Chapter 4 - Non-uniform Flow in Open Channel
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Transcript of Chapter 4 - Non-uniform Flow in Open Channel
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HYDRAULICS(BFC 21103)
Prepared by:-
MR WAN AFNIZAN BIN WAN MOHAMEDDEPT. OF WATER & ENVIRONMENTAL ENGINEERINGFAC. OF CIVIL & ENVIRONMENTAL ENGINEERINGe-mail: [email protected]
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CHAPTER 4
NON-UNIFORM FLOW IN OPEN CHANNEL
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CONTENT☻ HYDRAULIC JUMP
RAPIDLY VARIED FLOW
INTRODUCTION USAGE OF HYDRAULIC JUMP TYPES OF THE JUMP MOMENTUM EQUATIONS
MINIMUM FLOW FORCE
FLOW FORCE ON RECTANGULAR SECTION
JUMP EQUATION ( CHANNEL)
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CONTENT☻ HYDRAULIC JUMP
RAPIDLY VARIED FLOW
ENERGY / HEAD LOSSES POWER JUMP LOSSES
HEIGHT OF THE JUMP
LENGTH OF THE JUMP (IF BOTTOM SLOPE IS FLAT)
.... Cont ‘
LOCATION OF THE JUMP (IF SLOPE CHANGE EXIST)
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CONTENT☻ DEFINITION
GRADUALLY VARIED FLOW
.... Cont ‘
☻ TYPES OF SLOPE
☻ FLOW ZONE
☻ PROFILE TYPE
☻ OCCURRENCE EXAMPLES OF THE ACTUAL FLOW PROFILE
☻ GRADUALLY VARIED FLOW FORMULA
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CONTENT☻ METHOD OF GVF CALCULATION
GRADUALLY VARIED FLOW … Con’t
DIRECT INTEGRATION METHOD
NUMERICAL INTEGRATION METHOD
.... Cont ‘
STEP METHOD
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RAPIDLY VARIED FLOW
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HYDRAULIC JUMP
INTRODUCTION
Figure 4.1: Flow inside pipe
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INTRODUCTION .... Cont ‘
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INTRODUCTION .... Cont ‘
Figure 4.2: Hydraulic jump in the hydraulic laboratory
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INTRODUCTION .... Cont ‘
Figure 4.3: Hydraulic jump at the spillway toe (Itaipu Dam, Brazil)
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INTRODUCTION .... Cont ‘
Figure 4.4: Hydraulic jump at the downstream of sluice gate (Harran cannal, Turkey)
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INTRODUCTION .... Cont ‘
Figure 4.5: Waves that hit sea wall in Depoe Bay (Oregon U.S)
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INTRODUCTION .... Cont ‘
Figure 4.6: Surges waves (Tangjiasan, China)
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USAGE OF HYDRAULIC JUMP
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TYPES OF THE JUMP
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TYPES OF THE JUMP
Fr = 1.6
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MOMENTUM EQUATION
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MOMENTUM EQUATION
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MOMENTUM EQUATION
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MINIMUM FLOW FORCE & CRITICAL DEPTH
MOMENTUM EQUATION
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FLOW FORCE ON RECTANGULAR SECTION
MOMENTUM EQUATION
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FLOW FORCE ON RECTANGULAR SECTION
MOMENTUM EQUATION
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Using the force equation draw the specific force graph if the hyraulic jump was occured inside a rectangular channel and the discharge per unit width is 25 ft3/s.ft..
SOLUTION:Given:-
Rectangular channelq = 25 ft3/s.ft
Plot y vs F ???
EXAMPLE 4.1
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SOLUTION
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JUMP EQUATION - RECTANGULAR SECTION
MOMENTUM EQUATION .... Cont ‘
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JUMP EQUATION - RECTANGULAR SECTION
MOMENTUM EQUATION .... Cont ‘
4.2 …..
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JUMP EQUATION - RECTANGULAR SECTION
MOMENTUM EQUATION .... Cont ‘
4.3 …..
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ENERGY LOSS
Figure 4.7
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ENERGY LOSS
……. 4.4
……. 4.5
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POWER LOSS
……. 4.6
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HEIGHT OF THE JUMP
……. 4.7
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LENGTH OF THE JUMP
……. 4.8
……. 4.9
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A spillway discharges a flood flow at a rate of 7.75 m3/s per metre width. At the downstream horizontal apron the depth of flow was found to be 0.5 m. What tailwater depth is needed to form a hydraulic jump? If a jump is formed, find its (a) type, (b) length, (c) head loss, and (d) energy loss as a percentage of the initial energySOLUTION:
Given:-q = 7.75 m3/s.m y1 = 0.5 m
(i) y2 (ii) Jump type (iii) L (iv) EL (v) EL/Eo (%)
EXAMPLE 4.2
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SOLUTION
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SOLUTION .... Cont ‘
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SOLUTION .... Cont ‘
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A 25-m wide spillway has velocity of 30 m/s and flow depth of 1 m. Hydraulic jump occurs immediately downstream. Find the height of the jump and power loss in the jump
SOLUTION:Given:-
B = 25 m v1 = 30 m/s y1 = 1 m
(i) Hj (ii) PL
EXAMPLE 4.3
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SOLUTION
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SOLUTION .... Cont ‘
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LOCATION OF THE JUMP
IF THERE IS CHANGING OF SLOPE
Figure 4.8
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LOCATION OF THE JUMP
IF THERE IS CHANGING OF SLOPE
Figure 4.9
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LOCATION OF THE JUMP
IF THERE IS CHANGING OF SLOPE
Figure 4.10
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LOCATION OF THE JUMP
IF THERE IS CHANGING OF SLOPE
PROCEDURES OF CALCULATION
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LOCATION OF THE JUMP
IF THERE IS CHANGING OF SLOPE
PROCEDURES OF CALCULATION
.... Cont ‘
……. 4.10
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LOCATION OF THE JUMP
IF THERE IS CHANGING OF SLOPE
PROCEDURES OF CALCULATION
.... Cont ‘
And,
where,
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A rectangular channel 5 m width convey flow at a rate of 15 m3/s. At a point inside the channel there is abrupt slope change from 0.010 to 0.0015. Determine :- (i) Whether the jump will occur,(ii) Location of the jump (if occur) and(iii) Power loss inside the jump. Use Manning, n = 0.013
EXAMPLE 4.4
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SOLUTION:Given:-
Rectangular channel
B = 5 m Q = 15 m3/s y1 = 1 m
Sos = 0.010 Som = 0.0015
Find :-(i) Occurence of the jump(ii) Lj(iii) PL
EXAMPLE 4.4
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SOLUTION
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SOLUTION .... Cont ‘
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SOLUTION .... Cont ‘
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SOLUTION .... Cont ‘
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SOLUTION .... Cont ‘
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GRADUALLY VARIED FLOW
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GRADUALLY VARIED FLOW
DEFINITION
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GRADUALLY VARIED FLOW
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GRADUALLY VARIED FLOW
TYPES OF SLOPE
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GRADUALLY VARIED FLOW
FLOW ZONE
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GRADUALLY VARIED FLOW
PROFILE TYPE
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GRADUALLY VARIED FLOW
PROFILE TYPE
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GRADUALLY VARIED FLOW
OCCURRENCE EXAMPLES OF THE ACTUAL FLOW PROFILE
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GRADUALLY VARIED FLOW
OCCURRENCE EXAMPLES OF THE ACTUAL FLOW PROFILE
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GRADUALLY VARIED FLOW
OCCURRENCE EXAMPLES OF THE ACTUAL FLOW PROFILE
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GRADUALLY VARIED FLOW
OCCURRENCE EXAMPLES OF THE ACTUAL FLOW PROFILE
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GRADUALLY VARIED FLOW
OCCURRENCE EXAMPLES OF THE ACTUAL FLOW PROFILE
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GRADUALLY VARIED FLOW
OCCURRENCE EXAMPLES OF THE ACTUAL FLOW PROFILE
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GRADUALLY VARIED FLOW
OCCURRENCE EXAMPLES OF THE ACTUAL FLOW PROFILE
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GRADUALLY VARIED FLOW
OCCURRENCE EXAMPLES OF THE ACTUAL FLOW PROFILE
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Determine the flow profile type as shown in figures below.
EXAMPLE 4.4
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SOLUTION
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GRADUALLY VARIED FLOW
GRADUALLY VARIED FLOW FORMULA
Figure 4.8
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GRADUALLY VARIED FLOW
GRADUALLY VARIED FLOW FORMULA
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GRADUALLY VARIED FLOW
GRADUALLY VARIED FLOW FORMULA
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GRADUALLY VARIED FLOW
GRADUALLY VARIED FLOW FORMULA
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GRADUALLY VARIED FLOW
GRADUALLY VARIED FLOW FORMULA
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GRADUALLY VARIED FLOW
GRADUALLY VARIED FLOW FORMULA
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GRADUALLY VARIED FLOW
GRADUALLY VARIED FLOW FORMULA
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GRADUALLY VARIED FLOW
GRADUALLY VARIED FLOW FORMULA
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GRADUALLY VARIED FLOW
GRADUALLY VARIED FLOW FORMULA
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GRADUALLY VARIED FLOW
GRADUALLY VARIED FLOW FORMULA
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GRADUALLY VARIED FLOW
METHOD OF GVF CALCULATION
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GRADUALLY VARIED FLOW
METHOD OF GVF CALCULATION
……. 4.10
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GRADUALLY VARIED FLOW
METHOD OF GVF CALCULATION
……. 4.11
……. 4.12
……. 4.13
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GRADUALLY VARIED FLOW
METHOD OF GVF CALCULATION
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GRADUALLY VARIED FLOW
METHOD OF GVF CALCULATION
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GRADUALLY VARIED FLOW
METHOD OF GVF CALCULATION
Figure 4.9
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GRADUALLY VARIED FLOW
METHOD OF GVF CALCULATION
Figure 4.10
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Given:- A very wide channelyo = 3 m So = 0.0005n = 0.035
Water behind the weir rise to 1.5 m from the normal depth. Calculate the length, L from upstream starting 1% higher of the normal depth using the direct integration method.
EXAMPLE 4.5
![Page 89: Chapter 4 - Non-uniform Flow in Open Channel](https://reader036.fdocument.pub/reader036/viewer/2022081721/5695d0c61a28ab9b0293d218/html5/thumbnails/89.jpg)
EXAMPLE 4.5
![Page 90: Chapter 4 - Non-uniform Flow in Open Channel](https://reader036.fdocument.pub/reader036/viewer/2022081721/5695d0c61a28ab9b0293d218/html5/thumbnails/90.jpg)
SOLUTION
![Page 91: Chapter 4 - Non-uniform Flow in Open Channel](https://reader036.fdocument.pub/reader036/viewer/2022081721/5695d0c61a28ab9b0293d218/html5/thumbnails/91.jpg)
SOLUTION .... Cont ‘
0.8262.260
![Page 92: Chapter 4 - Non-uniform Flow in Open Channel](https://reader036.fdocument.pub/reader036/viewer/2022081721/5695d0c61a28ab9b0293d218/html5/thumbnails/92.jpg)
SOLUTION .... Cont ‘
0.8262.260
![Page 93: Chapter 4 - Non-uniform Flow in Open Channel](https://reader036.fdocument.pub/reader036/viewer/2022081721/5695d0c61a28ab9b0293d218/html5/thumbnails/93.jpg)
SOLUTION .... Cont ‘
0.8262.260
![Page 94: Chapter 4 - Non-uniform Flow in Open Channel](https://reader036.fdocument.pub/reader036/viewer/2022081721/5695d0c61a28ab9b0293d218/html5/thumbnails/94.jpg)
SOLUTION .... Cont ‘
0.8262.260
![Page 95: Chapter 4 - Non-uniform Flow in Open Channel](https://reader036.fdocument.pub/reader036/viewer/2022081721/5695d0c61a28ab9b0293d218/html5/thumbnails/95.jpg)
SOLUTION .... Cont ‘
0.8262.260
-ve = Against co-ordinate
![Page 96: Chapter 4 - Non-uniform Flow in Open Channel](https://reader036.fdocument.pub/reader036/viewer/2022081721/5695d0c61a28ab9b0293d218/html5/thumbnails/96.jpg)
SOLUTION .... Cont ‘
0.8262.260
Sketch of flow profile
![Page 97: Chapter 4 - Non-uniform Flow in Open Channel](https://reader036.fdocument.pub/reader036/viewer/2022081721/5695d0c61a28ab9b0293d218/html5/thumbnails/97.jpg)
EXAMPLE 4.6
![Page 98: Chapter 4 - Non-uniform Flow in Open Channel](https://reader036.fdocument.pub/reader036/viewer/2022081721/5695d0c61a28ab9b0293d218/html5/thumbnails/98.jpg)
EXAMPLE 4.6
Sketch of the problem
![Page 99: Chapter 4 - Non-uniform Flow in Open Channel](https://reader036.fdocument.pub/reader036/viewer/2022081721/5695d0c61a28ab9b0293d218/html5/thumbnails/99.jpg)
SOLUTION
![Page 100: Chapter 4 - Non-uniform Flow in Open Channel](https://reader036.fdocument.pub/reader036/viewer/2022081721/5695d0c61a28ab9b0293d218/html5/thumbnails/100.jpg)
SOLUTION
![Page 101: Chapter 4 - Non-uniform Flow in Open Channel](https://reader036.fdocument.pub/reader036/viewer/2022081721/5695d0c61a28ab9b0293d218/html5/thumbnails/101.jpg)
SOLUTION .... Cont ‘
0.8262.260
![Page 102: Chapter 4 - Non-uniform Flow in Open Channel](https://reader036.fdocument.pub/reader036/viewer/2022081721/5695d0c61a28ab9b0293d218/html5/thumbnails/102.jpg)
SOLUTION .... Cont ‘
0.8262.260
![Page 103: Chapter 4 - Non-uniform Flow in Open Channel](https://reader036.fdocument.pub/reader036/viewer/2022081721/5695d0c61a28ab9b0293d218/html5/thumbnails/103.jpg)
SOLUTION .... Cont ‘
0.8262.260
![Page 104: Chapter 4 - Non-uniform Flow in Open Channel](https://reader036.fdocument.pub/reader036/viewer/2022081721/5695d0c61a28ab9b0293d218/html5/thumbnails/104.jpg)
SOLUTION .... Cont ‘
0.8262.260
![Page 105: Chapter 4 - Non-uniform Flow in Open Channel](https://reader036.fdocument.pub/reader036/viewer/2022081721/5695d0c61a28ab9b0293d218/html5/thumbnails/105.jpg)
SOLUTION .... Cont ‘
0.8262.260
+ve = Follow the co-ordinate
![Page 106: Chapter 4 - Non-uniform Flow in Open Channel](https://reader036.fdocument.pub/reader036/viewer/2022081721/5695d0c61a28ab9b0293d218/html5/thumbnails/106.jpg)
SOLUTION .... Cont ‘
0.8262.260
Sketch of flow profile
![Page 107: Chapter 4 - Non-uniform Flow in Open Channel](https://reader036.fdocument.pub/reader036/viewer/2022081721/5695d0c61a28ab9b0293d218/html5/thumbnails/107.jpg)
GRADUALLY VARIED FLOW
METHOD OF GVF CALCULATION
……. 4.14
……. 4.15
![Page 108: Chapter 4 - Non-uniform Flow in Open Channel](https://reader036.fdocument.pub/reader036/viewer/2022081721/5695d0c61a28ab9b0293d218/html5/thumbnails/108.jpg)
GRADUALLY VARIED FLOW
METHOD OF GVF CALCULATION
……. 4.16
……. 4.17
![Page 109: Chapter 4 - Non-uniform Flow in Open Channel](https://reader036.fdocument.pub/reader036/viewer/2022081721/5695d0c61a28ab9b0293d218/html5/thumbnails/109.jpg)
GRADUALLY VARIED FLOW
METHOD OF GVF CALCULATION
Figure 4.11
![Page 110: Chapter 4 - Non-uniform Flow in Open Channel](https://reader036.fdocument.pub/reader036/viewer/2022081721/5695d0c61a28ab9b0293d218/html5/thumbnails/110.jpg)
GRADUALLY VARIED FLOW
METHOD OF GVF CALCULATION
……. 4.18
……. 4.19
![Page 111: Chapter 4 - Non-uniform Flow in Open Channel](https://reader036.fdocument.pub/reader036/viewer/2022081721/5695d0c61a28ab9b0293d218/html5/thumbnails/111.jpg)
GRADUALLY VARIED FLOW
METHOD OF GVF CALCULATION
……. 4.20
![Page 112: Chapter 4 - Non-uniform Flow in Open Channel](https://reader036.fdocument.pub/reader036/viewer/2022081721/5695d0c61a28ab9b0293d218/html5/thumbnails/112.jpg)
10 m3/s of water flows inside a rectangular channel with 3 m width, channel slope of 0.0016 and Manning n 0.013. A weir is constructed causing water level to rise as shown in the Figure. Using the numerical integration method, calculate the distance of L. Devide into 5 segments.
SOLUTION:Given:-
Rectangular channelQ = 10 m3/s B = 3.0 m So = 0.0016 n = 0.013yend limit = 1.85 m
EXAMPLE 4.7
![Page 113: Chapter 4 - Non-uniform Flow in Open Channel](https://reader036.fdocument.pub/reader036/viewer/2022081721/5695d0c61a28ab9b0293d218/html5/thumbnails/113.jpg)
EXAMPLE 4.7
Sketch of the problem
![Page 114: Chapter 4 - Non-uniform Flow in Open Channel](https://reader036.fdocument.pub/reader036/viewer/2022081721/5695d0c61a28ab9b0293d218/html5/thumbnails/114.jpg)
SOLUTION
![Page 115: Chapter 4 - Non-uniform Flow in Open Channel](https://reader036.fdocument.pub/reader036/viewer/2022081721/5695d0c61a28ab9b0293d218/html5/thumbnails/115.jpg)
SOLUTION .... Cont ‘
0.8262.260
![Page 116: Chapter 4 - Non-uniform Flow in Open Channel](https://reader036.fdocument.pub/reader036/viewer/2022081721/5695d0c61a28ab9b0293d218/html5/thumbnails/116.jpg)
SOLUTION .... Cont ‘
![Page 117: Chapter 4 - Non-uniform Flow in Open Channel](https://reader036.fdocument.pub/reader036/viewer/2022081721/5695d0c61a28ab9b0293d218/html5/thumbnails/117.jpg)
SOLUTION .... Cont ‘
![Page 118: Chapter 4 - Non-uniform Flow in Open Channel](https://reader036.fdocument.pub/reader036/viewer/2022081721/5695d0c61a28ab9b0293d218/html5/thumbnails/118.jpg)
SOLUTION .... Cont ‘
-ve = Against the co-ordinate
![Page 119: Chapter 4 - Non-uniform Flow in Open Channel](https://reader036.fdocument.pub/reader036/viewer/2022081721/5695d0c61a28ab9b0293d218/html5/thumbnails/119.jpg)
SOLUTION .... Cont ‘
Sketch of flow profile
![Page 120: Chapter 4 - Non-uniform Flow in Open Channel](https://reader036.fdocument.pub/reader036/viewer/2022081721/5695d0c61a28ab9b0293d218/html5/thumbnails/120.jpg)
Water flow at 5.25 m3/s inside a trapezoidal channel having side slopes of 1 vertical : 2 horizontal, base width of 3 m, bottom slope of 0.0005 and Manning n 0.017. A weir is constructed causing the water level to rise as shown in the. Using the numerical integration method, find L distance. SOLUTION:
Given:-Trapezoidal channelQ = 5.25 m3/s B = 3.0 mz = 2 So = 0.0005 n = 0.017yinitial limit = 1.5 m yend limit = 2.5 m
EXAMPLE 4.8
![Page 121: Chapter 4 - Non-uniform Flow in Open Channel](https://reader036.fdocument.pub/reader036/viewer/2022081721/5695d0c61a28ab9b0293d218/html5/thumbnails/121.jpg)
EXAMPLE 4.7
Sketch of the problem
![Page 122: Chapter 4 - Non-uniform Flow in Open Channel](https://reader036.fdocument.pub/reader036/viewer/2022081721/5695d0c61a28ab9b0293d218/html5/thumbnails/122.jpg)
SOLUTION
![Page 123: Chapter 4 - Non-uniform Flow in Open Channel](https://reader036.fdocument.pub/reader036/viewer/2022081721/5695d0c61a28ab9b0293d218/html5/thumbnails/123.jpg)
SOLUTION .... Cont ‘
0.8262.260
![Page 124: Chapter 4 - Non-uniform Flow in Open Channel](https://reader036.fdocument.pub/reader036/viewer/2022081721/5695d0c61a28ab9b0293d218/html5/thumbnails/124.jpg)
SOLUTION .... Cont ‘
0.8262.260
![Page 125: Chapter 4 - Non-uniform Flow in Open Channel](https://reader036.fdocument.pub/reader036/viewer/2022081721/5695d0c61a28ab9b0293d218/html5/thumbnails/125.jpg)
SOLUTION .... Cont ‘
0.8262.260
![Page 126: Chapter 4 - Non-uniform Flow in Open Channel](https://reader036.fdocument.pub/reader036/viewer/2022081721/5695d0c61a28ab9b0293d218/html5/thumbnails/126.jpg)
SOLUTION .... Cont ‘
0.8262.260
![Page 127: Chapter 4 - Non-uniform Flow in Open Channel](https://reader036.fdocument.pub/reader036/viewer/2022081721/5695d0c61a28ab9b0293d218/html5/thumbnails/127.jpg)
SOLUTION .... Cont ‘
0.8262.260
Calculation table :-
+ve = Follow the co-ordinate
![Page 128: Chapter 4 - Non-uniform Flow in Open Channel](https://reader036.fdocument.pub/reader036/viewer/2022081721/5695d0c61a28ab9b0293d218/html5/thumbnails/128.jpg)
SOLUTION .... Cont ‘
0.8262.260
![Page 129: Chapter 4 - Non-uniform Flow in Open Channel](https://reader036.fdocument.pub/reader036/viewer/2022081721/5695d0c61a28ab9b0293d218/html5/thumbnails/129.jpg)
SOLUTION .... Cont ‘
0.8262.260
Sketch of flow profile
![Page 130: Chapter 4 - Non-uniform Flow in Open Channel](https://reader036.fdocument.pub/reader036/viewer/2022081721/5695d0c61a28ab9b0293d218/html5/thumbnails/130.jpg)
GRADUALLY VARIED FLOW
METHOD OF GVF CALCULATION
……. 4.16
……. 4.17
Formula applied …
……. 4.18
![Page 131: Chapter 4 - Non-uniform Flow in Open Channel](https://reader036.fdocument.pub/reader036/viewer/2022081721/5695d0c61a28ab9b0293d218/html5/thumbnails/131.jpg)
GRADUALLY VARIED FLOW
METHOD OF GVF CALCULATION
……. 4.19
Formula applied … con’t
![Page 132: Chapter 4 - Non-uniform Flow in Open Channel](https://reader036.fdocument.pub/reader036/viewer/2022081721/5695d0c61a28ab9b0293d218/html5/thumbnails/132.jpg)
10 m3/s of water flows inside a rectangular channel having the width of 3 m, channel slope of 0.0016 and Manning’s n 0.013. A weir is built causing water to rise 1.85 m behind it. Using the step method, calculate the length of L.
SOLUTION:Given:-
Rectangular channelQ = 5.25 m3/s B = 3.0 m So = 0.0016 n = 0.013yend limit = 1.85 m
EXAMPLE 4.9
![Page 133: Chapter 4 - Non-uniform Flow in Open Channel](https://reader036.fdocument.pub/reader036/viewer/2022081721/5695d0c61a28ab9b0293d218/html5/thumbnails/133.jpg)
EXAMPLE 4.7
Sketch of the problem
![Page 134: Chapter 4 - Non-uniform Flow in Open Channel](https://reader036.fdocument.pub/reader036/viewer/2022081721/5695d0c61a28ab9b0293d218/html5/thumbnails/134.jpg)
SOLUTION
![Page 135: Chapter 4 - Non-uniform Flow in Open Channel](https://reader036.fdocument.pub/reader036/viewer/2022081721/5695d0c61a28ab9b0293d218/html5/thumbnails/135.jpg)
SOLUTION
![Page 136: Chapter 4 - Non-uniform Flow in Open Channel](https://reader036.fdocument.pub/reader036/viewer/2022081721/5695d0c61a28ab9b0293d218/html5/thumbnails/136.jpg)
SOLUTION
![Page 137: Chapter 4 - Non-uniform Flow in Open Channel](https://reader036.fdocument.pub/reader036/viewer/2022081721/5695d0c61a28ab9b0293d218/html5/thumbnails/137.jpg)
SOLUTION
Calculation table :-
+ve = Follow the co-ordinate
![Page 138: Chapter 4 - Non-uniform Flow in Open Channel](https://reader036.fdocument.pub/reader036/viewer/2022081721/5695d0c61a28ab9b0293d218/html5/thumbnails/138.jpg)
SOLUTION
![Page 139: Chapter 4 - Non-uniform Flow in Open Channel](https://reader036.fdocument.pub/reader036/viewer/2022081721/5695d0c61a28ab9b0293d218/html5/thumbnails/139.jpg)
SOLUTION .... Cont ‘
0.8262.260
Sketch of flow profile
![Page 140: Chapter 4 - Non-uniform Flow in Open Channel](https://reader036.fdocument.pub/reader036/viewer/2022081721/5695d0c61a28ab9b0293d218/html5/thumbnails/140.jpg)
TIME’S UP …
THANK YOU