Group 15 Report-2-Open-Channel (FLUID MECHANICS) PDF

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FLOW PROFILE IN OPEN CHANNEL GROUP OPEN ENDED LAB REPORT GROUP 15:NO. GROUP MEMBER MATRIC NUMBER 1. OOI LING JIE A166455 2. AZMIR NURHAQIM BIN JEFRIN ROZILEY A165451 3. RABIATUL ADAWIAH BINTI CHAIRUL ANWAR A165473 4. MUHAMMAD SYAMIM BIN MD SAAD A165584 5. WANG GUO XIAN A165556 LECTURER: Dr. SITI FAT...

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FLOW PROFILE IN OPEN CHANNEL GROUP OPEN ENDED LAB REPORT GROUP 15:NO.

GROUP MEMBER

MATRIC NUMBER

1. OOI LING JIE

A166455

2.

AZMIR NURHAQIM BIN JEFRIN ROZILEY

A165451

3.

RABIATUL ADAWIAH BINTI CHAIRUL ANWAR

A165473

4. MUHAMMAD SYAMIM BIN MD SAAD

A165584

5. WANG GUO XIAN

A165556

LECTURER: Dr. SITI FATIN BINTI MOHD RAZALI COURSE: FLUID MECHANICS (KKKH1223) DATE OF SUBMISSION: 21FIRST MAY 2018

PART A: THE HYDRAULIC JUMPS 1.0 INTRODUCTION . A hydraulic jump is a fluid shockwave created at the transition between laminar and turbulent flow. One common example of a hydraulic jump can be seen in the water radiating outward when the stream of tap water strikes the horizontal surface of a sink. The water initially flows in a smooth sheet with consistent current patterns. In this region, the speed of the water exceeds the local wave speed. Friction against the sink surface slows the flow until an abrupt change occurs. At this point, the depth increases as water piles up in the transition region and flow becomes turbulent. The motion of individual water molecules becomes erratic and unpredictable. The interruption of flow patterns also reduces the kinetic energy of the water. In addition to the kitchen sink example, hydraulic jumps are also typical features of river rapids where the water swirls and foams around rocks and logs. Some practical applications of hydraulic jump are: 1. To dissipate energy in water flowing over dams, weirs, and other hydraulic structures and thus prevent scouring downstream from the structures; 2. To recover head or raise the water level on the downstream side of a measuring flume and thus maintain high water level in the channel for irrigation or other water-distribution purposes; 3. To increase the discharge of a sluice by holding back tailwater, since the effective head will be reduced if the tail water is allowed to drown the jump; 4. To remove air pockets from water supply lines, thus prevent air locking. 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.



Find out the characteristic of standing wave produce when waters beneath an undershot weir thus observe the patterns flow.

3.0 THEORY

Figure 2.1:Hydraulic Jump When water flow 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 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 within the fluid on either side of a hydraulic jump of unit width it can be shown that (eq. 1):

v ΔH =d a +

a

2

2g

(

− d b+

v

b

2

2g

)

____________(eq. 1)

Where, ∆H is the total head loss across jump (energy dissipated) (m), v a is the mean velocity before jump (m/s), da is the depth of flow before hydraulic jump (m). Because of the working section is short, da ≈ d1 and db ≈ d3. Therefore, simplifying the above equation,

ΔH =

( d 3 −d 1) 4 d1 d3

3

____________(eq. 2)

Figure 2.2: Relationships between Hydraulic Jump with y/yc Hydraulic jump is analogous to a shock wave in aerodynamics. It is a turbulent, nonenergy-conserving process that passes from rapid flow to tranquil flow. Since energy is not conserved, the before and after states do not correspond to the intersection of a line E = const. with the specific energy curve. However, momentum is conserved, and this permits us to find the final state if we know the initial state. We will assume the channel is horizontal, since the hydraulic jump takes place in a limited length of channel, and gravitational energies will not be important. Let the water enter the jump with velocity V 1 and height y1, and leave with velocity V2 and height y2. In one second, the momentum lost by the fluid in passing through the jump is ρQ (V2 - V1), and this must be equal to the difference in pressure forces on the two cross sections. This difference is ρgh1A1 - ρgh2A2, where h is the distance from the surface to the centroid of area A. This is one place where the pressure plays an explicit role. Equating the rate of momentum change with the net force acting, we find that the combination hA + QV/g is conserved. For a rectangular channel, A = by and h = y/2m so a conserved quantity is f = q 2/yg + y2/2. We can write this in dimensionless form by setting x = y/y c, as we did for the specific energy above. Then, f/yc = 1/x + x2/2. We shall call a plot of f/yc vs y/yc the hydraulic jump curve. Curiously, it is just the specific energy curve with 1/x in place of x. The hydraulic jump curve for a wide rectangular channel is shown at the right. When the slope changes from steep to mild, a hydraulic jump occurs at a point such that y2 is the normal depth on the downstream end. Depending on the corresponding value of y 1, the jump may occur either after or before the break in slope, wherever the required value of y 1 can

be found. If y1 is greater than the depth on the steep slope, then the jump will occur on the mild slope when the height of the rapid flow reaches y 1. If it is not, then the jump will occur on the steep slope to a depth that becomes the normal depth on the mild slope.

Figure 2.3 The hydraulic Jump This may be illustrated by the jump shown in the figure at the left. The water comes through a gate with high velocity and a depth less than the critical depth, so the flow is rapid. The depth conjugate to y1' is y2'. Since this is greater than the normal depth on the mild slope, the jump will not occur immediately, because there is insufficient energy. The rapid flow decelerates on the mild slope and the water depth increases, bringing down the conjugate depth. Both the energy line and the water surface are slightly concave upwards in this region, though drawn as straight. When the depth reaches y1 conjugate to the normal depth on the mild slope, the jump begins, at the first point where it is possible. It does not occur at a single section, but extends for the length of the turbulent jump (foreshortened in the diagram). It is easy to see that the momentum on the left is balanced by the pressure force on the right. The energy line falls rapidly through the region of the jump, then resumes with the mild slope, S' = S. There is also a small energy loss at the gate, from the level of the slack water upstream. There are other cases, but this is the most common one. A good place to see a jump is at the overflow spillway of a dam, with a rapid flow down the inclined face of the spillway, and a jump on the apron beyond in the tail water.

4.0 EQUIPEMENTS 41

Self-contained Glass Sided Tilting Flume

42

Adjustable Undershot Weir

43

Instrument Carrier

44

Hook and Point Gauge

5.0 PROCEDURE 1. The machine was turned on and the water was pumped into the maximum water level fixed until the water about 80% of the channel. Thus, the water pressure of the flow rate should be not more than 0.018 m³. 2. Flume is levelled with the downstream tilting overshot weir at the bottom and actual breadth (m) of the undershot weir. Undershot weir is installed towards inlet end and securely clamped in position with plastacine provided. 3. Undershoot weir have been adjusted to position the sharp edge of the weir 20 mm above the bed of channel. The height of the tilting overshot weir was increased until the downstream level just start to rise.

4. The flow control valve gradually opens and the flow was adjusted until an undular jump was created with small ripple decaying towards the discharge end of the working section. 5. The height of the upstream of the undershot weir was increased by increasing the height of the tilting overshot weir to create and hydraulic jump. 6. With this, d0, d1, d3 and dg and also fixed q are measured. 7. The experiment was repeated with different height of gate opening, dg. 8. The graph is plot.

6.0 RESULT

Weir breadth, b = 0.310 m Table 1: Data recorded for hydraulic jumps experiment

Weir

Upstream

Flow

Flow

Flow

Opening

Flow

Depth

Depth

Rate

Depth

Above

Below

Jump

Jump

d1

d3

(m)

(m)

dg

(m)

d o (m)

ΔH

v1

ΔH d1

d3 d1

(m3/s)

0.020

0.3938

0.0128

0.1247

0.011

0.219

2.772

17.109

9.742

0.025

0.2550

0.0153

0.1105

0.011

0.128

2.232

8.366

7.222

0.030

0.2077

0.0188

0.1041

0.011

0.079

1.887

4.202

5.537

0.035

0.1840

0.0199

0.1011

0.011

0.067

1.783

3.367

5.080

0.040

0.1213

0.0253

0.0846

0.011

0.024

1.403

0.949

3.343

7.0 DATA ANALYSIS Calculation for velocity, V1

Given formula: V=

Q , w h ere A=b x d 1 A 1

V1 = Q / (b x d1) = 0.011 / (0.31 x 0.0128)

=

2.772 m/s

V2 = Q / (b x d1) = 0.011 / (0.31 x 0.0159)

=

2.232 m/s

V3 = Q / (b x d1) = 0.011 / (0.31 x 0.0188)

=

1.887m/s

V4 = Q / (b x d1) = 0.011 / (0.31 x 0.0199)

=

1.783 m/s

V5 = Q / (b x d1) = 0.011 / (0.3 x 0.0253)

=

1.403 m/s

Calculation for ratio of d3 and d1

Given formula: d3 d1 d3 1. d 1

=

0.1247 0.0128 = 9.742

d3 2. d 1

=

0.1105 0.0153 = 7.321

=

0.0846 0.0253 = 3.343

0.1041 =5.537 0.0188

d3 4. d 1

0.1011 = 0.0199

= 5.080

d3 5. d 1

d3 3. d 1

=

Calculation for total head loss across jump,

ΔH

Given formula:

v ΔH =d a +

a

2

2g

(

− db+

v

b

2

2g

Value for da = d1 and db = d3

)

therefore simplify the equation, ∆H = (d3-d1)3 / 4 d1d3

∆H1

= (d3-d1)3 / 4 d1d3 = (0.1247-0.0128)3/ 4(0.0128)(0.1247)

=

0.219

∆H2

= (d3-d1)3 / 4 d1d3 = (0.1105-0.0153)3/ 4(0.0153)(0.1105)

=

0.128

∆H3

= (d3-d1)3 / 4 d1d3 = (0.1041-0.0188)3/ 4(0.0188)( 0.1041)

=

0.081

∆H4

= (d3-d1)3 / 4 d1d3 = (0.1011-0.0199)3/ 4(0.0199)( 0.1011)

=

0.067

∆H5

= (d3-d1)3 / 4 d1d3 = (0.0846-0.0253)3/ 4(0.0253)(0.0846)

=

0.024

Calculation for ratio of ∆H and d1

Given formula: ΔH d1

∆H1/d1

=

0.219/ 0.0128

=

17.109

∆H2/d1

=

0.128 / 0.0153

=

8.366

∆H3/d1

=

0.079 / 0.0188

=

4.202

∆H4/d1

=

0.067 / 0.0199

=

1.521

∆H5/d1

=

0.024 / 0.0253

=

0.949

Calculation for dc, and verify d3: d1< dc< d3 A specific value is fixed for the flow rate for every testing to ensure there is no over flow happen on open channel equipment while carrying out the experiment. Q =

0.011m3/s

b =

0.31 m

q =

Q/b

dc

=

(q2/g)1/3

=

0.011/0.31

=

((0.035)2 / 9.81)1/3

=

0.035

=

0.0450 m

Based on the calculation, we concluded that the value of dc is within the stated interval. d1< dc< d3 = 0.0128 < 0.0450 < 0.1247.

8.0 DISCUSSION AND SUGGESTION 1. To 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 /d c , the solution are shown as below. From the graph of ∆H/d1 against d3/d1, Gradient = ∆H/d1 ÷ d3/d1 = (15.2 – 4) / (9 – 4.65) = 2.57 From 1st data, ∆H/dc

= 0.219 / 0.0450 = 4.87

Total head of upstream of weir crest, q² 2 gd ² = 0.3938 + (0.035)2 / (2 x 9.81 x 0.39382)

H o = do +

= 0.3942m Total head of downstream of weir crest, q² 2 gd з ² = 0.1247 + (0.035)2 / (2 x 9.81 x 0.12472)

H 3 = d3 +

= 0.1287m ∆H = Ho – H3 = 0.3942 – 0.1287 = 0.2655 < 4.56 (from theory)

Difference of the theoretical and experimental ∆H values may due to: a) Parallax error: During recording the data, the observer observed the wrong value. b) Defect of the experimental equipment: The plasticine is not strong enough to block the water from enter the free space of weir.

2. The most important engineering applications of the hydraulic jump is to dissipate energy in canals, dam spillways, and similar structures so that the excess kinetic energy does not damage these structures. The energy dissipation or head loss across a hydraulic jump is a function of the magnitude of the jump. The larger the jump as expressed in the fraction of final height to initial height, the greater the head loss.

Two methods of designing an induced jump are common: i.

If the downstream flow is restricted by the down-stream channel such that water backs up onto the foot of the spillway, that downstream water level can be used to identify the location of the jump.

ii.

If the spillway continues to drop for some distance, but the slope changes such that it will no longer support supercritical flow, the depth on the lower subcritical flow region is sufficient to determine the location of the jump.

In both cases, the final depth of the water is determined by the downstream characteristics. The jump will occur if and only if the level of inflowing (supercritical) water level (h 0) satisfies the condition:

Fr= Upstream Froude Number g = acceleration due to gravity (9.81 m/s2) h = height of the fluid from the bed of stream (h0 = initial height; h1 = final downstream height)

Dissipation of energy results in decreases of velocity and increases of flow depth.

water depth 0.25

0.2

0.15

0.1

0.05

0 0.06

0.08

0.1

0.12

0.14

0.16

0.18

0.2 energy2 0.22 Specific

0.24

Graph 1.1:

Graph shown above consists in water depth in the y-axis and Specific energy in the x-axis, with Dc (Critical depth) and Ec (Critical Specific energy) shown in red color.

Specific energy 0.25

0.2 0.15

0.1

0.05

0 0.2

0.4

0.6

0.8

1

Length along the channel 1.2 1.4 1.6

Graph 1.2: Graph shown above consists in specific energy in the y-axis and length along the channel in the x-axis,and is also shown the number of the location points and the water movement, it’s notice that the energy is loss along the channel with a big drop of energy when the hydraulic jump is formed.

Open Channel Flow Tank

Sub Critical

Sub Critical Critical depth

Super Critical

Fig 1.2The image shown Super Critical above represents the open channel with the critical depth line, above the line the water flow is sub-critical and below the line the water flow is supercritical. Critical depth

Fig 1.3- Image of a hydraulic jump with the Froude number between 4,5 to 9, jump consider a steady jump, information taken from B.S.Massey (1989) Mechanics of Fluids, 6th edition.

Fig 1.4- Drawing of the Lab hydraulic jump, Froude number 5.8, is assumed as “steady jump” by the use of previous studies, the jump above (Fig 1.3) was the jump more similar to the jump obtained in the experiment.

9.0 CONCLUSION As a conclusion, based on the investigation and analytical of result data, it shown that the degree of hydraulic jump are affected by the change of height of opening gate dg, level before jump d1, the velocity before jumps v1 and level of upstream do. This function of this agent can be determined by manipulated those variable and result with the observed transformation of flow hydraulic jump pattern. The rapidly varied hydraulic jump if the depth change abruptly over a comparatively short distance is classified as rapidly varies flow (RVP). In term of engineering, the objective of hydraulic jump is to perform an energy-dissipating device to reduce the excess energy of water flows. The relationship between the graphs of ΔH/d1 against d3/d1 is linearly proportional. When d3/d1 increases, value of ΔH/d1 increases. On the other hand, the graph of dg against v, when the value of weir opening, dg increases, the value of velocity of water decreases. When the gate is opened, the value of stream flow depth decreases and the downstream flow depth increases. Hydraulic jumps can be classified according to the Froude Number and ratio of d3/d1.

PART B: THE FORCE ON A SLUICE GATE

I.1 OBJECTIVE i.

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

ii.

To figure out the upstream and downstream of the sluice gate to calculate the hydrostatic pressure force.

iii.

To determine the relationship between the specific energy and flow depth in which can cause by water flowing under a sluice gate.

2.1 THEORY It can be shown that the resultant force on the gate is given by the equation,

F g=

1 2

ρ gd

12

(

d

02

d

)

−1 − 1

2

ρg bd 1

distribution is given by the equation, thrust (N),

FH

1−

d1 d0

1 FH=

) 2

. The gate thrust for a hydrostatic pressure ρg ( d 0 −d g )2

is resultant hydrostatic thrust (N),

density of fluid (kg/m3),

dg

(