Bncftk Stress.pdf

BNP20103 Hydraulics and Hydrology

Chapter 2. Uniform Flow in Open Channel

At the end of this chapter, students should be able to: i.

Understand the concept of uniform flow

ii. Calculate normal flow depth in variable channel sections using Chezy and Manning equations iii. Determine the best hydraulic/effective section of open channel

• Uniform flow is considered to be steady only, since unsteady uniform flow is practically does not exist. • Steady uniform flow is rare in natural streams, only happens in prismatic channels. • We adopt / assume uniform flow for most flow computations because uniform flow calculation is simple, practical and provide satisfactory solution.

The 132 km long All-American Canal links California's Imperial Valley to the Colorado River. This new concrete-lined section saves about 3.8 million of water a year over its leaky earthen forerunner

The concrete channel of Los Angeles River (NGM, 2010)

The Klang River, Kuala Lumpur & Selangor

• In uniform flow, the normal depth yo occurs when depth of water is the same along the channel. • Normal depth yo implies that the water depth, flow area, wetted perimeter, velocity and discharge at every section of the channel are constant within a prismatic channel. • Thus, in uniform flow, the energy line, water surface and channel bottom are parallel, i.e. the slopes are equal Sf = Sw = So = S.

V2 2g yo

1.1 Velocity Distribution Depends on the geometry of the channel and wetted boundary roughness y Vmax Vmax 0.84 0.82 0.80 0.76 0.70 0.62 0.48

Natural channel

0.2yo 0.53 yo

0.52 0.50 0.45 0.40 0.35

Rectangular channel

Vmax

0.6yo yo Vaverage V

Velocity distribution

1.2 Chezy and Manning Equations Two most common equations used in the uniform flow computations: 1. Chezy formula

1 2

V  CR S

1 2 o

C = Chezy roughness coefficient

2

1 3 12 2. Manning formula V  R So n

n = Manning roughness coefficient

Thus, the general uniform flow equation: V  constant R x Sox

1

1 6 Difference between Chezy and Manning formulae C  R n

Factors determining the roughness are surface roughness, vegetation, channel irregularity, channel alignment, silting and scouring, obstruction, size and shape of channel, stage and discharge, seasonal change, and suspended material and bed load.

Derivation of Chezy equation The Chezy two assumptions are: 1. The force resisting the flow per unit area of the channel bed is proportional to the square of the velocity: Ff  kV 2PL 2. The effective component of the gravity force causing the flow must be equal to the total force of resistance. This is also the basic principle of uniform flow where uniform flow will be developed if the resistance is balance by the gravity forces: Fg   ALsin

V2 2g

A yo W P

Datum

p1  Wsin  Ff  p2  M2  M1

Since for uniform flow, p1  p2 and M1  M2

kV 2PL   ALsin kV 2PL   ALSo

 A V  So k P 2

1 2

  2 2 V    R So  k  1 2

V  CR S

1 2 o

1

1

where C = Chezy coefficient

Example 2.1 A rectangular channel 2.0 m wide carries water at a depth of 0.5 m. The channel is laid on a slope of 0.0004. The Chezy coefficient is 73.6. Compute the discharge of the channel. Given B = 2.0 m, y = 0.5 m, So = 0.0004 and C = 73.6 A = By = 1 m2, P = B + 2y = 3 m, R = 1/3 m Q  AC RSo y

1 Q  1  73.6  0.0004 3 Q  0.850 m /s 3

B

Example 2.2 Water flows in a triangular channel with side slope 1.5(H) : 1(V), bottom slope 0.0002 and Chezy coefficient of 67.4. The depth of flow is 2.0 m. Find the flow rate and average velocity. Based on Froude number, determine the state of flow. Given y = 2.0 m, z = 1.5, So = 0.0002 and C = 67.4 A = zy2 = 6 m2, P = 2y = 7.211 m, R = A/P = 0.832 m, D = A/T = 6/2zy = 1 m V  C RSo

V  67.4 0.832 0.0002 V  0.869 m/s

1

y z

Q  AV Q  6 0.869 Q  5.217 m3 /s

Fr 

V gD

0.869 Fr  9.81  1

Fr  0.277  subcritical flow

Chezy resistance factor C The following two equations can be used to determine Chezy coefficient:

0.00155 1 23   So n n = Kutter coefficient 1. Ganguillet-Kutter C   0.00155  n  1   23  So  R 

2. Bazin C 

87 m 1 R

m = Bazin coefficient

Table 1.1a Values of Manning roughness coefficient n Surface characteristics

Range of n

(a) Lined channels with straight alignment Concrete i. formed, no finish

0.013 - 0.017

ii. trowel finish

0.011 - 0.015

iii. float finish

0.013 - 0.015

iv. gunite, good section

0.016 - 0.019

v. gunite, wavy section

0.018 - 0.022

Concrete bottom, float finish, sides as indicated i. dressed stone in mortar

0.015 - 0.017

ii. random stone in mortar

0.017 - 0.020

iii. cement rubble masonry

0.020 - 0.025

iv. cement rubble masonry, plastered

0.016 - 0.020

v. dry rubble (rip-rap)

0.020 - 0.030

Tile

0.016 - 0.018

Brick

0.014 - 0.017

Table 1.1b Values of Manning roughness coefficient n Surface characteristics Sewers (concrete, asbestos-cement, vitrified-clay pipes)

Range of n 0.012 - 0.015

Asphalt i. smooth

0.013

ii. rough

0.016

Concrete lined, excavated rock

i. good section

0.017 - 0.020

ii. irregular section

0.022 - 0.027

Laboratory flumes-smooth metal bed, glass or perspex sides

0.009 - 0.010

Manning roughness coefficient n = 0.020 - 0.022

Manning roughness coefficient n = 0.020 - 0.022

Manning roughness coefficient n = 0.022 - 0.024

Manning roughness coefficient n = 0.020

Table 1.1c Values of Manning roughness coefficient n Surface characteristics

Range of n

(b) Unlined, non-erodible channels Earth, straight and uniform i. clean, recently completed

0.016 - 0.020

ii. clean, after weathering

0.018 - 0.025

iii. gravel, uniform section, clean

0.022 - 0.030

iv. with short grass, few weeds

0.022 - 0.033

Channels with weeds and brush, uncut i. dense weeds, high as flow depth

0.050 - 0.120

ii. clean bottom, brush on sides

0.040 - 0.080

iii. dense weeds or aquatic plants in deep channels

0.030 - 0.035

iv. grass, some weeds

0.025 - 0.033

Rock

0.025 - 0.045

Table 1.1d Values of Manning roughness coefficient n Surface characteristics

Range of n

(c) Natural channels Smooth natural earth channels, free from growth, little curvature

0.020

Earth channels, considerably covered with small growth

0.035

Mountain streams in clean loose cobbles, rivers with variable section with some vegetation on the banks

0.040 - 0.050

Rivers with fairly straight alignment, obstructed by small trees, very little under brush

0.060 - 0.075

Rivers with irregular alignment and cross-section, covered with growth of virgin timber and occasional patches of bushes and small trees

0.125

Manning roughness coefficient n = 0.11

Manning roughness coefficient n = 0.20

Grassed swale

Table 2.2 Values of Manning roughness coefficient for grassed swale Surface cover

Manning n

Short grass

0.030 - 0.035

Tall grass

0.035 - 0.050

Table 1.3 Proposed values of Bazin coefficient m Description of channel

Bazin coefficient m

Very smooth cement of planed wood

0.11

Unplaned wood, concrete, or brick

0.21

Ashlar, rubble masonry, or poor brickwork

0.83

Earth channels in perfect condition

1.54

Earth channels in ordinary condition

2.36

Earth channels in rough condition

3.17

Example 2.3 Calculate the velocity and discharge in a trapezoidal channel having a bottom width of 20 m, side slopes 1(H) : 2(V), and a depth of water 6 m. Given Kutter's n = 0.015 and So = 0.005. Given B = 20 m, y = 6.0 m, z = 0.5, So = 0.005 and n = 0.015

A = By + zy2 = 138 m2, P = B + 2y = 33.42 m, R = A/P = 4.13 m

1

y z B

Ganguillet-Kutter

0.00155 1 23   So n C  0.00155  n  1   23  So  R  0.00155 1  0.005 0.015 C 0.00155  0.015 1   23   0.005  4.13  23 

C  76.769

Chezy velocity V  C RSo

V  76.769 4.13 0.005 V  11.03 m/s

Discharge Q  AV

Q  138 11.03 Q  1522.14 m3 /s

Example 2.4 Find the equivalent Bazin coefficient m for the question in Example 2.3 and compare the Chezy coefficients obtained from Kutter n & Bazin m. Known A = 138 m2, P = 33.42 m, R = 4.13 m Assume that for concrete with Kutter n = 0.015, Bazin m = 0.21

87 Bazin C  m 1 R 87 C 0.21 1 4.13 C  78.852 (from Bazin)  76.769 (from Ganguillet - Kutter)

Example 2.5 A trapezoidal channel is 10.0 m wide and has a side slope of 1.5(H) : 1(V). The bed slope is 0.0003. The channel is lined with smooth concrete n = 0.012. Compute the mean velocity and discharge for a depth of flow of 3.0 m. Given B = 10 m, y = 3.0 m, z = 1.5, So = 0.0003 and n = 0.012 A = By + zy2 = 43.5 m2, P = B + 2y 1  z 2 = 20.817 m, R = A/P = 2.090 m 1

y z B

2

1 3 12 Manning velocity V  R So n 2 1 1 V  2.090 3  0.00032 0.012 V  2.359 m/s

Discharge Q  AV  43.5  2.359

 102.625 m3 /s

Example 2.6 In the channel of Example 2.5, find the bottom slope necessary to carry only 50 m3/s of the discharge at a depth of 3.0 m. Given B = 10 m, y = 3.0 m, z = 1.5 and n = 0.012 and A = 43.5 m2, P = 20.817 m, R = 2.090 m 2

1 3 12 Manning discharge Q  AR So n 2 1 1 50   43.5  2.09 3  So2 0.012

So  0.0000712

So  7.12  105

Example 2.7 A triangular channel with an apex angle of 75 carries a flow of 1.2 m3/s at a depth of 0.80 m. If the bed slope is 0.009, find the roughness coefficient C and n of the channel. Given y = 0.80 m, So = 0.009,  = 75, and Q = 1.2 m3/s

 75    z  tan   tan   0.767 2  2  and A = zy2 = 0.491 m2, P = 2y 1  z 2 = 2.017 m, R = A/P = 0.2435 m

1

75

z

y

2

1 3 12 Using Manning equation Q  AR So n 2

1 1 1.2   0.491  0.24353  0.0092 n n  0.0151

1 2

Using Chezy equation Q  CAR S

1 2 o 1 2

1.2  C  0.491 0.2435  0.009 C  52.197

1 2

Conveyance Conveyance K of a channel section is a measure of the carrying capacity of the channel section per unit longitudinal slope. It is directly proportional to discharge Q. 1. Chezy formula

1 2

Q  CAR S 2

1 2 o

 K  CAR

1 2

2

1 3 12 1 2. Manning formula Q  AR So  K  AR 3 n n

1.4 Section Factor Section factor Z in the Manning formula is AR2/3, which is a function of the depth of flow. 2

1 3 12 In Manning formula Q  AR So n

Therefore,

2 3

AR 

Qn S

1 2 o

Section factor AR2/3 is normally used to compute the normal depth yo when the discharge Q, bottom slope So and Manning roughness coefficient n are provided. Computation of yo could be through either direct trial-and-error computation, based on graph, or through provided design chart.

Example 2.8 A trapezoidal channel 5.0 m wide and having a side slope of 1.5(H) : 1(V) is laid on a slope of 0.00035. The roughness coefficient n = 0.015. Find the normal depth for a discharge of 20 m3/s through this channel. Given B = 5.0 m, z = 1.5, So = 0.00035, n = 0.015, and Q = 20 m3/s A = By + zy2 = 5y + 1.5y2 P = B + 2y 1  z 2 = 5 + 2 3.25y

A 5y  1.5y 2  R  P 5  2 3.25y  1

y z B

2

1 3 12 Manning equation, Q  AR So n Arranging Manning equation as a function of section factor, 2 3

AR 

Qn S

1 2 o

2 3

 5yo  1.5yo2  20  0.015   5yo  1.5y  1 5  2 3 . 25 y  o  0.000352 2 o

5y

o  1.5y

5  2



5 2 3 o

3.25y o 

2 3

 16.036

Therefore, yo = 1.820 m

By trial-and-error: yo (m)

5y

o

5  2

 1.5y



5 2 3 o

3.25y o 

1

5.391

2

19.159

1.8

15.706

1.820

16.035

2 3

Graphically,

5y

o

 1.5y

5  2



5 2 3 o

3.25y o 

yo (m)

2 3

 16.036

5y

o

5  2

 1.5y

Therefore, yo = 1.820 m



5 2 3 o

2.5

3.25y o 

2 3

5.391

1.5

11.198

1.7

14.115

1.8

15.706

0.5

1.9

17.387

0

19.159

y o (m)

1

2

yo = 1.82 m

2 1.5 1

16.036 0

5

10

15 AR 2/3

20

25

Design Chart is available,

Rectangular (z = 0)

y y and B do

0.37 Circular

B

0.2194 AR B

8 3

2 3

and

AR d

8 3 o

2 3

2 3

At the x-axis, AR  16.036

AR B

8 3

2 3



16.036 5

8 3

 0.2194

Intersecting at z = 1.5 of trapezoidal channel gives

y  0.37 B y  0.37  5 Therefore, yo = 1.85 m

Example 2.9 A concrete-lined trapezoidal channel with n = 0.015 is to have a side slope of 1(H) : 1(V). The bottom slope is to be 0.0004. Find the bottom width of the channel necessary to carry 100 m3/s of discharge at a normal depth of 2.50 m. Given yo = 2.5 m, z = 1, So = 0.0004, n = 0.015, and Q = 100 m3/s A = By + zy2 = 2.5B + 6.25 P = B + 2y 1  z 2 = B + 7.071

A 2.5B  6.25 R  P B  7.071 1

y z B

Manning equation as a function of section factor, 2 3

AR 

Qn S

1 2 o

2 3

2.5B  6.25  100  0.015   1  B  7.071  0.00042

2.5B  6.25

2.5B  6.25 2 3

B  7.071

5 3

 75

By trial-and-error, B = 16.33 m

Activity 1

Water flows uniformly at 10 m3/s in a rectangular channel with a base width of 6.0 m, channel slope of 0.0001 and Manning's coefficient n = 0.013. Using trial-and-error method, find the normal depth.

Example 2.10 A sewer pipe of 2.0 m diameter is laid on a slope of 0.0004 with n = 0.014. Find the depth of flow when the discharge is 2 m3/s.

D2 Area A = 2  sin2  8 Perimeter P = D

r yo

2

D

2 3

Manning equation: AR 

Qn S

2 3

AR 

1 2 o

2 0.014 0.0004

For design chart:

AR D

8 3

AR D

2 3

2 3

8 3



1 2

1.4 2

8 3

 0.2205

Intersecting at circular section gives

yo  0.6 D yo  0.6  2 = 1.20 m

Design Chart:

Rectangular (z = 0)

y y and B do

0.6 Circular

B

0.2205

AR B

8 3

2 3

and

AR d

8 3 o

2 3

Simplification for Wide Rectangular Channel yo  0.02 Wide channel: B y For wide channel, o is small, therefore R  y o B Or simply, R  yo

Discharge per unit width Normally used in rectangular channels.

Q Discharge per unit width q  B

or

q  yV

Unit is m3/s/m. BFC21103 Hydraulics Tan et al. ([email protected])

Example 2.11 Water flows through a very wide channel at a rate of 2.5 m3/s/m. The channel has a base width of 60 m, channel slope of 0.005 and Manning's coefficient of 0.013. What is the normal depth? Given: q = 2.5 m3/s/m, B = 60 m, So = 0.005, n = 0.013

For a wide rectangular channel, R = y Manning equation: q  yoV 2

1 3 12 q  y o R So n 5

1 3 12 q  y o So n 5 1 1 2.5  y o3  0.0052 0.013 yo  0.6272 m

1.5 Best Hydraulic Section (Most Effective Section) A non-erodible channel should be designed for the best hydraulic efficiency. Best hydraulic section gives minimum area for a given discharge. Referring to the channel conveyance, 2

1 3 K  AR n for a constant flow area A, the conveyance increases with increase in hydraulic radius R or decrease in the wetted perimeter P. Simply, Qmax, Rmax and Pmin gives best hydraulic section. Pmin - reduces construction cost (less lining material), and - reduces friction force.

Table. Best hydraulic sections Cross section

Side slope z

Area A

Trapezoid

1 3

3y 2

Rectangle

-

2y2

Triangle

1

y2

Semicircle

-

y2

Parabola

-

 2

4 2 2 y 3

Wetted perimeter P

Hydraulic radius R

Top width T

y 2

4 3 y 3

y 2

2 2y

Hydraulic depth D

Section factor Z

3 y 4

3 2.5 y 2

2y

y

2y 2.5

2y 4

2y

y 2

2 2.5 y 2

y

y 2

2y

8 2 y 3

y 2

2 2y

2 3y 4y

 4

y

2 y 3

 4

y 2.5

8 3 2.5 y 9

What is the best hydraulic section for a rectangular channel?

For a rectangular channel,

A  By P  B  2y

For best hydraulic section

dP 0 dy

y B

Let's first assume A to be constant: A  2y y dP A   2 2 dy y P

For best hydraulic section

A  2 2  0 ye

A  2ye2

Bye  2y

2 e

B  2y e

P  B  2ye P  2y e  2y e P  4ye

A P 2ye2 R 4ye R

ye R 2

Show that the best hydraulic trapezoidal section is one-half of a hexagon.

For a trapezoid,

A  By  zy2 P  B  2y 1  z 2 1

dP 0 For best hydraulic section dy Let's first assume A and z to be constant: B 

z

A  zy y

60

1 3

Substituting B

A  zy  2y 1  z 2 y dP A   2  z  2 1  z2 dy y P

For best hydraulic section

A And, P   zy  2y 1  z 2 y



P  2ye 2 1  z 2  z



A  2  z  2 1  z2  0 ye A  2 1  z 2  z ye2



Therefore,



A R P ye2 2 1  z 2  z R 2y e 2 1  z 2  z

 

ye R 2

 





If z is allowed to vary, A  2 1  z 2  z ye2

A ye  2 1  z2  z



  2 z

Substitute into P, P  2ye 2 1  z 2  z

P 2

A 2 1  z2



P  2 A 2 1  z2  z When

dP 0 dz 1 ze  3

1  z2  z





1 When ze  , 3



P  2ye 2 1  z 2  z P  2 3ye P  B  2y 1  z 2 B  P  2y 1  z 2

2 ye 3

B





A  2 1  z 2  z ye2 A  3ye2



Example 2.12 A slightly rough brick-lined trapezoidal channel carrying a discharge of 25.0 m3/s is to have a longitudinal slope of 0.0004. Analyse the proportions of (a) an efficient trapezoidal channel section having a side of 1.5(H) : 1(V), (b) the most efficient-channel section of trapezoidal shape.

Rough brick-lined gives Manning roughness n = 0.017

(a) Fixed side slopes of 1.5(H) : 1(V),





For best hydraulic section A  2 1  z  z y 2

2 e

ye and R  2

A  2.1056ye2 2

1 3 12 From Manning equation, Q  AR So n 2 1 3 y 1   25   2.1056ye2   e   0.00042 0.017 2

A  By  zy2 2.1056ye2 B  1.5ye ye B  1.7137 m

ye  2.8298 m

1

2.830 m 1.5 1.714 m

(b) If the side slope is not fixed, the side slope and other channel characteristics for most-efficient trapezoidal section are

1 ze  3

ze  0.5774

A  3ye2

R

ye 2

2

1 3 12 From Manning equation, Q  AR So n 2 1 3 y 1 25   2.1056ye2   e   0.00042 0.017 2

ye  3.045m

B

2 ye 3

Be  3.516 m

1

3.045 m 0.5774 3.516 m