Chapter 4 - Non-uniform Flow in Open Channel

HYDRAULICS(BFC 21103)

Prepared by:-

MR WAN AFNIZAN BIN WAN MOHAMEDDEPT. OF WATER & ENVIRONMENTAL ENGINEERINGFAC. OF CIVIL & ENVIRONMENTAL ENGINEERINGe-mail: afnizan@uthm.edu.my

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 ‘

Figure 4.2: Hydraulic jump in the hydraulic laboratory

Figure 4.3: Hydraulic jump at the spillway toe (Itaipu Dam, Brazil)

Figure 4.4: Hydraulic jump at the downstream of sluice gate (Harran cannal, Turkey)

Figure 4.5: Waves that hit sea wall in Depoe Bay (Oregon U.S)

Figure 4.6: Surges waves (Tangjiasan, China)

USAGE OF HYDRAULIC JUMP

TYPES OF THE JUMP

Fr = 1.6

MOMENTUM EQUATION

MINIMUM FLOW FORCE & CRITICAL DEPTH

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 ‘

4.2 …..

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 ‘

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

LOCATION OF THE JUMP

IF THERE IS CHANGING OF SLOPE

Figure 4.8

Figure 4.9

Figure 4.10

PROCEDURES OF CALCULATION

.... Cont ‘

……. 4.10

.... Cont ‘

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

DEFINITION

TYPES OF SLOPE

FLOW ZONE

PROFILE TYPE

OCCURRENCE EXAMPLES OF THE ACTUAL FLOW PROFILE

Determine the flow profile type as shown in figures below.

EXAMPLE 4.4

SOLUTION

GRADUALLY VARIED FLOW FORMULA

Figure 4.8

METHOD OF GVF CALCULATION

……. 4.10

……. 4.11

……. 4.12

……. 4.13

Figure 4.9

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

SOLUTION

0.8262.260

-ve = Against co-ordinate

0.8262.260

Sketch of flow profile

EXAMPLE 4.6

Sketch of the problem

SOLUTION

0.8262.260

+ve = Follow the co-ordinate

0.8262.260

……. 4.14

……. 4.15

……. 4.16

……. 4.17

Figure 4.11

……. 4.18

……. 4.19

……. 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

SOLUTION

0.8262.260

-ve = Against the co-ordinate

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

SOLUTION

0.8262.260

Calculation table :-

0.8262.260

……. 4.16

……. 4.17

Formula applied …

……. 4.18

……. 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

SOLUTION

Calculation table :-

SOLUTION

0.8262.260

TIME’S UP …

THANK YOU

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