PART A : THE HYDRAULIC JUMPS1.0 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 lessthancritical 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. By considering the forces acting with the fluid on either side of a hydraulic jump of unit width it can be shown that : H = d a + 2 va 2g da . 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.d 1 )^3 / 4 d 1 d 3 .¿ + 2 vb 2g ) Where ΔH is the total head loss across jump (energy dissipated) (m). d a ≈ db and db ≈ d3. This phenomenon can be seen where water shooting under a sluice gate mixes with deeper water downstream. a hydraulic jump or standing wave is produced. . The surface of the water undulates in a series of oscillations. The large amount of energy loss produces a zone of extremely turbulent water before it settles to smooth tranquil flow. An undular jump occurs when the change in depth is small. Therefore.0 THEORY When water flowing rapidly changes to slower tranquil flow. vb is the mean velocity after hydraulic jump (m) and d b is the depth of flow after hhydraulic jump (m). A direct jump occurs when the change in depth is great. v a is the mean velocity before jump (m/s). simplifying the above equation.4. d a is the depth of flow before hydraulic jump (m). Because the working section is short. which gradually decay to a region of smooth tranquil flow. H ( d a . Figure 3: Control Panel 4) Hook and point gauge. .contained Glass Sided Tilting Flume. 2) Adjustable Undershot Weir Tilting Flume. 3) Instrument Carrier.5.0 EQUIPMENT 1) Self. 4. Install the undershot weir towards the inlet end of the flume and ensure that it is securely clamped in position. with the downstream tilting overshot weir. Measure and record the values of d1. 3. Measure and record the actual breadth b (m) of the undershot weir. 5. dg. and q. Observe and sketch the flow pattern.6. d3. Observe and sketch the flow pattern. repeat this for other flow rates q (upstream head) and heights of the gate dg . Ensure the flume is level. E at the bottom of its travel. 2. 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. Increase the height of the tilting overshot weir until the downstream level just stars to rise. 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.0 PROCEDURES 1. Adjust the undershot weir to position the sharp edge of the weir 20mm above the bed of the channel. 0814 0.0196 0.212 4. Calculate H/ d 1 and plot H/ d 1 against d 3 / d 1 3.698 7. Calculate dc and verify d 1 < d c < d 3 Weir breadth.0846 1.0995 0.1964 0.0879 1.300 m Weir opening.13 0.25 0.904 0.3520 Flow Depth above jump.0146 0.1735 .60 0.030 ∆H d1 d3 d1 9.43 6. d1 (m) 0.0815 1. b = 0.021 Upstream flow depth.111 4.012 0.37 5.667 5.07 0.1052 0.0913 0.0106 Flow depth below jump.012 0.0857 1.036 0.7.2545 0.012 0.2474 0.024 0.0188 0.512 6. dg (m) 0. Calculate v 1 and plot d g against v 1 2.033 0.012 0.333 1.012 0.1100 0. do (m) 0.795 6.012 0.027 0.0969 0.0158 0.61 Flow rate (m3/s) ΔH V1 0.3252 0.34 5.0164 0.68 0.0871 1.0 RESULT 1.1028 1.481 5. d3 (m) 0. 055 m The weir opening.0814 0. V = 0.021 x 0.021m H = (0.0814 – 0.1028 Calculation for v 1 .0 DATA ANALYSIS Calculation for opening weir.055 < 0.055 < 0. dg 21 24 27 d1 < dc < d3 0. dg = 0.904 m/s dc = √3 q2 /g q =Q/b = (0.0913 1.300 2 = 0.0969 .04 m /s Therefore.012 / 0. Q = AV V = Q/A A = dg x b = 0.012/ 0.0106) ^3 / 4(0.8.0063 m Therefore.0814) = 0. d c = √3 ( 0.04 )2 / g = 0.0146 < 0.0106 < 0.158 < 0.0063 = 1.300) 2 = 0.055 < 0.0106) (0. 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 / dc. the upstream values of E are different. The energy dissipates when .9.055 < 0.0188 < 0. Suggest application where the loss of energy in hydraulic jump would be desirable.055 < 0.0164 < 0.0196 < 0.0 30 0. Fbefore = Fafter 2.1100 QUESTION 1. How is the energy dissipated? The hydraulic jump flow process can be illustrated by use of the specific energy concept. About the graph. Equation loss energy can be written in term of the specific energy: E = do + V2/ 2g Where do and E are feet.0995 33 0.055 < 0.1052 36 0. (1) to state (2) the fluid does not proceed along the specific energy curve and pass through the critical condition. Because of the head loss across the jump. To mix chemicals used for water purification. 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.water flow at weir opening and the energy became 0 because d0 and d3 has are force from adverse. ∆H. weirs. From the experiment. we can get the force at weir opening. To raise water level on the downstream side for irrigation or other water distribution purposes. To increase weight on an apron and reduce uplift pressure under a structure by raising the water depth on the apron. Fbefore = Fafter. . 11. 10. 7. 6. Same like the equation. To aerate water for city water supplies. In the water channel. 5. 2. This phenomenon can be seen where water shooting under a sluice gate mixes with deeper water downstream.0 DISCUSSION Practical applications of hydraulic jump are: 1. 3. 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. 4. To dissipate energy in water flowing over hydraulic structures as dams. and others and prevent scouring downstream structures. water flowing rapidly changes to slower tranquil flow a hydraulic jump or standing wave is produced. To remove air pockets from water supply lines and prevent air locking. 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.0 CONCLUSION The conclusion from the experiment. A variety of gate-type structure is available for flow rate control at the crest of an overflow spillway.engineeringcivil. Inc. Reason the experiment perform because almost drain are open channel. Both graphs are sloping downward. we get the inverse line from graph g d1 against v1 and curve line from graph Δ H/d1 against d3/d1. Chaudhry. H.A. Hydraulics.com (serve on 19/11/2011) PART B: THE FORCE ON A SLUICE GATE 1. Open Channel Flow. L. 1988. REFERENCES John J. As the lower edge of the gate opening is flush with the floor of the channel. Side contraction will of course depend on the extent to which the opening spans the width of the channel. M.E. Inc. A.0 INTRODUCTION The Sluice gate is a device used to control the passage of water in an open channel. So the objective achieved and the experiments are success. Final result we can get the value of dc between d1 and d3. iii. Three . Introduction to Fluid Mechanics. contraction of the bottom surface of the issuing stream is entirely suppressed. 12. Prentice Hall. 1993. http://www. Simon. pp 302-408. we know about water flowing.0 i. Prentice-Hall.1997. pp 283-312. Prentice Hall. From the experiment. or at the entrance of an irrigation canal or river from a lake. pp 330-342. ii.From the result. When properly calibrated it may also serve as a means of flow measurement. Inc iv. 2. radial gate and drum gate. they . students should be able to apply the knowledge and skill have learned to: a) Understand the basic terms and concept of a sluice gate.typical types are vertical gate. In certain situation.0 LEARNING OUTCOMES At the end of the course. 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.0 OBJECTIVE To determine the relationship between upstream head and trust on a sluice gate (undershot weir) for water flowing under the sluice gate. Picture 1: Drowned outflow from a sluice gate. b) Understand on the characteristics of the force on a sluice gate. 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. 3. 4. q is volume flowrate (m/s).0 EQUIPMENTS 1. b is breadth of gate (m).81m/s2).d g )^2 where Fg is the resultant gate thrust (N). Hook and Point Gauge . FH is the resultant hydrostatic thrust (N). g is the gravitational constant (9. Adjustable Undershot Weir 3. Instrument Carrier 4. Self-contained Glass Sided Tilting Flume 2.4.0 THEORY It can be shown that the resultant force on the gate is given by the equation : Fg = 1 2 pg d 1 ^2 ( d 0 ^2 / d 1 ^2-1 ) – pg / b d 1 (1- d1 d0 ) The gate thrust for hydrostatic pressure distribution is given by the equation : FH = ½ pg ( d 0 . dg is height of upstream opening (m). d0 is upstream depth of flow (m) and d1 is downstream depth of flow (m). ρ is density of fluid (kg/m3). At each level of the weir. The undershot weir adjusted to set its bottom edge 20mm above the bed of the channel. d1 (m) Flow Rate q (m3/s) Gate Thrust Fg (N) Hydrostatic Thrust. with the downstream tilting overshot weir at the bottom of its travel its travel. The procedure with a constant flow q. dg (m) Upstream flow depth. 2. the readings for q and d1 are taken. do (m) Downstream Flow Depth. d1 and q. Introduce water into the flume until do = 200mm. with do at this position. Ensure the flume is level. Raise the undershot weir in increments of 10mm. 4. record the values of dg. allowing do to vary are repeated and the value of do and d1 are recorded. maintaining constant do by varying q.6. 7. 3. b = 300 m Weir opening.0 PROCEDURES 1. FH (N) Fg/ FH dg /do .0 RESULT Weir breadth. The actual breadth b (m) of the undershot weir are measured and recorded. 036 0.16 x106 650. (Refer graph) 2. Based on the graph.024 0.55 0.0100 0.59 x 106 117. Comment of the graph obtained.97 x 106 270.d g ) ^2 = 650.96 N 9.055 0. Plot graph of the ratio Fg / FH against the ratio dg / do.021m.176 0.074 0.42 -7284.18 0.22 0.1876 0.24 13561. FH (N) FH = ½ pg ( d 0 .96 -4854.012 -0.2348 0.012 -2.0166 0.0185 0. Gate Thrust.12 x 106 438.207 8.033 0.0195 0. the value of Fg / FH also increased.51 -4834.027 0.73 -2819.012 1.021 0.3853 0.3230 0..41 15951.37 0. Fg (N) Fg = 1 2 pg d 1 ^2 ( d 0 ^2 / d 1 ^2-1 ) – pg / b d 1 (1- d1 d0 ) 6 = -3.0 QUESTIONS 1.128 0.2618 0.0.49 x 106 93.16x 10 N Hydrostatic Thrust. .1740 0.012 1.103 0.96 0.0147 0. the pattern of the graph is linear line and an increased. When the value of dg /do are increased .030 0.92 0.012 -1.012 -3.58 x 106 205.0156 0.0 DATA ANALYSIS Calculation for weir opening 0. What is the effect of flow rate on the result obtained? From the result. . The flow is from left to right and enters at a velocity Vo. Fg = ½ ρgd12 [ (d02 / d12 ) . Since most of these devices operate by controlling the water surface elevation being stored or routed. the more flow rate will give the less thrust for both of the gate and the hydrostatic. river. to adjust flow rates in sluices and canals. They may be designed to set spillway crest heights in dams.0 DISSCUSSION Floodgates are adjustable gates used to control water flow in reservoir. 11. we get the value in positive (+ve). We can conclude that before the water is flow to the sluice gate. This obstruction is called a sluice gate (see Figure 1). the force are F H are in positive (+ve) because is follow the direction of the water flow. and exits the upstream section under the gate of height b.0 CONCLUSION The flow through a channel in which a gate partially obstructs the flow will be used for this measurement of total force. they are also known as crest gates. 10.1 ] – ρg / bd1 [ 1 – (d1 / d0 )]. The fluid attains a higher velocity V1 as it passes under the gate and a shallower free surface height y1 downstream. or levee systems. Compared your calculated values for Fg and FH and comment on any differences. stream. 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 basin when the main river or canal is approaching a flood stage.3. or they may be designed to stop water flow entirely as part of a levee or storm surge system. In the case of flood bypass systems. 4.dg )2 . Fg is resultant gate thrust (N) and FH is resultant hydrostatic thrust (N). This is because of the decreasing pressure at both of them. The fluid in the upstream section builds up against the gate to a level y0. we get the value are in negative (ve) and when we calculated FH with FH = ½ ρg ( d0 . The force are happened after sluice gate are Fg in negative (-ve) because the resultant force of the flow is opposite the direction. pp 302-408. Prentice Hall. 1959. John J. Prentice-Hall. 1993. Chaudhry. Figure 1.E. V. Introduction to Fluid Mechanics. . Flow under a vertical sluice gate. Open Channel Flow. 1988. 3. Chow. pp 608-710. Inc.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.A. 2. pp 330-342. M. T. 3) Flow at upstream and downstream sections is uniform and the effect of the sidewalls is negligible. McGraw-Hill. Inc. H. Open Channel Hydraulics. 12. Inc.0 REFERENCES 1. 2) The flow is steady and has a uniform velocity distribution at the inlet and outlet sections.