CHAPTER 2: OPEN CHANNEL FLOW



OPEN CHANNEL FLOW
- A branch of hydraulics and fluid mechanics, is a type of liquid flow within a conduit with a free surface, known as a channel. The other type of flow within a conduit is pipe flow. These two types of flow are similar in many ways but differ in one important respect: the free surface. Open-channel flow has a free surface, whereas pipe flow does not.


4 CLASSIFICATION OF OPEN CHANNEL:
1. Classification 1
- Natural Channels [natural open channels include all channels that exists naturally on the earth]
- Artifical Channels [the channels develops by men]

2. Classification 2
- Prismatic Channels [unvarying cross-section channel and constant bottom slope]
- Non-prismatic Channels [unconstant bottom slope]

3. Classification 3
- Rigid Boundary Channels [a channel with imnovable bed and sides]
- Mobile Boundary Channels [composed of loose sedimentary particles moving]


4. Classification 4
- Large Slope Channels [having bottom slope greater than 1 in 10]
- Small Slope Channels [artifical channel like drops and chutes have for more than 1 in 10]

TYPES OF FLOW IN CHANNEL:
1. Steady State and Unsteady State
- Steady state have no changes in velocity patterns and magnitude with time at a given channel cross-section.
Unsteady state have changing velocity with time at a given cross-section.

2. Laminar Flow and Turbulent Flow
- Laminar flow having lowest speed flow of velocity.
Turbulent flow having highest speed flow of velocity.

3. Uniform Flow and Non-uniform Flow
- Uniform flow have constant flow rate.
- Non-uniform flow have to reaches of channel, bottom slope, cross-section slope, and cross-section size change.

UNIFORM STEADY FLOW:

  • MANNING EQUATION
Q=VA=[1.49/n]×A×R⅔×√S {ft}

Q=AV=[1.00/n]×A×R⅔×√S {unit}

• A=Area of flow
• S=Slope

EXAMPLES:
Water flow in a rectangular 6 ft-wide timber flume with n=0.013, what channel slope is needed to convey water uniformly at 20 ft/s when the depth is 3 ft?

Solution:
A= 6×3=12
V= 20 ft/s
n= 0.013
P= 3+6+3=12
S=?

Q=VA=[1.49/n]×A×R⅔×√S

R=A/P=18/12=1.5

Q  =(20)(18)=[1.49/0.013](18)(1.5)⅔(√S)
     = 360=2703.40 √S
√S =360/2703.40
S½ =0.1332
S    =(0.1332)²
S    =0.0177
NON-UNIFORM STEADY FLOW

  • Gradually Varied Flow
H= Z+(P/pg)+(V²/2g)=Z+y+(V²/2g)

• Z= elevation head
• y= gage pressure head
• V²/2g=dynamic head


  • Taking datum Z=0 as the bottom
Es= y+ (V²/2g)          Es= y+(Q²/2gA²) 

q= kadar air per unit lebar
q= Q/B
Q= qB


  • For a channel with constant width, b
Q=AcV=ybv


  • Plot of Es vs y for constant V and b
Es= y + (Q²/2gb²y²)


  • The three parameter can be correlated in two situation:
E value is constant
Yc= 2E/3
qm= V(gyc)³

 q value is constant
Yc= (q²/g)⅓
E min= 1.5 yc


  • Non-uniform flow can be caused by:
  1. Differentiate in depth of channel
  2. Differentiate in width of channel
  3. Differentiate in the nature of bed
  4. Differentiate in slope of channel
  5. Obstruction in the equation of flow


  • Frounde Number
Fr= V/(√gh)


Rapidly Varied Flow

  • Depth ratio:
y2/y1= 0.5 ( -1 + √8Fr²)


  • Head loss associated with hydraulics jump
hL= y1-y2 + [V1² - V2²/2g] = y1-y2 + [y1Fr²/2] [1-(y1²/y2²)]


  • Dissipation ration:
hL/Es= hL/(y+V²/2g)

EXAMPLE :
Sebuah saluran yang panjang 3m lebar membawa aliran pada kadar 12m³/s. Pada satu titik dalam saluran ini, terdapat perubahan cerun dengan mengejut, daripada 0.015 kepada 0.0016. Gunakan formula manning,n=0.013 dan tentukan:

  1. Sama ada lompatan hidraulik akan berlaku
  2. Kedudukan lompatan jika berlaku
  3. Kuasa yang terlesap dalam lompatan

SOLUTION:
b= 3m
Q= 12 m³/s
n= 0.013

A=by=3y
P= b+2y
   = 3+2y
R= by/b+2y
   = 3y/3+2y

So= 0.015
(1)

Qn/So½= AR⅔
(12)(0.013)/(√0.015)= (3y)(3y/3+2y)⅔
1.2737= (3y)(3y/3+2y)⅔

Try and Success!
1. y= 0.812
      = (3)(0.812) [(3)30.812)/ 3+2(0.812)]⅔
      = 1.589

2. y= 0.7
      = (3)(0.7) [(3)(0.7)/ 3+2(0.7)]⅔
      =1.28


So= 0.0016

Qn/So½= AR⅔
(12)(0.013)/(√0.0016)= (3y)(3y/3+2y)⅔
3.9= (3y)(3y/3+2y)⅔

Try and Success!
1. y=1
      =(3)(1) [(3)(1)/3+2(1)]⅔
      =2.13

2. y=1.5
      =(3)(1.5) [(3)(1.5)/3+2(1.5)]⅔
      =3.71

3. y=1.6
       =(3)(1.6) [(3)1.6)/3+2(1.6)]⅔ 
       =3.9~4.0

(2) L=6y
       =6(1.6)
       =9.6 m

(3) kuasa yang terlesap

Q=AV

V1= Q/A1
     = 12/(3)(0.7)
     = 5.714

V2= Q/A2
     = 12/(3)(1.6)
     = 2.5

Specific Energy
Example :
Sebuah saluran segi empat tepat 3m lebar membawa air pada kadar alir 12m³/s mempunyai nombor Froude = 0.8.
1. Berapakah ukur dalam gentingnya?
2. Berapakah ketinggian air dalam saluran ?
3. Apakah keadaan aliran dalam saluran ?
4. Berapakah kecerunan saliran jika pekali Manning, n = 0.035 ?
5. Berapakah tenaga tentu aliran sungai ?
6. Dapatkan ukur dalam sela bagi aliran ?
7. Berapakah nilai nombor Froude pada ukur dalam sela ?
8. Apakah keadaan aliran dalam saluran pada ukur dalam sela ?

Solution for Question 1 :
Q = 12m³/s
 
     






        


Solution for Question 2 :

Solution for Question 3 :

 







Solution for Question 4 :
Pekali Manning, n = 0.035


Solution for Question 5 :


Solution for Question 6 :




1st try, y=1
2nd try, y=0.8
 
3rd try, y=1.1
 
4th try, y=1.02









Solution for Question 7 :
 


Solution for Question 8:

Answer : Supercritical flow.













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