American University of Madaba

Fluid Mechanics and Hydraulics Lab

Impact of a jet

Bilal Khalil
Objective:
The purpose of this experiment is to measure the force developed by a jet of
water deflected on a fixed impact object and compare it to the force predicted by
the momentum theory.

Introduction:
The force exerted by a jet coming from a nozzle striking an object with a particular
shape depends on the shape of the same object.
Two kinds of objects struck by a water jet
a. Flat object: the outlet angle, B of the jet is at 90 in respect to the inlet direction.
b. Hemispherical object: the outlet angle, B of the jet is at 180 in respect to the
inlet

The jet impact force is given by the different the jet inlet and outlet momentum:
F X = ρ. Q . ( V 1−V 1 cos β ) =ρ .Q . V 1 ( 1−cosβ )

Where:
Kg
ρ=liquid density , liquid is water ρ water=1000 ¿
m3

Q = liquid flow rate m3/s

V = liquid speed m/s

1. In case of impact on a flat object β=90 ¿

F X = ρ. Q .V 1

2. In case of impact on a hemispherical object β=180 ¿

F X =2. ρ . Q. V 1
3. In case of a conical object (45) β=45 ¿
F X = ρ. Q .V 1 ¿

The jet outlet speed from the nozzle, V0, can be measured from the flow rate Q as
it follows:
V0 = Q / area of the nozzle

π d2
A=
4

Where d = nozzle diameter, m

The jet at the nozzle, has vertical direction but where it impact the objects the
object it is deflected of an angle B depending on the shape of the object

Applying the Bernoulli equation between the nozzle and the impact object
P n V 20 P p v 21
+ + Z n= + + Z p =constant
γ 2g γ 2g

Where pn is the pressure at the nozzle outlet

Zn is the position of the nozzle outlet
V0 the outlet speed from the nozzle
Pp is the pressure after the impact on the support
Zp is the impact position
V1 is the impact speed
Now, considering the nozzle as well as the impact objective is at atmospheric
pressure, it is given by:
Pn Pp
− =0
γ γ

Besides, being
Z n−Z p =s

For example equal to the distance between nozzle and impact, we obtain:
V 21=V 20 −2 gs

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.
S: distance between the nozzle and the impact object = 20mm

Apparatus:
1. Plexiglas diameter cylindrical tank.
2. 8mm-diameter nozzle.
3. 5mm-diameter nozzle.
4. Impact object of flat shape having a 30mm diameter.
5. Impact object of hemispherical shape having a 30mm diameter.
6. Impact object of conical shape (45°) having a 30mm diameter.
7. Nozzle distance – impact object (s) = 20mm.
8. Set of stainless steel weights.

Procedure:
1. Remove the cover from the equipment by unscrewing the fixing screws.
2. Screw the wished object to the support stem.
3. Connect the nozzle with the wished diameter to the bottom pipe
4. Set the cover and screw the fixing screws.
5. Adjust the pins so that the equipment is perfectly leveled.
6. Set the pointer besides the weight stem assembly to the red level (balance
position without jet).
7. Shut off the flow control valve V1 of the hydraulic bench mod.HB/EV
 8. Open the exhaust valve V2 of the volumetric tank of the hydraulic bench
mod. HB/EV.
9. Switch the pump G1 of mod. HB/EV on and slowly opens the valve V1 until
you reach the weight flow value.
10.Due to the jet, the impact object will push the support stem upward; add
the weights provided with the stem until the plate will be taken back to the
pointer height and note the value.
11.Note the flow value indicated by the variable area flow-meter of mod.
HB/EV and the one calculate using the volumetric tank of mod. HB/EV and a
timer.
12.Repeat the test with different flow values.
13.At the end of the experiment shut the valve V1 of the hydraulic bench
HB/EV off and switch the pump G1 off. In case the water supply network is
used. Just shut off the valve to the water supply network.

Results:
Flat impact object
Nozzle diameter: 5mm Area: π(0.005)2/4
Flow Q Mass Speed V0 Speed v1 Force Force Error
3
m /hr g m/s m/s Ftheo, N Fexp, N %
0.9 450 12.7 12.68 3.17 4.41 39.12%
0.8 320 11.2 11.18 2.46 3.14 27.64%
0.5 130 7.1 7.07 0.99 1.26 27.27%
Q 0.9/3600
v 0= = =12.7 m/s
A π∗0.0052 /4

v1 =√ v 02−2∗g∗s=√12.7 2−2∗9.81∗0.02=12.68 m/s

1000∗0.9
F=ρ∗Q∗v 1= ∗12.68=3.17 N
3600
F exp=m∗g=0.45∗9.81=4.41 N

|Ftheo −F exp| 4.41−3.17

Error= ∗100 %= ∗100 %=39.12 %
F theo 3.17
 45

40

35

30

25
Error %

20

15

10

0
6 7 8 9 10 11 12 13 14
V1 m/s

Hemispherical impact object

Nozzle diameter: 5mm Area: π(0.005)2/4
Flow Q Mass Speed V0 Speed v1 Force Force Error
3
m /hr g m/s m/s Ftheo, N Fexp, N %
0.9 450 12.7 12.68 6.34 5.59 11.83%
0.8 320 11.2 11.18 4.92 5.10 3.66%
0.5 130 7.1 7.07 1.98 2.94 48.48%
Q 0.9/3600
v 0= = =12.7 m/s
A π∗0.0052 /4

v1 =√ v 02−2∗g∗s=√12.7 2−2∗9.81∗0.02=12.68 m/s

2∗1000∗0.9
F=2∗ρ∗Q∗v 1= ∗12.68=6.34 N
3600
F exp=m∗g=0.57∗9.81=5.59 N

|Ftheo −F exp| 6.34−5.59

Error= ∗100 %= ∗100 %=11.83 %
F theo 6.34
 60

50

40
Error %

30

20

10

0
6 7 8 9 10 11 12 13 14
V1 m/s

Conical impact object

Nozzle diameter: 5mm Area: π(0.005)2/4
Flow Q Mass Speed V0 Speed v1 Force Force Error
3
m /hr g m/s m/s Ftheo, N Fexp, N %
0.9 450 12.7 12.68 0.93 1.47 58.06%
0.8 320 11.2 11.18 0.72 1.08 50%
0.5 130 7.1 7.07 0.29 0.39 34.48%
Q 0.9/3600
v 0= = =12.7 m/s
A π∗0.0052 /4

v1 =√ v 02−2∗g∗s=√12.7 2−2∗9.81∗0.02=12.68 m/s

0.2929∗1000∗0.9
F=0.2929∗ρ∗Q∗v 1= ∗12.68=0.93 N
3600
F exp=m∗g=0.15∗9.81=1.47 N

|Ftheo −F exp| 1.47−0.93

Error= ∗100 %= ∗100 %=50.06 %
F theo 0.93
 70

60

50

40
Error %

30

20

10

0
6 7 8 9 10 11 12 13 14
V1 m/s

Flat impact object

Nozzle diameter: 8mm Area: π(0.008)2/4
Flow Q Mass Speed V0 Speed v1 Force Force Error
3
m /hr g m/s m/s Ftheo, N Fexp, N %
0.9 130 5 4.96 1.24 1.27 2.83%
0.8 100 4.4 4.35 0.96 0.98 2.38%
0.5 30 2.8 2.73 0.38 0.29 22.97%
Q 0.9/ 3600
v 0= = =5 m/s
A π∗0.0082 /4

v1 =√ v 02−2∗g∗s=√5 2−2∗9.81∗0.02=4.96 m/s

1000∗0.9
F=ρ∗Q∗v 1= ∗4.96=1.24 N
3600
F exp=m∗g=0.13∗9.81=1.27 N

|Ftheo −F exp| 1.27−1.24

Error= ∗100 %= ∗100 %=2.83 %
F theo 1.24
 25

20

15
Error %

10

0
2.5 3 3.5 4 4.5 5 5.5
V1 m/s

Hemispherical impact object

Nozzle diameter: 8mm Area: π(0.008)2/4
Flow Q Mass Speed V0 Speed v1 Force Force Error
3
m /hr g m/s m/s Ftheo, N Fexp, N %
0.9 130 5 4.96 2.48 2.75 10.74%
0.8 100 4.4 4.35 1.92 2.06 7.51%
0.5 30 2.8 2.73 0.76 0.98 28.38%
Q 0.9/ 3600
v 0= = =5 m/s
A π∗0.0082 /4

v1 =√ v 02−2∗g∗s=√5 2−2∗9.81∗0.02=4.96 m/s

2∗1000∗0.9
F=2∗ρ∗Q∗v 1= ∗4.96=2.48 N
3600
F exp=m∗g=0.28∗9.81=12.75 N

|Ftheo−F exp| 2.75−2.48

Error= ∗100 %= ∗100 %=10.74 %
F theo 2.48
 30

25

20
Error %

15

10

0
2.5 3 3.5 4 4.5 5 5.5
V1 m/s

Conical impact object

Nozzle diameter: 8mm Area: π(0.008)2/4
Flow Q Mass Speed V0 Speed v1 Force Force Error
3
m /hr g m/s m/s Ftheo, N Fexp, N %
0.9 130 5 4.96 0.36 0.49 35.03%
0.8 100 4.4 4.35 0.28 0.29 4.87%
0.5 30 2.8 2.73 0.11 0.98 12.34%
Q 0.9/ 3600
v 0= = =5 m/s
A π∗0.0082 /4

v1 =√ v 02−2∗g∗s=√5 2−2∗9.81∗0.02=4.96 m/s

0.2929∗1000∗0.9
F=0.2929∗ρ∗Q∗v 1= ∗4.96=0.36 N
3600
F exp=m∗g=0.05∗9.81=0.49 N

|Ftheo −F exp| 0.49−0.36

Error= ∗100 %= ∗100 %=35.03 %
F theo 0.36
 40

35

30

25
Error %

20

15

10

0
2.5 3 3.5 4 4.5 5 5.5
V1 m/s

Discussion:
In this experiment we can see that we have high error values. This error is mostly
because we didn’t level the pointer to the wright level (balance level).
From the graphs (v1 vs. error) we can see that in most cases the error decrease
when the velocity decreases, that maybe because with low velocity less water
accumulates in the tank which makes the experiment more accurate.
When a jet of water flowing with a steady velocity strikes a solid surface the
water is deflected to flow along the surface. 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.

Conclusions:
From the results above we can see that the hemisphere has the highest impact
force, then the flat shape, and the smallest is the impact force from the conical
shape.