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.

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 andular 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 within the fluid on either side of a hydraulic jump of unit

width it can be shown that:

ΔH=d a+va2

2 g−(db+ vb22g )

Where, ΔH is the total head loss across jump (energy dissipated) (m), va is the mean

velocity before jump (m/s), da is the depth of flow before hydraulic jump (m), vb is the

mean velocity after hydraulic jump (m) and db is the depth of flow after hydraulic jump

(m). Because the working section is short, da≈d1 and db≈d3 . Therefore, simplifying

the above equation, ΔH=(d3−d1)3 /4 d1d3 .

5.0 EQUIPMENTS USED

Clear-acrylic rectangular open channels Sump tank

supported by steel frames (0.3m width)

Switch pump with water meter

Rectangular sluice gate (0.3m width) Control valve & pump

2 units measurement gauges 1 meter long steel ruler

Plasticine

6.0 EXPERIMENTAL METHODS

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 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 20 mm above the

bed of the channel. Increase the height of the tilting overshot weir until the

downstream level just start to rise.

3. Gradually open the flow control valve and adjust the flow until an andular 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 tilting overshot weir to create a hydraulic

jump in the center of the working section. Observe and sketch the flow pattern.

5. Measure and record the values ofd1 , d3 , d g and q . Repeat this for other flow

rates q (upstream head) and heights of the gated g .

7.0 SAMPLE DATA

Channel Width, b=0.30m

Weir Opening

,

Upstream Flow Depth,

Flow Depth Above Jump,

Flow Depth Below Jump,

Flow Rate,

∆H V1 ∆H/d1 d3/d1

dg (m) do (m) d1 (m) d3 (m) Q (m3/s)

0.20 0.3145 0.0173 0.0944 0.011 0.0702 0.1833 4.0554 5.4566

0.21 0.3184 0.0155 0.0918 0.011 0.0780 0.1746 5.0351 5.9226

0.22 0.3085 0.0166 0.0943 0.011 0.0749 0.1667 4.5131 5.6807

0.23 0.2625 0.0171 0.0941 0.011 0.0709 0.1594 4.1479 5.5029

0.24 0.2508 0.0210 0.0938 0.011 0.0490 0.1528 2.3318 4.4667

Table 7.1. The Hydraulic Jumps

7.1 Calculate V1 and plot dg against V1

7.2 Calculate ∆H/d1 and plot ∆H/d1 against d3/d1

7.3 Calculate dc and verify d1<dc<d3

8.0 ANALYSIS OF RESULTS, EQUATIONS USED

8.1 Equation used for V1, Q=AV

V=Q/A

A=dg x b =0.20 x 0.30 =0.060m2

V=Q/A =0.011/0.06 =0.1833m/s

A=dg x b =0.21 x 0.30 =0.063m2

V=Q/A =0.011/0.063 =0.1746m/s

A=dg x b =0.22 x 0.30 =0.066m2

V=Q/A =0.011/0.066 =0.1667m/s

A=dg x b =0.23 x 0.30 =0.069m2

V=Q/A =0.011/0.069 =0.1594m/s

A=dg x b =0.24 x 0.30 =0.072m2

V=Q/A =0.011/0.072 =0.1528m/s

0.15 0.200.20

0.21

0.22

0.23

0.24

0.25

Graph 8.1. dg against V1

dg/V1

, V1

dg ,

8.2 Equation used for ∆H = (d3-d1)3 / 4d1d3

∆H = (0.0944-0.0173)3

4(0.0173)(0.0944)

= 0.0702m

∆H = (0.0918-0.0155)3

4(0.0155)(0.0918)

= 0.0780m

∆H = (0.0943-0.0166)3

4(0.0166)(0.0943)

= 0.0749m

∆H = (0.0941-0.0171)3

4(0.0171)(0.0941)

= 0.0709m

∆H = (0.0938-0.0210)3

4(0.0210)(0.0938)

= 0.0490m

8.3 Calculations for ∆H / d1 ,

= 0.0702

0.0173

= 4.0554

= 0.0780

0.0155

= 5.0351

= 0.0749

0.0166

= 4.5131

= 0.0709

0.0171

= 4.1479

= 0.0490

0.0210

= 2.3318

8.4 Calculations for d3 / d1 ,

4 5 6 72.0

2.5

3.0

3.5

4.0

4.5

5.0

5.5

Graph 8.2. ∆H/d1 against d3/d1

d3/d1

, d3/d1

∆H/d1 ,

8.5 Calculations for Critical depth, dc = (q2/g)1/3 , where q = Q/b

= (0.03672/9.81)1/3 = 0.011/0.3

= 0.0516m = 0.0367m3/s/m

and acceleration due to grativy, g = 9.81m2/s

dc

Verifications, if d1<dc<d3

0.20m0.0173<0.0516<0.0944

0.21m0.0155<0.0516<0.0918

0.22m0.0166<0.0516<0.0943

0.23m0.0171<0.0516<0.0941

= 0.0944

0.0173

= 5.4566

= 0.0918

0.0155

= 5.9226

= 0.0943

0.0166

= 5.6807

= 0.0941

0.0171

= 5.5029

= 0.0938

0.0210

= 4.4667

0.24m0.0210<0.0516<0.0938

9.0 QUESTIONS

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 ∆H/dc.

Fbefore = Fafter

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

concept. Equation loss energy can be written in term of the specific energy:

E = do + V2/ 2g

Where do and E are feet. Because of the head loss across the jump, the

upstream values of E are different. About the graph, (1) 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 d0 and d3 has are force from adverse. Same like the equation,

Fbefore = Fafter.

Practical applications of hydraulic jumps:

Dissipation of energy of water flowing over dams and weirs – to prevent

possible erosion and scouring due to high velocities.

Raising water levels in canals to enhance irrigation practices and reduce

pumping heads.

Reducing uplift pressure under the foundations of hydraulic structures.

Creating special flow conditions to meet certain special needs at control

sections – gaging stations, flow measurement, flow regulation.

10.0 COMPARISON WITH THEORY, DISCUSSION

11.0 CONCLUSIONS