FACULTY OF PETROLEUM AND RENEWABLE ENERGY ENGINEERING
UNIVERSITI TEKNOLOGI MALAYSIA FLUID MECHANICS LABORATORY TITLE OF EXPERIMENT JET IMPACT (E3) Name
Bong Woei Shen (A13KP0021) Kumaresan a/l Sinathurai (A13KP0038) Ahmed Gamal Mahmoud Moteir (A12KE4016)
Group / Section
2/Section 11
Supervisor
Associate Professor Issham bin Ismail
Date of Experiment
3 March 2014
Date of Submission
9 March 2014
Marks obtained (%) TECHNICIANS: 1. EN. MAHMOOD RASIDON 2. PN. ZAIMARHAMAH ZAINUDDIN
1
1.0
Objective
The objective of this experiment is to measure the force exerted by a fluid jet impinging upon a flat plate or a hemispherical surface and to compare the results with the theoretical values.
2.0
Introduction
Water jet from a small opening, with a high velocity, when exerted on a surface of plate will produce force that gives power to move a system. The principle of jet impact is the basis for the understanding of liquid flow in turbines. This principle is used in designing impulse turbines. In these turbines part of the fluid energy is transformed into kinetic energy in a nozzle ( or a set of nozzles) which issues a jet of fluid at high speed. The jet strikes the moving blades, mounted on the turbine wheel, producing the force required to drive it. 3.0
Theory
A jet of fluid when impinging upon a flat or a curved surface generates a force due to change of momentum of the fluid according to Newton’s second law of motion. For example, when water of a velocity is forced out from a jet nozzle with diameter d on a plate, the rate of change of momentum produced and its magnitude is the same with the force exerted on the surface of the plate to support the water jet.
Force = Rate of momentum change of water jet = (mass of fluid/time) x change of velocity The force (F) generated by a jet of water as it strikes depends on the shape of the plate surface, e.g. flat plate or a curved (hemispherical) surface. Diagrams 1-2 show a jet of fluid issuing from a nozzle of diameter ‘d’, and moving vertically upward with velocity v and strikes a stationary surface. The jet is deflected by the vane through an angle θ and the fluid leaves the vane with velocity v’ The force generated is
2
Fth m v v'
ρav v v cos
(1)
ρav 2 ρav 2 cos
where, Fth a θ v v’
= Theoretical force exerted on the plate (Newton) = Cross-sectional area of nozzle (m2) = Density of water (kg/m3) Angle of water flow after impact on the plate surface = Velocity of water jet before impact on the plate surface = Velocity of water jet after impact on the plate surface
Water jet Nozzle d
Diagram 1- Flat Surface
Water jet Nozzle d Diagram 2 – Hemispherical Surface
For flat plate (diagram 1) , θ = 90o, therefore cos θ = 0 , so Fth ρav 2
(2)
3
For hemispherical plate (diagram 2), θ = 180o, therefore cos θ = -1, So Fth av v v 2 av 2
(3)
4
4.0
Apparatus
The apparatus consists of an upward discharging jet surrounded by a clear Plexiglas tube provided with levelling screws. The plate located directly over the jet is mounted on a stainless steel spindle, which passes through the top plate of the apparatus. A weight pan is mounted on the upper end of this spindle Water is supplied from the lab faucet (supply valve) to the inlet of the apparatus via a hose. Water flowing through the nozzle strikes the flat plate and deflects from the flat plate and falls to the base of the clear Plexiglas tube where it exit and drain in the sink.
JET IMPACT
Weight Mass Plate Apparatus Spring Coils
Standard Indicator Jet Impact Plate
Control Valve
Water Volume Scale
Pump Switch Water Tank Valve
5
5.0
Experimental Procedure Notice:The weight cannot be reentered to make up the total required weight
1. Be sure the flat or hemispherical plates are fixed. (please be careful, do not turn it too tight since it would be difficult to open) 2. 700g was the standard weight for each plate apparatus. Weights were put on the spring plat. Standard weight for both plates must be similar. 3. Standard indicator was adjusted to be level with the position of the plate containing the weight as standard mark. (zero velocity of water, V = 0) 4. The water control valve was sure to be closed (clockwise). Both switches of the pump were turn on and water control valve was opened slowly (anticlockwise) until maximum. 5. The plate with standard weight height will increase above standard mark. More weights were added until it returns to the standard mark. The total maximum weight for the first reading of the load is taken. 6. The valve of the water tank is closed (clockwise). Time started to be taken as the volumes at 2 liter until it reached 7 liter. 5 liter of water was accumulated. 7. The total weight load was reduced; the plate will rise above the standard mark. The flow of water jet was reduced slowly (clockwise) until the plate apparatus returns to the standard mark level. Step 6 was repeated. 8. Step 7 and 6 was repeated for next readings until the last total weight load is the same with the standard weight load 9. Control valve was closed (clockwise) and the pumps were switched off after the experiment was finished. The equipments were cleaned and dry.
6.0
Experimental data and analysis
6
Water density,
= 1000 kg/m3
Water velocity of the jet by the nozzle with diameter d = 5 mm Q A
V
m3 m liter 1 x x 2 3 s s m 10 liter
where V
= Water velocity (m/s)
Q
= Volumetric flow rate of water
A
= Area of nozzle with diameter “d” (m2)
A
x d 2 ... m 2 4 m π 2 x d mm x 3 4 10 mm
2
Hence, the force measured is Fmea
RESULTS Flat plate Standard Weight
weight (gram) x 9.81 m/s 1000 gram/kg gram x 9.81 x 10 3 Newton
= 700 (g)
Maximum Weight
= 1300 (g)
7
Weight Load (Gram)
Actual Weight (Gram)
Fmea (Newton)
Time (Second)
Q (L/S)
V (m/s)
Fth (Newton)
1300 1200
600 500
5.89 4.90
14.38 16.16
0.348 0.309
17.67 15.74
6.13 4.86
1.247
0.770
1.197
0.690
12.07 0.82
1100
400
3.92
18.71
0.267
13.60
3.63
1.134
0.593
7.99
1000
300
2.94
21.31
0.235
11.97
2.81
1.078
0.468
4.63
900
200
1.96
28.10
0.178
9.12
1.63
0.960
0.292
20.25
800
100
0.98
42.78
0.117
5.96
0.70
0.775
-0.009
40.00
700
0
0
-
0
0
0
0
0
0
Log Fmea
Percentage of relative error
Log V
Log Fmea
Percentage of relative error
%
Hemispherical plate Standard Weight = 700 (g) Time (Second)
Q (L/S)
Maximum Weight V (m/s)
Fth (Newton)
Fmea Fth x100 Fth
= 1700 (g)
Weight Load (Gram)
Actual Weight (Gram)
Fmea (Newton)
1700
1000
9.81
13.38
0.374
19.05
14.25
1.280
0.992
31.16
1600
900
8.83
14.3
0.350
17.83
12.48
1.251
0.946
29.25
1500
800
7.85
15.36
0.326
16.60
10.82
1.220
0.895
27.45
1400
700
6.87
16.66
0.300
15.28
9.17
1.184
0.837
25.08
1300
600
5.89
17.56
0.285
14.51
8.27
1.162
0.770
28.78
1200
500
4.91
19.47
0.257
13.09
6.73
1.117
0.691
27.04
1100
400
3.92
21.75
0.230
11.71
5.38
1.069
0.593
27.14
1000
300
2.943
23.15
0.216
11.00
4.75
1.041
0.469
38.04
900
200
1.962
25.70
0.195
9.93
3.87
0.997
0.293
49.30
800
100
0.981
28.63
0.175
8.91
3.12
0.950
-0.008
68.56
700
0
0
-
0
0
0
0
0
0
Log V
%
Note: (1) Nozzle size, d = 5 mm
Fmea Fth x100 Fth
(2) Actual Weight = Load Weight – Standard Weight
8
7.0 RESULT AND DISCUSSION Calculation for each data for both flat and hemisphere plate is same. Then we take the calculation for first data of each plate as example. For Flat Plate, Actual weight =
Fmeasured =
=
1400 g - 800 g
=
600 g
weight (g) / (1000g/kg) x 9.81 m/s2 =
(600/1000) kg x 9.81 m/s2
=
5.886 N
Flow rate, Q
Velocity, v
Ftheory
Log V Log Fmeasured
Maximum weight (g) – Standard weight (g)
=
Volume (L)/Time (s)
=
5 L/14.13 s
=
0.35386 L/s
=
(Q/1000)/A
=
0.35386 L/s x 1 m3/103 L ÷ [(π/4) x (5/1000)2] m2
= =
18.02179 m/s ρAV2
=
1000 kg/m3 x [(π/4) x (5/1000)2] m2 x (18.02179 m/s)2
= =
1000 kg/m3 x 0.00001963 m2 x 324.785 m2/s2 6.3771 N
=
log (18.02179)
=
1.256
=
log (5.886)
=
0.770
Percentage of relative error, % = │(Fmea – Fth) / Fth│ x 100% = │(5.886 – 6.3771) / 6.3771│ x 100% = 7.70%
9
For Hemisphere Plate, Actual weight = Maximum weight (g) – Standard weight (g)
Fmeasured =
=
1700 g - 800 g
=
900 g
weight (g) / (1000g/kg) x 9.81 m/s2 =
(900/1000) kg x 9.81 m/s2
=
8.829 N
Flow rate, Q
Velocity, V
Ftheory
=
Volume (L) / Time (s)
=
5L / 15.90s
=
0.31447 L/s
=
(Q/1000)/A
=
0.31447 L/s x 1 m3/103 L ÷ [(π/4) x (5/1000)2] m2
= =
16.01559 m/s 2ρAV2
=
2 x 1000 kg/m3 x [(π/4) x (5/1000) 2] m2 x (16.01559
=
10.0727 N
=
log (16.01559)
=
1.205
=
log (8.829)
=
0.946
m/s)2 Log V Log Fmeasured
Percentage of relative error, % = │(Fmea – Fth) / Fth│ x 100% = │(8.829 – 10.0727) / 10.0727│ x 100% = 12.35%
10
11
12
A.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:
Log
Log
=
av2
=
Log av2
=
Log + Log a + Log v2
=
2Log v + A
……………… (1)
……………… (2) ,
Where A is constant, A= Log + 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:
Log
Log
=
2av2
=
Log 2av2
=
Log 2 + Log + Log a + Log v2
=
2Log v + B
……………… (1)
………………. (2)
Where B is constant, B= Log 2 + Log + Log a (The same goes to
.)
The slope of Fmea on flat plate is 2.2718 while its Fth is 1.9977. The differences of slope is only 0.2741, 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 2.2989 while it’s Fth 1.9976. The difference is also 0.301 thus3 can be considered 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. 13
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 = -1.8142/-2.029 = 0.8236 (Fmean) Ratio= -1.4037/1.7044 = 0.8236 (Ftheory) The ratio of y-intercept of hemispherical to flat plate for (
) is 0.8236
: 1 while the ratio of y-intercept of hemispherical to flat plate for(
is
0.8236 : 1. Both values for gradient and y-intercept for both graphs are identical, thus the result obtained is 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 behaviour 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.
14
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 7.70% for flat plate and 12.35% for hemispherical plate. The percentage of error ranged from around 0.52% to around 31%. If we have less percentage relative error, so it means jet velocity is more constant. Some of the percentages of 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. 15
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.
8.0
CONCLUSIONS From our experiment, we found that the force produced by the jet is directly proportional to the square of the velocity of water for both flat plate and the hemisphere, F 2V The force produced by the hemisphere plate is greater than the flat plate that is approximately two-fold. This happen due to the structure of hemisphere plate that 16
curve, resulting fountain out of the water jet nozzles experienced rate of change of momentum is higher compared to flat plate structures.
Surface area nozzle jet large flow rates will slow the water. This will reduce water flow velocity and lower the rate of change of momentum flow. With this, the power produced will also be less. From question 5, the relationship with the power nozzle diameter can be described as follows: Fth
9.0
REFERENCES
Fluid Mechanics (Fundamental and Application) Second Edition in SI Units
by Yunus A. Cengel and John M. Cimbala Fundamental of Fluid Mechanics Bruce R. Munson, Donald F. Young, Theodore H. Okiishi
10.0
APPENDICES Data table for graph of Log Fmea vs
Hemispherical plate Log V Log Fmea 1.205 0.946 1.183 0.895 1.142 0.837 1.115 0.77 1.091 0.691 1.045 0.594 1.009 0.469 0.917 0.293 0.782 -0.008
by
Log
V
17
18
