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Impact of Jet
Swinburne University of Technology
School of Engineering (Sarawak Campus)
HES 2340 Fluid Mechanics 1
Semester 2, 2008
Lab Sheet: IMPACT OF JET
Name: Student ID: Group Number: Date performed experiment: Lab supervisor:
OBJECTIVE 1. To determine the reaction force produced by the impact of jet of water on to variety type of target vanes. 2. To experimentally determine the force required to keep a target at a datum level while it is subjected to the impact of water jet. 3. The experimentally measured force is compare with the theoretical calculated force APPARATUS REQUIRED
Impact of jet apparatus with hydraulic bench
1. 1 Parts Identification
W e ig h t C a r rie r B ra ss W e ig h ts P o in te r
W e ig h t P la tfo r m I n te rc h a n g e a b le T arg e t V a n e In te rc h a n g e a b le N o z z le
D ra in h o le s in b a se W a te r S u p p ly C o n n e c tio n
Figure 1: Impact of Jets Apparatus
120?
60?
O 30
O 30
O 30
O 30
A) Flat Target
B) Conical Target
C) Semi-Spherical Target
D) 60° Plate Target
Figure 2: Interchangeable Target Vanes
2. 0
SUMMARY OF THEORY 2. 1 General Analysis When a jet of water flowing with a steady velocity strikes a solid surface, the water is deflected to flow along the surface. Unlike the impact of solid bodies, there is no rebound and unless the flow is highly turbulent, there will be no splashing. If friction is neglected by assuming an inviscid fluid and it is also assumed that there are no losses due to shocks then the magnitude of the water velocity is unchanged, the pressure exerted by the water on the solid surface will everywhere be at right angles to the surface. Newton’s second law of motion states that a mass that is accelerated required a force that is equal to the product of the mass and acceleration. In fluid mechanics, whenever fluid are forced to go through a restriction or change direction. The analogy to Newton’s second law in fluid mechanics is known as the momentum equation.
FX
Vi
?
Vi
Impact Velocity, Vi
V i c o? s Vi
Height, h
Vi
V i s ?in
Exit Velocity, Vn
Figure 3: Impact of a Jet Consider a jet of water which impacts on to a target surface causing the direction of the jet to be changed through and angle ? as shown in Figure 3 above. In the absence of friction, the magnitude of the velocity across the surface is equal to the incident velocity Vi. The impulse force exerted on the target will be equal and opposite to the force which acts on the water to impart the change in direction.
Applying Newton’s Second law in the direction of the incident jet
Force - FX
= Mass ? Accelerati on = Mass Flow Rate ? Change in Velocity = = M ?V M (VX, out - VX, in )
. .
. .
- FX FX
= M ( Vi cos ? - Vi ) = M Vi (1 - cos ? )
.
But M
= ?Q
therefore
? Q V i (1 ? cos ?)
.
And dividing trough by ? Q V which is the incident momentum i 4
F ? Q Vi
.
1 ? cos?
2. 2 Application to Impact of Jet Apparatus In each case it is assumed that there is no splashing or rebound of the water from the surface so that the exit angle is parallel to the exit angle of the target. a) Effect of Height The jet velocity can be calculated from the measured flow rate and the nozzle exit area.
Q Vn = A However, as the nozzle is below the target, the impact velocity will be less than the nozzle velocity due to interchanges between potential energy and kinetic energy.
.
Applying the Bernoulli equation between nozzle and plate:
? Pn ? ? ? ? Vn2 ? ? Pi ?+? ? ? 2g ? + ( Z n ) = ? ? ? ? ? ? ? ? ? ? Vi 2 ? ?+? ? ? 2g ? + ( Zi ) ? ? ? ?
Since the jet is open to the atmosphere,
? Pn ? ? ? ? Pi ? ? ? ? ? ?=0 ? ?
And
( Z n ) ? ( Zi ) = h
Therefore,
Vi 2 = Vn2 ? 2 gh
Where h is the height of target above the nozzle exit. b) Impact on Normal Plane Target For the normal plane target ? is 90?. Therefore cos ? = 0 F = 1 ? cos ? = 1 . ? Q Vi
c)
Impact on Conical and 30? Plate Target The cone semi-angle ? is 120?. Therefore cos ? = 0. 5 F = 1 ? cos ? = 0. 5 . ? Q Vi Impact on Semi-Spherical Target The target exit angle is 180?. Therefore cos ? = - 1 F = 1 ? cos ? = 2 . ? Q Vi By using the above equation, we can compare the theoretical and experimental of force value of target with different angle. Theoretically,
F = mg
d)
Experimentally,
F = ? Q Vi ? (1 - cos ? )
3. 0
EXPERIMENTAL PROCEDURES 3. 1 General Start-up Procedures
Figure 4: Impact of a Jet Apparatus with Hydraulic Bench The Impact of Jet (Model: FM 31) is supplied ready for use and only requires connection to the Hydraulic Bench (Model: FM 110) as follows: 1. The apparatus is located on top of the Hydraulic Bench with the left hand support feed of the Impact of Jets Apparatus located on the two left hand locating pegs of the Hydraulic Bench so that the apparatus straddles the weir channel. 2. A spirit level is about to attached to baseboard and level the unit on top of the bench by adjusting the feet. 3. The feed tube is connected from the Hydraulic Bench to the base of the Impact of Jets Apparatus by using a hose. 4. Water is filled into the volumetric tank of the hydraulic bench until approximately 90% full. 5. Fully close the bench flow control valve, V1 then switch on the pump. 6. Open V1 gradually and allow the piping to fill with water until all air has been expelled from the system. 7. The actual flow of water can be measured using the volumetric tank with a stopwatch.
3. 2
Experiment: Reaction force Determination
Objective: 1. To determine the reaction force produced by the impact of a jet of water on to various target vanes. 2. To experimentally determine the force required to keep a target at a datum level while it is subjected to the impact of a water jet. 3. To compare the experimentally measured force with the theoretically calculated force Procedures: 1. The weight carrier is positioned on the weight platform. The spring tension adjuster is adjusted to a distance of 20 mm between the nozzle and the target, then record this value as h. The pointer is to be moved so that it is aligned to the weight platform that is floating in mid position. 2. The pump is started and the water flow is established by steadily opening the bench regulating valve until it is fully open. 3. The vane will now be deflected by the impact of the jet. Weights are added onto the weight carrier until the weight platform is again floating in mid position. 4. The flow rate is measured and the result is recorded on the test sheet, together with the corresponding value of weight on the tray. The form of the deflected jet is observed and its shape is noted. 5. The weight on the weight carrier is reduced in steps and balance of weight platform is maintained by regulating the flow rate in about eight or ten even steps, each time recording the value of flow rate and weight on the weight carrier. 6. The control valve is closed and the pump is switched off. 7. The experiment is repeated with different target vanes and nozzles. Results and analysis: 1. The results are recorded on the result sheets. 2. The flow rate and the nozzle exit velocity are calculated. The nozzle velocity for the height of the target is corrected above the nozzle to obtain the impact velocity. 3. The experimental force and the theoretical force are calculated, then to compare.
Discussion: 1. In the installation of this apparatus, it’s crucial to make sure the placement of the nozzle head is at the centre under the vane. The displacement of it causing a loss in water velocity due to splashing by the rebound water. If the vane and the nozzle shaft are placed in series and centered, there will be no water rebound as jet water exerted will be deflected to flow along the surface to the surrounding shield when it hits the target vane. Due to this displacement also, it will cause an uneven force impact on the target vane hence decreasing the reaction force produced on the vane. 2. Higher water jet velocity will produce a higher force exerted onto the target vane. The amount of weight can be supported indicate the force exerted by the jet. Table for 120° Conical Target Weight (g) 100 150 200 Flow Rate (LPM) 12. 8 14. 8 18. 3
Flow Rate, Q (m? s) 4 2. 13 x10 ? ? 4 2. 47 x10 4 3. 05 x 10 ?
Flow Rate, Q (m? s)
4 2. 13 x10 ? 4 2. 47 x10 ? 4 3. 05 x 10 ?
Exit Velocity, Vn (m s) 10. 85 12. 58 15. 53
h, (mm) 25 25 25
Impact Velocity, Vi (m s) 10. 83 12. 56 15. 51
Experimental Theoretical Force, F(N) Force, Fn(N) 0. 98 1. 15 1. 47 1. 55 1. 96 2. 37
Error (%) 17. 23 5. 44 20. 92
Graph for 120° Conical Target
Calculation for 120° Conical Target Average Time, t (s) = 60s Flow Rate, Q (m? s) = V t = ( 12. 8l x 0. 001 m? l ) 60s 4 = 2. 13 x 10 ? m? s Nozzle Diameter: 5 x 10? ? m Area, A = ? D? 4 = ? ( 5 x 10? ? ) 4 5 =1. 9635 x 10 ? m? Exit Velocity, Vn (m s) = Q A 4 5 = ( 2. 13 x 10 ? m? s ) 1. 9635 x 10 ? m? = 10. 85 m s Impact Velocity, Vi (m s)
= V n
2
?2 gh
= (10 . 85 ) 2 ?( 2 x9. 81 x 0. 025 ) =10 . 83 m s
Experimental Force, F(N)
10
= ? V (1 ?cos ?) Qi =1000 x 2. 13 x10 ?4 x10 . 83 x (0. 5) =1. 15 N
Theoretical Force, Fn(N) = mg
= (100g 1kg 1000g) x 9. 81 m s = 0. 98 N Error (%)
= = Theoretica l ? exp erimenta Theoretica l 0. 92 ?1. 15 x100 % 0. 98 x100 %
= 17. 23%
11
Table for Flat Surface Target Weight 150 200 250 Flow Rate, Q (m3 s) 1. 833 x 10-4 2. 167 x 10-4 2. 333 x 10-4 Flow (LPM) 11 13 14 Exit H Velocit (mm) y, Vn (m s) 9. 343 25 11. 036 11. 88 25 25 Rate Flow (m3 s) 1. 833 x 2. 167 x 2. 333 x Impact Velocit y Vi 9. 333 11. 025 11. 87 Rate 10-4 10-4 10-4 Theoret ical Force, Fn (N) 1. 472 1. 962 2. 45 ERROR (%) 16. 2 21. 8 11. 6 Q
Experime ntal Force, F (N) 1. 71 2. 39 2. 73
12
Calculations for Flat Target surface Average Time, t (s) = 60s Flow Rate, Q (m? s) = V t = ( 13 x 0. 001 m? l ) 60s 4 = 2. 167 x 10 ? m? s Nozzle Diameter: 5 x 10? ? m Area, A = ? D? 4 = ? ( 5 x 10? ? ) 4 5 =1. 9635 x 10 ? m? Exit Velocity, Vn (m s) = Q A 4 5 = ( 2. 167 x 10 ? m? s ) 1. 9635 x 10 ? m? = 11. 036 m s Impact Velocity, Vi (m s)
= V n
2
?2 gh
= (11 . 036 ) 2 ?( 2 x9. 81 x 0. 025 ) =11 . 025 m s
Experimental Force, F(N) = ? V (1 ?cos ?) Qi
=1000 x 2. 167 x10 ?4 x11 . 025 x (1) = 2. 39 N
Theoretical Force, Fn(N) = mg
= (200g 1kg 1000g) x 9. 81 m s = 1. 962 N Error (%)
= = Theoretica l ? exp erimenta Theoretica l 1. 962 ? 2. 39 x100 % 1. 962 x100 %
= 21. 8%
13
Table for Hemisphere Target Weight 150 250 300 Flow Rate, Q (m3 s) 1. 4 x -4 10 1. 717 x 10-4 1. 867 x 10-4 Flow (LPM) 8. 4 10. 3 11. 2 Exit H Velocit (mm) y, Vn (m s) 7. 13 25 8. 75 9. 51 25 25 Rate Flow Rate (m3 s) 1. 4 x 10-4 1. 717 x 10-4 1. 867 x 10-4 Impact Velocit y Vi 7. 11 8. 74 9. 50 Experime ntal Force, F (N) 1. 71 2. 39 2. 73 Q
Theoret ical Force, Fn (N) 1. 4715 2. 45 3. 55
ERROR (%) 35. 29 22. 5 20. 63
14
Calculations for Flat Target surface Average Time, t (s) = 60s Flow Rate, Q (m? s) = V t = ( 13 x 0. 001 m? l ) 60s 4 = 2. 167 x 10 ? m? s Nozzle Diameter: 5 x 10? ? m Area, A = ? D? 4 = ? ( 5 x 10? ? ) 4 5 =1. 9635 x 10 ? m? Exit Velocity, Vn (m s) = Q A 4 5 = ( 1. 717 x 10 ? m? s ) 1. 9635 x 10 ? m? = 8. 75 m s Impact Velocity, Vi (m s)
= V n
2
?2 gh
= (8. 75 ) 2 ?( 2 x9. 81 x 0. 025 ) = 8. 74 m s
Experimental Force, F(N) = ? V (1 ?cos ?) Qi
=1000 x1. 717 x10 ?4 x8. 74 x ( 2) = 3. 001 N
Theoretical Force, Fn(N) = mg
= (250g 1kg 1000g) x 9. 81 m s = 2. 45 N Error (%)
= = Theoretica l ? exp erimenta Theoretica l 2. 45 ?3. 001 x100 % 2. 45 x100 %
= 22. 5%
15
Table for 30° Plate Target Weight (g) 100 150 200 Volum e (L) 12. 5 13. 0 16. 1 Time (s) T2 60 60 60 Average Time (s ) 60 60 60
T1 60 60 60
T3 60 60 60
Flow Rate, Q (m? s)
4 2. 0830 x10 ? 4 2. 1667 x 10 ? ? 4 2. 6830 x10
Exit Velocity, Vn (m s) 10. 6090 11. 0347 13. 6670
h, (mm) 25 25 25
Impact Velocity, Vi (m s) 10. 5870 11. 0125 13. 6490
Experimental Force, F(N)
1. 10263 1. 19304 1. 83101
Theoretical Force, Fn(N) 0. 981 1. 4715 1. 9620
Error (%) 12. 40 18. 92 6. 68
Graph for 30° Plate Target
16
Calculation for 30° Plate Target Average Time,t (s) = ( T1 + T2 + T3 ) 3 = ( 60 + 60 + 60 ) 3 = 60 s Flow Rate, Q (m? s) = V t = ( 12. 5l x 0. 001 m? l ) 23s 4 = 2. 083 x 10 ? m? s Nozzle Diameter: 5 x 10? ? m Area, A = ? D? 4 = ? ( 5 x 10? ? ) 4 5 =1. 9635 x 10 ? m? Exit Velocity, Vn (m s) = Q A 4 5 = ( 2. 083 x 10 ? m? s ) 1. 9635 x 10 ? m? = 10. 609 m s Impact Velocity, Vi (m s)
= Vn
2
?2 gh
= (10 . 609 ) 2 ?( 2 x9. 81 x 0. 025 ) =10 . 587 m s
Experimental Force, F(N) = ? Vi (1 ?cos ?) Q
=1000 x 2. 083 x10 ?4 x10 . 587 x (1 ?cos 60 ) =1. 10263 N
Theoretical Force, Fn(N) = mg
= (100g 1000g) x 9. 81 m s = 0. 981 N Error (%)
= = Theoretica l ? exp erimental Theoretica l 0. 981 ?1. 10263 x100 % 0. 981 x100 %
= 12. 40% 4. 0 O BSERVATIONS
AND
D ISCUSSION
17
•
When the graphs of Theoretical Force vs Experimental Force were plotted all the vanes except the hemispherical one gave a gradient very close to 1. The hemisphere gave a gradient of 1. 92 Gradient of graph = Theoretical Force Experimental Force Therefore we can see most of the experimental results were very close to the theoretical results.
•
•
This can be further analyzed by using the percentage errors calculated.
Type of Vane used 120 Conical Flat Plate Hemisphere 30 Plate Maximum experimental error % 20. 92 21. 8 35. 29 18. 92
• • •
Most of the experimental errors above are below 25% which although are not within the usually range of about 10-15 percent are not totally unacceptable. It was also observed that the experimental force was at all instances higher than the theoretically required force. Possible Sources of Error 1. The height between the nozzle and the target of the spring tension should be a constant value - This value can fluctuate due to parallax errors and also inaccuracy of measuring instruments 2. The height between the nozzle and the vane can also change due to the change of vanes as all vanes do not have equal heights. 3. At all instances the nozzle and the vane have to be concentric – In practice this does not always happen as there is a slight play between the weight platform and the cylinder that holds it and it can move around slightly due to the action of the force of the water. 4. There could also be a frictional force between the weight platform and where it is fixed – This could be one reason why a higher force than the calculated was required to support the vane. 5. The reason the hemispherical vane gives a higher discrepancy than the others could be because once the water hits its center the only way it can travel is downwards and hence come in the way of the water coming from the jet 6. Bubbles present in the water can be a reason to get inaccurate readings as well. 7. The water which hits the Vane could flow downwards and hit the jet again which will give a momentum in the opposite direction and hence give false values. C ONCLUSION 18
5. 0
•
For orifices having a sharp edge, A, has been found to be approximately 62% of the orifice area (pg 117, Kundu) – Therefore the area used for the calculations can be one reason for the discrepancies. Although assumed as uniform throughout the jet during calculation, the velocity of the water in the jet is not. To account for this a Momentum-Flux correlation factor(Beta) has to be used where
•
(pg 155, White) • The elasticity of spring acted on the weight platform is one of the main cause to the errors occurred in the experiment when weight is been added. To obtain a theoretical force, a suitable formulae is: f = mg – kx where k is the constant of elasticity and x is the length of the spring. • The experimental results and the theoretically calculated values are similar within experimental error and proves the law of conservation of momentum. R EFERENCES • • Fluid Mechanics,2nd Edition,2002, Kundu and Cohen Fluid Mechanics,4th Edition,Frank M White
19
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Impact of Jet Derry Brownfield Funeral GAAP for Non-Profit Organizations