Class Name: Hydraulics Lab
Class Number: CEE312
Report Title: Impact of a jet
Name: Qusai Alsaadat
ID: 20171205043
Section: 2
Date: 17/03/2020 Tuesday
Introduction:
In this experiment we learned how decreasing the diameter of pipe can create a high velocity
jet of water that can implicate a high amount of momentum which can be used in many fields ,
moreover we learned how different shapes of plates can deflect and absorb energy from the
water jet .
Objectives:
This experiment enables us to measure the force developed by a jet of water deflected on a
fixed impact object by comparing it to the force predicted by the moment theory.
Equipment and apparatus:
1) Impact of jet apparatus
2) Hydraulic bench
3) Weights
Notes:
The equipment is composed of a transparent cylinder inside which water is fed from
bottom pipe provided with a nozzle. The jet produced by the nozzle strikes an object
supported on a stem which extends through the cover.
To counterbalance the force exerted by the water jet, weights are applied to the upper
end of the stem When there is jet of water, the force exerted by the last causes the
stem to rise and this will be balanced by adding weights until the stem returns to its
starting position .In this way it is possible to determine the force exerted by the jet of
water on the object under test .
3 kinds of objects are available: a flat one, a conical one at 4so and a hemispherical one.
Procedure:
1) The cover was removed from the equipment by unscrewing the fixing Screws.
2) The wished impact object was screwed to the support stem.
3) The cover was set, and the fixing screws was screwed.
4) The pins were adjusted so that the equipment is perfectly levelled.
5) The pointer was set besides the weight stem assembly to the red level (Balance position
without jet). flow rate step by step by operating.
6) A nominal mass was placed on the weight pan, then the water was allowed to now increase
the control valve on the bench, and switch on the pump.
7) The now rate was adjusted until the weight pan is adjacent to the level gauge.
8) The readings were taken as volume and time to find the flow rate.
9) The mass was written down on the weight pan.
10) Steps 7-10 were repeated with additional masses on the weight pan. conical one at 45 and,
11) The above steps were repeated by using the three different types of objects provided (flat
one, a hemispherical one).
Data and data analysis:
Data acquired from the lab:
Flat plate:
Weight (kg) Water volume Time (S)
(m3)
0.030 0.003 30.94
0.050 0.003 18.00
0.070 0.003 16.96
0.080 0.003 15.67
0.100 0.003 14.16
Conical plate:
Weight (kg) Water volume Time (S)
(m3)
0.030 0.003 27.80
0.050 0.003 12.36
0.070 0.003 9.82
0.080 0.003 9.56
0.100 0.003 9.02
Hemispherical plate:
Weight (kg) Water volume Time (S)
(m3)
0.030 0.003 36.12
0.050 0.003 27.80
0.070 0.003 25.12
0.080 0.003 21.55
0.100 0.003 20.44
Data of flat plate:
Weight Water Time Flow rate Mass flow Velocity of Force due Force
(kg) volume (s) (m3/s) rate (kg/s) water (m/s) to due to
(m3) momentum gravity
(N) (N)
0.030 0.003 30.94 0.00009696 0.09696186 1.92821884 0.18696368 0.2943
2 2 8
0.050 0.003 18.00 0.00016666 0.16666666 3.31439393 0.55239899 0.4905
7 7 9
0.070 0.003 16.96 0.00017688 0.17688679 3.51763507 0.62222318 0.6867
7 2 7 6
0.080 0.003 15.67 0.00019144 0.19144862 3.80721703 0.72888647 0.7848
9 8 3 7
0.100 0.003 14.16 0.00021186 0.21186440 4.21321263 0.89262979 0.981
4 7 5 6
Data of the conical plate:
Weight Water Time Flow rate Mass flow Velocity of Force due to Force
(kg) volume (s) (m3/s) rate (kg/s) water (m/s) momentum due to
(m3) (N) gravity
(N)
0.030 0.003 27.80 0.000107914 0.107913669 2.146010464 0.067829292 0.2943
0.050 0.003 12.36 0.000242718 0.242718447 4.82678729 0.343138886 0.4905
0.070 0.003 9.82 0.000305499 0.305498982 6.07526384 0.543605576 0.6867
0.080 0.003 9.56 0.000313808 0.313807531 6.240490681 0.573576162 0.7848
0.100 0.003 9.02 0.000332594 0.332594235 6.614089901 0.644308415 0.981
Data of the hemispherical plate:
Weight Water Time Flow rate Mass flow Velocity of Force due Force
(kg) volume (s) (m3/s) rate (kg/s) water (m/s) to due to
(m3) momentum gravity
(N) (N)
36.12 0.00008305 0.08305647 1.65169133 0.27436733 0.2943
0.030 0.003 7 8 2 1
27.80 0.00010791 0.10791366 2.14601046 0.46316772 0.4905
0.050 0.003 4 9 4 6
25.12 0.00011942 0.11942675 2.37496381 0.56726842 0.6867
0.070 0.003 7 2 6
21.55 0.00013921 0.13921113 2.76840329 0.77078513 0.7848
0.080 0.003 1 7 9
0.100 0.003 20.44 0.00014677 0.14677103 2.91874221 0.85677364 0.981
1 7 7 5
Relationship between the force due to gravity and momentum:
flat plate
1.2
force due to momentum (N)
1
y = 0.9781x + 0.0639
0.8
0.6
0.4
0.2
0
0 0.2 0.4 0.6 0.8 1
force due to gravity (N)
Original equation:
y = 0.6173x + 0.1957
conical plate
1.2
force due to momentum (N)
1
y = 1.0878x + 0.1748
0.8
0.6
0.4
0.2
0
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7
force due gravity (N)
Original equation:
y = 0.7046x + 0.1777
hemispherical plate
1.2
force due to momentum
1
y = 1.1103x - 0.0037
0.8
0.6
0.4
0.2
0
0 0.2 0.4 0.6 0.8 1
force due to gravity
Original equation:
y = 1.1953x - 0.2731
Discussion:
Errors in this experiment may occur due to wrong setting of the arrow that the
equilibrium is read from , or by not reading from eye level .also the valve may be not
closed completely which may change the time , and regarding the use of a stop watch
the time reading may be not so accurate .
If we changed the flow rate very slowly down to zero the curve of momentum and
gravity forces will pass threw zero because we are expecting equilibrium on the
theoretical part moreover the leaner equation if it has no (b) component , it will pass
through zero because it’s (Y intercept) is zero. While on the practical side it will not pass
through zero because the linear equation always has a (b) component that make it
impossible of a point to have a (0,0) coordinates.
Slope of the momentum vs gravity forces curve:
Plate Theoretical slope Theoretical Practical Practical
slope value slope slope value
Flat plate Β =0 1 From the 0.9781
1-cosβ graph
1- 0=1
Conical plate Β =45 0.292893 From the 1.0878
1-cosβ graph
1-07071=0.292893
Hemispherical Β =180 2 From the 1.1103
plate 1-cosβ graph
1- -1=2
The difference in these values is because of the high error that occurred in experiment.
(the errors are mention up)
The shapes of the water formed after hitting the plate:
Flat: gives a 90 degree
Conical: gives a 45 degree
Hemispherical: gives a 180 degree
Conclusion:
After the experiment we how jet deflection prosses works on the plates, and how to calculate
the impact of the water jet.
References:
Hydraulics lab (ECIV 3122) lab report no.3 done in Islamic university – (GAZA) IUG
Hydraulics lab report d one in Sultan Alzan Shah polytechnic (PSAS)
