To measure the force developed by a jet of water defected on a fixed impact object by comparing it to the force predicted by the momentum theory.

1. ‫ﻋﻠﻲ‬ ‫اﻟدﯾن‬ ‫ﺳﯾف‬
Baghdad University
College of Engineering
Department of Mechanical Engineering
Name of Experiment
" Impact of Jet"
Preparation:
Saif al-Din Ali Madi
The second phase
Group A ""

2. ‫ﻋﻠﻲ‬ ‫اﻟدﯾن‬ ‫ﺳﯾف‬
1. Experiment name:
"Impact of Jet"
2. objec ve:
To measure the force developed by a jet of water defected on
a fixed impact object by comparing it to the force predicted by
the momentum theory.
3. Apparatus
Impact of jet apparatus as shown in Fig. 1
 Plexiglas diameter cylindrical tank.
 10 mm-diameter nozzle.
 Impact object of flat shape having a30mm diameter.
 Impact object of hemispherical shape having a 30mm
diameter
 Nozzle distance- impact object (S): 37mm.
 Stainless steel weight
3. Theory
The force executed by a jet coming from a nozzle striking an
object with a particular shape depends on the shape of the
same object. Fig. 2. Shows two kinds of objects stuck by a water
jet.
a) Flat object: the outlet angle, ^B^ of the jet is at 90oin respect
to the inlet direction.
b) Hemispherical object: the outlet angle, B of the jet is at
180 ° in respect to the inlet direc on.

3. ‫ﻋﻠﻲ‬ ‫اﻟدﯾن‬ ‫ﺳﯾف‬
Fig 1. arrangement of apparatus
1 2
Fig 2 . impact of a jet on a : 1 )flat object : 2)hemispherical object

4. ‫ﻋﻠﻲ‬ ‫اﻟدﯾن‬ ‫ﺳﯾف‬
Theoretical Force:
The jet impact force is given by the difference between the jet inlet
( ) and outlet ( ) momentum :
= ∗
( − )
The impulse force exerted on the target will be equal and opposite to
the force which acts on the water to fraction the change in direction.
Applying Newton's Second Law of Motion to the problem of the impact
of a jet un a stationary target, for an ideal fluid a general theoretic
relationship can be derived for the theoretical force produced
Where:
V=Liquid speed
= (If we neglect the effect of change of elevation on jet speed,
and the loss of speed due to friction over the surface of the vane)
^
-Mass flow rate
∗
= = =
.
In case of impact on a flat object, =90 ° (cos - 0)
= ∗
In case of impact on a hemispherical object, =180 ( = − )
= ∗

5. ‫ﻋﻠﻲ‬ ‫اﻟدﯾن‬ ‫ﺳﯾف‬
Calculate velocity by :
the jet outlet speed from the nuzzle, can be measured from the flow
rate Q as it follows:
= =
∗
Where
= ( )
= ( ) = . ∗
The jet, at the nozzle, has vertical direction but when it affects the
object it is deflected of an angle A depending on the shape of the
object
Fig. 3: outlet speed and impact speed of the jet.
Applying the Bemoulli equation between the nozzle and the impact
object:
+ + = + + =

6. ‫ﻋﻠﻲ‬ ‫اﻟدﯾن‬ ‫ﺳﯾف‬
: is the pressure at the nozzle outlet
:is the position of the nozzle outlet
Vo: the outlet speed from the nozzle
:is the pressure after the impact on the support
: is the impact position
: is the impact speed
Now, considering the nozzle as well as the impact object is at
atmospheric pressure, it is given by
− =
: Between, being
− =
i.e., equal to the distance between nozzle and impact object, we
obtain:
= −
= −
So, it is possible to obtain the jet impact speed on the object from the
jet speed at the nozzle outlet and from the distance between the
nozzle and the impact object.
Where:
S 37 mm (height of vane above nozzle)

7. Experimental force :
Without jet:
With jet :
‫ﻋﻠﻲ‬ ‫اﻟدﯾن‬ ‫ﺳﯾف‬

8. ‫ﻋﻠﻲ‬ ‫اﻟدﯾن‬ ‫ﺳﯾف‬
+ 1 + = 0 ……(without jet)
− ∗ 1 + ( 1 − 2) + = 0 …….. (with jet)
=
2
1
Where:
X1= 150 mm (Distance from center of vane to pivot of lever)
mj= 0.610 kg (Mass jockey weight)
g =9.81 m s accelerate
w=mg= 0.610x 9.81= 5.89 N (weight of jockey weight)
x2= jockey weight is moved a distance from its zero position
=
=
−

9. ‫ﻋﻠﻲ‬ ‫اﻟدﯾن‬ ‫ﺳﯾف‬
Accounts:
" Flat plate "
= . = ∗
= . ∗ =
1.
= =
∗
= =
7.5
60
= 0.125
=
∗
∗
=
0.125
0.0785
= 1.25923 /
= − = ( . ) − ( . ) ∗ ( ∗ )
= . /
= ∗
∗ = 0.125 ∗ 1.3451 = 0.1681
= 4 2 = 4(9.81) ∗ 10 ∗ 10 = 0.3924
= =
. .
.
= 133.2%

10. ‫ﻋﻠﻲ‬ ‫اﻟدﯾن‬ ‫ﺳﯾف‬
= . = ∗
= . ∗ =
2.
= =
∗
= =
7.5
40
= 0.1875
=
∗
∗
=
0.1875
0.0785
= 2.388535 /
= − = ( . ) − ( . ) ∗ ( ∗ )
= . /
= ∗
∗ = 0.1875 ∗ 2.2314 = 0.418388
= 4 2 = 4(9.81) ∗ 22 ∗ 10 = 0.86328
= =
. .
.
= 106.33%

11. ‫ﻋﻠﻲ‬ ‫اﻟدﯾن‬ ‫ﺳﯾف‬
= . = ∗
= . ∗ =
3.
= =
∗
= =
7.5
= 0.3
=
∗
∗
=
0.3
0.0785
= 3.82165 /
= − = ( . ) − ( . ) ∗ ( ∗ )
= . /
= ∗
∗ = 0.3 ∗ 3.7254684 = 1.11764
= 4 2 = 4(9.81) ∗ 36 ∗ 10 = 1.41264
= =
. .
.
= 26.3948%

12. ‫ﻋﻠﻲ‬ ‫اﻟدﯾن‬ ‫ﺳﯾف‬
= . = ∗
= . ∗ =
4.
= = .
∗
= =
7.5
.
= 0.36372
=
∗
∗
=
0.36372
0.0785
= 4.633433 /
= − = ( . ) − ( . ) ∗ ( ∗ )
= . /
= ∗
∗ = 0.36372 ∗ 4.55442 = 1.65655
= 4 2 = 4(9.81) ∗ ∗ 10 = 1.962
= =
. .
.
= 18.43%

13. ‫ﻋﻠﻲ‬ ‫اﻟدﯾن‬ ‫ﺳﯾف‬
= . = ∗
= . ∗ =
5.
= =
∗
= =
7.5
= 0.441176
=
∗
∗
=
0.441176
0.0785
= 5.6200824 /
= − = ( . ) − ( . ) ∗ ( ∗ )
= . /
= ∗
∗ = 0.441176 ∗ 5.555122 = 2.450789
= 4 2 = 4(9.81) ∗ ∗ 10 = 2.86452
= =
. .
.
= 16.881%

14. ‫ﻋﻠﻲ‬ ‫اﻟدﯾن‬ ‫ﺳﯾف‬
"Hemispherical"
= . = ∗
= . ∗ =
1.
= = .
∗
= =
7.5
.
= 0.12376
=
∗
∗
=
0.12376
0.0785
= 1.5765908 /
= − = ( . ) − ( . ) ∗ ( ∗ )
= . /
= 2 ∗
∗ = 2 ∗ 0.12376 ∗ 1.326536 = 0.32835
= 4 2 = 4(9.81) ∗ ∗ 10 = 0.3924
= =
. .
.
= 19.5064%

15. ‫ﻋﻠﻲ‬ ‫اﻟدﯾن‬ ‫ﺳﯾف‬
= . = ∗
= . ∗ =
2.
= =
∗
= =
7.5
= 0.170454
=
∗
∗
=
0.170454
0.0785
= 2.1713954 /
= − = ( . ) − ( . ) ∗ ( ∗ )
= . /
= 2 ∗
∗ = 2 ∗ 0.170454 ∗ 1.9972526 = 0.68088
= 4 2 = 4(9.81) ∗ ∗ 10 = 0.7848
= =
. .
.
= 15.2623%

16. ‫ﻋﻠﻲ‬ ‫اﻟدﯾن‬ ‫ﺳﯾف‬
= . = ∗
= . ∗ =
3.
= =
∗
= =
7.5
= 0.25
=
∗
∗
=
0.25
0.0785
= 3.184713 /
= − = ( . ) − ( . ) ∗ ( ∗ )
= . /
= 2 ∗
∗ = 2 ∗ . ∗ . = 1.534312
= 4 2 = 4(9.81) ∗ ∗ 10 = 1.64808
= =
. .
.
= 17.414885%

17. ‫ﻋﻠﻲ‬ ‫اﻟدﯾن‬ ‫ﺳﯾف‬
= . = ∗
= . ∗ =
4.
= = .
∗
= =
7.5
.
= 0.32923
=
∗
∗
=
0.32923
0.0785
= 4.194091 /
= − = ( . ) − ( . ) ∗ ( ∗ )
= . /
= 2 ∗
∗ = 2 ∗ . ∗ . = 2.7041
= 4 2 = 4(9.81) ∗ ∗ 10 = 2.43288
= =
. .
.
= 10.0301%

18. ‫ﻋﻠﻲ‬ ‫اﻟدﯾن‬ ‫ﺳﯾف‬
= . = ∗
= . ∗ =
5.
= =
∗
= =
7.5
= 0.416666
=
∗
∗
=
0.416666
0.0785
= 5.307855 /
= − = ( . ) − ( . ) ∗ ( ∗ )
= . /
= 2 ∗
∗ = 2 ∗ . ∗ . = 4.365854
= 4 2 = 4(9.81) ∗ ∗ 10 = 3.3354
= =
. .
.
= 23.602%

19. ‫ﻋﻠﻲ‬ ‫اﻟدﯾن‬ ‫ﺳﯾف‬
Flat plate
No X2(mm) t(sec) m (kg/s) V1(m/s)
1 10 60 0.125 . 0.1681 0.3924 133.2
2 22 40 0.1875 . 0.418388 0.86328 106.33
3 36 25 0.3 . 1.11764 1.41264 26.3948
4 50 20.62 0.3637 . 1.65655 1.962 18.43
5 73 17 0.4411 . 2.45 2.864 16.881
"Hemispherical"
No X2(mm) t(sec) m (kg/s) V1(m/s)
1 10 60.6 0.12376 . 0.32835 0.3924 19.5064
2 10 44 0.170454 . 0.68088 0.7848 15.2623
3 42 30 0.25 . 1.53431 1.64808 7.4141
4 62 22.78 0.32923 . 2.7041 2.4328 10.0301
5 85 18 0.4166 . 4.3658 3.3354 23.602

20. ‫ﻋﻠﻲ‬ ‫اﻟدﯾن‬ ‫ﺳﯾف‬
"Flat plate"
0
1
2
3
4
5
6
0 20 40 60 80 100 120 140
v1(m/s)
error
0
0.5
1
1.5
2
2.5
3
0 0.5 1 1.5 2 2.5 3 3.5
fy(th)
fy(ex)

21. ‫ﻋﻠﻲ‬ ‫اﻟدﯾن‬ ‫ﺳﯾف‬
"Hemispherical"
0
1
2
3
4
5
6
0 5 10 15 20 25
v1(m/s)
error
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
0 0.5 1 1.5 2 2.5 3 3.5 4
fy(th)
fy(ex)

22. ‫ﻋﻠﻲ‬ ‫اﻟدﯾن‬ ‫ﺳﯾف‬
3.What suggestion have you to improving the apparatus
First, add water measuring device
Second, replace the ruler balance with another kind of newer as in the
picture
Thirdly, replacing the water pump with another one is more efficient, in
which the flow is dependent
Fourthly, a circuit breaker is connected with a sensitive electrical circuit
if an electrical contact occurs to observe the conditions of safety
Fifth, the iron basins are replaced with plastic ones because we deal with
water and electricity
In the other, I put pictures of his devices

23. ‫ﻋﻠﻲ‬ ‫اﻟدﯾن‬ ‫ﺳﯾف‬