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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]
CHAPTER 4
NON-UNIFORM FLOW IN OPEN CHANNEL
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)
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)
CONTENT☻ DEFINITION
GRADUALLY VARIED FLOW
.... Cont ‘
☻ TYPES OF SLOPE
☻ FLOW ZONE
☻ PROFILE TYPE
☻ OCCURRENCE EXAMPLES OF THE ACTUAL FLOW PROFILE
☻ GRADUALLY VARIED FLOW FORMULA
CONTENT☻ METHOD OF GVF CALCULATION
GRADUALLY VARIED FLOW … Con’t
DIRECT INTEGRATION METHOD
NUMERICAL INTEGRATION METHOD
.... Cont ‘
STEP METHOD
RAPIDLY VARIED FLOW
HYDRAULIC JUMP
INTRODUCTION
Figure 4.1: Flow inside pipe
INTRODUCTION .... Cont ‘
INTRODUCTION .... Cont ‘
Figure 4.2: Hydraulic jump in the hydraulic laboratory
INTRODUCTION .... Cont ‘
Figure 4.3: Hydraulic jump at the spillway toe (Itaipu Dam, Brazil)
INTRODUCTION .... Cont ‘
Figure 4.4: Hydraulic jump at the downstream of sluice gate (Harran cannal, Turkey)
INTRODUCTION .... Cont ‘
Figure 4.5: Waves that hit sea wall in Depoe Bay (Oregon U.S)
INTRODUCTION .... Cont ‘
Figure 4.6: Surges waves (Tangjiasan, China)
USAGE OF HYDRAULIC JUMP
TYPES OF THE JUMP
TYPES OF THE JUMP
Fr = 1.6
MOMENTUM EQUATION
MOMENTUM EQUATION
MOMENTUM EQUATION
MINIMUM FLOW FORCE & CRITICAL DEPTH
MOMENTUM EQUATION
FLOW FORCE ON RECTANGULAR SECTION
MOMENTUM EQUATION
FLOW FORCE ON RECTANGULAR SECTION
MOMENTUM EQUATION
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
SOLUTION
JUMP EQUATION - RECTANGULAR SECTION
MOMENTUM EQUATION .... Cont ‘
JUMP EQUATION - RECTANGULAR SECTION
MOMENTUM EQUATION .... Cont ‘
4.2 …..
JUMP EQUATION - RECTANGULAR SECTION
MOMENTUM EQUATION .... Cont ‘
4.3 …..
ENERGY LOSS
Figure 4.7
ENERGY LOSS
……. 4.4
……. 4.5
POWER LOSS
……. 4.6
HEIGHT OF THE JUMP
……. 4.7
LENGTH OF THE JUMP
……. 4.8
……. 4.9
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
SOLUTION
SOLUTION .... Cont ‘
SOLUTION .... Cont ‘
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
SOLUTION
SOLUTION .... Cont ‘
LOCATION OF THE JUMP
IF THERE IS CHANGING OF SLOPE
Figure 4.8
LOCATION OF THE JUMP
IF THERE IS CHANGING OF SLOPE
Figure 4.9
LOCATION OF THE JUMP
IF THERE IS CHANGING OF SLOPE
Figure 4.10
LOCATION OF THE JUMP
IF THERE IS CHANGING OF SLOPE
PROCEDURES OF CALCULATION
LOCATION OF THE JUMP
IF THERE IS CHANGING OF SLOPE
PROCEDURES OF CALCULATION
.... Cont ‘
……. 4.10
LOCATION OF THE JUMP
IF THERE IS CHANGING OF SLOPE
PROCEDURES OF CALCULATION
.... Cont ‘
And,
where,
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
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
SOLUTION
SOLUTION .... Cont ‘
SOLUTION .... Cont ‘
SOLUTION .... Cont ‘
SOLUTION .... Cont ‘
GRADUALLY VARIED FLOW
GRADUALLY VARIED FLOW
DEFINITION
GRADUALLY VARIED FLOW
GRADUALLY VARIED FLOW
TYPES OF SLOPE
GRADUALLY VARIED FLOW
FLOW ZONE
GRADUALLY VARIED FLOW
PROFILE TYPE
GRADUALLY VARIED FLOW
PROFILE TYPE
GRADUALLY VARIED FLOW
OCCURRENCE EXAMPLES OF THE ACTUAL FLOW PROFILE
GRADUALLY VARIED FLOW
OCCURRENCE EXAMPLES OF THE ACTUAL FLOW PROFILE
GRADUALLY VARIED FLOW
OCCURRENCE EXAMPLES OF THE ACTUAL FLOW PROFILE
GRADUALLY VARIED FLOW
OCCURRENCE EXAMPLES OF THE ACTUAL FLOW PROFILE
GRADUALLY VARIED FLOW
OCCURRENCE EXAMPLES OF THE ACTUAL FLOW PROFILE
GRADUALLY VARIED FLOW
OCCURRENCE EXAMPLES OF THE ACTUAL FLOW PROFILE
GRADUALLY VARIED FLOW
OCCURRENCE EXAMPLES OF THE ACTUAL FLOW PROFILE
GRADUALLY VARIED FLOW
OCCURRENCE EXAMPLES OF THE ACTUAL FLOW PROFILE
Determine the flow profile type as shown in figures below.
EXAMPLE 4.4
SOLUTION
GRADUALLY VARIED FLOW
GRADUALLY VARIED FLOW FORMULA
Figure 4.8
GRADUALLY VARIED FLOW
GRADUALLY VARIED FLOW FORMULA
GRADUALLY VARIED FLOW
GRADUALLY VARIED FLOW FORMULA
GRADUALLY VARIED FLOW
GRADUALLY VARIED FLOW FORMULA
GRADUALLY VARIED FLOW
GRADUALLY VARIED FLOW FORMULA
GRADUALLY VARIED FLOW
GRADUALLY VARIED FLOW FORMULA
GRADUALLY VARIED FLOW
GRADUALLY VARIED FLOW FORMULA
GRADUALLY VARIED FLOW
GRADUALLY VARIED FLOW FORMULA
GRADUALLY VARIED FLOW
GRADUALLY VARIED FLOW FORMULA
GRADUALLY VARIED FLOW
GRADUALLY VARIED FLOW FORMULA
GRADUALLY VARIED FLOW
METHOD OF GVF CALCULATION
GRADUALLY VARIED FLOW
METHOD OF GVF CALCULATION
……. 4.10
GRADUALLY VARIED FLOW
METHOD OF GVF CALCULATION
……. 4.11
……. 4.12
……. 4.13
GRADUALLY VARIED FLOW
METHOD OF GVF CALCULATION
GRADUALLY VARIED FLOW
METHOD OF GVF CALCULATION
GRADUALLY VARIED FLOW
METHOD OF GVF CALCULATION
Figure 4.9
GRADUALLY VARIED FLOW
METHOD OF GVF CALCULATION
Figure 4.10
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
EXAMPLE 4.5
SOLUTION
SOLUTION .... Cont ‘
0.8262.260
SOLUTION .... Cont ‘
0.8262.260
SOLUTION .... Cont ‘
0.8262.260
SOLUTION .... Cont ‘
0.8262.260
SOLUTION .... Cont ‘
0.8262.260
-ve = Against co-ordinate
SOLUTION .... Cont ‘
0.8262.260
Sketch of flow profile
EXAMPLE 4.6
EXAMPLE 4.6
Sketch of the problem
SOLUTION
SOLUTION
SOLUTION .... Cont ‘
0.8262.260
SOLUTION .... Cont ‘
0.8262.260
SOLUTION .... Cont ‘
0.8262.260
SOLUTION .... Cont ‘
0.8262.260
SOLUTION .... Cont ‘
0.8262.260
+ve = Follow the co-ordinate
SOLUTION .... Cont ‘
0.8262.260
Sketch of flow profile
GRADUALLY VARIED FLOW
METHOD OF GVF CALCULATION
……. 4.14
……. 4.15
GRADUALLY VARIED FLOW
METHOD OF GVF CALCULATION
……. 4.16
……. 4.17
GRADUALLY VARIED FLOW
METHOD OF GVF CALCULATION
Figure 4.11
GRADUALLY VARIED FLOW
METHOD OF GVF CALCULATION
……. 4.18
……. 4.19
GRADUALLY VARIED FLOW
METHOD OF GVF CALCULATION
……. 4.20
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
EXAMPLE 4.7
Sketch of the problem
SOLUTION
SOLUTION .... Cont ‘
0.8262.260
SOLUTION .... Cont ‘
SOLUTION .... Cont ‘
SOLUTION .... Cont ‘
-ve = Against the co-ordinate
SOLUTION .... Cont ‘
Sketch of flow profile
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
EXAMPLE 4.7
Sketch of the problem
SOLUTION
SOLUTION .... Cont ‘
0.8262.260
SOLUTION .... Cont ‘
0.8262.260
SOLUTION .... Cont ‘
0.8262.260
SOLUTION .... Cont ‘
0.8262.260
SOLUTION .... Cont ‘
0.8262.260
Calculation table :-
+ve = Follow the co-ordinate
SOLUTION .... Cont ‘
0.8262.260
SOLUTION .... Cont ‘
0.8262.260
Sketch of flow profile
GRADUALLY VARIED FLOW
METHOD OF GVF CALCULATION
……. 4.16
……. 4.17
Formula applied …
……. 4.18
GRADUALLY VARIED FLOW
METHOD OF GVF CALCULATION
……. 4.19
Formula applied … con’t
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
EXAMPLE 4.7
Sketch of the problem
SOLUTION
SOLUTION
SOLUTION
SOLUTION
Calculation table :-
+ve = Follow the co-ordinate
SOLUTION
SOLUTION .... Cont ‘
0.8262.260
Sketch of flow profile
TIME’S UP …
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