1.0 TITTLEImpact of a Jet2.0 OBJECTIVEThe purpose of this experiment is to demonstrate and verify the integral momentum equation. The force generated by a jet of water deflected by an impact surface is measured and compared to the change of the jet.3.0 EQUIPMENTThe following equipment is required for the experiment:3.1 impact of a Jet Apparatus3.2 Steady water supply with a flow control valve3.3 Flow meter3.4 A flat plate and a hemispherical cup3.5 Set of calibrated weight

4.0 EQUIPMENT DESCRIPTIONFigure 1 shows the arrangement, in which water supplied from the Hydraulic Bench is fed to a vertical pipe terminating in a tapered nozzle. This produces a jet of water which impinges on a vane, in the form a flat plate or a hemispherical cup.The nozzle an vane are contained within a transparent cylinder, and at the base of the cylinder there is an outlet from which the flow is directed to the measuring tank of the bench. As indicated in Figure 1, the vane is supported by a lever which carries a weight pan, and which is restrained by alight spring. The lever may be set to a balanced position (as indicated by a tally supported from it) by placing the weight pan at its zero position, and then adjusting the knurled nut above the spring. Any force generated by impact of the jet on the vane may now be measured by moving the weight pan along the lever until the tally shows that it has been restored to its original balanced position.

5.0 IMPACT OF A JET THEORYA theoretical model for the force necessary to hold the impact surface stationary is obtained by applying the integral forms of the continuity and momentum equations. The details of the model depend on whether, or not the fluid stream leaving the impact surface is symmetric relative to the vertical axis of the surface.The control volume, bounded by the dashed lines, is chosen so that it crosses the jet streams at right angles. To proceed with the analysis make the following assumptions: Friction between the impact surface and the water jet is negligible The magnitude of the jet velocity does not change as the jet is turned Velocity profiles are uniform where the floe crosses the control surface The jet exit is circumferentially symmetricalIf any of the impact surfaces used in the experiment cause the flows that violate these assumptions, the formula for reaction forces given below will not match the measured reaction forces.Applying the conservation of mass to the jet streams givesV A - V2 A2 = 0where V is the average velocity at a given cross-section, and A is the cross-sectional area normal to the direction of the average velocity. The subscripts 1 and 2 refer to the inlet and outlet of the control volume, respectively. Since the magnitude of the velocity is assumed to not change, Equation (1) simplifies toA = A2 = AThe integral equation for momentum conservation in the x-direction is

Rh = Where Rh is the reaction force in the x-direction necessary to hold the impact surface stationary, and is the angle between the horizontal and the velocity vector of the fluid leaving the control volume. Equation (4) shows that Fh = 0 if the flow leaving the impact surface is symmetric about the vertical axis of the impact surface. If there is any disruption to the symmetry, e.g., variation in or around the periphery of the exit, Rh will not be zero.

Figure 2 : Nomenclature for control volume analysis of the jet. Ideally, theApparatus and jet are symmetric about the center lining of the jet.

Applying the y-direction integral momentum equation gives

- = (-) + (- Where is the reaction force in the y-direction. Using the simplification A = A2 = A and V = V2 = V, Equation (6) reduces to = Where = =

6.0 PROCEDURE

6.1 The flat plate in the apparatus was installed6.2 Note the no load position of the weight tray by aligning the pointer on top of the apparatus to the weight pan.6.3 The appropriate masses added to the way tray until it returns to the no load position6.4 The flow rate and masses were recorded6.5 The water supply was reduced and the procedure (steps 4-5) were repeated for each flow rate6.6 The flat plate was replaced with an hemispherical cup and the procedure was repeated

7.0 RESULTSFluid Dynamics (Impact Jet) Data SheetApparatus design geometry:Diameter of the nozzle: 5mmCross sectional area of nozzle: = = 1.963495 Height impact above nozzle tips: 3.5 cmExperiment results:

7.1 Plate

For mass, m = 800g, time, t = 26.12s, Q = 0.005 / 26.12 = 1/5224 /s V = Q / A = = 9.749156 m/s= mg = (0.8-0.75)(9.81) = 0.4905 N=QV = (1000)(1/5224)(9.749156) = 1.8662 N For mass, m = 850g, time, t = 22.62, Q = 0.005 / 22.62 = 1/4524 /s V = Q / A = = 11.25765 m/s= mg = (0.85-0.75)(9.81) = 0.981 N=QV = (1000)(1/4524)( 11.25765) = 2.4888 N For mass, m = 900g, time, t = 20.02s, Q = 0.005 / 20.03 = 1/4006 /s V = Q / A = = 12.71333 m/s= mg = (0.9-0.75)(9.81) = 1.4715 N =QV = (1000)(1/4006)( 12.71333) = 3.1736 N

For mass, m = 950g, time, t = 18.31, Q = 0.005 / 18.31 = 1/3662 /s V = Q / A = = 13.9076 m/s= mg = (0.95-0.75)(9.81) = 1.962 N=QV = (1000)(1/3662)( 13.9076) = 3.7978 N

For mass, m = 1000g, time, t = 17.96s Q = 0.005 / 17.96 = 1/3592 /s V = Q / A = = 14.17862 m/s= mg = (1.00-0.75)(9.81) = 2.4525 N=QV = (1000)(1/3592)( 14.17862) = 3.9473 N

CaseMass (g)Time (s)Q (m/s)V(m/s) (N)Rv = (N)Comment (Diff)

180026.1299.74920.49051.86221.3757(smallest)

285022.6211.25770.98102.48881.5078

390020.0312.71331.47153.17361.7021

495018.3113.90761.96203.79781.8358(largest)

5100017.9614.17862.45253.94731.4948

7.2 Hemisphere

For mass, m = 800g, time, t = 34.81s, Q = 0.005 / 34.81 = 1/6962 /sV = Q / A = = 7.31537 m/s= mg = (0.8-0.75)(9.81) = 0.4905 N=QV = (1000)(1/6962)( 7.31537) = 1.0508N

For mass, m = 900g, time, t = 26.75s, Q = 0.005 / 26.75 = 1/5350 /sV = Q / A = = 9.51955 m/s= mg = (0.9-0.75)(9.81) = 1.4715 N=QV = (1000)(1/5350)( 9.51955) = 1.7794 N

For mass, m = 1000g, time, t = 23.80s,Q = 0.005 / 23.80 = 1/4760 /sV = Q / A = = 10.69949 m/s= mg = (1.00-0.75)(9.81) = 2.4525 N=QV = (1000)(1/4760)( 10.69949) = 2.2478 N

For mass, m = 1100g, time, t = 22.50s,Q = 0.005 / 22.50 = 1/4500 /sV = Q / A = = 11.31769 m/s= mg = (1.10-0.75)(9.81) = 3.4335 N=QV = (1000)(1/4500)( 11.31769) = 2.5150N

For mass, m = 1200g, time, t = 19.39s,Q = 0.005 / 19.39 = 1/3878 /sV = Q / A = = 13.13295 m/s= mg = (1.20-0.75)(9.81) = 4.4145 N=QV = (1000)(1/3878)( 13.13295) = 3.3865 N

CaseMass (g)Time (s)Q (/s)V (m/s) (N)Rv = (N))Comment (Difference)

180034.817.31540.49051.05080.5603

290026.759.51961.47151.77940.3079

3100023.8010.69952.45252.2478-0.2047 (smallest)

4110022.5011.31773.43352.5150-0.9185

5120019.3913.13304.41453.3865-1.028 (largest)

8.0 ANALYSIS

8.1 For each setting of the control volume8.1.1 The flow rate converted to average velocity for the jet and tabulated in table in Results part. 8.1.2 The theoretical reaction force given V was computed from the flow rate measurement as shown in the Results part.8.1.3 was computed from the applied weight as shown in Results part.

8.2 On the same axes, and versus V plotted for each impact surface.

This graph shows that the is linear proportional to V. is also linear proportional to V except for when V = 14.1786m/s, the is decreased a bit. This also show that the experimental force and theoretical force has significant difference between them for plate surface. Both and increases with V.

This graph show that is linear proportional to V. However, do not have linear relationship with V. Both and increase when V increase. For hemisphere surfaces, the theoretical force and experimental force is less difference if compare to plate surfaces.

8.3 The discrepancy - versus V plotted.8.3.1 For plate surface

- V (m/s)

1.37579.7492

1.507811.2577

1.702112.7133

1.835813.9076

1.494814.1786

This graph show that the Discrepancy between and versus V for plate surface. The discrepancy increases until it reaches V = 13.9017m/s and decreases lastly.

8.3.2 For hemisphere surface

- V (m/s)

0.56037.3154

0.30799.5196

0.204710.6995

0.918511.3177

1.02813.1330

This graph shows the discrepancy between and versus V for hemisphere surface. The discrepancy has a dramatically increase when reached V = 11.3177m/s.

9.0 REPORT

9.1 When velocity is high, the flow rate of the water will getting higher, hence the reaction forces will also increase. This trend is consistent with the theory. From the equation V = Q/A, Hence, V and Q have linear relationship. Also, = QV, also linear proportional to Q, Hence, the trends in the reaction forces versus jet velocity will form a linear graph because V Q Rv. This relation can be seen clearly from the four plots accompanied with this report. This result was already predicted from the change in momentum equation of calculating the force.

9.2 The theoretical model does not predict the measured force for the plate and hemispherical surface well. It is because there are significant large difference between the measured forces and theoretical forces. The density of air was not measured causing there is discrepancy between experimental and theoretical results in both surfaces cases. The error arose in the experimental velocity because of the area measurements taken from the nozzle as well as the air flow rates measured from the experimental apparatus.

9.3 Yes, measured values point to differences between the assumptions and actual flow conditions. We assume that the surfaces are smooth and very small friction exists, but in the actual flow conditions, there is friction between the water and the nozzle and the surfaces. Negligible variation in elevation of the incoming and outgoing jets. Uniform distribution of velocity throughout. But in actual flow conditions, not all the surfaces get the uniform distribution of velocity because the water tap is closed during the experiments.

10.0 PRECAUTIONS

10.1 Apparatus should be in leveled condition.10.2 Reading must be taken in steady conditions.10.3 Discharge must be varied very gradually from a higher to smaller value.

11.0 CONCLUSION

As a conclusion, the theoretical force is correlated with the measured force. Both of the forces will have directly proportional relation. The flow rate for the hemisphere is found to be the lowest and thus require a longer time for the volumetric tank to rise 5 liters. Theoretically, the theoretical force should be the same as the measured for, but this cannot be achieved experimentally due to the errors made during conduct this experiment, such as parallax error due to human and servicing factors. This error occur during observer captured the value of the water level. Besides, error had occurred during adjusting the level gauge to point at the white line on the side of the weight pan and also maybe because of the water valve. The water valve was not completely close during collecting the water and may affect the time taken for the water to be collected. Hence, recommendation to overcome the errors is to ensure that the position of the observers eyes must be 90 perpendicular to the position.

12.0 REFERENCES

YUNUS A. CENGEL, JOHN M. CIMBALA (2010), Fluid Mechanics Fundamentals and Applications (Second Edition in SI Units), Mc Graw Hill.

BRUCE R. MUNSON, DONALD F. YOUNG (2009), Fundamental of Fluid Mechanics (Sixth Edition), John Wiley & Sons, Inc.