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:

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1. To determine determine the the reaction reaction force force produced produced by the the impact impact of jet jet of water water on to to variety type of target vanes. 2. To experimenta experimentally lly determine determine the the force requir required ed to keep a target target at a datum datum level while it is subjected to the impact of water jet. 3. The experiment experimentally ally measured measured force force is compare compare with with the theoreti theoretical cal calculated calculated force

Impact of jet apparatus with hydraulic bench

1.1 Parts Identification Identification

W e i g h t C a r r iei e r Pointer

B r a ss ss W e i g h t s

W e i g h t P l a t fo fo r m Interchangeable Target V ane Interchangeable N ozz le

D r a in h o l e s i n b a s e W a te r S u p p ly Connection

Figure 1: Impact of Jets Apparatus

2

120º

Ø 30

60º

Ø 30

Ø 30

Ø

Figure 2: Interchangeable Target Vanes

2.1 General General Analysis 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.  F  X  

3

Vi

Vi

θ

V i c oθ s

Impact Velocity, Vi

Vi

Height, h

V i s iθ n

Vi

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 θ 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

=

Mass

=

Mass Flow Rate

×

Accelerati on × Change

in Velocity

.

- FX

=

M ∆V

=

M (VX, out - VX, in )

.

.

- FX

=

M( Vi cos cos θ  - Vi ) .

FX .

But M

=

.

= ρQ

F

M Vi (1 - cos cos θ )

therefore .

=

ρ Q V i (1 cos θ) −

.

And dividing trough by ρ Q Vi which is the incident momentum

4

F

1 cosθ

=

.

−

ρ Q Vi 2.2 Application Application to Impact of Jet Apparatus 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. .

Vn

=

Q

A However, as the nozzle is below the target, the impact velocity will   be less less than than the the nozzl nozzlee velo veloci city ty due to inte interc rchan hange gess betwe between en  potential energy and kinetic energy.

Applying the Bernoulli equation between nozzle and plate: V    V i     P n       P i              ( )  Z  +  n   + = + + ( Z i ) n            g   g  γ   2 γ   2                 2

2

Since the jet is open to the atmosphere,

  P n    P i       −     = 0   γ       γ     And

( Z n ) − ( Z i ) = h Therefore, V i 2

=

V n2

−

2 gh

Where h is the height of target above the nozzle exit.

b)

Impact on Normal Plane Target For the normal plane target θ is 90º. Therefore cos θ = 0 F = 1 − cos θ = 1 . ρ Q Vi

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c)

Imp Impact on Conic nical and 30˚ 30˚ Plate ate Targe rget The cone semi-angle θ is 120º. Therefore cos θ = 0.5 F = 1 − cos θ = 0.5 . ρ Q Vi

d)

Impact on on Se Semi-Spherical Ta Target The target exit angle is 180º. Therefore cos θ = - 1 F = 1 − cos θ = 2 . ρ 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

Experimentally, .

cos θ )  F  = ρ Q Vi × (1 - cos

3.1 General Start-up Procedures Procedures

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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 o f 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: Experiment: Reaction Reaction force Determination Determination

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Objective:

1. To determine the reaction force produced by the impact of a jet of  water on to various target vanes. 2. To experimentally experimentally determine determine the force required required to keep a target target at a datum level while it is subjected to the impact of a water jet. 3. To comp compar aree the the expe experi rime ment ntal ally ly meas measur ured ed forc forcee with with the 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  pointer is to be moved so that it is aligned to the weight platform 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 resul results ts are are recorde recorded d on the resul resultt sheets sheets.. 2. The The flow flow rate rate and the nozzl nozzlee exit exit veloci velocity ty are calcul calculat ated ed.. The The nozzle velocity for the height of the target is corrected above the nozzle to obtain the impact velocity. 3. The experim experiment ental al force force and the theoret theoretica icall force force are calculat calculated, ed, then to compare.

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 Discussion:

1. In the install installati ation on of this apparat apparatus, us, it’s it’s crucial 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 Higher water water jet veloci velocity ty will will produce produce a higher force force exerte exerted d onto the target vane. The amount of weight can be supported indicate the force exerted by the jet.

Weight (g) 100 150 200

Flow Rate, Q (m³ /s)

Flow Rate (LPM) 1 2 .8 1 4 .8 1 8 .3

Exit Velocity, Vn (m/s)

h, (mm)

Flow Rate, Q (m³ /s) 2.13 x10 −4 2.47 x10 −4 3.05 x 10 −4

Impact Velocity, Vi

Experimental Force, F(N)

(m/s) 2.13 x10

Theoretical Force,

Error (%)

Fn(N)

4

10.85

25

10.83

1.15

0.98

17.23

4

12.58 15.53

25 25

12.56 15.51

1.55

1.47 1.96

5.44 20.92

− 2.47 x10 − 3.05 x 10 −

4

9

2.37

Graph for 120° Conical Target

Calculation for 120° Conical Target Average Time, t (s) = 60s Flow Rate, Q (m³/s) = V / t = ( 12.8l x 0.001 m³/l ) / 60s = 2.13 x 10 −4 m³/s  Nozzle Diameter: 5 x 10ˉ ³ m Area, A = ∏ D² / 4 = ∏ ( 5 x 10ˉ ³ ) / 4 =1.9635 x 10 −5 m² Exit Velocity, Vn (m/s) = Q / A = ( 2.13 x 10 −4 m³/s ) / 1.9635 x 10 −5 m² = 10.85 m/s Impact Velocity, Vi (m/s) =

=

Vn

2

2 gh

−

(10 .85 ) 2

( 2 x9.81 x0.025 )

−

10 .83 m /  s

=

Experimental Force, F(N)

10

 ρ  QVi (1

=

cos θ  )

−

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

0.92

1.15

−

0.98

 x100

l 

 x100

%

%

= 17.23%

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Table for Flat Surface Target

Weight

Flow (LPM) 11 13 14

150 200 250

Flow Rate, Q (m3/s) 1.833 x 10-4 2.167 x 10-4 2.333 x 10-4

Exit H Velocit (mm) y, Vn (m/s) 9.343 25 11.036 11.88

Rate Flow (m3/s) 1.833 x 2.167 x 2.333 x Impact Velocit y Vi

Rate

Q

10 -4 10 -4 10 -4

9.333

Experime ntal Force, rce, F (N) 1.71

Theoret ical Force, Fn (N) 1.472

16.2

25

11.025

2.39

1.962

21.8

25

11.87

2.73

2.45

11.6

12

ERROR (%)

Calculations for Flat Target surface Average Time, t (s) = 60s Flow Rate, Q (m³/s) = V / t = ( 13 x 0.001 m³/l ) / 60s = 2.167 x 10 −4 m³/s  Nozzle Diameter: 5 x 10ˉ ³ m Area, A = ∏ D² / 4 = ∏ ( 5 x 10ˉ ³ ) / 4 =1.9635 x 10 −5 m² Exit Velocity, Vn (m/s) = Q / A = ( 2.167 x 10 −4 m³/s ) / 1.9635 x 10 −5 m² = 11.036 m/s Impact Velocity, Vi (m/s) =

=

Vn

2

2 gh

−

(11 .036 ) 2

( 2 x9.81 x0.025 )

−

11 .025 m /  s

=

Experimental Force, F(N) QVi (1 cos θ    ρ  ) =

−

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

Theoretica

1.962

−

2.39

1.962

 x100

erimenta l 

 x100

%

%

= 21.8%

13

Table for Hemisphere Target

Weight

Flow (LPM) 8.4 10.3 11.2

150 250 300

Flow Rate, Q (m3/s) 1.4 x -4 10 1.717 x 10-4 1.867 x 10-4

Exit H Velocit (mm) y, Vn (m/s) 7.13 25 8.75 9.51

Rate Flow Rate 3 (m /s) 1.4 x 10 -4 1.717 x 10 -4 1.867 x 10 -4 Impact Velocit y Vi

Q

7.11

Experime ntal Force, rce, F (N) 1.71

Theoret ical Force, Fn (N) 1.4715

35.29

25

8.74

2.39

2.45

22.5

25

9.50

2.73

3.55

20.63

14

ERROR (%)

Calculations for Flat Target surface Average Time, t (s) = 60s Flow Rate, Q (m³/s) = V / t = ( 13 x 0.001 m³/l ) / 60s = 2.167 x 10 −4 m³/s  Nozzle Diameter: 5 x 10ˉ ³ m Area, A = ∏ D² / 4 = ∏ ( 5 x 10ˉ ³ ) / 4 =1.9635 x 10 −5 m² Exit Velocity, Vn (m/s) = Q / A = ( 1.717 x 10 −4 m³/s ) / 1.9635 x 10 −5 m² = 8.75 m/s Impact Velocity, Vi (m/s) =

=

Vn

2

2 gh

−

(8.75 ) 2

( 2 x9.81 x0.025 )

−

8.74 m /  s

=

Experimental Force, F(N)  ρ QVi (1 cos θ  ) =

−

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 =

−

exp

Theoretica

2.45 3.001 2.45 −

=

l 

 x100

erimenta l 

 x100

%

%

= 22.5%

15

Weight (g) 100 150 200

Volum e (L) 1 2 .5 1 3 .0 1 6 .1

Time (s) T2 60 60 60

T1 60 60 60

T3 60 60 60

Average Time (s ) 60 60 60

Flow Rate, Q (m³ /s)

Exit Velocity, Vn (m/s)

h, (mm)

Impact Velocity, Vi (m/s)

Experimental Force, F(N)

Theoretical Force, Fn(N)

2.0830 x10 −4 2.1667 x 10 −4

10.6090 11.0347

25 25

10.5870 11.0125

1.10263 1.19304

0.981 1.4715

12.40 18.92

4

13.6670

25

13.6490

1.83101

1.9620

6.68

2.6830 x10 −

Graph for 30° Plate Target

16

Error (%)

Calculation for 30° Plate Target Average Time,t (s) = ( T1 + T2 + T3 ) / 3 = ( 60 + 60 + 60 ) / 3 = 60 s Flow Rate, Q (m³/s) = V / t = ( 12.5l x 0.001 m³/l ) / 23s = 2.083 x 10 −4 m³/s  Nozzle Diameter: 5 x 10ˉ ³ m Area, A = ∏ D² / 4 = ∏ ( 5 x 10ˉ ³ ) / 4 =1.9635 x 10 −5 m² Exit Velocity, Vn (m/s) = Q / A = ( 2.083 x 10 −4 m³/s ) / 1.9635 x 10 −5 m² = 10.609 m/s Impact Velocity, Vi (m/s) =

=

Vn

2

2 gh

−

(10 .609 ) 2

( 2 x9.81 x0.025 )

−

10 .587 m /  s

=

Experimental Force, F(N)  ρ QVi (1 cos θ  ) =

−

1000  x2.083  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 =

−

exp

Theoretica

0.981 1.10263 0.981 −

=

l 

erimental  l 

 x100

 x100

%

%

= 12.40% 4.0

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•

•

When the graphs of Theoretical Force vs v s 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 Maximum experimental error % 120 Conical 20.92 Flat Plate 21.8 Hemisphere 35.29 30 Plate 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 height between between the nozzle nozzle and and the target target of the spring spring tension tension should  be a constant value - This value can fluctuate due to parallax errors and also inaccuracy of measuring instruments 2. The height height between between the the nozzle nozzle and the vane vane can also also change change due to the change of vanes as all vanes do not have equal heights. 3. At all instances instances the the nozzle and and the vane vane have to to be concentri concentricc – In practice 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 could also also be a fricti frictional onal force force between between the weight weight platfor platform m 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 reason the hemispher hemispherical ical vane vane gives a higher discrepancy discrepancy than 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 Bubbles present present in the the water water can be a reason reason to get inaccurat inaccuratee readings readings as well. 7. The water water which which hits the the Vane could could flow flow downwards downwards and and hit the the jet again again which will give a momentum in the opposite direction and hence give false values.

5.0

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•

For orifices orifices having a sharp edge, A, has been found to be approximately approximately 62% of  the orifice area (pg 117, Kundu) – Therefore the area used for the calculations can  be one reason for the discrepancies.

•

Although Although assumed assumed as uniform uniform throughout the jet during calculation, calculation, the velocity 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.

•

Fluid Mechanics,2nd Edition,2002, Kundu and Cohen

•

Fluid Mechanics,4th Edition,Frank M White

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