174048087 Flow in Open Channel

FACULTY OF CIVIL AND ENVIRONMENTAL ENGINEERING DEPARTMENT OF WATER & ENVIROMENTAL ENGINEERING WATER ENGINEERING LABORATORY

LAB REPORT Subject Code Code & Experiment Title Course Code Date Section / Group Name Members of Group

Lecturer/Instructor/Tutor Received Date

Comment by examiner

BFC 21201 MKA – MKA – 02 02 ; FLOW IN OPEN CHANNEL 2 BFF/1 14 NOVEMBER 2011 2/5 AFANDI BIN ABD WAHID 1.MUHAMMAD IKHWAN BIN ZAINUDDIN 2.MOHD HASIF BIN AZMAN 3.MUHAMMAD HUZAIR BIN ZULKIFLI CIK AMNANI BIN ABU BAKAR EN JAMILULLAIL BIN AHMAD TAIB 21 NOVEMBER 2011

Received

(DF100122) (DF100018) (DF100079) (DF100040)

STUDENTS’ ETHICAL CODE

(SEC) DEPARTMENT OF WATER & ENVIRONMENTAL ENGINEERING FACULTY OF CIVIL & ENVIRONMENTAL ENGINEERING UNIVERSITI TUN HUSSEIN ONN MALAYSIA BATU PAHAT, JOHOR

“I declare that I have prepared this report with my own efforts. I also declare not receive or give any assistance in preparing this report and make this affirmation in the belief that nothing is in, it is true”

………………………………………. (STUDENT SIGNATURE) NAME : AFANDI BIN ABD ABD WAHID MATRIC NO. : DF100122 DATE : 21 NOVEMBER 2011




………………………………………. (STUDENT SIGNATURE) NAME : MUHAMMAD IKHWAN BIN ZAINUDDIN MATRIC NO. : DF100018 DATE : 21 NOVEMBER 2011




………………………………………. (STUDENT SIGNATURE) NAME : MOHD HASIF BIN AZMAN MATRIC NO. : DF100079 DATE : 21 NOVEMBER 2011




………………………………………. (STUDENT SIGNATURE) NAME : MUHAMMAD HUZAIR BIN ZULKIFLI MATRIC NO. : DF100040 DATE : 21 NOVEMBER 2011

PART A : THE HYDRAULIC JUMPS

1.0

INTRODUCTION

The concept of the hydraulic jump when the hydraulic drop that occurs at a sudden drop in the bottom of a channel, and the free surface flow around obstructions like bridge piers. A hydraulic jump forms when a supercritical flow changes into a subcritical flow. The change in the flow regime occurs with a sudden rise in water surface. Considerable turbulence, energy loss and air entrainment are produced in the hydraulic jump. A hydraulic is used for mixing chemicals in water supply systems, for dissipating energy below artificial channel controls, and as an aeration device to increase the dissolved oxygen in water.

In a hydraulic jump there occurs a sudden change in liquid depth from less-thancritical to greater-than-critical depth. The velocity of the flow changes from supercritical to subcritical as a result of the jump. This transition takes place over a relatively short distance, usually less than 5 times the depth of flow after the jump, over which the height of the liquid increase rapidly, incurring a considerable loss of energy. An example of a hydraulic jump can be observed when a jet of water from a faucet strikes the horizontal surface of the kitchen sink. The water flows rapidly outward and a circular jump occurs.

We shall restrict the derivation of the basic equation of the hydraulic jump to rectangular horizontal channels. First, we shall determine the downstream depth of the jump by using the momentum and continuity equations for one-dimensional flow. Then the energy loss due to the jump will be evaluated, using the energy equation.

2.0

OBJECTIVE

To investigate the characteristic a standing wave (the hydraulic jump) produced when waters beneath an undershot weir and to observe the flow patterns obtained.

3.0

LEARNING OUTCOMES

At the end of the course, students should be able to apply the knowledge and skills they have learned to: a) Understand the concept and characteristics of hydraulic jump. b) Understand the factors which influence the hydraulic jump.

4.0

THEORY

When water flowing rapidly changes to slower tranquil flow, a hydraulic jump or standing wave is produced. This phenomenon can be seen where water shooting under a sluice gate mixes with deeper water downstream. It occurs when a depth less than critical changes to a depth which is greater than critical and must be accompanied by loss of energy. An undular jump occurs when the change in depth is small. The surface of the water undulates in a series of oscillations, which gradually decay to a region of smooth tranquil flow. A direct jump occurs when the change in depth is great. The large amount of energy loss produces a zone of extremely turbulent water before it settles to smooth tranquil flow.

By considering the forces acting with the fluid on either side of a hydraulic jump of unit width it can be shown that : 2  v b    H  d a    d b  2g  2g 

va

2

Where ΔH is the total head loss across jump (energy dissipated) (m). v a is the mean velocity before jump (m/s), d a is the depth of flow before hydraulic jump (m). v b is the mean velocity after hydraulic jump (m) and d b is the depth of flow after hhydraulic jump (m). Because the working section is short, d simplifying the above equation, H  d 3  d1  / 4d1d 3 . 3

a

≈ d b and d b ≈ d3. Therefore,

5.0

EQUIPMENT

Figure 1: Self-contained Glass Sided Tilting Flume.

Figure 3: Control Panel

Figure 5: Hook and Point Gauge

Figure 2: Adjustable Undershot Weir

Figure 4: Instrument Carrier

Figure 6: Slope Adjustment

6.0

PROCEDURES

1. Ensure the flume is level, with the downstream tilting overshot weir, E at the bottom of its travel. Measure and record the actual breadth b (m) of the undershot und ershot weir. Install the undershot weir towards the inlet end of the flume and ensure that it is securely clamped in position. 2.

Adjust the undershot weir to position the sharp edge of the weir 20mm above the bed of the channel. Increase the height of the tilting overshot weir until the downstream level just stars to rise.

3.

Gradually open the flow control valve and adjust the flow until an undular jump is created with small ripple decaying towards the discharge end of the working section. Observe and sketch the flow pattern.

4.

Increase the height of water upstream of the undershot weir by increasing the flow rate and increase the height of the titling overshot weir to create a hydraulic jump in the centre of the working section. Observe and sketch the flow pattern.

5. Measure and record the values of d 1 , d 3 , d g , and , and q. repeat this for other flow rates q (upstream head) and heights of the gate d g

7.0

RESULT

Weir breadth, b = 0.300 m

Weir opening, dg (m) 0.020

Upstream flow depth, do (m) 0.416

Flow Depth above jump, d1 (m) 0.013

Flow depth below jump, Flow rate 3 d3 (m /s) (m) 0.124 0.012

0.030

0.228

0.017

0.101

0.040

0.146

0.024

0.094

1. Calculate V1 and plot dg against V1 2. Calculate ΔH / d1 and plot ΔH / d1 against d3 / d1 3. Calculate dc and verify d1 < dc < d3

ΔH

V1

ΔH d1

d3 d1

0.2121

2.000

16.32

9.54

0.012

0.0863

1.333

5.08

5.94

0.012

0.0380

1.000

1.58

3.92

8.0

DATA ANALYSIS

Calculation for :

∆H

= (0.124 - 0.013) 4 (0.013)(0.124) = 0.2121 m

Calculation for :

=

= (0.101 - 0.017) 4 (0.017)(0.101) = 0.0863 m

V1 ,

= (0.094 - 0.024) 4 (0.024)(0.094) = 0.0380 m

Q = AV V= Q A

A = dg x b = 0.020 x 0.300 2 = 0.006 m

A = dg x b = 0.030 x 300 2 = 0.009 m

Therefore, V = 0.012 0.006 = 2.000 m/s

Therefore, V = 0.012 0.009 = 1.333 m/s

Calculation for :

3

(d3 – d d1) 4d1d3

A = dg x b = 0.040 x 300 2 = 0.012 m Therefore, V = 0.012 0.012 = 1.000 m/s

∆H

d1

= 0.2121 0.013 = 16.32

Calculation for :

= 0.124 0.013 = 9.54

= 0.0863 0.017 = 5.08

= 0.0380 0.024 = 1.58

= 0.101 0.017 = 5.94

= 0.094 0.024 = 3.92

d3 d1

Calculate for: dc = q2 g

⅓

q=Q b = 0.012 0.300 3 = 0.040 m /s m Therefore, 2 ⅓ dc = 0.040 9.81 = 0.055 m Therefore weir opening:

9.0

dg 20

d1 < dc < d3 0.013 < 0.055 < 0.124

30

0.017 < 0.055 < 0.101

40

0.024 < 0.055 < 0.094

QUESTION

1.

Verify the force of the stream on either side of the jump is the same and that the specific energy curve predicts a loss equal to ΔH / ΔH / dc. F before before = F after after

2.

Suggest application where the loss of energy in hydraulic jump would be desirable. How is the energy dissipated? The hydraulic jump flow process can be illustrated by use of the specific energy e nergy concept. Equation loss energy can be written in term of the specific energy: 2

E = do + V / 2g

Where d o and E are feet. Because of the head loss across the jump, the upstream values of E are different. About the graph, (1) to to state (2) the fluid does not proceed along the specific energy curve and pass through the critical condition. The energy dissipates when water flow at weir opening and the energy became 0 because d 0 and d 3 has are force from adverse. Same like the equation, F before before = F after after .

10.0

DISCUSSION

Practical applications of hydraulic jump are:

1. To dissipate energy in water flowing over hydraulic structures as dams, weirs, and others and prevent scouring downstream structures. 2. To raise water level on the downstream side for irrigation or other water distribution purposes. 3. To increase weight on an apron and reduce uplift pressure under a structure by raising the water depth on the apron. 4. To indicate special flow conditions such as the existence of supercritical flow or the presence of a control section so that a gaging station maybe located. 5. To mix chemicals used for water purification. 6. To aerate water for city water supplies. 7. To remove air pockets from water supply lines an d prevent air locking.

11.0

CONCLUSION

The conclusion from the experiment, we can investigate the characteristic a standing wave (the hydraulic jump) produced when waters beneath an undershot weir and to observe the flow patterns obtained. From the experiment, we can get the force at weir opening, ∆H. In the water channel, water flowing rapidly changes to slower tranquil flow a hydraulic jump or standing wave is produced. This phenomenon can be seen where water shooting under a sluice gate mixes with deeper water downstream. It occurs when a depth less than critical changes to a depth which are greater than critical and must be accompanied by loss of energy.

From the result, we get the inverse line from graph gd1 against v1 and curve line from graph Δ H/d1 against d3/d1. Both graphs are sloping downward. Final result we can get the value of dc between d1 and d3. So the objective achieved and the experiments are success. Reason the experiment perform because almost drain are open channel. From the experiment, we know about water flowing.

12.0

REFERENCES

i. John J.E.A. 1988. Introduction to Fluid Mechanics, pp 330-342. Prentice Hall, Inc. ii. Chaudhry, M. H. 1993. Open Channel Flow, pp 302-408. Prentice-Hall, Inc. iii. Simon, A. L.1997. Hydraulics, pp 283-312. Prentice Hall, Inc iv. http://www.engineeringcivil.com (serve on 19/11/2011)

PART B: THE FORCE ON A SLUICE SLUICE GATE

1.0

INTRODUCTION

The Sluice gate is a device used to control the passage of water in an open channel. When properly calibrated it may also serve as a means of flow measurement. As the lower edge of the gate opening is flush with the floor of the channel, contraction of the bottom surface of the issuing stream is entirely suppressed. Side contraction will of course depend on the extent to which the opening spans the width of the channel.

A variety of gate-type structure is available for flow rate control at the crest of an overflow spillway, or at the entrance of an irrigation canal or river from a lake. Three typical types are vertical gate, radial gate and drum gate. The flow under a gate is said to be free outflow when the fluid issues as a jet of supercritical flow with a free surface open to the atmosphere. atmosphere.

In certain situation, the depth downstream of the gate is controlled by some downstream obstacle and the jet of water issuing from under the gate is overlaid by a mass of water that is quite turbulent.

Picture 1: Drowned outflow from a sluice gate.

2.0

OBJECTIVE

To determine the relationship between upstream head and trust on a sluice gate (undershot weir) for water flowing under the sluice gate.

3.0

LEARNING OUTCOMES

At the end of the course, students should be able to apply the knowledge and skill they have learned to: a) Understand the basic terms and concept of a sluice gate. b) Understand on the characteristics of the force on a sluice gate.

4.0

THEORY

It can be shown that the resultant force on the gate is given by the equation :

  g  d 1  1 2  d   F g   gd 1  0 2  1   d  bd 1  d  2 1  0   1  2

The gate thrust for hydrostatic pressure distribution is given by the equation : F H 

1 2



 g d 0

 d g 

2

where Fg is the resultant gate thrust (N), F H is the resultant hydrostatic thrust (N), q is 3

volume flowrate (m/s), ρ is density of fluid (kg/m ), g is the gravitational constant 2

(9.81m/s ), b is breadth of gate (m), d g is height of upstream opening (m), d0 is upstream depth of flow (m) and d1 is downstream depth of flow (m).

5.0

EQUIPMENT

Figure 1: Self-contained Glass Sided Tilting Flume.

Figure 2: Adjustable Undershot Weir

Figure 3: Control Panel

Figure 6: Instrument Carrier

Figure 4: Hook and Point Gauge

Figure 6: Slope Adjustment

6.0

PROCEDURES

1. Ensure the flume is level, with the downstream tilting overshot weir at the bottom of its travel its travel. The actual breadth b (m) of the undershot weir are measured and recorded. 2. The undershot weir adjusted to set its bottom edge 20mm above the bed of the channel. 3. Introduce water into the flume until do = 200mm. with do at this position, the readings for q and d1 are taken. Raise the undershot weir in increments of 10mm, maintaining constant do by varying q. At each level of the weir, record the values of dg, d1 and q. 4. The procedure with a constant flow q, allowing do to vary are repeated and the value of do and d1 are recorded.

7.0

RESULT

Weir breadth, b = 300 m

Weir opening, dg (m) 0.025

Upstream flow depth, do (m) 0.292

Downstream Flow Depth, d1 (m) 0.016

0.030

0.225

0.035 0.040

Flow Rate 3 q (m /s)

Gate Thrust Fg (N)

Hydrostatic Thrust, FH (N)

Fg FH

dg do

0.012

-1.931 x10

349.67

-5.52

0.086

0.018

0.012

-1.58 x 10

186.51

-8.47

0.133

0.167

0.022

0.012

-1.48 x 10

85.46

-17.32

0.210

0.138

0.024

0.012

-1.12 x 10

47.11

-23.77

0.290

8.0

DATA ANALYSIS

Calculation for Gate Thrust at Weir Opening = 0.025 m

 d   g  d 1  1 Gate Thrust = F g   gd 1 2  0  1     d 2  bd 1  d  2 1 0    1  2

2  1 1000(9.81)  0.016  2  0.292   1  F g  (1000)(9.81)(0,016)  1   2  0.016  (0.300)(0.016)  0.292 

6

F g = = -1.931 x 10 N Calculation for Gate Thrust at Weir Opening = 0.030 m

 d   g  d 1  1 Gate Thrust = F g   gd 1 2  0  1     d 2  bd 1  d  2 1  0   1  2

2  1 1000(9.81)  0.018  2  0.225 F g  (1000)(9.81)(0,016)   1  1   2 0 . 018 ( 0 . 300 )( 0 . 018 ) 0.225    

6

F g = -1.58 x 10 N Calculation for Gate Thrust at Weir Opening = 0.035 m

  g  d 1  1 2  d 0     gd 1  1  2   bd 1  d  2 d 1  0   1  2

Gate Thrust = F g 

2  1 1000(9.81)  0.022  2  0.167  1  F g  (1000)(9.81)(0,016)  1   2 0 . 022 ( 0 . 300 )( 0 . 022 ) 0 . 167    

6

F g = -1.48 x 10 N Calculation for Gate Thrust at Weir Opening = 0.040 m

 d   g  d 1  1 Gate Thrust = F g   gd 1 2  0  1     d 2  bd 1  d  2 1  0   1  2

2  1 1000(9.81)  0.024  2  0.138  1  F g  (1000)(9.81)(0,016)  1   2 0 . 024 ( 0 . 300 )( 0 . 024 )  0.138   

F g = -1. -1.12 12 x 10 N

Calculation for Hydrostatic Thrust at Weir Opening = 0.025 m Hydrostatic Thrust , , F H 

1

F H 

1

2 2



 g d 0

 d g 

2

(1000)(9.81)0.292  0.025

2

F H = 349.67 N


1

F H 

1

2 2



 g d 0

 d g 

2

(1000)(9.81)0.225  0.030

2

F H = 186.51 N


1

F H 

1

2 2



 g d 0

 d g 

2

 g 0.167  0.035

2



F H = 85.46 N


1

F H 

1

2 2



 g d 0

 d g 

2

(1000)(9.81)0.138  0.040

F H = 47.11 N

2

Calculation for F g /F H ratio at Weir Opening = 0.025 m 6

Fg = - 1.931 x 10 FH 349.67 = - 5.52


Fg = - 1.580 x 10 FH 186.51 = - 8.47


Fg = - 1.48 x 10 FH 85.46 = - 17.32


Fg = - 1.12 x 10 FH 47.11 = - 23.77

Calculation for d g /d o ratio at Weir Opening = 0.025 m dg = 0.025 do 0.292 = 0.086




9.0

QUESTIONS

1. Plot graph of the ratio Fg / FH against the ratio dg / do. (Refer graph)

2. Comment of the graph obtained. Based on the graph, the pattern of the graph is linear line and an increased. When the value of d g /d o are increased , the value of F also increased. g / F H also

3. Compared your calculated values for Fg and FH and comment on any differences. After

we 2

are 2

calculated

Fg

with

used

this

formula,

2

F g = ½ ρgd 1 [ (d 0 / d 1 ) - 1 ] – ] – ρg ρg / bd 1 [ 1 – 1 – (d (d 1 / d 0 )], we get the value are in negative 2

(-ve) and when we calculated F H with F H = ½ ρg ( d 0 - d g ) , we get the value in positive (+ve). F g is resultant gate thrust (N) and F H is resultant hydrostatic thrust (N). We can conclude that before the water is flow to the sluice gate, the force are F H are in positive (+ve) because is follow the direction of the water flow. The force are happened after sluice gate are Fg in negative (-ve) because the resultant force of the flow is opposite the direction.

4. What is the effect of flow rate on the result obtained? From the result, the more flow rate will give the less thrust for both of the gate and the hydrostatic. This is because of the decreasing pressure at b oth of them.

10.0

DISSCUSSION

Floodgates are adjustable gates used to control water flow in reservoir, river, reservoir, river, stream, stream, or levee systems. They may be designed to set spillway set spillway crest heights in dams, in dams, to to adjust flow rates in sluices and canals, and canals, or or they may be designed to stop water flow entirely as part of a levee or storm storm surge system.

Since most of these devices operate by controlling the water surface elevation being stored or routed, they are also known as crest gates. In the case of flood bypass systems, floodgates sometimes are also used to lower the water levels in a main river or canal channels by allowing more water to flow into a flood bypass or detention detention basin when the main river or canal is approaching a flood stage. Picture below shown the floodgates.

Tokyo floodgates created to protect from typhoon from typhoon surges

Another example of floodgate or sluice gate that we can explore is as shown below :

A sluice gate on the Harran the Harran canal

Tainter gate diagram

Drum gates are controlled with valves .

Tainter gates and spillway

Drum gates on a diversion dam

11.0

CONCLUSION

The flow through a channel in which a gate partially obstructs the flow will be used for this measurement of total force. This obstruction is called a sluice gate (see Figure 1). The flow is from left to right and enters at a velocity Vo. The fluid in the upstream section builds up against the gate to a level y level y0 0 , and exits the upstream section under the gate of height b. The fluid attains a higher velocity V 1 as it passes under the gate and a shallower free surface height y height y1 1 downstream.

Three assumptions will be made in this derivation of the equation for horizontal force on a sluice gate: 1) The viscous force at the bottom of the channel and the energy dissipation at the gate are negligible. 2) The flow is steady and has a uniform velocity distribution at the inlet and outlet sections. 3) Flow at upstream and downstream sections is uniform and the effect of the sidewalls is negligible.

Figure 1. Flow under a vertical v ertical sluice gate.

12.0

REFERENCES

1. John J.E.A. 1988. Introduction to Fluid Mechanics, pp 330-342. Prentice Hall, Inc. 2. Chaudhry, M. H. 1993. Open Channel Flow, pp 302-408. Prentice-Hall, Inc. 3. Chow, V. T. 1959. Open Channel Hydraulics, pp 608-710. McGraw-Hill, Inc.

174048087 Flow in Open Channel

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