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 tailwater 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):

ΔH=d a+

va

2

2 g−(db+

vb

2

2 g )____________(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=(d3−d1 )3

4d1d3 ____________(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, non-

energy-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 V1 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 y1 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 y1. 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

Opening

d g (m)

Upstream

Flow Depth

do (m)

Flow

Depth

Above

Jump

d1 (m)

Flow

Depth

Below

Jump

d3 (m)

Flow

Rate

(m3/s)

ΔH v1 ΔHd1

d3

d1

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

Weir breadth, b = 0.310 m

Table 1: Data recorded for hydraulic jumps experiment

7.0 DATA ANALYSIS

Calculation for velocity,

V1

Given formula:

V=QA

,w h ere A=b xd 11

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

1.

d3

d1 =

0.12470.0128= 9.742 2.

d3

d1 =

0.11050.0153= 7.321 3.

d3

d1 =

0.10410.0188=5.537

4.

d3

d1 =

0.10110.0199 = 5.080 5.

d3

d1 =

0.08460.0253= 3.343

Calculation for total head loss across jump,

ΔH

Given formula:

ΔH=d a+v

a2

2 g−(db+

vb

2

2 g )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:

ΔHd1

∆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 toΔH /dc , 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,

Ho = do + q ²

2 gd ⃘² = 0.3938 + (0.035)2 / (2 x 9.81 x 0.39382)

= 0.3942m

Total head of downstream of weir crest,

H3 = d3 + q ²

2 gd з ² = 0.1247 + (0.035)2 / (2 x 9.81 x 0.12472)

= 0.1287m

∆H = Ho – H3

= 0.3942 – 0.1287= 0.2 655 < 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 (h0) 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.

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.0 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.

2.0 THEORY

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

Fg=

12

ρ gd12 (d02

d1

2

−1)− ρgbd1

(1−d1

d0). The gate thrust for a hydrostatic pressure

distribution is given by the equation, FH=1

2ρg (d0−d g)2

where, Fg is resultant gate thrust

(N), FH is resultant hydrostatic thrust (N), q is volume flowrate (m/s), ρ is density of fluid

(kg/m3), g gravitational constant (9.81m/s2), 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).

In its simplest form, a control structure is designed to change the water depth to (or through) the

critical depth (dc), so that the discharge is fixed relative to the depth. In practice, most control

structures accelerate a subcritical flow, through the critical regime, to produce a shallow fast

supercritical flow.

Open channel control structures.

To include more than one effective control within an open channel, the supercritical flow

produced by an upstream control must be reconverted to a subcritical flow. This is usually

achieved by a hydraulic jump (or stationary bore) in which the characteristics of the subcritical

flow are determined by a second downstream control. These events are associated with large

energy losses, and are often used as an effective means of dissipating unwanted kinetic energy

downstream of an overflow (spillway) or underflow (sluice gate) structure. The hydraulic jump

is in many respects analogous to a shock wave arising within a compressible flow. For example,

whereas the hydraulic jump provides a transition from supercritical to subcritical flow, the shock

wave involves a transition from supersonic to subsonic flow. In both cases there is a critical

velocity below which these transitions cannot occur, and both processes involve an increase in

entropy. Indeed, in the case of a hydraulic jump the increase in entropy per unit mass is

proportional to the cube of the depth change; whereas in a shock wave this increase is

proportional to the cube of the pressure difference (provided this is small).

iii.0 EQUIPMENTS

a) Self-contained Glass Sided Tilting Flume.

b) Adjustable Undershot Weir.

c) Hook and Point Gauge.

4.0 PROCEDURE

1. Flume was levelled with the downstream tilting overshot weir at the bottom of its travel. The

actual breadth b (m) of the undershot weir was measured and recorded. Then the undershot

weir towards the inlet end of the flume was installed and it is securely clamped in position.

2. The undershot weir was adjusted and set its bottom edge 20mm above the bed of the

channel.

3. Water introduced into the flume until d0 = 200mm. Reading for q and d1 taken with d0 at

this position, the undershot weir raised in increments of 10mm, constant d0 maintained with

vary, q . The values of d g , d1 and q were recorded at each level of the weir.

4. The procedure was repeated with a constant flow q , allowing do to vary. The values of do

and d1 was recorded.

6.0 RESULT

Weir breadth, b = 0.310 m

Density of fluid, ρ = 1000 kg/m3

Gravitational constant, g = 9.81 m/s2

Weir Openingd g (m)

Upstream Flow Depth

do (m)

Downstream Flow Depth

d1 (m)

Flow Rate

q (m3/s)

Gate Thrust

Fg

( x106N)

Hydrosatic ThrustFH (N)

Fg

F H

x103

dg

do

0.021 0.3210 0.0151 0.012 -1.977 441.45 -4.523 0.065

0.022 0.3704 0.0183 0.012 -1.643 595.38 -2.760 0.059

0.023 0.2984 0.0169 0.012 -1.766 372.02 -4.747 0.077

0.024 0.2454 0.0172 0.012 -1.711 240.43 -7.116 0.098

0.025 0.2499 0.0163 0.012 -1.815 248.09 -7.316 0.100

Table 2: Data recorded for force on sluice gate Lab

7.0 DATA ANALYSIS

Calculation of Gate Resultant Force, Fg

Given the formula of

Fg=1

2ρ gd

12 (d02

d1

2

−1)− ρgbd1

(1−d1

d0)

where, ρ = density of the fluid in the experiment (1000kg/m3)

g = gravitational constant (9.81 m/s2)

b = width breadth of the undershot weir (unit of meter)

d0 = Upstream depth of flow (unit of meter)

d1 = Downstream depth of flow (unit of meter)

By applying the formula above with substitution of achieved result, Fg can be obtained as below

shown.

Fg 1 = ½ (1000)(9.81)(0.0151)2[(0.3210)2/ (0.0151)2-1)-

(1000)(9.81)/(0.310X0.0151)[1-(0.0151)/0.3210)]

= - 1.977 x 10 6 N

Fg 2 = ½ (1000)(9.81)(0.0183)2[(0.3704)2/ (0.0183)2-1)-

(1000)(9.81)/(0.310X0.0183)[1-(0.0195)/0.3704)]

= - 1.643 x 10 6 N

Fg 3 = ½ (1000)(9.81)(0.0169)2[(0.0.2984)2/ (0.0169)2-1)-

(1000)(9.81)/(0.310X0.0169)[1-(0.0169)/0.2984)]

= - 1.766 x 10 6 N

Fg 4 = ½ (1000)(9.81)(0.0172)2[(0.2454)2/ (0.0172)2-1)-

(1000)(9.81)/(0.310X0.0172)[1-(0.0172)/0.2454)]

= - 1.711 x 10 6 N

Fg 3 = ½ (1000)(9.81)(0.0163)2[(0.2499)2/ (0.0163)2-1)-

(1000)(9.81)/(0.310X0.0163)[1-(0.0163)/0.2499)]

= - 1.815 x 10 6 N

Calculation for Hydrostatic Pressure Distribution, Fh

Given the formula of

Fh=1

2ρg (d0−dg )2

where, ρ = density of the fluid in the experiment (1000kg/m3)

g = gravitational constant (9.81 m/s2)

d0 = Upstream depth of flow (unit of meter)

dg = Height of weir opening (unit of meter)

By applying the formula above with substitution of achieved result, Fh can be obtained as below

shown.

Fh1 = ½ (1000)(9.81) (0.3210 – 0.021)2 = 441.45 N

Fh2 = ½ (1000)(9.81) (0.3704 – 0.022)2 = 595.38 N

Fh3 = ½ (1000)(9.81) (0.2984 – 0.023)2 = 372.02 N

Fh4 = ½ (1000)(9.81) (0.2454 – 0.024)2 = 240.43 N

Fh5 = ½ (1000)(9.81) (0.2499 – 0.025)2 = 248.09 N

Calculation for Ratio of Resultant Force (Fg) to Hydrostatic Pressure Distribution (Fh)

With the calculation of resultant force (Fg) and hydrostatic pressure distribution (Fh) that we

have analyzed earlier on, the ratio

Fg

F H can be obtained as below shown.

Fg/Fh (1) = (- 1.977 x106)/ (441.45) = -4.523x10 3

Fg/Fh (2) = (-1.643 x106)/ (595.38) = -2.760 x10 3

Fg/Fh (3) = (-1.766 x106)/ (372.02) = -4.747 x10 3

Fg/Fh (4) = (-1.711 x106)/ (240.43) = -7.116x10 3

Fg/Fh (5) = (-1.815 x106)/ (248.09) = -7.316x10 3

Calculation for Ratio of dg to d0

With the attachment of table 2 in section result, the value of dg and d0 are substituted to get the

value of ratio

dg

do .

dg/do (1) = (0.021 /0.3210) = 0.065

dg/do (2) = (0.022 /0.3704) = 0.059

dg/do (3) = (0.023 /0.2984) = 0.077

dg/do (4) = (0.024 /0.2454) = 0.098

dg/do (5) = (0.025 /0.2499) = 0.100

8.0 DISCUSSION

1. Plot a graph of the ratio Fg/FH against the ratio dg/do .

2. Comment on the graph obtained.

The value of the ratio Fg/FH is increase when the value of the ratio dg/do is decrease

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

Value for the gravity force, Fg is negative (-) because the resultant force of the flow is opposite

of the direction of water flow on the gate. The value for the hydrostatic force, FH is positive (+)

because the resultant force is as the direction a flow of the water.

4. What is the effect of flow rate on the results obtained?

The effect of flow rate on the results obtained is the flow rate value will increase when weir

opening increase. This is because flow rate value is based on weir opening.

In this experiment, two sides of the weir have been leaking because the plasticine

which is used to block the water at both side of the weir is damaged. So, it will give impact to

the flow of rate water. Besides that, value of flow rate is not too accurate because we use manual

method to set the value of flow rate. The adjustable key of the equipment to set the value of flow

rate is locked. The results of this experiment are based on some elements such as the equipment

that have been used. In this experiment

1. Common errors and suggested precautions that are taken places during the experiment to

minimize the inaccuracy of result.

i. Water may leak at the sides of the weir that are not designed in the exact shape

according to the breadth of weir. Porcelain is introduced to stick of the sides of

the weir as well as the bolt and each of every small gap that may caused leakage.

ii. Parallax error occurs when the steel rule is placed inside the water channel to

measure the height of the opening weir. The refraction of water may cause a

slight different of reading. Therefore, we suggest that measurement must be read

while the reader’s eye is exact on the level of meniscus and have the best short

distance of reading to minimise the refraction due to the water.

iii. Random error happens when adjusting the opening weir takes a period of time

which may causes the depth of upstream flow decreased. The longer the time

taken to adjust the opening weir, the more water loss from upstream to

downstream since the water flow rate that we set constant earlier on is real small.

The lower the water level in the upstream will influence the pressure of water

flow through the opening weir. Therefore, we are advised to work in fast action

in adjusting the height of the opening weir.

9.0 CONCLUSION

The conclusion for this experiment A variety of gate-type structures 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. The relationship between upstream head and thrust on a sluice gate (undershot weir) for

water flowing under the sluice gate can be determined by the experiment that has been done.

When upstream head is high, thrust on a sluice gate also high

For the sluice gate, when the weir opening increases, the force at the gate thrust will also

increases but the force for hydrostatic thrust decreases.

Hydraulic jumps application for real life are downstream of hydraulic structure such as

spillways, sluice gate, venturi flumes, downstream of bridge piers during flooding and many

more while sluice gate is used at the end of a pipe line or to cover an opening in a wall and is not

an in live valve.

A hydraulic jump is a sudden rise in water level due to decreasing velocity. In flowing

water, the water before the jump is moving much faster than the water after the jump. When the

velocity of the water decreases to critical flow speed a jump occurs.

APPENDIX

REFERENCES

Chaudhry, M. H. (1993). Open Channel Flow. Prentice-Hall, Inc.

Richard H. French. (1994). Open Channel Hydraulics. McGraw-Hill, Inc.

Janusz M.G, John F.D, John A.S.(2001). “Fluid Mechanics”.(4th ed). London. Pitman

Publishing.

http://en.wikipedia.org/wiki/Open-channel_flow

http://www.engineeringcivil.com/what-are-the-effects-of-hydraulic-jump.html