Civil Engineering Department
CC501 Hydraulics 2
EXPERIMENT
:
3 HM150.08 – IMPACT IMPACT OF JET
OBJECTIVE
:
To investigate jet force impacting against stationary deflectors.
APPARATUS
:
Hydraulic bench Impact of jet apparatus i.
Loading
ii.
Deflectors
iii.
Stop watch
Unit Description
Civil Engineering Department
THEORY
CC501 Hydraulics 2
:
The theoretical jet force is calculated from the principle of linear momentum. The density (ρ) of the water is 1000 kg/m
Force =
Where
3
Fluid Mass × Variable of velocity
=
(ρ.A.V )×(∆V
=
( ρ Q ) × ( V 1 – V2 )
:
ρ
=
density of fluid
A
=
area of nozzle
V1
=
first velocity
V2
=
second velocity
1. For plate Fth = ρ Q × ( V1 – V2 )
If V2 = 0 then, Fth = ρ Q . V1
Civil Engineering Department
CC501 Hydraulics 2
2. For hemisphere Fth = ρ Q × ( V1 – V2 )
If V2 = -V1 then, Fth = ρ Q . 2V1
3. For slope
Fx = ρ Q × ( V1 )
Fth = Fx , with α = 60ᵒ Fth = ρ Q . V1
Civil Engineering Department
CC501 Hydraulics 2
PROCEDURES
i.
The test set-up was placed on the HM150 so that the drain routes the water into the channel.
ii.
The connecting hose between HM150 and unit was fitted.
iii.
HM150 drain was opened.
iv.
Deflector’s were assembled [1], (plate, hemisphere, slope or clone ). The 3 screws [3] were loosened on the cover [4] and the cover was removed together with lever mechanism. The deflector was fitted appropriately. Do not forget to tighten lock nut[2] on rod. The cover was screwed back onto the vessel.
v.
Adjusting screw was used [5] to set pointer to z ero (zero notch [7]). When doing so, do not place any loading weights on measurement system when doing [8].
vi.
Desired loading weight were applied [8] 0.5 ; 1N ; 1.5N ; 2N ; 2.5N ; 3N or combination thereof.
vii.
The main HM150 cock was closed.
viii.
The HM150 pump was switched on.
ix. x. xi.
The main cock was opened carefully until the pointer was on zero again. HM150 drain cock was closed.
Volumetric flow was then determined. Recording time,t required to fill up the volumetric tank of the HM150 from 20 to 30 litres was involved.
xii.
Loading weights were added and time, t taken for 10 litres was jotted down.
xiii.
The pump was switch off, the drain was opened.
Civil Engineering Department
CC501 Hydraulics 2
RESULTS OF MEASUREMENT 1. Measured Value for Plate Force F (N)
Measuring Time (s)
Flow Rate (m /s)
0.5
45
2.222 × 10
1.0
35
2.857 × 10
2.0
24
4.167 × 10
3.0
21
4.762 × 10
7.0
13
7.692 × 10
Force F (N)
Measuring Time (s)
Flow Rate (m /s)
0.5
61
1.639 × 10
1.0
48
2.083 × 10
2.0
36
2.778 × 10
3.0
28
3.571 × 10
6.0
21
4.762 × 10
-
Measured volume : 10 liter / 0.01 m
2. Measured Value for Hemisphere
-
Measured volume : 10 liter / 0.01 m
NOTE :
Nozzle diameter = 7mm Cross Sectional Area of Nozzle = Discharge, Q =
Velocity, V =
2
Gravity, g = the gravitational acceleration (9.81m/s ).
Civil Engineering Department
CC501 Hydraulics 2
RESULT 1. Plate Flow Rate, Q 3
(m /s)
3
Velocity, V (m /s)
Calculated Force
Measured Force,F
Fth, (N)
(N)
-
5.774
1.283
0.5
-
7.425
2.129
1.0
-
10.829
4.512
2.0
-
12.375
5.893
3.0
-
19.990
15.37
7.0
Calculated Force
Measured Force,F
Fth, (N)
(N)
2.222 × 10 2.857 × 10 4.167 × 10 4.762 × 10 7.692 × 10
2. Hemisphere Flow Rate, Q 3
(m /s)
3
Velocity, V (m /s)
-
4.259
0.698
0.5
-
5.413
1.128
1.0
-
7.219
2.005
2.0
-
9.280
3.314
3.0
-
12.375
5.893
6.0
1.639 × 10 2.083 × 10 2.778 × 10 3.571 × 10 4.762 × 10
Civil Engineering Department
CC501 Hydraulics 2
CALCULATION 1. PLATE
3
F low Rate (m /s)
Measured volume
3
=
0.01 m
3
Q= V/t (m /s)
0.5N Q
1.0N = =
Q
-4
2.222 × 10
3
m /s
2.0N Q
=
-4
m /s
-4
m /s
2.857 × 10
3
3.0N = =
Q
-4
m /s
-4
m /s
4.167 × 10
3
7.0N Q
=
= =
7.692 × 10
3
= =
4.762 × 10
3
Civil Engineering Department
CC501 Hydraulics 2
Velocity
V
=
(m/s)
0.5N V
1.0N = =
5.774 m/s
2.0N V
= =
7.425 m/s
3.0N = =
10.829 m/s
7.0N V
V
= =
19.990 m/s
V
= =
12.375 m/s
Civil Engineering Department
CC501 Hydraulics 2
F orce, (N)
Fth
=
ρQ × v1,
ρ
0.5N Fth = =
=
-4
1000 (2.222 × 10 ) × 5.774
=
Fth
1.283 N
-4
=
1000 (2.857 × 10 ) × 7.425
=
2.129 N
=
1000 (4.762 × 10 ) × 12.375
=
5.893 N
3.0N -4
1000 (4.167 × 10 ) × 10.829 4.512 N
7.0N Fth =
3
1000kg/m 1.0N
2.0N Fth =
=
-4
1000 (7.692 × 10 ) × 19.990 15.37 N
Fth
-4
Civil Engineering Department
CC501 Hydraulics 2
2. HEMISPHERE (180 ᵒ DEFLECTION)
3
F low Rate (m /s)
Measured volume
3
=
0.01 m
3
Q= V/t (m /s)
0.5N Q
1.0N = =
Q
-4
1.639 × 10
3
m /s
2.0N Q
=
-4
m /s
-4
m /s
2.083 × 10
3
3.0N = =
Q
-4
m /s
-4
m /s
2.778 × 10
3
6.0N Q
=
= =
4.762 × 10
3
= =
3.571 × 10
3
Civil Engineering Department
CC501 Hydraulics 2
Velocity
V
=
(m/s)
0.5N V
1.0N = =
4.259 m/s
2.0N V
= =
5.413 m/s
3.0N = =
7.219 m/s
6.0N V
V
= =
12.375 m/s
V
= =
9.280 m/s
Civil Engineering Department
CC501 Hydraulics 2
F orce, (N)
Fth
=
ρQ × v1,
ρ
0.5N Fth = =
=
-4
1000 (1.639 × 10 ) × 4.259
=
Fth
0.698 N
-4
=
1000 (2.083 × 10 ) × 5.413
=
1.128 N
=
1000 (3.571 × 10 ) × 9.280
=
3.314 N
3.0N -4
1000 (2.778 × 10 ) × 7.219 2.005 N
6.0N Fth =
3
1000kg/m
1.0N
2.0N Fth =
=
-4
1000 (4.762 × 10 ) × 12.975 5.893 N
Fth
-4
Civil Engineering Department
CC501 Hydraulics 2
QUESTION
1. Comparison of the two bodies with completing the table below.
Loading (N)
Calculated Force, Fth (N) Plate
Hemisphere
0.5
1.283
0.698
1.0
2.129
1.128
2.0
4.512
2.005
3.0
5.893
3.314
6.0
-
5.893
7.0
15.37
-
Civil Engineering Department
CC501 Hydraulics 2
DISCUSSION
In this experiment 10 liter of water and a nozzle of diameter 7mm were used. The forces were obtained by calculating the time recorded and the information, and the 3
percentage of error. The unit of flow rate, Q was converted from L/min to m /s for an easier calculation. The force and percentage of error were calculated according to the given formula.
In this experiment, when the measurement increased, the time needed for the level in the volume metric tank to rise 20 to 30 liter decreased. This result in a decrease in the flow rate. The flow rate for hemisphere is found to be the lowest, when comparing deflectors.
CONCLUSION
The two bodies used have a different geometry. This is the reason why the energy of water jet is used differently by each body. The size of the extent of utilisation can be seen based on the amount of water needed in order to raise a body against the force.
As the volume metric rate of flow ‘Q’ increased, the force resulted from the impact of the jet on both the flat plate and the hemispherical cup, is increased to for the predicted ‘F1’ and the measured ‘F2’ value of the force. This relation can be seen clearly from the four plots accompanied with this report. This result was alread y predicted based from the changes in momentum equation of calculating the force.
The predicted value of the jet force showed larger value than the measure one.This may caused by error in taking the reading while conducting the experiment. According to the last point the hemispherical cup is more efficient for using in the turbine than the flat plate, but only one point to be considered is that the water exit the cup is going to collide with the water entering the cup which will be reduce the force, for that reason the cup is made angles less than 180ᵒ.
Civil Engineering Department
CC501 Hydraulics 2
The impact of the jet force on the flat plate where half of it on the hemispherical cup. This result is proven based from the graph plotted .This was already predicted from the change in momentum equation of calculating the force.
REFERENCES
Hydraulic report lab sheet www.eng.ucy.ac.cy/EFM/manual/HM%2015008/HM15008E-ln.pdf (HM150.08 Impact of Jet)
