Experiment #4

Impact of a Jet of Water

Stephen Mirdo

Performed on September 30, 2010

Report due October 7, 2010

Table of Contents Object ………………………………………..………………………….………….…. p. 1 Theory ………………………………………………………………..……….....…pp. 1 -3 Procedure …………………………………………………………………….………...p. 4 Results ………………………………………………………..……...……………pp. 5 - 6 Discussion and Conclusion …………………………………………………….………p. 7 Appendix ……………………………………………………..…….…....………pp. 8 - 10

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Object The object of this experiment was to determine the force exerted by a jet of water on a stationary vane and compare the experimental results to the theoretical results.

Theory

A moving stream of fluid carries momentum. Momentum is defined as the product of the mass of a body and its velocity. When a moving stream of fluid, or jet, is deflected by a surface, a change in linear momentum occurs. This change in linear momentum results in the water jet exerting a force on the surface it is impacting.

ΣF = ∫∫CS V ρ Vn dA (Equation 1)

Equation 1 is the general equation for linear momentum for a deformable control

volume. The components of this equation are V, the velocity of the stream of fluid, ρ, the density of the fluid, and Vn, the velocity of the fluid normal to the impact surface. These components are integrated with respect to the impact surface area and yield the force the jet exerts.

To calculate the theoretical force a jet exerts on a surface, first a control volume

must be defined. For this experiment, a Pelton bucket will be used for the vane. A diagram of the force exerted on the Pelton bucket by the jet of fluid is seen in Figure 1 below.

Figure 1: Diagram of Pelton bucket, reaction forces and angle of velocity of

escaping water.

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To calculate the force the jet of fluid exerts on the Pelton bucket, a force sum calculation is used. A control volume is defined such that the fluid exits at 90 degrees, normal to the defined control volume. This definition of the control volume allows for no fluid to escape in the y direction, thereby simplifying the force calculation. It is assumed that to the left is positive on the x axis. The velocity of the fluid escaping the Pelton bucket is calculated as follows:

V = Vcos(19o) = V(cos161o) (Equation 2)

The force balance equation for this system, with to the left on the x axis being positive is as follows:

ΣFx = Fwater – ρAV(Vout – Vin) = 0 (Equation 3)

Simplifying the equation by factoring the velocity term and substituting Equation 3 into the force equation for the velocity of the escaping jet of fluid acting in the x direction is as follows:

ΣFx = Fwater – ρAV2(1 – cos(161o)) = 0 (Equation 4)

The force the Pelton bucket experiences due to the change in momentum of the jet of water is equivalent to the force of the water. Therefore, the equation for the theoretical force of the jet of fluid exerted on the Pelton bucket is as follows:

F = ρAV2(1 – cos(161o)) (Equation 5)

To calculate the area term, A of Equation 5, calculate the cross sectional area of

the exit of the nozzle. The V term, velocity, is the fluid jet’s velocity at the exit of the nozzle. To calculate the exit velocity, divide the volumetric flow rate of the nozzle, Q, by the cross sectional area A.

Q = VA V = Q / A (Equation 6)

A method of determining the actual force exerted by a stream of fluid is to

evaluate the moment the force exerts on a beam. As seen in Figure 1 below, a nozzle emits a jet of fluid at a constant volumetric flow rate that impacts the vane. The vane is connected to beam that will rotate about a pivot as seen to the left on the beam. The impacting jet of fluid, having its direction of momentum altered, will exert a force on the vane at a fixed distance from the pivot, do, and cause the beam to rotate counterclockwise.

To assess the force exerted by the jet, F, a mass jockey is positioned on the beam so as to counteract the jet’s force with a force that is equal in magnitude and opposite in direction. The mass jockey will exert a force due to weight, Fw, in the downward direction at a distance, dw, from the pivot, thereby exerting a clockwise moment about the

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pivot on the beam. The tally, or leveling device, of the apparatus will register in the neutral position when the mass jockey is positioned at the appropriate distance from the pivot to counteract the counterclockwise moment exerted by the jet of fluid.

Figure 2: Jet impact testing apparatus diagram

(Source: University of Delaware, Department of Mechanical Engineering)

Any subsequent increase in the volumetric flow rate, Q, will cause a proportional

increase to the impact force the jet exerts on the vane. The increased force equates to an increase in the moment exerted by the jet around the beam’s pivot. To counteract the increased force, and therefore increased moment, the mass jockey must be shifted further to the right on the beam to a position dwi to counteract the increase. To account for the force at each flow rate, multiply the weight of the mass jockey by the difference in each displacement with reference to the initial position and divide the product by the distance from the fluid jet to the pivot.

F = [ Fw (dwi – dw1) ] / do (Equation 7)

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Procedure Equipment: TQ H1D Volumetric Bench (SN: A7771/2)

Pelton Bucket Plate Stopwatch Experiment:

1) Close the drain in the catch basin of the jet impact apparatus. Open the flow control valve, turn the pump on and set the flow rate to a desired value. Using the sight glass on the side of the device, measure the time taken for the volume in the catch basin to rise from zero to an arbitrary volume. Divide this volume by the time taken to achieve it. This value is a flow rate, Q.

2) Obtain a theoretical velocity by dividing the flow rate Q by the cross-sectional area of the nozzle. This obtained value is the theoretical velocity of the fluid exiting the nozzle.

3) Use the theoretical velocity to calculate a theoretical moment exerted on the beam by the jet.

4) Turn off the pump of the apparatus. Ensure that the flow control valve has been closed and open the drain of the catch basin so as to let the fluid in the system return to the pump.

5) Move the mass jockey to the zero position and calibrate the lever arm of the testing apparatus by adjusting the spring via the thumb screw.

6) Turn the pump back on and open the flow control valve. Adjust the flow rate until a deflection of the lever arm is noticed. Leave the pump running and move the mass jockey to the right until the moment created by the water jet has been balanced. Record this distance. Do not adjust the spring.

7) Increase the flow rate of the water jet by an increment of two liters per minute. Again, slide the jockey mass to the right until the moment has been balanced. Record this value.

8) Repeat Step 7 for a desired number of data points.

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Results

Table 1: Known values used to determine theoretical and actual values of force exerted by a jet of water

Jockey Mass 0.600 kg Nozzle Diameter 0.010 m Density of Water 1000 kg/m3 Distance from Pivot to Vane 0.150 m

Table 2: Theoretical values of the force exerted by a jet of water

obtained by using Equation 2 and Equation 3. Volumetric

Flow (LPM)

Volumetric Flow (m3/s)

Velocity (m/s) Force (N)

10 1.67E-04 2.12 0.6054 12 2.00E-04 2.55 0.8718 14 2.33E-04 2.97 1.1866 16 2.67E-04 3.40 1.5499 18 3.00E-04 3.82 1.9615 20 3.33E-04 4.24 2.4217 22 3.67E-04 4.67 2.9302 24 4.00E-04 5.09 3.4872 26 4.33E-04 5.52 4.0926 28 4.67E-04 5.94 4.7465 30 5.00E-04 6.37 5.4487

Table 3: Experimental values of the force exerted by a jet of water obtained by

using Equation 2 and Equation 4. Volumetric

Flow (LPM)

Volumetric Flow (m3/s)

dw (mm)

dwi-dw1 (m)

Velocity (m/s) Force (N)

10 1.667E-04 12 0.012 2.12 0.4709 12 2.000E-04 19 0.007 2.55 0.2747 14 2.333E-04 26 0.014 2.97 0.5494 16 2.667E-04 35 0.023 3.40 0.9025 18 3.000E-04 45 0.033 3.82 1.2949 20 3.333E-04 58 0.046 4.24 1.8050 22 3.667E-04 72 0.06 4.67 2.3544 24 4.000E-04 88 0.076 5.09 2.9822 26 4.333E-04 100 0.088 5.52 3.4531 28 4.667E-04 113 0.101 5.94 3.9632 30 5.000E-04 130 0.118 6.37 4.6303

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0.0000

1.0000

2.0000

3.0000

4.0000

5.0000

6.0000

10 15 20 25 30 35

Flow Rate (LPM)

Forc

e (N

)

Actual Theoretical

Figure 3: Theoretical and actual force exerted by a jet of water on a Pelton bucket

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Discussion and Conclusion

The Pelton bucket is a component of the Pelton wheel that is found in impulse turbines. The raised ridge between the two hollows of the device essentially splits a jet of water into two and deflects each nearly 180o. This deflection prevents the jet from spraying out of and away from the device. By controlling the velocity vector of the fluid jet, the Pelton bucket is able to extract more energy from the moving fluid by changing its linear momentum. The fluid that leaves the Pelton bucket has a small fraction of the velocity it came in with, concluding that it has a very efficient design. It was that the theoretical and experimental forces have a significant percentage of error between them, as seen in Table 4 below. Most of this error is due to the theoretical calculations neglecting the force of gravity on the jet of water. When the jet leaves the nozzle, it will have the calculated velocity obtained by using Equation 3. However, after the fluid has obtained any height above the nozzle, the force of gravity acts on it and decreases the velocity. The reduction in velocity can be determined by using Bernoulli’s equation, where V is the respective velocity, g is gravitational acceleration and Δz is the distance between the nozzle and the vane.

V2vane = V2

nozzle – 2gΔz

Table 4: Percent error of experimental and theoretical force of a jet of water Experimental

Force Theoretical

Force % Error

0.4709 0.6054 22.22% 0.2747 0.8718 68.49% 0.5494 1.1866 53.70% 0.9025 1.5499 41.77% 1.2949 1.9615 33.98% 1.8050 2.4217 25.46% 2.3544 2.9302 19.65% 2.9822 3.4872 14.48% 3.4531 4.0926 15.63% 3.9632 4.7465 16.50% 4.6303 5.4487 15.02%

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Appendix Data Usage Sample Calculation of Velocity at 10 LPM

V = Q/A = [(10 LPM * 1.5e-2 m3/L * min/60 s)] / [(π/4) * (0.010m)2] = 2.12 m/s

Sample Calculation of Experimental Force at 12 LPM

F = 0.600 kg * 9.81 m/s2 * (0.007 m) / 0.15 m = 0.2747 N

Sample Calculation of Theoretical Force at 20 LPM

F = 1000 kg/m3 * (π/4) * (0.010m)2 * (4.24 m/s)2 * (1-cos(161o) = 2.4217 N

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Bibliography

Introduction to Fluid Mechanics, 3rd Edition William S. Janna (1993)

Fundamentals of Material Science and Engineering: An Integrated Approach

W.D. Callister, Jr and D.G. Rethwish (2008)

A Manual for the Mechanics of Fluid Laboratory William S. Janna (2008)