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Impact Jet

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Impact Of a Water Jet

• Introduction:

Over the years, engineers have found many ways to utilize the force that can be imparted by a jet of fluid on a surface diverting the flow. For example, the pelt on wheel has been used to make flour. Further more, the impulse turbine is still used in the first and sometimes in the second stages of steam turbine. Firemen make use of the kinetic energy stored in a jet to deliver water above the level in the nozzle to extinguish fires in high-rise buildings. Fluid jets are also used in industry for cutting metals and debarring. Many other applications of fluid jets can be cited which reveals their technological importance. This experiment aims at assessing the different forces exerted by the same water jet on a variety of geometrical different plates. The results obtained experimentally are to be compared with the ones inferred from theory through utilizing the applicable versions of the Bernoulli and momentum equations.

• Objectives:

1. To determine the force produced by a water jet when it strikes a flat vane and a hemispherical cup. 2. To compare the results measured with the theoretical values calculated from the momentum flux in the jet.

• Theory:

For the general case shown in figure (1) the momentum flux in the jet is (muo)
Where:
m is the mass flow rate
Uo is the jet velocity just upstream of the vane.
After being deflected through the angle (, the momentum flux is (mu1cos() in the x-direction. The force on the fluid is therefore (mu1cos(-mu() in the x-direction. Thus the force F in the x-direction on the vane is:
F = m (uo-u1cos() (1)
Now in the case of the plate figure 3a (=90( so cos(=0 and equation (1) reduces to:
F=muo
For a hemispherical cup, figure 2b (=180( so cos( = -1 and equation (1) reduces to:
F=m (uo-u1)
Furthermore if there is a negligible reduction of a speed so that (u0=u1) then
F=2muo
In the experiment it is not possible to measure directly just upstream of the vane. However the velocity u at the exit of the nozzle can be determined. The velocity uo is somewhat less this due to declamation caused by gravity and can be calculated from the Bernoulli equation that is:
(U^2/2g)+(Z)+(P/(g) = (u0^2/2g)+(Zo)+(Po/(g)
Now from figure 2a Z=0 Po=P Zo=s and this yields to: u^2= uo-2gs which can be written as: uo^2=u^2-2gs where s is the distance between the nozzle exit and the surface of the valve. In order to calculate the force on the valve due to the jet we take the moment about the pivot of the weighing beam figure 3 and substitute known values into the equation we get:
F*0.1525 = 0.610*g*(X
Or
F = 4g(X • Apparatus: Hydraulic bench, water jet apparatus, stopwatch. The water is supplied to the jet apparatus in a closed loop by a pump. The flow rate is determined with the use of a weighing tank, and a stopwatch. The water issues vertically upwards into the air, through a nozzle. Two objects are available: ( A flat plate ( Hemispherical cup

Each object can be mounted on a horizontal lever above the water jet and receive its impact. The force on the object can be determined with the use of weights that can be hung at different positions on the lever, see figure 4.

• Procedure: 1. Stand the apparatus on the hydraulic bench, with the drainpipe immediately above the hole leading to the weighing tank, see figure 4. Connect the bench supply hose to the inlet pipe on the apparatus, using a hose-clip to secure the connection. 2. Fit the flat plate to the apparatus. If the cup is fitted, remove it by undoing the retaining screw and lifting it out, complete with the loose cover plate. Take care not to drop the cup in the plastic cylinder. 3. Fit the cover plate over the stem of the flat plate and hold it in position below the beam. Screw in the retaining screw and tighten it. 4. Set the weigh-beam to its datum position. First set the jockey weight on the beam so that the datum groove is at zero on the scale, figure 5. Turn the adjusting nut, above the spring, until the grooves on the tally are in line with the top plate as shown in figure 6. This indicates the datum position to which the beam must be returned, during the experiment, to measure the force produced by the jet. 5. Switch on the bench pump and open the bench supply valve to admit water to the apparatus. Check that the drainpipe is over the hole leading to the weighing tank. 6. Fully open the supply valve and slide the jockey weight along the beam until the tally returns to record the reading on the scale corresponding to the groove on the jockey weight. 7. Measure the flow rate by limiting the collection of 30Kg of water in the bench-weighing bank. 8. Move the jockey weight inwards by 10 to 15mm and reduce the flow rate until the beam is approximately level. Set the beam to exactly the correct position (as indicated by the tally) by moving the jockey weight, and record the scale reading. Measure the flow rate. 9. Repeat step 8 until you have about 6 sets of readings over the range flow. For the last set, the jockey should be set at about 10mm from the zero position. At the lower flow rates you can reduce the mass of water collected in the weighing tank to 15Kg. 10. Switch off the bench pump and fit the hemispherical cup to the apparatus using the method in steps 2 and 3. Repeat step 4 to check the datum setting. 11. Repeat steps 5 to 9, but this time move the jockey in steps of about 25mm and take the last set of readings at about 20mm. 12. Switch of the bench pump and record the mass m of the jockey weight, the diameter d of the nozzle, and the distance s of the vanes from the outlet of the nozzle.

• Technical data:

Mass of jockey weight: m = 0.610Kg

Distance from centerline of vane to weight-beam pivot: 0.1525m

Diameter of nozzle: d = 0.01m

Height of vane above nozzle outlet: s = 0.04m

Diameter of hemispherical cup: 0.06m

• Tables- Data and Results:

Table 1: Results for Flat Plate

|QTY |T |(X |m |u |uo |muo |F |
|(Kg) |(Sec) |(m) |(Kg/S) |(m/s) |(m/s) |(N) |(N) |
|7.5 |16.52 |0.06 |0.454 |5.78 |5.71 |2.59 |2.35 |
|7.5 |18.35 |0.05 |0.409 |5.21 |5.13 |2.10 |2.96 |
|7.5 |20.16 |0.04 |0.372 |4.74 |4.66 |1.73 |1.57 |
|7.5 |22.77 |0.03 |0.329 |4.19 |4.10 |1.35 |1.17 |
|7.5 |27.76 |0.02 |0.270 |3.44 |3.32 |0.90 |0.78 |
|7.5 |33.33 |0.01 |0.225 |2.86 |2.72 |0.083 |0.392 |

Table 2: Results for Hemispherical Cup

|QTY |T |(X |m |U |uo |muo |F |
|(Kg) |(Sec) |(m) |(Kg/S) |(m/s) |(m/s) |(N) |(N) |
|7.5 |17.29 |0.12 |0.434 |5.53 |5.46 |2.37 |4.74 |
|7.5 |18.63 |0.10 |0.403 |5.13 |5.05 |2.04 |4.08 |
|7.5 |19.79 |0.08 |0.379 |4.83 |4.75 |1.80 |3.60 |
|7.5 |22.77 |0.06 |0.329 |4.19 |4.10 |1.35 |2.70 |
|7.5 |27.09 |0.04 |0.277 |3.53 |3.42 |0.95 |1.90 |
|7.5 |37.93 |0.02 |0.198 |2.52 |2.36 |0.47 |0.94 |

• Sample Of Calculations:

1. m = mw/T [ in table 1: 7.5/16.52=0.454 Kg/S] [ in table 2: 7.5/17.29=0.434 Kg/S]

2. m = ρuA ⇒u= m/ρA ρA= 1000*(/4*0.001=0.0785 [in table 1: 0.454/0.0785=5.78m/s] [in table 2: 0.434/0.0785=5.53m/s]

3. From Bernoulli’s equation: (u^2/2g)+(Z1)+(P1/ρg)= (uo^2/2g)+(Z2)+(P2/ρg) Z1=0 Po=P Z2=s=40mm ⇒ uo^2= u^2-2gs [in table 1: (5.78^2)-(2*9.81*.04)]^2=5.71m/s [in table 2: (5.53^2)-(2*9.81*.04)]^2=5.46m/s

4. By summing the moment around the pivot in figure (3) we obtain:

Fw*0.1525-m1g(X=0 ⇒ Fw=4g(X

5. For flat plate:

F = m (uo-u1cos() Where (=90(

⇒ F = muo [so F for (X = 0.06m = 2.59N]

6. For the hemispherical vane:

F = m (uo- u1 cos() Where (=180(

⇒ F = m (uo+ u1) ⇒ F = 2muo [so F for (X = 0.012m = 2*2.37=4.74N]

• Source of Errors:

1. Turning the adjusting nut above the spring until the grooves on the tally are in the line with the top plate as shown in figure 6. 2. Recording the reading on the scale corresponding to the groove on the jockey weight. 3. Starting timer and adding weights when beam moves to horizontal. 4. Stopping timer when beam moves to horizontal again.

The values of F theoretical (calculated from 4g(x) are close to those found experimentally. So we connect these points with a straight line.

Also from this graph we see that the calculated F (4g(x) is equal to the double of mu ( 2mu

• Conclusions:

1. The results obtained theoretically are close to those obtained experimentally.

2. Accuracy = (muo-4g(X /4g(X) *100%

For flat plate:

(2.59-2.35/2.35)=10.2% (2.10-1.96/1.96)=7.14% (1.73-1.57/1.57)=10.2% (1.35-1.18/1.18)=14.4% (0.9-0.78/0.78)=15.4%

For hemispherical cup:

(4.74-4.71/4.71)=0.64% (4.08-3.92/3.92)=4.08% (3.6-3.14/3.14)=14.6% (2.7-2.35/2.35)=14.9% (1.90-1.57/1.57)=21.0% (0.94-0.78/0.78)=20.5%

3. If the line didn’t pass through the origin that means that there is an error, because if the force is zero ( the jet doesn’t touch the vane) the should be placed at the origin which means (X=0 so F=0

4. F = m (uo = u) u ( uo because we neglect reduction of speed so that u=uo fo = 2muo but the force on the hemispherical cup less than twice that on the flat plate.

5. to find the thickness of the water film at the circumference of the hemispherical case: mo = (.uo.A m1 for the water film is equal to (.u1.A where A = 2(rt and u0 = u1 r = 0.03m and the mass flow is constant then t = ((*0.01*0.01)/(8*(*0.03)= 0.4167mm

6. The effect on the calculated force on the flat plate if the jet was assumed to leave the plate at 1( upward will be a moment in the x-direction which will decrease the moment in the y-direction F=m (1.99uo) and it won’t effect the results too much.

7. a. The effect will be:

F = [(0.610 ( 0.001) g(X]/0.1525

b. The effect will be: F = (0.610*g(X)/(01525(0.001)

c. The effect will be: F = m [{(m)/( ((¼(D(0.0001)²)}² -2gs]^1/2

8. If the velocity in the jet wasn’t uniform we will divide the jet into a two separate regions in which we solve for each one separately.

♣ in our experiment we have learned lots of things, we have had a close look on how the hydraulic bench and how a water jet work. We also have learned to calculate the force caused by a water jet experimentally using different sets of data and also learned how to calculate the mass flow rate and speed of water using the mass of weights and their (X using time intervals measured experimentally.

♣ there is for sure a difference between our results and the theoretical ones because of different sources of errors that have been discussed earlier in this report.

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Jet Copies

...completed only one simulation, they might have received a value of $9,870 and determined that they didn’t need to purchase a back up copier. If the actual results generated something similar, they would have made out better than they would have if they chose to purchase a backup copier. Likewise, if they had used the simulation above and any of the component variables behaved differently than modeled, they could have lost a substantial amount of revenue. In many cases large losses of revenue can have a detrimental impact on a business both in reputation and income used to service debt. The potential of variance in each of the simulated variables causes some risk in the calculation that is not captured fully through the use of a single simulation. It is therefore good business practice to utilize several simulations that will better incorporate variances in the component variables and provide an average of all observed possibilities. Based on my simulations, JET Copies would chose to purchase a backup copier because the average revenue lost was above $12,000. This is likely a good choice as having a backup copier will insult the company somewhat from more dire scenarios. Even using the number of simulations calculated in my example (50), there were some variables that when 2 standard deviations were added to the mean the result was a value outside minimum and maximum values. This would signify that the values didn’t likely fall within a normal distribution all each of......

Words: 1407 - Pages: 6

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