance to civil engineers, deals with flows having a free surface in channels constructed for water supply, irrigation, drainage, and hydroelectric power generation; in sewers, culverts, and tunnels flowing partially full; and in natural streams and rivers.

Pipe system

Open Channel

Classification

• Steady Flow– Not time

dependent

• Unsteady Flow– Is time dependent

Typical situations Uniform flow

Gradually Varied Flow Rapidly Varied Flow

Open channel section types

Uniform Flow

SRCV h

Chezy equation

C is the Chezy C, a dimensional factor which characterizes the resistance to flow

P wetted

A wetted Radius hydraulic

slope bedS

Rh

Manning equation

2/13/21SR

nV h

tCoefficien M

P wetted

A wetted Radius hydraulic

anningn

slopebedS

Rh

Example 1

1.5m

3.0m

2

1

open channel of width = 3m as shown, bed slope = 1:5000, d=1.5m find the flow rate using Manning equation, n=0.025.

sVAQ

V

P

AR

P

A

SRn

V

h

h

/m 84.49538.0

m/s 538.050001927.0

025.0

1

927.0708.9

9

9.708 35.132

m 95.1935.0

1

3

3

2

22

2

3

2

Example 2open channel as shown, bed slope = 69:1584, find the flow rate using Chezy equation, C=35.

sVAQ

V

P

AR

P

A

SRCV

h

h

/m 84.11352.1627.0

m/s 7.01584

69.0917.035

917.018.177

52.162

m 177.18 04.552.28.166.38.115072.0

m 52.16215072.06.32

52.272.08.1652.2

2

04.552.2

3

2222

2

Most Efficient Sections

During the design stages of an open channel, the channel cross-section, roughness and bottom slope are given.

The objective is to determine the flow velocity, depth and flow rate, given any one of them. The design of channels involves selecting the channel shape and bed slope to convey a given flow rate with a given flow depth. For a given discharge, slope and roughness, the designer aims to minimize the cross-sectional area A in order to reduce construction costs

Most Efficient Sections

The most ‘efficient’ cross-sectional shape is determined for uniform flow conditions. Consideringa given discharge Q, the velocity V is maximum for the minimum cross-section A. According to the Manning equation the hydraulic diameter is then maximum.

It can be shown that:1. the wetted perimeter is also minimum,2. the semi-circle section (semi-circle having its

centre in the surface) is the best hydraulic section

Most Efficient Sections

Because the hydraulic radius is equal to the water cross section area divided by the wetted parameter, Channel section with the least wetted parameter is the best hydraulic sectionRectangular section

DBA

BD2P

D

A 2DP

0 dD

dP

222 2 0 2 D

DB

D

A

D

A

dD

dP

D

B 2

2

B D

Trapezoidal section D)Dk(BA

212 kDBP

Dk D

A B

212 kD ) kD D

A(P

0 dD

dP

0 12 22

kkD

A

dD

dPk

D

Ak12

22

D

DkBk

D

DkD)(Bk

2 12

22

2

Dk2Bk1D 2

ork

k

Other criteria for economic Trapezoidal section

DOF

The best side slope for Trapezoidal section

3

1 k 60

0 dk

dP

k

Circular section

2sin8

4

22 ddA

drP 2

dD 95.0154 Maximum Flow using Manning

Maximum Flow using Chezy

dD 94.0151

dD 81.075.128

Maximum Velocity using Manning or Chezy

Example 3Circular open channel as shown d=1.68m, bed slope = 1:5000, find the Max. flow rate & the Max. velocity using Chezy equation, C=70.

sVAQ

V

mP

AR

dP

ddA

SRCV

h

h

/m 496.117.269.0

m/s 69.05000

1485.070

485.05.4

17.2

m 4.5 68.1180

154

m 17.21542sin8

68.1

180154

4

68.12sin

84

3

22222

154Max. flow rate

m/s 748.05000

157.070

57.03775.3

93.1

m 378.368.1180

75.128

m 93.175.1282sin8

68.1

18075.128

4

68.12sin

842

2222

V

mP

AR

dP

ddA

SRCV

h

h

75.128Max. Velocity

Variation of flow and velocity with depth in circular pipes

Circular open channel as shown Q=10m3/s, velocity =1.5m/s, for most economic section. find wetted parameter, and the bed slope n=0.014.

Example 4

mD

DDDA

DkDBA

mV

QA

BD

DBD

kDBkD

78.1

667.6)2

36055.0(

667.65.1

10

6055.02

232

231

2

21

2

2

2

mP

kDDP

kDBP

49.72

3178.12)78.1(6055.0

126055.0

12

2

2

2

To calculate bed Slope

6.1941:1

5.189.0014.0

1

89.049.7

667.6

m 49.7

m 667.6

1

3

2

2

3

2

S

SV

P

AR

P

A

SRn

V

h

h