FACULTY OF PETROLEUM AND RENEWABLE ENERGY ENGINEERINGUNIVERSITI TEKNOLOGI MALAYSIA FLUID MECHANICS LABORATORYTITLE OF EXPERIMENT JET IMPACT (E3)

Name

MUHAMMAD KHAIRIL IKRAM(A13KP0047) AKMAL FAIZ BIN ABDUL RAHIM (A13KP0008) ABDUL WAHAB (A13KP4006)KSATRIYA ANANTAYUTYA (A13KP4001)

Group / Section

1/Section

Supervisor

Associate Professor Issham bin Ismail

Date of Experiment

10/3/2014

Date of Submission

16/3/2014

Marks obtained (%)

OBJECTIVE:The objectives of this experiment are:1. To measure the force produced by a jet on flat and curved surfaces.2. To compare the experimental results with the theoretically calculated values

INTRODUCTION:Water turbines are widely used throughout the world to generate power. In the type of water turbine referred to as a Pelton wheel, one or more water jets are directed tangentially on to vanes or buckets that are fastened to the rim of the turbine disc. The impact of the water on the vanes generates a torque on the wheel, causing it to rotate and to develop power. Although the concept is essentially simple, such turbines can generate considerable output at high efficiency. Powers in excess of 100 MW, and hydraulic efficiencies greater than 95%, are not uncommon. It may be noted that the Pelton wheel is best suited to conditions where the available head of water is great, and the flow rate is comparatively small. For example, with a head of 100 m and a flow rate of 1 m3/s, a Pelton wheel running at some 250 rev/min could be used to develop about 900 kW. The same water power would be available if the head were only 10 m and the flow were 10m3/s, buta different type of turbine would then be needed

SUMMARYIn this experiment the force generated by the jet of water striking a flat plate or a hemispherical cup is measured and results are compared to those of the computed flow rate in the jet. Through the use of simple moments an equation was derived so that the force exerted by the jet using the jockey weight , the graduated length of the pivot arm, and the distance from the pivot to the jet acting at zero point of the jockey weight will be measured. The force acting on the surfaces for various flow rates is also measured. Results are represented in a table and in form of graphs.

Theory

Force can be defined as the rate at which momentum changes in a system, When a fluid is deflected by a solid surface the fluid momentum changes because the direction changes and it therefore exerts a force on that surface. In the same way, fluid which is discharged from a reservoir through a nozzle exerts a reaction force on the nozzle and attached hose. Although real fluids are viscous and viscous interaction is responsible for some force, most of the force exerted in these cases is due to the change in fluid momentum. A control volume analysis can be used to determine the impact force of the jet, as shown in equation [1] for flat plate and equation [2] for hemispherical plate.

Force = Rate of change in momentum

For flat plate (diagram 1), = 90o, therefore cos = 0 , soeqn.[1]For hemispherical plate (diagram 2), = 180o, therefore cos = -1, Soeqn.[2]where,Fth= Theoretical force exerted on the plate (Newton)a= Cross-sectional area of nozzle (m2)r= Density of water (kg/m3)= Angle of water flow after impact on the plate surfacev= Velocity of water jet before impact on the plate surfacev= Velocity of water jet after impact on the plate surface

Equation [1] and [2] also predicts the reaction force opposite to the flow direction in the reaction of a jet experiment, which is the focus of this analysis.

PROCEDUREThe flat plate was fixed in its place. Extra care was carried out when we are fixing the plate to make the replacing plate at the end of first set of experiment easier.Standard weights for each plate apparatus was putted by putting weights on the spring plat until the coils will touch each other.The level of the standard indicator was adjusted with the position of the plate containing the weight. This was taken as the standard. The total standard weight was noted.The water control valve was closed. Both switches of the pump and open the water control valve were then opened slowly until it reached its maximum. Then, plate with the standard weight was increase above the standard mark. At this point, more weights were added until the indicator plate returned to the standard mark. The total maximum weight for the first reading of the load was taken. The valve of the water tank was closed and when the volume reaches 2 liter, the time for it to reaches 7 liter was taken. The total volume of accumulated water was 5 liter. The total weight load then reduced, plate apparatus will change from the standard mark, the flow of the water jet was also reduced by closing the control valve slowly until the plate apparatus returned to the standard mark level. Step 6 was then repeated. For the next reading, step 7 and 6 was repeated until the last total weight load is the same with the standard weight load. Steps 1 to 8 were then repeated using the hemispherical surface plate.When the experiment is completed, the control valve was closed. Both pumps were switched off All equipment used in this experiment were then cleaned.

6.0 Data and Analysis

Water density, r = 1000 kg/m3

Water velocity of the jet by the nozzle with diameter d = 5 mm

Q V =

m A liter

1 m 3 = x 2 x 3

where s s m 10 liter

V = Water velocity (m/s)

Q = Volumetric flow rate of water

A = Area of nozzle with diameter d (m2)

A = p x d 24 ... m 2

2 = x(d mm)2 x m 4 103 mm

Hence, the force measured is

Fmea = weight (gram) x 9.81m/s1000 gram/kg= gram x 9.81x10-3 NewtonRESULTS

Flat plateStandard Weight = 700 (g) Maximum Weight = 1300 (g)WeightLoad(Gram)ActualWeight(Gram)Fmea(Newton)Time(Second)Q(L/S)V(m/s)Fth(Newton)Log VLog FmeaPercentage of relativeerrorF - F% = mea th x100Fth13006005.8914.380.34817.676.131.2470.77012.0712005004.9016.160.30915.744.861.1970.6900.8211004003.9218.710.26713.603.631.1340.5937.9910003002.9421.310.23511.972.811.0780.4684.639002001.9628.100.1789.121.630.9600.29220.258001000.9842.780.1175.960.700.775-0.00940.0070000-000000

Hemispherical plate

Standard Weight = 700 (g) Maximum Weight = 1700 (g)

WeightLoad(Gram)ActualWeight(Gram)Fmea(Newton)Time(Second)Q(L/S)V(m/s)Fth(Newton)Log VLog FmeaPercentage of relativeerrorF - F% = mea th x100Fth170010009.8113.380.37419.0514.251.2800.99231.1616009008.8314.30.35017.8312.481.2510.94629.2515008007.8515.360.32616.6010.821.2200.89527.4514007006.8716.660.30015.289.171.1840.83725.0813006005.8917.560.28514.518.271.1620.77028.7812005004.9119.470.25713.096.731.1170.69127.0411004003.9221.750.23011.715.381.0690.59327.1410003002.94323.150.21611.004.751.0410.46938.049002001.96225.700.1959.933.870.9970.29349.308001000.98128.630.1758.913.120.950-0.00868.5670000-000000

1.200 Graph of Log Fmea Vs Log V

y = 2.7744x - 2.4792R = 0.9369

1.000

0.800 y = 1.6585x - 1.299R = 0.9985

0.600

0.400 Flat plate Hemispherical plate Linear (Flat plate)Linear (Hemispherical plate)

0.200

0.0000.000 0.200 0.400 0.600 0.800 1.000 1.200 1.400

Log Fmea (N)

-0.200 Log V (m/s)18

DISCUSSION AND DISCUSSION

Estimate the slope of the graph for each plate and compare with the theoretical value as shown in eq. 1 and eq. 2, respectively. Comment on the difference.

The slope of the graph 1 for flat plate is a linear graph. By logging both side of the theoretical equation for flat plate we are able to get:Fth = rav2 (1)

Log Fth = Log rav2

= Log r + Log a + Log v2

Log Fth = 2Log v + A (2), Where A is constant, A= Log r + Log a

While the slope of graph 2 for hemispherical plate is also a linear graph. By logging both side of the theoretical equation for hemispherical plate we are able to get:

Fth = 2rav2 (1)

Log Fth = Log 2rav2

= Log 2 + Log r + Log a + Log v2

Log Fth = 2Log v + B . (2) Where B is constant, B= Log 2 + Log r + Log a (The same goes for Fmea)The slope of Fmea on flat plate is 1.6585 while its Fth is 1.9868. The differences of slope is only 0.3283, and slightly deviated from the theoretical value but still result can be considered acceptable. The difference might be caused by the height between the nozzle and the vane due to the change of vanes as all vanes do not have equal heights. As for hemispherical plate, the Fmea slope is 2.7744 while its Fth 1.9997. The difference is also 0.7747 thus it can be considered as acceptable. The difference might be cause by error such as bubbles present in the water can be a reason to get inaccurate readings as well.A. Estimate the y-intercept ratio of hemispherical to flat plate and compare with the theoretical ratio, as deduced from eq. 1 and eq. 2. Comment on the difference.

Y- Intercept ratio of hemispherical plate to flat plate

Ratio = -2.4792/-1.299 = 1.9085 (Fmean)Ratio= -1.4057/1.5921 = 0.8835 (Ftheory)

The ratio of y-intercept of hemispherical to flat plate for (Fmea) is 1.9085:1 while the ratio of y-intercept of hemispherical to flat plate for (Fth) is 0.8835:1. Both values for gradient and y-intercept for both graphs are not identical, but the result obtained is still near to the theoretical value. So, the data that we calculated and recorded can be considered acceptable. The differences might be due to errors when taking the measurement and might be due to systematic errors while handling experiment apparatus.

B. Comparing the force exerted on the hemispherical vane with the one on the flat plate, which one is greater? Why?

Comparison of force on both plates

Force exerted on both hemispherical plates and the flat plate was totally different. Force exerted on hemispherical plate greater than flat plate because it lies on the behavior of water jet when it strikes the flat surface. It forms a radial sheet which impinges on the inner wall of the surrounding cylinder, and then divides, some of the water flowing down the cylinder wall and the rest flowing upwards. Although visibility is impaired by the spray which is generated, it does seem that some water falls on to the top side of the vane. This would have the effect of producing a small momentum force in the downwards direction, so reducing the net upwards force on the vane.

1) Comparing the percentage of relative error for the two plates as function of jet velocity. Comment on the analysis. Can one deduce sources of error due to the shape of the plates? Explain your reason. State other possible sources of error.

Comparison of the percentage of relative error for the two plates

Based on our data, the percentage of relative error for both plates different, that is 12.07% for flat plate and 31.16% for hemispherical plate. The percentage of error ranged from around 0.82% to around 68.56%. If we have less percentage relative error, so it means jet velocity is more constant. Some of the percentages error are large due to several errors made during the experiment.

The shape of the plate can be as sources of error, because the equation using the angle where the impact of the velocity from water to the surface of the plate, so if the plate is not in perfect shape , in case got incomplete sphere , the angle will be different which will get a different force. Then possible source of error could be is spring coil. The shape of coil must be in a standard position which is straight. If not, the velocity that applied by the water is not accurate.

i. Briefly discuss factors contributing to errors or inaccuracy in experimental data and propose recommendation to improve the results.

While conducting the experiment several errors may have been made which affected accuracy of our data. Firstly, parallax error occurred when we were taking the reading of 5 L water in the water tank and when we were synchronizing the height of weight with standard height. Secondly, the control valve may not be open to maximum. Thirdly, the time reading for increasing of 5 L water may not be accurate. The contact angle between water and the plates also may not be the same as stated in experimental procedure. The spring that was used to balance the weights may not be able to be compressed to its full potential

There are some precautionary steps that we must follow in order to obtain data with high degree of accuracy. First of all, make sure that all the apparatus is in good condition and do some repetition in the experiment so that the reading will accurate and precise. Secondly, always remember to open the control valve to its maximum so steady flow rate of water can be achieved. Next, tally the standard height carefully so that the weight height and the standard height is equal. Parallax error can be avoided via placing our eye position perpendicular to the meniscus of water. Furthermore, the surfaces of plates also should be examined before carrying the experiment to eliminate possibilities of defect surfaces. The control valve should be handled carefully and slowly to avoid disturbance in the water flow rate. The person who taking the time reading should remained focus and alert while taking the time do that better data can be obtained.

Conclusion:There are two important points that the graphs and results highlight,The rate on flow, Q, is directly proportional to the force resulted from the impact of the jet on the plates. This relation was shown clearly by the two plots. This proportionaliy was already predicted by the equation of momentum for calculating the force.There is higher efficiency in using the hemispherical cup for turbines. Keeping in mind that the angles made for the watering exiting the cup should be less than 180 degrees so to prevent the reduction of force due to collision of the water entering and water exiting the cup.

REFERENCESFluid Mechanics (Fundamental and Application) Second Edition in SI Units by Yunus A. Cengel and John M. CimbalaFundamental of Fluid Mechanics by Bruce R. Munson, Donald F. Young, Theodore H. Okiishi(2) Munson, B.R., D.F. Young, and T.H. Okiishi (2006). Fundamentals of Fluid Mechanics. 5th Edition. London: John Wiley and Sons.Douglas, J.F., J.M. Gasiorek, and J.A. Swaffield (1981). Fluid Mechanics. 3rd

Edition. London: Longman Scientific and Technical. (Munson, B.R., D.F. Young, and T.H. Okiishi (2006). Fundamentals of Fluid

Mechanics. 5th Edition. London: John Wiley and Sons.The vol

AppendicesCalculation