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بسم الله الرحمن الرحيم

Open Channel Flow)3- )الجريان في القنوات المفتوحةrd

Class Dr. Sataa A. Al-Bayati(09-10)

Introduction:

What is the difference between the flow in open-channel & closed pipe?

Open channel flow is flow of a liquid in a conduit in which upper surface of

liquid (free surface) is in contact with atmosphere.

Natural channels: rivers, & streams.

Manmade channels: irrigation canals, sewer lines, storm drains, & street

gutters.

What are the important parameters to be calculated in open-channel?

Fig.(1) Flow in an open-channel

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Fig.(2) Uniform flow in an open-channel

Calculation of Q using Moody diagram:

The discharge can be calculated as follows,

oRSf

gAAVQ

8 ------- (1)

Where:

A = cross section area

ƒ = friction factor

R = hydraulic radius = A/P

P = wetted perimeter

So= channel slope

What is Moody Diagram Fig.(3)?

V2/2g

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Fig.(3) Moody diagram: friction factors for commercial pipe (Featherstone,

R. E. & Nalluri, C., 1995)

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Example (1):

Determine the discharge in a long rectangular concrete channel that is 5ft

wide, that has a slope of 0.002, & in which the water depth is 2ft. Use

K=5×10-3

ft, ν = 1.22 × 10-5

ft2/s.

Solution:

What are the relation between D & R?

Given: rectangular ch., b = 5ft, So = 0.002, y = 2ft, K = 5x10-3

ft, &

ν = 1.22×10-5

ft2/s.

Q =?

522

52

P

AR

= 1.11ft

as D = 4R

11.14

105

4

3

R

K s

= 1.13 x 10-3

Assume ƒ = 0.020

Eq. (1)

oRSf

gAAVQ

8

002.011.102.0

2.32852

= 53.5ft3/s

V = Q/A = 53.5 / 10

= 5.35ft/s

RVRe

4

51022.1

11.1435.5

eR

= 1.95 × 106

From Moody diagram, Fig. (3), ƒ = 0.02 OK

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Stop & take Q = 53.5cfs.

Note: if ƒ is not equal repeat the calculation with the new ƒ.

*************

Calculation of Q using Manning Equation:

Manning Equation

What is Manning equation? Used for what?

2/13/249.1oSAR

nQ (EI) ----------- (2-a)

2/13/21oSAR

nQ (SI) ----------- (2-b)

Where:

n = Manning coefficient

Dividing the actual parameters of Eqs.(2) by the flowing full parameters we

get the following ( o or f = flowing full):

2/13/2

2/13/2

oo

o

o SnR

SnR

V

V

2/13/2

2/13/2

ooo

o

o SRnA

SnAR

Q

Q

From these two forms Fig.(4) is plotted.

Fig. (4) Flow characteristics of a circular section, Q/Qf vs. z=y/df.

V/Vf vs. z=y/df

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Fig. (4) Flow characteristics of a circular section

For quick solution Eqs.(2) is arranged in Fig. (5).

Fig. (5) Chart for flow of water in pipes flowing full, (Q, D, n, V, & S).

Need to know 3-parameters → find the other 2-parameters.

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Fig. (5) Chart for flow of water in pipes flowing full.

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Example (2):

Determine the discharge in a 3ft sewer pipe if the depth of flow is 1ft, &

pipe slope is 0.0019. Assume n = 0.012.

Solution:

Given: pipe ch., d = 3ft, y = 1ft, S = 0.0019, & n = 0.012.

Q =?

Use Fig. (5).

Draw a straight line through the points for n = 0.012 & S = 0.0019.

This line intersects match line.

Draw a line through intersection point & the 3ft diameter point.

Full discharge, Qf = 20mgd.

y/df = 1/3 = 0.333 → Fig. (4) → Q/Qf = 0.2

Q = 0.2Qf

= 0.2 × 20

= 4mgd.

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Flow in Channel of Trapezoidal Cross Section)الجريان في القنوات شبه منحرف(

Fig. (6) Determining the normal depth.

A R2/3

/ b8/3

vs. y/b (trapezoidal)

A R2/3

/ do8/3

vs. y/do (circular)

Where:

b = bottom channel width,

do = pipe diameter.

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Fig. (6) Curves for the computation of normal depth, s=z (Chow 1959)

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Example (3):

Determine the normal depth for a trapezoidal channel with side slopes of 1

vertical to 2 horizontal, a bottom width of 8ft, discharge of 200cfs, channel

bottom slope of 1ft in 1000ft, & n = 0.012.

Solution:

Given: trapezoidal ch., 1:z = 1:2, b =8ft,Q =200cfs, So= 1/1000, &n = 0.012.

y =?

We need (A R2/3

)/ b8/3

, so we do the following steps:

2/13/249.1oSAR

nQ

2/1

3/2

49.1 oS

nQAR

Divide the two sides by b8/3

,

3/82/13/8

3/2

49.1 bS

nQ

b

AR

o

3/82/1 )8()001.0(49.1

012.0200

= 0.199

Go to Fig. (6) with (A R2/3

)/ b8/3

= 0.199 → y/b = 0.33

y = 0.33 × b

= 0.33 × 8

= 2.64ft.

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Design of Erodible Channels)القنوات القابلة للتآكل(

They are channels constructed in erodible material e.g. soil or fine gravel.

These channels may be eroded. Methods for designing erodible channels are:

1. Permissible velocity method.

2. Tractive force method.

Permissible velocity method:

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Table (1) Permissible side slopes for different materials

Material Side slopes

Rock Nearly vertical

Stiff clay or earth with concrete lining 1: ½ to 1:1

Firm soil 1:1

Loam sandy soil 1:2

Sandy loam 1:3

Table (2) Maximum permissible velocities for different materials, with n

values

Material V, ft/s n

Fine sand 1.50 0.020

Sandy loam 1.75 0.020

Silt loam 2.00 0.020

Firm loam 2.50 0.020

Stiff clay 3.75 0.025

Fine gravel 2.50 0.020

Coarse gravel 4.00 0.025

Example (4):

An unlined irrigation canal is to be constructed in a firm loam soil. The slope

is to be 0.0006, & it is to carry a water flow of 100cfs. Determine an

appropriate cross section for this canal.

Solution:

Given: Firm loam soil, So = 0.0006, & Q = 100cfs

y& b =?

From Table (1), firm soil → 1:1

From Table (2): max. permissible V = 2.5ft/s & n = 0.02

2/13/249.1oSR

nV

or

2/3

2/149.1

oS

VnR

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2/3

2/1)0006.0(49.1

5.202.0

R

= 1.6ft

A = Q/V

= 100/2.5

= 40ft2

R = A/P

P = A/R

= 40/1.6 = 25ft

A = by + zy2 = by + y

2 = 40

_____ _

P = b + 2y√1 + z2 = b + 2√2 y = 25

Solve these two Eqs. simultaneously

b = 19.8ft & y = 1.85ft

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Best Hydraulic Section)أفضل مقطع هايدروليكي(

Section factor = A R2/3

= A (A/P)2/3

For given n & So in Manning eq.

Q = (1/n) A R2/3

So1/2

= (1/n) A (A/P)2/3

So1/2

Q increase with A → Q α A

decrease with P, → Q α 1/P

For given A & shape e.g. rectangular, there will be a certain

y/b → A (A/P)2/3

= max. → Qmax

Such a ratio establishes the best hydraulic section.

Best hydraulic section: is channel proportion that yields a Pmin. & Qmax for a

given A.

Given A: y/b → Pmin.

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Example (5):

Determine the best hydraulic section for a rectangular channel.

Solution:

A = by --- (1)

P = b + 2y --- (2)

Let A be constant, then from Eq.(1), b = A/y

Take Eq.(2),

P = b + 2y

= A/y + 2y

By differentiation,

022

y

A

dy

dp

A/y2 = 2

But A = by

Then, b/y = 2 → y = b/2

Therefore, best hydraulic section for a rectangular channel occurs when the

depth is one half the widths (½ square).

***********************

The best hydraulic section for;

Trapezoidal channel = 1/2 (hexagon)

Circular = 1/2 (circle)

Triangular = 1/2 (square)

Best hydraulic section → min. excavation → min. cost

Specific Energy)الطاقة النوعية(

Specific Energy = depth + velocity head

E = y + V2/2g

E1 = E2 = ---

2

2

2

22

1

2

122 gA

Qy

gA

Qy ----- (1)

Fig. (7) Relation of y vs. E

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Fig. (7) Specific energy curve

Characteristics of Critical Flow)خواص الجريان الحرج(

Critical flow occurs at E for a given Q

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2

c

c

gA

TQ --- (2)

Multiply by Ac & divided by Tc ,

2

2

cc

c

gA

Q

T

A --- (3)

Q = VA,

g

V

T

A c

c

c

2

---- (4)

Where:

T = channel top width

Example (6):

Determine the critical depth in a trapezoidal channel for a discharge of

500cfs. The width of the channel bottom is 20ft, & the sides slope upward at

an angle of 45o.

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Solution:

Given: Q =500cfs, b = 20ft, & 1:z = 1:1.

yc =?

Eq.(2), 13

2

c

c

gA

TQ

or

g

Q

T

A

c

c

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= (500)2 / 32.2 = 7764ft

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But the A & T of trapezoidal channel are,

A = y (b + y) = y (20 + y)

T = b + 2y = 20 + 2y

Therefore,

7764220

20323

y

yy

T

A

c

c

By iteration: choose different y & place in the left side of the equation to

satisfy the right side =7764.

yc = 2.57ft

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For rectangular channel:

c

c

cT

Ay

&

2

2

2

2

cc y

q

A

Q

Where: T

Qq = discharge / unit width

Eq. (3), 2

2

cc

c

gA

Q

T

A

becomes,

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3

2

g

qyc -----(5)

Eq. (4)

g

V

T

A c

c

c

2

yc = Vc2 /g

1c

c

gy

V Frc (critical Froud number) ----(6)

__

What is Froud number? Fr = V / √g y

Critical flow is unstable character of flow. Therefore, flow in ordinary canals

is subcritical. Critical flow location is useful in discharge measurement,

why?

Eq. (3)

2

2

cc

c

gA

Q

T

A

Rearrangement

2/1

2/3

T

A

g

Q = Zc

_

For each Q /√g & shape → one yc

Fig. (8) curves for determining yc

5.2

oBg

Q

vs. yc/Bo (trapezoidal)

5.2

oDg

Q

vs. yc/Do (circular)

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Fig.(8) Curves for determining the critical depth, s=z (Chow, 1959)

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Example (7):

Solve example (6) using Fig. (8).

Solution:

5.2bg

Q

=

5.2)20(2.32

500

= 0.0493

Fig. (8) with z = 1 → yc/Bo = 0.13

yc = 0.13 * 20

= 2.6ft

References:

- Chow, V. T., 1959, “ Open-channel Hydraulics” Japan.

- Roberson, J.A., et.al., “ Hydraulic Engineering”, 2nd Ed, John Wiley & Sons. Inc., New York.

- Featherstone, R. E. & Nalluri, C., 1995, "Civil Eng. Hydraulics", 3rd

Ed,

Blackwell Sc. Ltd.

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H.W: 05-06

M E

Ex.1: wide = 5 + 0.1(no.) = 5.1ft depth = 2 + 0.1(no.) = 2.1ft

Ex.2: Dia. = 3 + 0.1(no.) = 3.1ft D = 6ft & y = 1 + 0.1 = 1.1ft

Ex.3 b = 8 + 0.2(no.) = 8.2ft Q = 200 + 5(no.) = 205cfs

Ex.4 stiff clay sandy loam

Q = 100 + 5(no.) = 105cfs Q = 100 – 2(no.) = 98cfs

Ex.6 Q = 500 + 10(no.) = 510cfs Q = 500 – 10(no.) = 490cfs

Ex.7

H.W: 06-07

M E

Ex.1: wide = 6 + 0.1(no.) = 6.1ft depth = 3 + 0.1(no.) = 3.1ft

Ex.2: D = 4 + 0.1(no.) = 4.1ft D = 6ft & y = 1.5 + 0.1 = 1.6ft

Ex.3 b = 9 + 0.2(no.) = 9.2ft Q = 220 + 5(no.) = 225cfs

Ex.4 Q = 90 - 5(no.) = 85cfs Q = 110 + 5(no.) = 115cfs

Ex.6 Q = 450 - 10(no.) = 440cfs Q = 550 + 10(no.) = 560cfs

Ex.7

H.W: 07-08

E M

Ex.1: width = 6 + 0.1(no.) = 6.1ft depth = 3 + 0.1(no.) = 3.1ft

Ex.2: D = 4 + 0.1(no.) = 4.1ft D = 6ft & y = 1.5 + 0.1 = 1.6ft

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Ex.3 b = 9 + 0.2(no.) = 9.2ft Q = 220 + 5(no.) = 225cfs

Ex.4 Q = 90 - 5(no.) = 85cfs Q = 110 + 5(no.) = 115cfs

Ex.6 Q = 450 - 10(no.) = 440cfs Q = 550 + 10(no.) = 560cfs

Ex.7