Impact force of a water jet upon a target (fluid mechanics)

In summary, the conversation discusses a lab assignment where a jet of water is shot at different targets with various geometries. The force of the jet is determined by measuring the mass of weights needed to push the target back down and by calculating the theoretical value using fluid density, nozzle cross-sectional area, volume of water, and time. The conversation also includes a discussion about a formula given by the professor and the attempts to solve the problem using different methods. The conversation ends with a request for help and an attached Excel spreadsheet.
  • #1
jjack99645
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This one is actually for a lab assignment. In the experiment you shoot a jet of water upward vertically at a variety of targets with different geometries (a flat plate, various cups). The target is suspended from a rod, and on top of the rod is a cup where you can add weights, held up by a spring. With the jet impacting the cup you determine the force two ways, first, by measuring the mass of the weight it takes to push the target apparatus back down to y=0 against the force of the water. Second, by calculating the theoretical value of the force given the density of water, nozzle cross-sectional area, volume of water through flowmeter, and time.

I think the formula given by the professor on the lab sheet is incorrect, or perhaps I don't understand it enough and am missing something.

In all we repeated the process 12 times, so I'll just post the data for one repetition. The process should be the same for each one.

Homework Statement



Problem:
1. Assuming the impact velocity is the same as nozzle exit velocity, calculate the theoretical impact velocity.
2. Using the theoretical equation for the impact force of a jet, calculate the impact force.
3. Compare the theoretical and measured values.

Data:
ρ = 1000 kg/m^3
θ = 90° (flat plate)
m = 159g (mass required to "zero" the target against the force)
vol = 0.018927 m^3
t = 56.40 s
D = 1cm

Homework Equations



F = ρA(V^2)cos(θ+1)
F = mg

where V is jet velocity, A is nozzle cross-sectional area, ρ is fluid density, θ is the angle of the target (given by professor), and g is the constant of gravitation.

The Attempt at a Solution



Experimental force = m*g = 159g(.001kg/g)(9.8m/s^2) = 1.5582 N

Jet area: A = ∏r^2 = ∏((1cm*.01 m/cm)/2)^2 = 7.854e-5 m^2

Jet velocity: V = ( vol / t ) / A = (0.018927 m^3 / 56.4 s) / (7.854e-5 m^2) = 4.2728 m/s

Impact force: F = ρA(V^2)cos(θ+1) = (1000kg/m^3)(7.854e-5 m^2)(4.2728 m/s)^2(cos(90° + 1)

- or -

= (1000kg/m^3)(7.854e-5 m^2)(4.2728 m/s)^2(cos(∏/2 + 1))

I have tried it both ways -- radians and degrees, it does not specify in the lab sheet -- and both of them are incorrect. For this one I get a value of -1.2065 N. This in itself wouldn't trouble me, except that for the next experiment, a 0° "dome" target I get a positive 0.46963N theoretical value versus a measured value of 1.7444N. The values alternate between positive and negative as we go across the chart.

I'm guessing the problem lies in the cos(θ+1). +1 what? It doesn't make sense to me. I've tried Google to see if an alternative form of the equation exists but no luck so far. For the 90° target, which is just a flat plate, the jet force should be the full force of the jet...or F = ρA(V^2). That means the cos() would have to go to 1, which would mean cos(∏) or, in this case, cos(θ + ∏/2). Don't see how +1 plays in.

Thanks for any help on this one in advance. I'm trying to get the data done so I can pass it along to my lab partners for the lab write-up but I've been scratching my head on this all day and they have no suggestions either, and, sine it's Saturday, the professor is not in his office. Hopefully I've explained it well enough to pinpoint where I've gone wrong.
 
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  • #2
Excel spreadsheet attached
 

Attachments

  • lab #4 data.xls
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FAQ: Impact force of a water jet upon a target (fluid mechanics)

1. What factors affect the impact force of a water jet?

The impact force of a water jet is affected by the velocity of the jet, the density of the water, the surface area of the target, and the angle at which the jet hits the target. These factors can all be manipulated to increase or decrease the impact force.

2. How is the impact force of a water jet calculated?

The impact force of a water jet can be calculated using the equation F = ρQV, where ρ is the density of the water, Q is the volume flow rate of the jet, and V is the velocity of the jet. This equation takes into account the mass and velocity of the water to determine the impact force.

3. Can the impact force of a water jet be controlled?

Yes, the impact force of a water jet can be controlled by adjusting the velocity and flow rate of the jet, as well as the angle at which it hits the target. By manipulating these factors, the impact force can be increased or decreased as desired.

4. How does the impact force of a water jet affect different types of targets?

The impact force of a water jet can have different effects on different types of targets. For softer targets, such as tissue or wood, a high impact force can cause damage or penetration. But for harder targets, such as metal or concrete, a high impact force is needed to cause any significant damage or penetration.

5. Are there any safety concerns when working with high impact forces of water jets?

Yes, working with high impact forces of water jets can be dangerous and proper safety precautions should be taken. The high velocity and force of the jet can cause bodily harm, and protective gear should be worn when operating or working near water jets with high impact forces.

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