This PPT covers introduction of all the part of open channel flow with numerical problem you can use it for memorizing the open channel flow

1. OPEN CHANNEL FLOW
1

2. TOPIC UNDER OPEN CHANNEL FLOW
Introduction
Types of flow in OCF
Determination of discharge
Different formulas
Most economical channel section
2

3. TOPIC UNDER OPEN CHANNEL FLOW
Specific energy
Hydraulic jump
Back water curve
Serge
3

4. INTRODUCTION
• Open channel flow is defined as the flow under the atmospheric
condition or flow of liquid with free surface.
• Best example of open channel flow is river flow
4

5. TYPES OF OPEN CHANNEL
Fig: Artificial open channel
5

6. TYPES OF OPEN CHANNEL
Fig: Natural Channel
6

7. TYPES OF FLOW IN OCF
 Steady and unsteady flow
Uniform and non- uniform flow
Laminar and turbulent flow
Sub- critical, critical and super critical flow
7

8. TYPES OF FLOW IN OCF
• STEADY FLOW:
𝑑𝑦
𝑑𝑡
= 0
𝑑𝑉
𝑑𝑡
= 0
• UNSTEADY FLOW:
𝑑𝑦
𝑑𝑡
≠ 0
𝑑𝑉
𝑑𝑡
≠ 0
8

9. TYPES OF FLOW IN OCF
• UNIFORM FLOW:
𝑑𝑦
𝑑𝑠
= 0
𝑑𝑉
𝑑𝑠
= 0
• NON-UNIFORM FLOW:
𝑑𝑦
𝑑𝑠
≠ 0
𝑑𝑉
𝑑𝑠
≠ 0
9

10. • Non uniform flow is further divided in
• Rapidly Varied Flow ( R. V. F.):
Suddenly change of depth in small length
• Gradually Varied Flow ( G. V. F.):
Gradually change of depth in small length
TYPES OF FLOW IN OCF
10

11. • Laminar and Turbulent flow
It id depends on Reynolds no. Which is written as
𝑅𝑒 = 𝜌𝑉𝑅/𝜇
Re < 600 = Laminar flow
Re < 500 − 200 = Transition flow
Re > 2000 = Turbulent flow
TYPES OF FLOW IN OCF
11

12. Sub-critical, Critical and Super-Critical:
It is depend on Froude number.
𝐹𝑒 =
𝑉
𝑔𝐷
𝐹𝑒 < 1 = 𝑆𝑢𝑏 − 𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙
𝐹𝑒 = 1 = 𝐶𝑟𝑖𝑡𝑖𝑐𝑎𝑙
𝐹𝑒 > 1 = 𝑆𝑢𝑏 𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙
12

13. DETERMINATION OF DISCHARGE
• CHEZY’S FORMULA FOR DISCHARGE
𝑽 = 𝑪 𝒎 𝒊
𝑸 = 𝑨 𝑿 𝑽
𝑸 = 𝑨 𝑿𝑪 𝒎 𝒊
13

14. • BAZIN’S FORMULA
𝑪 =
𝟏𝟓𝟕. 𝟔
𝟏. 𝟖𝟏 +
𝑲
𝒎
S. No. Nature of Channel inside surface Value of K
1 Smooth cemented or planned wood 0.11
2 Brick or concrete or unplanned wood 0.21
3 Rubble masonry or ashlar or poor brick work 0.83
4 Earthen channel of very good surface 1.54
5 Earthen channel or ordinary surface 2.36
6 Earthen channel or rough surface 3.17
14

15. • MANNING FORMULA
𝑪 =
𝟏
𝑵
𝒎 𝟏 𝟔
S. No. Nature of Channel inside surface Value of N
1 Very smooth surface of glass, plastic or brass 0.010
2 Smooth surface of concrete 0.012
3 Rubble masonry or ashlar or poor brick work 0.017
4 Earthen channels neatly excavated 0.018
5 Earthen channel or ordinary surface 0.027
6 Earthen channel or rough surface 0.030
7 Natural streams, clan and straight 0.030
8 Natural streams with weeds, duppools etc. 0.075 to 0.15
15

16. • Problem:
Find the discharge through a rectangular channel 2.5 m wide, having depth of water 1.5m and bed
slope as 1 in 2000. The channel is Earthen channel of ordinary surface. Use Bazin’s and Manning’s
formula.
Solution:
b=2.5 m
d= 1.5 m
A= b X d
A=2.5 X 1.5
A= 3.75 sqm.
P = d+b+d
P = 1.5 + 2.5 + 1.5
P = 5.5 m
Hydraulic Mean Depth = m = A/P = 3.75/5.5 = 0.682
Slop = i = 1/2000
Value of k = 2.36
Value of N = 0.027 16

17. • BY BAZIN’S FORMULA
𝑸 = 𝑨 𝑿𝑪 𝒎 𝒊
𝑸 = 𝑨 𝑿𝑪 𝒎 𝒊
𝑪 =
𝟏𝟓𝟕. 𝟔
𝟏. 𝟖𝟏 +
𝑲
𝒎
𝑪 =
𝟏𝟓𝟕. 𝟔
𝟏. 𝟖𝟏 +
𝟐. 𝟑𝟔
𝟓. 𝟓
𝑪 = 𝟑𝟑. 𝟕𝟔
𝑸 = 𝟑. 𝟕𝟓 𝑿𝟑𝟑. 𝟕𝟔 𝟎. 𝟔𝟖𝟐 𝑿 𝟏/𝟐𝟎𝟎𝟎 = 𝟐. 𝟑𝟑𝟕 𝒎 𝟑/𝒔
17

18. Manning’s Formula:
𝑸 = 𝑨 𝑿𝑪 𝒎 𝒊
𝑸 = 𝑨 𝑿𝑪 𝒎 𝒊
𝑪 =
𝟏
𝑵
𝒎 𝟏 𝟔
𝑪 =
𝟏
𝟎. 𝟎𝟐𝟕
𝟎. 𝟔𝟖𝟐 𝟏 𝟔
𝑪=34.74
𝑸 = 𝟑. 𝟕𝟓 𝑿𝟑𝟒. 𝟕𝟒 𝟎. 𝟔𝟖𝟐 𝑿 𝟏/𝟐𝟎𝟎𝟎 = 𝟐. 𝟒𝟎 𝒎 𝟑/𝒔
18

19. MOST ECONOMICAL CHANNEL SECTION
• A section of channel is said to be most economical when the cost
of construction of the channel is minimum. Which is depends on
the excavation and lining. If this two thing become cheap than the
section will be the most economical section.
• In short the section which gives more discharge in less perimeter
is known as most economical channel section
19

20. • Condition for the Most economical channel section
• Rectangular section
• b=2d
• Hydraulic width m= d/2
• Trapezoidal section
• Sloping side length = 𝑑 𝑛2 − 1
• Hydraulic width m= d/2
• Circular Section
• Perimeter = 2R𝜃
• Hydraulic Mean Depth =
𝑅
2𝜃
𝜃 − sin 2𝜃
• Triangle Section
• N = 1 : 1
• Vertex angle = 2 = 90 degree
20

21. A rectangular channel carries water at the rate of 400 liters/s when bed slope is l in 2000. Find the most economical
dimension of the channel if C = 50.
SOLUTION: - Given
Discharge ( Q ) = 400 liters / s = 0.4 m3/s
Bed Slop i = 1 / 2000
Chezy’s Constant C = 50
For the rectangular channel to be most economical
( i ) Width b = 2d
(ii) Hydraulic mean depth = m = d /2
Area of flow A= b X d = 2d X d = 2𝑑2
Using Equation of discharge
𝑸 = 𝑨 𝑿𝑪 𝒎 𝒊
0.4 = 2𝑑2
𝑿 𝟓𝟎
𝒅
𝟐
𝟏
𝟐𝟎𝟎𝟎
𝑑 = 0.577 m
𝑏 = 2d = 2 X 0.577 = 1.154 m
21

22. HYDRAULIC JUMP
• When water strike with any object or obstruction in the flow than in down
stream the fluid strike to the surface and again subjected to the rise of level this
rise of level known as hydraulic jump
22

23. BACK WATER CURVE
• The reservoir constructed at the down stream side of the river and this side is at the lower level
to the level of the river. In this process a angle is formed between actual level of dam and actual
level of river that curve known as back water curve.
23

24. THANK YOU
24