Abstract
When a body is in a state of motion, the total momentum is maintained, such that the product of mass and the velocity vector, until that moment when an external force is applied to it. This law is called the conservation of linear momentum, this principle somewhat corroborates the conservation of energy principle which states that energy is neither created or destroyed, but can be transformed from one form to the other, like mechanical energy transformed to electrical energy etc. When a body is in a motion is possesses an energy within itself, but with an impact or when there is a change of the course of this motion, there’s often energy change (The Editors of Encyclopaedia Britannica, 2019). A falling body has an amount of energy that is constant, this energy changes from a potential to kinetic energy after impact, but the theory of relativity equates energy and mass. Still at the law of conservation of momentum, we know that the universe is not impacted by any external motion, therefore its momentum is conserved. As a result, its components that are in any direction are conserved as well. Applying this law is relevant to solving problems that goes with collisions. Like operations of rockets or the ones that creates energy like the Pelton Wheel Turbine, where a high moving water moves and impacts the cupped shapes of the wheel to create movement, which in turns results into electricity after transfer of energy to the gearboxes or motors.
This paper specifically deals with the impact of a high-speeding water on different shapes and how the movement is at the point of impact. There are different shapes, the cupped, conical, and the flat shape being used for the experiment.
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Introduction
Water turbines are important and is used throughout the world today, especially for power generation. The vanes of the turbines are subjected to a Pressurized fluid that strikes them into rotation and mechanical work by the turbines begins from this impact. In simple terms, the impact that is caused by water on the vanes, creates a torque on the wheel which in turns makes it to rotate and power is developed. This force exerted on the surface plane of the vanes is dependent on the fluid density at a given temperature. Such that the force generated also determines the mass flow-rate of the fluid. The Pelton wheel produces an output that is expressed easily where its optimum speed of rotation can be obtained. It therefore becomes possible to understand the jet’s deflection and how it causes power at the vane and the force that relates to the rate of momentum flow that is in the jet. The vane is propelled from the due to the shape of the surface that the water hits. It is in this that the experiment is conceived, where the experiment determines how different, if any, is the force created when water hits the surface. In the experiment there are different shapes that are subjected to a speeding water. In the experiment there are three types of geometrical shapes, which are conical, cupped-shape, and a flat surface, and so the aim is to assess the different forces by the same fluid, water jet. The reaction force that hits the objects shall be determined and compared and a conclusion drawn. The instrument to be used shall be impact of jet apparatus. The experiment also helps us determine and understand how the forces can affect the change of the flow of momentum in a jet flow. Another lesson is understanding the working principle of turbines in water when subjected under the fluid pressure.
Theory
When water flows with a stable velocity and hits a solid surface, there will be a deflection which takes the pattern of that shape, since it tends to flow along that surface. Assuming the fluid is inviscid and therefore there is no frictional force, another assumption is that there is no loss caused by shocks, the water velocity magnitude would be not be changed. The water that is exerted at the surface hits the surface at right-angle. Consider a jet of water under pressure impacting a solid surface causing the water to go at and angle θ. Assuming the frictional force is negligible or there’s non, the velocity magnitude across the surface is equal to the incident velocity Vi. The water force that impacts the surface is equal to the force of water after the impact that moves in the opposite direction.
Based on the second Newton’s Law, also known as the momentum equation, states that the sum of external forces that are exerted to control the volume of the fluid moving in any given direction is equal to the rate at which the momentum changes in those directions. The external forces referred to here are the weight of the component of the fluid and the exerted forces acting externally on the boundary surface that controls the volume. The impact of jet apparatus enables a vertical movement of water at a velocity V, and is directed towards a target, a free vane moving in a vertical direction, the force that the is made to impact on the target of the jet.
The Newton’s second law that moves in vertical direction, which is the incident jet
So, the resultant force that acts on the fluid in motion’s direction, shall be found as shown in (1)
Where: is the density, Q is the flow rate, is the velocity of the fluid after the impact, and is the velocity of the fluid before the impact.
The force F in the equation, comprises of the force exerted in the direction of the fluid by a solid body that touches the volume control, that is , force applied in the direction of the fluid by the force of the body, like gravity etc. , and the force that is applied in a given direction of the pressure of the fluid that is not in the volume control, .
The third law of Newton states that for every action force exerted on a body, there’s a reaction force acting on the opposite of that force (Lucas, 2017). The fluid exerts an equal reaction force that acts on the opposite direction to the surrounding. This means that the force that the fluid exerts on the solid material touches the volume control shall be equal and in the opposes , this force shall be expressed as:
When (1) is substituted, the expression shall be:
Which shall be simplified as:
The impact jet apparatus
This apparatus is able to have different shapes of deflectors or where water impacts, and investigates the jet forces upon these targets. When using the apparatus in figure 1, it is assumed that there’s no fluid splashing or there’s no rebound of the fluid, in this case water, from the surface such that the angle to which water exits is parallel to the angle the water exits at the target. The apparatus has an upward discharging jet which is inside a clear glass, cylindrical in shape called Plexiglas cylinder. There are screws at the bottom of the apparatus, they are called levelling screws. There’s a weight pan on top of the apparatus is attached at the spindle’s top and is passing through the lid of the cylinder through to the plate which is above the water jet exit. Water goes through the inlet from the lab spout through a hose. This water that oozes from the nozzle hits the surface (any surface that is placed) and get deflected downwards at the clear Plexiglas cylinder’s base, where it then drains out to the sink.
Figure 1:Water jet apparatus and its parts
The apparatus standard dimensions are as follows:
(4) Nozzle 10 mm,
(4) Nozzle’s cross-sectional area, A, 78.5 mm2,
(1) The jokey weight (loading weight), w = 0.6 kg,
Vane to pivot distance 0.15 m,
Height of the vane which is at above the nozzle exit, s =35 mm.
Figure 2:The pictures of the apparatus
Once the flow rate has been determined and the area of the nozzle from which water exits, the jet velocity can be calculated as follows:
According to the arrangement on the apparatus, the nozzle is below the target (which is replaced each time), the velocity of the impact will be less than the velocity of the nozzle since there will be interchange between the kinetic and potential energy, this means:
Where is the target’s height from the nozzle below it.
Considering the different targets, where there is the normal flat surface, hemispherical, and conical shapes. There shall be theoretical calculations of the jet force based on the shapes of the targets and the principle of this calculation shall be based on linear momentum. The density of water is 1kg/m2, and the force equation is as equation (1).
(i. For Flat surface we have:
The following parameters shall be considered: , that means
(ii. Hemispherical surface we have:
The change of parameter here becomes , therefore,
(iii. Conical surface we have:
In the case of the conical shape, we have the angle being 45O, therefore, we have , but
The procedure of the experiment
- Ensure that the apparatus is at a level, so that you can set the lever to a position that is balanced, the balancing is achieved by adding weight until the target’s top is clear. Ensure that the platform is at a mid-position.
- Move the pointer to be aligned with the platform weight and record the value of the weight carrier.
- Pump the water so that water flows in steadily and fully open the regulating valve
- There will be a deflection due to the impact of the jet. While this is happening, add weights on the carrier till the platform floats in the mid position.
- Now measure and record the results of the flow rate, alongside the weights that corresponds to those values. Observe the nature and the shape of the water getting deflected from the solid.
- Take off the weight on the weight carrier piece by piece to maintain the balance of the platform by controlling the flow rate in three steps, thereabout, each time record the values of the flow rate alongside the weight on the carrier.
- Close the regulating valve and put off the pump allowing the apparatus to drain.
- Replace the normal vane with hemispherical, 45o conical, and a flat vane and repeat the test with the nozzle dimeters and record the values.
Results
The results shall be recorded in the sheets from table 1 to table 3, remember we have the parameters of the nozzle identified above, which are:
Nozzle diameter and its cross-sectional area are 10 mm or 0.001m and 0.0007855m2 respectively, and Height of the vane which is at above the nozzle exit, s =35 mm, and the vane to pivot distance 0.15 m. From the considered parameters in the table, to calculate the forces we need to calculate the velocity of the nozzle of each shape, remember equation (3)
The density of water is 1kg/m3,
Flat surface:
Flow rate in a mass of 0.01808 kg was measured to be 0.000659 m/s, therefore,
We know that the velocity of the water jet that strikes the vane is as below:
Where h, is the height and g are the gravitational acceleration given as 9.81m2/s
Also, remember that;
The second force shall be calculated as:
Area
Distance
mass(g)
t (s)
Q (m3/s)
V(nozzle)
V(in)
F1
F2
0.0007855
0.15
0.01808
27.44
0.000659
0.838819
0.016917
-0.0000111464
0.026605
0.0007855
0.15
0.01335
33.6
0.000397
0.50582
-0.43085
0.000171184
0.019645
0.0007855
0.15
0.00978
40.29
0.000243
0.309026
-0.5912
0.000143509
0.014391
0.0007855
0.15
0.01195
48.87
0.000245
0.3113
-0.58979
0.00014422
0.017584
For the hemispherical surface
Flow rate in a mass of 0.01496 kg was measured to be 0.000659 m/s, therefore,
Where h, is the height and g are the gravitational acceleration given as 9.81m2/s
The second force is calculated as:
Area
Distance
mass(kg)
t (s)
Q (m3/s)
V(nozzle)
V(in)
F1
F2
0.0007855
0.15
0.01496
23.44
6.38X10-7
0.000813
-0.6867
8.76538X10-7
0.022014
0.0007855
0.15
0.01345
39.65
3.39X10-7
0.000432
-0.6867
6.5882X10-7
0.019792
0.0007855
0.15
0.00987
47.56
2.08X10-7
0.000264
-0.6867
2.85018X10-7
0.014524
0.0007855
0.15
0.00963
91.31
1.05X10-7
0.000134
-0.6867
1.44845X10-7
0.014171
For the conical surface
The flow rate in a mass of 0.01513 kg was measured to be 0.000762217 m/s, therefore,
Where h, is the height and g are the gravitational acceleration given as 9.81m2/s
Second force
Area
Distance
mass(g)
t (s)
Q (m3/s)
V(nozzle)
V(in)
F1
F2
0.0007855
0.15
15.13
19.85
0.0007622
0.9703585
0.2548957
-0.0001943
22.2637950
0.0007855
0.15
13.94
26.06
0.0005349
0.6809923
-0.2229495
-0.0001193
20.5127100
0.0007855
0.15
11.81
43.06
0.0002743
0.3491642
-0.5647844
0.0003098
17.3784150
0.0007855
0.15
8.37
84.44
0.0000991
0.1261918
-0.6707756
-0.0000665
12.3164550
We find the average force and the flow rate then compare in a chart, we have
For flat we have:
Q (m3/s)
F
0.000658892
0.013296787
0.000397321
0.009907855
0.00024274
0.007267389
0.000244526
0.008864322
For hemispherical surface (cup) we have
Q (m3/s)
F
0.00000063823
0.011007258
0.00000033922
0.00989607
0.00000020753
0.007261995
0.00000010546
0.007085345
For conical surface we have
Q (m3/s)
F
0.0000008
0.0111322
0.0000005
0.0102562
0.0000003
0.0086894
0.0000001
0.0061582
Average shape per shape, we have the following
Surface
Force
Flat
0.009834
Hemi
0.009059
Con
0.008813
Discussions
The three bodies have different geometry, that is why the force exerted from the water jet having impacted by them is different. The main parameter is the flow rate, since the flow rate tells the velocity of water and therefore the force it exerts on the surface. It is also noticeable that the higher the flow rate, the more the force exerted on each body. The force is also dependent on the height from which the jet is and the nozzle, where the shorter the distance, the greater the force, but in this case, we had the same the same distance, and the same diameter. The results also show the efficiency of the shape, in terms of giving greater force, this is shown in the chart, where the conical shaped solid doesn’t give a lot of force compared to the other three shapes. The longer the time, the smaller the flow rate, this can also be seen on the tables shown.
The results obtained shows that the force is dependent on the shape of what it strikes, in that the water striking the cupped-shaped solid diverts less water outside, that is why it was assumed that it takes 180O. It is the water that leaves the surface that determines the force, such that the more this water is diverted the less the force is exerted on the solid. This fact is why the cupped surface performs well in giving force than the conical surface.
Nomenclature
A – cross-sectional area of the nozzle
F – Force (N or kg)
D – distance from the nozzle to the target (m)
G – acceleration due to gravity (m2/s)
– mass flow rate (kg/s)
Q – Flow rate (m3/s)
R – Reaction force (N or kg)
T – time in seconds (s)
Vi – Velocity of input (water) (m/s)
Vn – Velocity at the nozzle (m/s)
– fluid density (kg/m3)
W – weight (kg)
Conclusions
The volumetric flow rate increased as the resulting force caused by the impact on the surface increased, therefore through the experiment, it has been proven that the flow rate is directly proportional to the force resulting from the impact. From the experiment we were able to find out that the different shapes give different magnitude of force. In the experiment the shapes that were used were flat, hemispherical, and conical shaped surface. They had an exposed surface of 90O, 180O, and 45O respectively. From this the hemispherical shape gave the best result in exhibiting force, and the conical surface gave the worst. Therefore, we can say that the larger the angle of water diverted, the more the force exhibited, and the reverse is true. The force exhibited is the reactionary force, therefore, the Newton’s third law of motion has been proven too. We can infer that the diameter of the nozzle also plays a bigger role in the force exerted on the surfaces. Where the nozzle needed to be replaced from the 10 mm to 8 mm and 5 mm diameter and the test is repeated. We also can infer that the best surface to use for a turbine or a water jet engine, should have a cupped surface.
References
- Lucas, J. (2017, September 26). Equal & Opposite Reactions: Newton's Third Law of Motion. Retrieved from Live science : https://www.livescience.com/46561-newton-third-law.html
- The Editors of Encyclopaedia Britannica. (2019). Physics: Conservation law. Retrieved from Encyclopedia Britannica: https://www.britannica.com/science/energy
Appendix
Flat surface:
Flow rate in a mass of 0.01808 kg was measured to be 0.000659 m/s, therefore,
We know that the velocity of the water jet that strikes the vane is as below:
Where h, is the height and g are the gravitational acceleration given as 9.81m2/s
Also remember that;
The second force shall be calculated as:
Hemispherical surface
Flow rate in a mass of 0.01496 kg was measured to be 0.000659 m/s, therefore,
Where h, is the height and g are the gravitational acceleration given as 9.81m2/s
Second force is calculated as:
Conical surface
Flow rate in a mass of 0.01513 kg was measured to be 0.000762217 m/s, therefore,
Where h, is the height and g are the gravitational acceleration given as 9.81m2/s
Second force
Flat Hemi Con 9.8340883026020835E-3 9.058979962070255E-3 8.8126671604147194E-3
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