Introduction
The purpose of this lab is to introduce a method by which internal pressure energy
is transformed to kinetic, and then mechanical energy. Water-jet impact forces will
be measured against a flat plate and a hemispherical cup, and prove that twice as
much mechanical energy is transferred to the cup, due to its shape. These
fundamentals are largely applied when making hydraulic turbines, as it is necessary
to maximize energy transfer from kinetic to mechanical
Theory
To determine the amount of force transferred to either the plate or cup, an
experiment can demonstrate a relationship between fluid velocity and shape of
impact plate; the angle of deflection has the greatest impact on the force
generated. The greater the angle of deflection, more kinetic energy is transferred
from the water into mechanic lift. See figure 1. The contact angle () for a flat plate
is 90o and for a cup it is 180o. Where uo and u1, is water velocity leaving the nozzle
and striking the plate; the fluid decelerates after leaving the nozzle due to gravity.
The equation for transfer of force into impact zone is:
F=W ( uo u1 cos ( ) ) ( Newtons )
(1)
Where:
F = Force (N)
U0/1 = velocity (m/s)
W = mass rate (kg/s)
For the plate the angle is 90 and 180 for the cup, which simplifies to:
Fcup=W ( u o+ u1 ) ( Newtons )
Fcup max=2 W u 0 ( Newtons )
(2)(3)
Fplate=W u0 ( Newton)
(4)
We can see that we should expect twice as much force generated into the cup.
For the experimental derivation of force, a sum of moments is used. This will be
explained in the experiment procedure. For an in-depth explanation of the
derivations and theory, please refer to pages 83 and 84 of 261Fluid mechanics lab
U
U
manual.
1
1
U
0
Fig 1
Equipment and Procedure:
For the full and original procedure, please refer to the 261 lab manual, page 84
First level the lever to the balanced position, indicated by tallies, with the
jockey at zero position
Allow water into the apparatus by opening the supply valve.
Increase the supply to maximum, and balance the lever by moving the
jockey. Record the distance on the lever scale
Record the mass flow rate using the water bench and stopwatch method
Decrease the flow rate slightly, and rebalance with the jockey.
Record around 10 points each with lower flow rates
Switch out the plate or cup. And repeat the same amount of measurements,
stating at the maximum flow and incrementally decreasing.
The procedure give us the ability to calculate upward force, by finding the amount
of torque applied by the jockey, to balance the lever.
F So=m g y
Where
F= Impact force ( N)
So = distance from center of vane to pivot of lever : 0.150 m
provided in manual
Equipment used:
m= mass of jockey : 0.600 kg provided in manual
g = gravitational acceleration : 9.81 m/s2
y = distance of jockey from pivot of lever. measured value
Stop
measurement).
Flat plate and hemispherical plate (supplied in lab)
Jet Impact machine (see page 80 in lab manual); make: TecQuipment, model:
H8 Impact
Hydraulics bench (see page 21 in lab manual); make: TecQuipment, model:
h1 Mk1V.
Thermometer.
watch
(make
sure
the
precision
is
consistent
for
each
time
Results
Diameter of nozzle
=
10mm
Cross-Sectional area of nozzle, ao
=
78.5mm2
Mass of Jockey weight
=
0.600kg
Distance from center of vane to pivot of lever=
0.150m
Table 1 : Data for hemispherical cup
balanc
tes e mass
t
(kg)
resultant
water
mass
(balance
3X) (kg)
time
(s)
distan
ce (m)
18
34.07
0.200
18
34.12
0.186
18
36.25
0.166
18
38.60
0.148
18
40.15
0.134
12
28.35
0.122
12
32.32
0.100
12
35.65
0.078
9
10
4
2
12
6
50.12
28.36
0.050
0.036
W,
mass
rate
uo
Wuo
F
(kg/s)
u (m/s) (m/s)
(N)
(N)
0.5283 6.7302 6.6790 3.5286
24
43
32
93
7.848
0.5275 6.7203 6.6690 3.5182
7.298
5
8
93
79
64
0.4965
6.2709 3.1138
6.513
52 6.3255
84
68
84
0.4663 5.9403 5.8823 2.7430
5.807
21
98
15
48
52
0.4483 5.7110 5.6506 2.5332
5.258
19
68
28
83
16
0.4232 5.3921 5.3280 2.2552
4.787
8
07
5
59
28
0.3712 4.7297 4.6566 1.7289
87
72
13
41
3.924
0.3366 4.2879 4.2071 1.4161
3.060
06
73
38
48
72
0.2394 3.0500 2.9352 0.7027
25
05
73
79
1.962
0.2115 2.6951 2.5645 0.5425
1.412
11
2
Example Calculation:
51.55
0.010
66
0.1163
92
For the hemispherical cup, row one:
u=
u=
W
103ao
0.528324 kg/ s
=6.730243 m/s
10378.5106
Velocity striking cup:
u o =u 2 gs
u o =u 29.81 m/s 0.035 mm
u o= (6.730243 m/s)229.81 m/s 0.035 mm
u o=6.679032m/ s
Force Upward
F So=m g y
F=
m g y
So
F=
0.600 kg 9.81 m/s 0.200 m
0.035m
F=7.848 N
03
1.4826
99
43
1.2295
1
69
0.1431
05
64
0.392
4
The same calculation procedure is used for the flat plat.
Table 2 : Data for flat plate
tes
t
balanc
e mass
(kg)
resultant
water
mass
(balance
3X) (kg)
time
(s)
distan
ce (m)
18
32.08
0.106
18
34.27
0.100
18
34.66
0.090
18
35.74
0.080
12
28.43
0.070
12
29.99
0.060
12
33.00
0.050
12
38.08
0.040
12
42.13
0.030
10
12
46.98
0.020
11
33.03
0.010
mass
rate(kg/
uo
Wuo
F
s)
u (m/s) (m/s)
(N)
(N)
0.5610 7.1477 7.0995 3.9835
4.159
97
36
37
31
44
0.5252 6.6909 6.6394 3.4873
41
65
51
1
3.924
0.5193 6.6156 6.5635 3.4086
3.531
31
77
72
64
6
0.5036 6.4157 6.3620 3.2041
3.139
37
63
21
52
2
0.4220 5.3769 5.3126 2.2424
2.746
89
34
94
32
8
0.4001 5.0972 5.0294 2.0124
2.354
33
4
29
43
4
0.3636 4.6323 4.5575 1.6573
36
1
87
04
1.962
0.3151 4.0143 3.9278 1.2377
1.569
26
45
83
78
6
0.2848 3.6284 3.5325 1.0061
1.177
33
42
47
85
2
0.2554 3.2538 3.1465 0.8037
0.784
28
58
68
21
8
0.1816 2.3140 2.1605 0.3924
0.392
53
51
87
77
4
Force on Vane vs Rate of Momentum
9
8
7
f(x) = 2.14x
cup
Force on Vane (N)
Linear (cup)
plate
f(x) = 1.08x
Linear (plate)
3
2
1
0
0.5
1.5
2.5
3.5
4.5
W uo (N)
Discussion
From the above graph we can see the force exerted on the vane is proportional to
the rate of momentum of the fluid. Since the trend lines both pass through the
origin, its indicative that our jockey was properly balanced and calibrated.
The equation of force on the flat plate is:
Fp=1.08(W u o)
The equation of force on the cup is:
Fc=2.14(W u o)
If the experiment was done accurately the ratio of the slopes should be equal to, as
the cup is expected to produce twice as much force.
R=
2.14
1.08
R=1.98
The true ratio is 1.98, which is extremely close to the predicted value of 2.
The efficiency of energy exchange can be calculated by comparing the ratio of
measured force, to the theoretical maximum. Evaluated at the 2 nd data point for the
hemispherical cup.
7.29864 N
2 W u0
7.29864 N
6.669093 m
2 (0.52755 kg /s)(
)
s
=1.03
Because we did not account for temperature when assuming water density, we
have an inaccurate mass rate, giving our efficiency an impossible value. If this was
corrected, the efficiency would be below 1, as all real systems never transfer 100%
of its energy.
Questions for discussion
1. To improve the accuracy of the measurements, it is suggested that the jockey
be weighed rather than assume the given weight. Also the addition of an
automatically adjusting jockey and calibration device, would help eliminate
human error in measurements.
2. a) The effect of 0.001kg less than actual would make the calculated force
greater and efficiency greater. The opposite is true if its mass was less than
actual by 0.001kg.
b) If the distance from center of vane to pivot of lever in error was 1mm
smaller, it would increase efficiency, and decrease it if the distance was
larger by 1mm.
c) If the measured diameter was smaller by 0.1mm the cross section error
would decrease, exit velocity would increase, and theoretical max force would
increase, resulting in a decreased efficiency. Opposite if error is greater than
actual.
Conclusion
In this experiment we proved the flat plate will have half of the energy transfer of a
spherical cup. Our experimental results agree with the theory. There was a slight
difference in actual and theoretical values and an efficiency greater that unity,
which is likely due to human error. We observed there was energy loss since the jet
was decelerated from the force of gravity.
Work Cited
Milton, Dean. Engineer Fluid Mechanics. Regina: University of Regina, n.d.
Enev 261
Lab #3 Impact of Jet
Section: 093
Alexander Calkins
SID: 200347607
