# Can someone explain the theory of impact of jets?

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1. Mar 18, 2017

### Brenda

So I did this lab on momentum on jets and the theory is that "a force is applied on the plate due to change of momentum" the jet both impacted on a flat plate and on a hemispherical plate; now i understand that the greater the angle of deflection of the water jet, the greater the change of momentum. I also understand that the force acting on the hemispherical plate will be double of that acting on the flat plate. I also understand that the force acting on the plate will be equal to the rate of momentum. Now having calculate that change of momentum i got values for both the plate and the hemispherical shell but the results that I obtained show that the rate of change of momentum for the flat is actually double that of the hemispherical shell which is not what the theory predicted and this is where i get confused as the shell deflect the water by 180 degrees whereas the flat plate deflects it by 90 degrees. I then used this formula Fv = Mjgb/a where Mj is the mass of the jockey weight, g is acceleration due to gravity, b the distance of the jockey to the neutral position and "a" the distance from the neutral position to the pivot. And that force Fv is meant to be the vertical force acting on the fluid due to the jockey. The aim to compare the forces acting on the two plates but I'm not sure how to do when the theory is being defied.
PS: i cheated with my other mates and their results showed a similar pattern

2. Mar 18, 2017

### JBA

Without seeing a clear and complete description the problem(s) and your attempted solution(s) there is no way to evaluate your difficulties.

3. Mar 18, 2017

### Brenda

Introduction
This experiment uses water jets impacting on metal plates to look at Newton’s second law.
Aim
The aim of the experiment is to compare the forces acting on the two metal plates.
Background and Analysis
Newton’s second law of motion states that the force acting on a body is equal to the rate of change of momentum. You may be familiar with the special case , which is only true when the mass is constant.
In this experiment, we have fluid flowing, and so there is not a fixed mass. Newton’s second law, in this case, can be written
where: is the (vertical) force acting on the fluid, is the mass flow rate,
and are the initial and final fluid velocities vectors (positive for upwards, negative for downwards), is a momentum flux.
Suppose the water exits the nozzle with a vertical velocity of , i.e. .
In the case of the flat plate, the water spreads out horizontally; the final vertical velocity is zero, . The force acting on the fluid is . Note the minus sign means that the force acting on the fluid is downwards.
In the case of the hemispherical shell, the water comes off the plate travelling vertically downwards. If we assume that the fluid speed is constant ( ) and only changes direction, then final velocity is ; note the minus sign because the fluid is going downwards. The force acting on the fluid in this case is double that of the flat plate.
Procedure

•  Starting with the flat plate, move the jockey weight 2cm along the lever arm.

•  Open the control valve and change the flow rate until the lever arm is level.

•  Measure the mass flow rate of water by measuring the length of time taken to collect a volume of water.

•  Repeat each measurement moving the jockey weight further down the lever arm in 2cm increments.

•  Each time note the distance of the jockey weight, the mass of water collected, and the time taken (there is a

table at the back of the lab sheet for you to record your measurements).

•  You can repeat each measurement several times to find an average.

•  Repeat the test for the hemispherical shell.

Analysis of Results

 For each measurement, calculate the mass flow rate dot/t

Where: m dot is the mass flow rate
m is the mass of water collected
t is the time of collection
• If you took multiple measurements, find the average mass flow rate (add them up and divide by the number of values)

•  Calculate the force acting downwards due to the jockey weight
Fv = mjgb/a
https://www.physicsforums.com/file:///page2image3816 [Broken]
where:

Fv is the vertical force acting on the fluid
mj = 0.61kg is the mass of the jockey weight,

g = 9.81 m/s^2 is acceleration due to gravity,

a = 152.4 mm is the distance of the neutral position from the pivot,
b is the distance of the jockey weight from the neutral position

I can't insert the image but is that of the setup of the hydraulic bench (jockey and plate part)
Figure 1: The forces acting on the balance beam in equilibrium when (a) there is no force on the plate and (b) the water jet produces a force on the plate and the jokey mass is moved to re-establish equilibrium. Note that in figure a) the torque produced from the spring force, S, cancels the torque from the jockey mass and the pass of the beam.

Presentation of results

 On the same graph, plot against for both sets of results  Find the line of best fit for each set of results

Discussion

How do the lines of best fit compare with the theory above?
What are the sources of error in the experiment? Could these be reduced or eliminated? Quantify the effect of the errors on your results.

Conclusions

Report your best fit equations and summarise the important points that arose in your discussion.

that's the problem I just need to understand why is that the results from the calculation of the average rate of change of momentum for the flat place are double that of the hemispherical plate when the theory says the opposite.

Last edited by a moderator: May 8, 2017
4. Mar 18, 2017

### Nidum

We need that image to understand what the problem is . Did you use the UPLOAD button bottom right of reply box ?

5. Mar 18, 2017

### Brenda

There you go that's the overall set up and the jockey part

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6. Mar 18, 2017

### JBA

I don't see in the problem statement the requirement to calculate the change in momentum, only the forces resulting from it for each case and then plotting those forces. Below is a sample of an experiment essentially identical to yours and equation 4.5 is what relates the impact force to the fluid momentum.
Unless your test measurements indicate that the flow rates for the flat plate are lower than those for the hemisphere for identical jockey weight positions I think this should help resolve your problem.

http://web.cecs.pdx.edu/~gerry/class/EAS361/lab/pdf/lab4_impactOfJet.pdf

7. Mar 18, 2017

### Brenda

What I don't understand is that it says that the force acting on the fluid for the hemispherical she'll should be twice that of the plate so (2mu) but the results I got shows the inverse the force acting in the flat plate (mu) are twice the one acting on the hemispherical plate and like I said I checked with some classmates and they got the same pattern I have attached the first page of the report and my results

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8. Mar 18, 2017

### JBA

Your posted data results page is basically not readable you might try photographing 90° from what you used and that may help but a simple plot of flow rate vs calculated force would be much easier to review.

If I understand what you are saying, your test results indicate that for the same jockey weight the flow rate for the hemisphere is twice that as for the flat plate.

9. Mar 19, 2017

### JBA

This morning I managed, with a bit more effort, to review the data sheet you posted and the the sheet shows that the fluid flow and therefore the momentum of the fluid flow for a given jockey position is lower for the hemisphere than for the flat plate, which is exactly how it should be because (as per the equation I referred you to indicates) the reaction force for a given fluid momentum is greater for the hemisphere than for for the flat plate.